text:
pKa of imidazoles
I was looking for the $\mathrm{p}K_\mathrm{a}$ of imidazole and found this $\mathrm{p}K_\mathrm{a}$ table [1]:
\begin{array}{ll}
\text{Compound} &\mathrm{p}K_\mathrm{a} \\
\hline
\text{Imidazolium} &7.38 \\
\text{2-Aminoimidazolium} &8.46 \\
\text{1-Methyl-2-aminoimidazolium} &8.65 \\
\text{4,5-Dimethyl-2-aminoimidazolium} &9.21
\end{array}
Wouldn't you predict totally inverse order of $\mathrm{p}K_\mathrm{a}$ values?
For example, $\ce{-NH2}$ is a donor by resonance, it will make the cation more stable than imidazole, as far as I know. So, how to understand those values of $\mathrm{p}K_\mathrm{a}$?
I upload the first two compound structures:
IMIDAZOLE

AMINOIMIDAZOLE

Storey, B. T.; Sullivan, W. W.; Moyer, C. L. The Journal of Organic Chemistry 1964, 29 (10), 3118–3120. DOI 10.1021/jo01033a537.

Edit
Only for completeness I should add that pka values are of the protonates species, I mean imidazolium, 2-aminoimidazolium, and so on..

Amino groups being "donors" is not nearly specific enough. Is it via resonance or inductive effects?

Right, I will edit. It is by resonance @Zhe

My point was that it is inductively withdrawing. Give the nature of the cation, resonance donating is probably not super helpful.

Another good question is what resonance of the amino lone pair in the guanidine moiety does to the aromaticity of the ring. Interesting competing effects for sure.

Everything's all right. You've got confused but what you pointed out in your edit. Cation becomes weaker acid - "more stable" just like you say. Higher pKbH+ means weaker acid - stronger neutral base.

I don’t understand where the problem lies. A higher $\mathrm pK_\mathrm{a}$ means less likely to get deprotonated. The more electron-donating groups, the more a positive charge is stabilised and the less likely it is for a proton to leave — exactly what the data tells you. Am I missing something?

@Jan right. No, I was missing the final part..thanks a lot.

@Mithoron yes you are right,thanks for the comment.

Ah. Do you want to write up a self-answer to your question? =)

@Jan I can try.., please pay attention to correct me if am wrong..(and my English is quite bad also)

That’s what we’re here for =)

As pointed out by @Jan and @Mithoron in comments, part of the answer was on the OP. I will complete it.
If we compare acidity of imidazolium and 2-aminoimidazolium (the cations), we find that 2-aminoimidazolium is a weaker acid because of the bigger $\mathrm{p}K_\mathrm{a}$ value.
The reason is that the amino group ($\ce{-NH2}$) is donor by resonance, stabilizing the cation more than imidazolium is, because it doesn't have any extra donor group.
The same reasoning can be done for the remainder compounds of the table. The more inductive or resonance donors, the bigger stability of the cation and weaker acid (bigger $\mathrm{p}K_\mathrm{a}$).
One last thing
As @Zhe observed, the pKa values of all those compounds are really different of guanidine($\mathrm{p}K_\mathrm{a}$=13.6). In the paper below, authors make the same observation. I haven't found an easy answer for that. Maybe it can be done with computational calculations.

Storey, B. T.; Sullivan, W. W.; Moyer, C. L. The Journal of Organic Chemistry 1964, 29 (10), 3118–3120. DOI 10.1021/jo01033a537.

With regard to the comparison to guanidine, see "The pKa Values of Some 2-Aminomidazolium Ions"J. Org. Chem. 1964, 29, 10, 3118–3120 https://doi.org/10.1021/jo01033a537


text:
In a jet or internal combustion engine that burns hydrogen as a fuel, what would happen if you replaced the hydrogen with deuterium?
Would it increase the power of the jet engine due to heavier atoms being exhausted out the back?
For the same reason would it slow down an internal combustion engine?

I thought jet engines burn kerosene, and hydrogen is more suited for rockets.

@andselisk There are several hydrogen powered jets and cars that have been produced.

I guess you need to specify the exact types of engines you are talking about, and provide some references. I still cannot imagine a jet reaction engine using hydrogen (maybe you meant a turboshaft?), I must be living under a rock. Also, you might want to split the question in two as an internal combustion engine and a reaction engine have fundamentally different operating principles.

Unfortunately I don't think it's possible to answer your question satisfactorily. Simply mentioning "What will happen..." is rather vague, for there are several things that could happen.

@Pritt Balagopal Its not very complicated I don't think. Deuterium is a hydrogen atom with double the mass.  I just want to know what effect in theory that should have on an engine burning it.

Deuterium is not a gas. It is a isotope. Normally it comes as HD gas simply because Deuterium is rather rare is that what you mean. There is not much DD gas.

Its not very complicated I don't think. Deuterium is a hydrogen atom with double the mass. An extra neutron can change many things. An extra neutron made heavy water colorless, while normal water is very light blue. I just want to know what effect in theory that should have on an engine burning it. First of all, you must realize that engines are designed based on the fuel it uses. Using $\ce{D2}$ instead of $\ce{H2}$ can cause several problems, and you would need several books that explain the engines functioning to know why. There will be effects of course, just way too many effects.

related https://chemistry.stackexchange.com/questions/23831/would-there-be-a-difference-if-deuterium-is-embedded-instead-of-protium-regular

Possible duplicate of Chemical properties of isotopes

Very roughly there is nothing to gain. D2 will double the mass of the rocket while providing less energy content (D2 dissociation energy is higher than that of H2).

Sorry. You asked about jet or IC engines :)) but it should work the same. Momentum and energy counts, at least for jet. In IC again you have a heavier vehicle and less energy inside,   you should both have littler acceleration and lower final attainable speed.

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sort: votes


text:
Sulphuric Acid in Esterification Reaction
Why is it that adding the sulphuric acid slowly to a solution of alcohol and carboxylic acid (making an ester) will increase the yield of the product? It seems that not only does the rate at which it is dispensed affects the yield of ester, but really does change the colour of the solution when heated. For example, my team used it slowly and the colour was a more faint peach colour whilst everyone else, who added it quickly, produced pure black solutions!
What is happening? Why does adding the acid slowly increase yield and purity of the ester?

Sulfuric acid of higher concentration acts as a dehydrating agent.

It would be really nice if this was all documented, for educational purpose, especially for those who are forced to read the theory without the seeing the experiments.

The esterification is not the only thing catalysed by sulphuric acid. Basically, it’s catalytic activity is protonation of whatever feels happy enough to be protonated. In your desired esterification mechanism, we want to protonate the acid:
$$\ce{R-C(=O)-OH + H2SO4 <=> R-C(=\overset{+}{O}H)-OH + HSO4-}\tag{1}$$
This will ease the nucleophilic attack of the alcohol onto the carboxy carbon. However, we also have other atoms that can be protonated.  For example, the hydrogen-carrying oxygen may also accept the proton:
$$\ce{R-C(=O)-OH + H2SO4 <=> R-C(=O)-\overset{+}{O}H2 + HSO4-}\tag{2.1}$$
This could then lead to the removal of water and the formation of a very reactive acyl cation:
$$\ce{R-C(=O)-\overset{+}{O}H2 -> R-C#\overset{+}{O} + H2O}\tag{2.2}$$
This in itself isn’t bad as it could also react with the alcohol in the desired fashion. However, the alcohol also can be protonated:
$$\ce{R-CH2-CH2-OH + H2SO4 <=> R-CH2-CH2-\overset{+}{O}H2 + HSO4-}\tag{3.1}$$
And depending on the type of alcohol you have, a number of side reactions may occur. For instance, you could simply eliminate $\ce{H3O+}$ from the protonated alcohol to give an alkene:
$$\ce{R-CH2-CH2-\overset{+}{O}H2 -> R-CH=CH2 + H2O + H+}\tag{3.2}$$
If you have a suitable neighbouring group that can stabilise positive charges, semi-pinacol-type rearrangements could occur. I won’t draw that with the simple primary alcohol I drew because it would be unlikely in e.g. propanol. It could be a problem in other cases, though.
Finally, the intermediates I drew up can react even further and unpredictably with sulphuric acid still present.
If you add sulphuric acid slowly, the local acid concentration will remain rather low and the reaction mixture is rapidly homogenised. This means that side reactions caused by sulphuric acid will be suppressed. If, however, you add it quickly, there will be areas of your solution with very high local sulphuric acid concentrations. In these, these side reactions are accelerated. Furthermore, concentrated sulphuric acid acts as a dehydrating agent which further speeds up the reactions we don’t want (they all include losing a water molecule).
These side reactions — especially the dehydrating ability of sulphuric acid — tend to end up in ‘black tar’ which would give a more or less uniform black solution. What exactly happened is hard to say except that some polymerisation probably took place.

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text:
How to determine fire hazard of heating ethylene glycol to dissolve borax and boric acid
I am looking to replicate instructions provided in the last paragrph of this article that requires dissolving borax (Sodium Tetraborate decahydrate ($\ce{Na2B4O7·10H2O}$)) and boric acid in ethylene glycol, for reasons of practicality and cost efficiency supplied in the impure form of concentrated automotive anti-freeze.
For reasons out of scope of this question I've modified the ratio provided as follows:

5 kg anti-freeze
2.5 kg Sodium Tetraborate decahydrate
1 kg Boric acid

In order to dissolve the borax completely it is necessary to boil off the water of crystallisation plus any water impurities in the anti-freeze.
While I have water in the system, the temperature of the mix is expected to be stable at around 100 °C, but once the water is removed the temperature will rise to potentially dangerous levels if the heat source is not removed in time.
I would like to understand the risks of this operation. The flash point of (pure) ethylene glycol is, according to Wikipedia, at 111 °C. The autoignition temperature is listed as 410 °C and seems to be safely above the boiling point of ethylene glycol (it would boil off before it auto ignites). However this hazard sheet for ethylene glycol states it 'causes ignition' at 100 °C in combination with (amongst others) Sodium Chlorite (no more details given).
Can I assume that Borax and Boric acid are 'safe' as they are not listed on the hazard sheet?
Is there an (online) reference where I can find out how Borax and Boric acid modulate the flash point and auto ignition temperature of Ethylene Glycol?

Chlorates, perchlorates, permanganates and other strong oxidants conflagrate even at room temperature with ethylene glycol.
Boric acid and borax, however, are not considered strong oxidants and should probably be safe to dissolve below the boiling point of water. That said, I would test a small batch in a safe place, to be sure there is not a large gain in temperature on dissolving.
BTW, you might be interested in reading of the preservation of the Swedish warship Vasa in polyethylene glycol, and long-term issues with acidification. N.B. Polyethylene glycol is generally considered innocuous, but ethylene glycol is toxic if ingested.

I think the ultimate answer here is 'test a smal batch in a safe place'. Probably with planned 'catastrophic failure' to have an idea of a worst case scenario. Then I'll repeat this with sodium chloride added, as this is always present in marine environments. Although I think they might have a typo on that safety sheet and meant to say sodium chlorite.


text:
How would the crystal structure of sodium stearate look like?
In my lessons of solid state, I was taught about Bravais Lattices, and Face Centered, Body Centered, and Primitive cubic lattices, to name a few. All these crystal structures had this assumption, that ions/atoms are considered as hard spheres.
Considering sodium stearate, the stearate anion isn't a hard sphere, not by a long shot. Since Bravais lattices assume spherical atoms, I am led to believe that sodium stearate doesn't follow Bravais lattice structures.
So my question is, what is the crystal structure of sodium stearate?

There's quite a bit of difference between atoms and molecules...

@Mithoron I think I already know that. Can you please be a bit more direct and tell me what I'm missing?

Bravais lattices have nothing to do with atoms as solid spheres, or with atoms at all, for that matter.

Lattices don't assume spherical atoms. There is no assumption of what is placed at each lattice point.

@orthocresol Perhaps my assumption was wrong. However, all the pictures of these lattices show the lattice points as spheres. How would you accommodate a long chain fatty acid's conjugate base in such a lattice point?

Stretch the lattice. Note that there's also no assumption about the lattice parameters $a, b, c, \alpha, \beta, \gamma$. Anyway, nobody also said the alkyl chain has to be fully extended; maybe it is, maybe it isn't.

Maybe take a look at the refined structure of any stearate, say, this one: http://pubs.acs.org/doi/suppl/10.1021/jp992279f (supplementary info would be enough). Pay attention how the structure is obtained, what symmetry it has and what the unit cell looks like.

@orthocresol Usually it is. Then again, lattice points have nothing to do with atoms; if they coincide, that's merely a coincidence.


text:
I (might) have accidentally combined these mold-killers at home: Any adverse reactions?
A trailer is a small enclosed space. About 2 months ago we had some mold and I applied some mold killer to a few spots that I could not remediate. Below I've listed 2 products I used. (I definitely used product #1, and it's possible I applied product #2 to the same 1 x 2 foot area.)
Air testing shows the mold is gone, but there's still something in the air that one can feel isn't good to breath in. It tickles your throat, tightens up throat and sinuses, headache, and sometimes makes you cough. (Again, this isn't a medical question. I don't live in the trailer, it's sitting.) The only way to find out what that is though, is to disect what 2 products were sprayed. Would mixing these 2 products create something super nasty? (Like formaldehyde, Chlorine, Chloroform, etc., the list goes on.)
We did a Volatile Organic Compound (VOC) test a few weeks ago and it turned out to be pretty much negative. It did pick up some very low levels of toluene, hexane, ethyl benzene, and xylene, but far from anything alarming. The levels were far below the EPA's acceptable exposure levels, but the reason I mention these chemicals is in case they are valuable information and tip you off on anything.
Also, in case this is a tip for anyone; All my fabrics smell and they really make me cough (etc.). After I moved out of the trailer to remediate it, I noticed that all of my cloths, towels, bedding, etc. (i.e. all fabrics) had a weird chemical smell until they were washed. It sort of smelled like a smokey chemical smell... but not really smokey. Hard to explain. But what's really interesting is; I had a pile of clothes I'd left in a closet. The other day I put them in my car to drive to my new house, and I started coughing in the car, the headache set in again, irritated throat... all the same symptoms, only now its some kind of gassing coming off my clothes that were in the trailer. Weird! It makes me wonder if you're aware of a specific chemical that can stick to clothes like that?

I have 2 questions:
QUESTION 1: Is/was it dangerous to mix these 2 products? Did I create an even worse chemical in doing so? (I've listed their chemical composition at the bottom of this post)
QUESTION 2: Is there a chemical method to 'pull' chemicals like these out of the trailer? For example, I've heard about people putting ammonia in a bowl and leaving in their trailer to pull out formaldehyde. Not sure if that is a valid chemical method?

Product #1: Mold Armor Mold Remover & Disinfectant
Ingredients:
Ethanol is the main ingredient. These ingredients below are the small/active ingredients

Quaternary ammonium compounds, benzyl-C12-16-alkyldimethyl, chlorides (< 0.1 %)
Decanaminium, N,N-Dimethyl-N-octyl-chloride {Decyldimethyloctylammonium chloride} (< 0.1 %)
Octanaminium, N,N-Dimethyl-N-octyl-chloride (< 0.1 %)
Decanaminium, N-Decyl-N,N-dimethyl-chloride (< 0.1 %)

See product documentation (SDS), here.

Product #2: Concrobium Mold control
Ingredients:
Water is the main ingredient. These 3 ingredients below are the small/active ingredients

Trisodium phosphate
Sodium carbonate (.95%)
Sodium bicarbonate

See product documentation (SDS), here.

Edit
I reworded my question. Again, not a medical question. This is a chemistry question about mixing 2 products, and their reactions.

Perhaps is the mold itself.

I don't  believe closing this question as a "Personal Medical Question" is a good idea. For starters, the OP hasn't asked for advice on medication he (might) need to use. Paraphrasing what the OP said in questions (1) and (2):  "Do the two liquids react to produce any substance(s) that are harmful/irritants? Are the symptoms I display indicative of any of these substances." Question (4) seems on-topic too (albeit a wee bit broad, maybe?). Question (3) might do with some back-of-the-envelope advice; or should be asked at HomeImprovement.SE. I, for one, really like the detail put in the question

{Continuing here since I didn't have enough space in the last comment} Surely. being able to identify whether a substance is a (known) health hazard counts as being "on-topic"? To work out possible reactions that may take place upon combining two (non-pharmaceutical) household chemicals and identify any (known) hazardous substances produced...is really more of a Chemist's issue than a Doctor's.

could also be stirred up mold, or an allergic reaction to one of the chemicals...

This is an issue for a GP or an environmental health officer.  All of these post-event problems could be anything from an allergic reaction to psychosomatic.  It's way beyond the scope of this forum.

It's not psychosomatic, nor is it mold as a mold test was negative after remediation. Almost everyone entering the trailer notices it after about 15 minutes. It is purely a chemistry question. Doctors and air quality experts are clueless. If you need, ignore the entire post except the real question, which is; "Do the two liquids react to produce any substance(s) that are harmful/irritants? Are the symptoms I display indicative of any of these substances." (How can a non-chemist can hope to answer this?) I'd appreciate y'all not bowing out on this. Any way you can take the question off hold?

@Sydney It takes 5 reopen votes to reopen a question. As of posting this comment, your question has already garnered 3 reopen votes (invisible to new users). You must also understand that everyone here is not a chemist. We have chemists, students and plenty of inquisitive people from other disciplines. Someone with sufficient grounding in organic chemistry can help you here. You just have to wait for one (users Jan, ron, Klaus, Loong, ortho, etc) who's willing to spend some time to answer this (organic reactions are not easy...even for organic chemists). Good luck!

I have reopened, as I do not believe this is a personal medical question. If this is to be closed as outside of the scope of SE, it should be done with a custom comment. However, I also strongly recommend getting info from places apart from the Internet. Your comments seem to indicate you have already done so, but the truth is that if you have already spoken to professionals and given them all the necessary information, and they still don't know the answer, then it's highly unlikely that anybody here can offer anything better.

Completely understood. We'll see what happens. ;-)

Chemically speaking there cannot be anything reacting by simply mixing (room T, not irradiation, nothing extreme) the salts listed in your question!

I wouldn't be too worried about any chemicals that came from either of the products, or from reactions between the products. Any of those would have been volatile enough that they're long gone by now.
From your description, it sounds like you may be having a reaction to either residual mold or another allergen present in the trailer.
Mold is very difficult to fully eradicate. Even if you cleaned it off of all visible surfaces, it could very easily be growing behind something and you'd never know it without ripping out pretty much everything that's not glass or metal. Also, even if you've managed to kill all the most that was originally present, there could still be residual mold spores or proteins that get kicked up into the air very easily. If you're sensitive to mold, it doesn't take much for your immune system to react, giving you eye and upper respiratory tract irritation.
If this is the case, it may be easier to get rid of the trailer. To fully decontaminate, you'd want to go in and do a full tear out of everything that could have mold on it. After you're down to bear metal, you'd need to wet-wipe everything with a good biocide, making sure that you don't spread possible contamination with your wiping cloth. The point isn't only to kill the mold - it's to get rid off every possible protein fragment that may have been generated by the mold or mold spores.
While it's true that some people are sensitive to mold generated mycotoxins, more common is that people have immune reactions to the mold proteins, and it is not a small job to fully get rid of them.
Now, before you go full bore, keep in mind that it could also be something other than mold. Dust mites and roaches are other common allergens that come to mind. It may be worth going to an allergist to get tested for these and other common environmental allergens. You may find out that you're actually allergic to some kind of pollen (like poppies!) and it just happens that the trailer was driven through a poppy field at some point...

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text:
Stereoelectronic effects in torquoselectivity of 4π electrocyclic opening
Thermal 4π electrocyclic ring openings of cyclobutenes to form butadienes are known to proceed in a conrotatory manner, consistent with the Woodward–Hoffmann rules. However, there are two different conrotatory modes which can lead to isomeric products if the cyclobutene is suitably unsymmetrically substituted. Torquoselectivity refers to selectivity arising due to the preference of one conrotatory mode over the other.
The Houk group has reported some examples of counterintuitive 4π ring openings1,2 (simply considering steric interactions, the larger group generally wants to move outwards). In both cases only one isomer of the product is formed.

Why is this selectivity seen?

References
1. Houk, K. N.; Spellmeyer, D. C.; Jefford, C. W.; Rimbault, C. G.; Wang, Y.; Miller, R. D. Electronic control of the stereoselectivities of electrocyclic reactions of cyclobutenes against incredible steric odds. J. Org. Chem. 1988, 53 (9), 2125–2127. DOI: 10.1021/jo00244a058.
2. Rudolf, K.; Spellmeyer, D. C.; Houk, K. N. Prediction and experimental verification of the stereoselective electrocyclization of 3-formylcyclobutene. J. Org. Chem. 1987, 52 (16), 3708–3710. DOI: 10.1021/jo00392a047.

Related: Torquoselectivity: Disrotatory ring opening

The torquoselectivity of the above reactions is dictated by stereoelectronic effects, which (in this case) are significant enough to take precedence over steric effects. In general, electron-donating substituents (e.g. OMe, OTMS, ...) prefer to move outwards, and electron-accepting substituents (e.g. CHO) prefer to move inwards.
These can be understood by considering the frontier orbitals in the transition state of the conrotatory opening, as explained by the Houk group in earlier publications1,2 (the reactions given above were, in fact, carried out in order to provide experimental verification of their theoretical predictions). A more accurate orbital correlation diagram for the reaction is given below (modified from ref 2 to include diagrams of the relevant MOs):

A brief explanation is as follows. As the reaction proceeds, the C–C σ bond in the reactant is stretched, and consequently the σ* orbital is initially strongly destabilised (its energy increases rapidly). At a certain point near the transition state, though, it becomes sufficiently close in energy to the π* orbital such that orbital mixing can occur; this mixing leads to stabilisation and eventual "evolution" into the HOMO of butadiene.
Therefore, in the transition state, the HOMO is mostly σ-like, and the LUMO σ*-like; this was also verified computationally (details in ref 2). These frontier orbitals are of the correct symmetry to interact with pi-type orbitals of substituents. For donor substituents such as OMe, these orbitals are filled (lone pair on oxygen); for acceptor substituents such as CHO, these orbitals are empty (low-lying C=O π*).
The dashed orbitals in the following diagram represent the positions of these orbitals in the TS for both cases of outward and inward rotation. In the case of outward rotation, the substituent orbital can only interact with the lobe on C–3 which is in close spatial proximity; however, in the case of inward rotation, the substituent orbital can interact with the lobes on both C–3 and C–4.

In general, the interaction of a filled orbital and an unfilled orbital leads to stabilisation, whereas the interaction of two filled orbitals leads to destabilisation. The torquoselectivity is determined by maximisation of the former and minimisation of the latter. If the interaction with the lobe of C–4 leads to destabilisation, then inward rotation will be disfavoured, and vice versa.
For a donor substituent:

Inward rotation decreases the stabilising interaction between the donor orbital and the LUMO of the TS, as the lobe on C–4 is of the opposite phase as the lobe on the donor orbital, which leads to poorer overlap between the two orbitals.
Inward rotation increases the destabilising interaction between the donor orbital and the HOMO of the TS, as the lobe on C–4 is of the same phase as the lobe on the donor orbital, which leads to stronger overlap.

Based on these reasons, donor substituents generally prefer to rotate outwards. In the case of OH, this preference was calculated to be a whopping 16.4 kcal/mol.3
For acceptor substituents, the interaction with the LUMO is not relevant as the interaction of two unfilled orbitals does not lead to any stabilisation. However, inward rotation maximises the stabilising interaction with the HOMO, since this is enhanced by overlap with the lobe on C–4. Hence, acceptor substituents generally prefer to rotate inwards. This preference, however, is markedly smaller than in the case of donor substituents - perhaps because there is only one factor that biases the course of the reaction (as opposed to two in the case of donor substituents). Experimentally there are fewer examples of torquoselectivity with electron-accepting substituents, and for the second reaction above, the preference for the inward rotation of the formyl group was calculated to be a much more modest 4.5 kcal/mol.4
Interestingly enough, for the cyano group (-CN), a preference of 4.3 kcal/mol was calculated - for outward rotation.4 It turns out that the donor π orbitals are more important than the acceptor π* orbitals, which are relatively high in energy. The formyl group only rotates inwards because it has an exceptionally low-lying π* orbital!

References
1. Kirmse, W.; Rondan, N. G.; Houk, K. N. Stereoselective substituent effects on conrotatory electrocyclic reactions of cyclobutenes. J. Am. Chem. Soc. 1984, 106 (25), 7989–7991. DOI: 10.1021/ja00337a067.
2. Rondan, N. G.; Houk, K. N. Theory of stereoselection in conrotatory electrocyclic reactions of substituted cyclobutenes. J. Am. Chem. Soc. 1985, 107 (7), 2099–2111. DOI: 10.1021/ja00293a046.
3. Houk, K. N.; Spellmeyer, D. C.; Jefford, C. W.; Rimbault, C. G.; Wang, Y.; Miller, R. D. Electronic control of the stereoselectivities of electrocyclic reactions of cyclobutenes against incredible steric odds. J. Org. Chem. 1988, 53 (9), 2125–2127. DOI: 10.1021/jo00244a058.
4. Rudolf, K.; Spellmeyer, D. C.; Houk, K. N. Prediction and experimental verification of the stereoselective electrocyclization of 3-formylcyclobutene. J. Org. Chem. 1987, 52 (16), 3708–3710. DOI: 10.1021/jo00392a047.


text:
Hydrogenation of pyridine compared to other heteroaromatic rings
Why does pyridine require the lowest temperature for hydrogenation, even though it has it has the highest resonance energy amongst pyridine, pyrrole and furan? Is it just because of the increase in the absolute pressure or does the HCl also effect the reaction somehow?

There's higher pressure though, and different catalyst, Pt is better I think.

In your example, you're not reducing pyridine, but pyridinium hydrochloride. The difference is quite large, pyridine requires 70 bar and up to 100 degrees Celsius to form piperidine at an effective rate, using nickel.

When we reduce pyridine using the $\ce{H2/Pt/HCl}$ reagent-catalyst combination, we require a significantly lower temperature and pressure compared to other similar reductions (but without $\ce{HCl}$), because $\ce{H+}$ (and other Lewis acids for that matter) will go and add to the nitrogen atom, forming pyridinium salts.
This makes the compound more reactive towards reduction (as a partial positive charge will develop in the ring).


text:
Why doesn't a buffer solution change ph(Appreciably?)
Consider a buffer solution containing weak acid $\ce{CH3COOH}$ and its salt $\ce{CH3COONa}$:
\begin{align}
\ce{CH3COOH &<=> CH3COO- + H+} \tag{1}\\
\ce{CH3COONa &-> CH3COO- + Na+} \tag{2}
\end{align}
These two reactions take place and now we are majorly left with $\ce{H+}$, $\ce{CH3COOH}$, $\ce{CH3COO-}$, $\ce{Na+}$ in the solution. Since, $\ce{Na+}$ is a spectator ion, we can ignore it.
We know that buffers resist change in pH on adding small quantities of $\ce{H+}$ and $\ce{OH-}$. Let us add $\ce{H^+}$ to this solution.
A buffer solution must contain some things that could remove these $\ce{H+}$ ions, otherwise the pH would change. As we add $\ce{H+}$ to this buffer solution, it will react with $\ce{CH3COO-}$ to form $\ce{CH3COOH}$.
But here's the catch.
$\ce{CH3COO-}$ is getting removed and $\ce{CH3COOH}$ is getting formed. According to Le Chatelier's principle (LCP), the equilibrium should shift towards in the forward reaction (in reaction (1)). Shifting of equilibrium in the forward reaction should increase the concentration of $\ce{H+}$ and hence decrease the pH of the solution.
Now, one might say that $\ce{CH3COOH}$ is a weak acid and it would not decompose much. So, we can actually ignore the amount of $\ce{H+}$ so generated.
Let's "assume" that's true for a while and move ahead.
Let us add $\ce{OH-}$ ions to the solution.
Remember there's $\ce{H+}$, $\ce{CH3COOH}$, $\ce{CH3COO-}$, $\ce{Na+}$ present in the solution. $\ce{OH-}$ can react with either (a) $\ce{H+}$, or (b) $\mathrm{CH_3COOH}$.
Here comes the tricky part.
Let us only consider hydroxide ions reacting with $\ce{H+}$
If $\ce{OH-}$ reacts with $\ce{H+}$, water will be formed.
Here although hydroxide ions are getting removed, $\ce{H+}$ are also being removed from the solution. Now, because of the removal of $\ce{H+}$ shouldn't the solution become less acidic and the pH should increase.
Now, some might say as soon as $\ce{H+}$ is removed, the equilibrium in the first reaction shift towards the right.
But wait, didn't we just ignore that (in the case when we added $\ce{H+}$ in the solution)?
Can anyone explain why is there a contradiction?
P. S.: I have ignored the $\ce{H+}$ and $\ce{OH-}$ from water because it is irrelevant to the discussion.

pH will change all right, only way less than it would change in pure water.

we can't ignore the forward reaction in the second case as the net equilibrium(of water formation and decompostion of CH3COOH) would be around 10^10.(10^15 of water and 10^(-5) of acetic acid)

@Ayushmaan and we can ignore it in the first casE?

yes, because in the first case the equilibrium constant would only be 1. ( 10^5*10^(-5)). So, if initial concentration of acetic acid is large in comparison to H+ we have put, the pH would be approx. same.

Having problems with that concept myself, above answers seem mighty complicated, this is how I think it works (warning- only my understanding, could well be wrong)
Point One:
You have a lot of $\ce{CH3COOH}$ (not much decomposition) and a lot of $\ce{CH3COO−}$ (from the salt) and very little $\ce{H+}$
Point Two:
pH works on the amount of $\ce{OH-}$ and $\ce{H+}$ in the system, there are usually only very very small amounts (when looking at weak acids and bases) present compared to the reactant (in water, there is about 500 million molecules to ever one that dissociates) So the point I'm making is adding a few $\ce{OH-}$ or $\ce{H+}$ ions will have a big impact on the pH.
Conclusion drawn:
As you have so much $\ce{CH3COO−}$ to begin with, removing a very very small amount by adding a few $\ce{H+}$ ions and forming a very very small amount of $\ce{CH3COOH}$, is not going to have a noticeable impact on the concentrations, in comparison to the amount already in the solution (both $\ce{CH3COO−}$ and $\ce{CH3COOH}$). You could even say that it is a negligible amount and can be ignored, therefore can say this will not have an effect on the equilibrium.
As the $\ce{H+}$ is removed, the pH is not changed; however as there is so little $\ce{H+}$ ions to begin with, not removing them will have an impact on the pH (see point two).
As already stated, there are so little $\ce{H+}$ ions in the system, removing them (by adding $\ce{OH-}$) will have an impact on equilibrium, you cannot say the amount of $\ce{H+}$ removed is negligible, you cannot ignore Le Chatelier's principle, the equilibrium will shift right to replace the lost $\ce{H+}$.
Note: I'm an A-level student, don't take this as 100% right or even 1% right, this, at best is a calculated guess as to why equilibrium can be ignored when you add $\ce{H+}$ ions and equilibrium can not be ignored when you add $\ce{OH-}$.

The characteristic feature of a buffer is that the equilibrium is balanced at an inflection point with $\mathrm{HA}$ and $\mathrm{A^-}$ set approximately equal. The $[\mathrm{H}^+]$ concentration changes very little here with a change in the ratio $\mathrm{[A^-]/[HA]}$.
Let's look at an imaginary case, with numbers: $K_{\mathrm{a}}$ of acid $\mathrm{HA} = 10^{-6}$. Then
$~~~~~~ K_{\mathrm{a}} = 10^{-6} = \mathrm{\dfrac{[H^+][A^-]}{[HA]}}$
For a 1 M solution of $\mathrm{HA}$, $\mathrm{[H^+] = [A^-] = 0.001~}M$. The pH will be 3.0.
These are very small amounts, reflecting the fact that weak acids are weakly ionized. If you add 0.1 mole of strong acid to 1 L of solution, the pH will change to ~1.0, a significant change.
If you add 0.1 mole of strong base, the pH increases:
$~~~~~~ K_{\mathrm{a}} = 10^{-6} = \mathrm{\dfrac{[H^+][\sim 0.1]}{[\sim 0.9]}}  \;\; \therefore \;\; [\mathrm{H}^+] = 9.1\times 10^{-6}$ and pH goes to ~5.04.
If we make 1 L buffer solution with HA = 0.5 M plus 0.5 M of an alkali salt of HA,
$~~~~~~ K_{\mathrm{a}} = 10^{-6} = \mathrm{\dfrac{[H^+][\sim 0.5]}{[\sim 0.5]}} \;\; \therefore \;\; [\mathrm{H}^+] = 1 \times 10^{-6}$ and pH = 6.
Now add 0.1 mole of strong acid ($\mathrm{A^-}$ is consumed, $\mathrm{HA}$ increases):
$~~~~~~ K_{\mathrm{a}} = 10^{-6} = \mathrm{\dfrac{[H^+][\sim 0.4]}{[\sim 0.6]}}  \;\; \therefore \;\; [\mathrm{H}^+] = 1.5\times 10^{-6}$ and pH drops to 5.82.
This is a relatively small change in pH.
If you add 0.1 mole of strong base, $\mathrm{HA}$ is consumed and $\mathrm{A^-}$ increases:
$~~~~~~ K_{\mathrm{a}} = 10^{-6} = \mathrm{\dfrac{[H^+][\sim 0.6]}{[\sim 0.4]}}  \;\; \therefore \;\; [\mathrm{H}^+] = 0.66\times 10^{-6}$ and pH rises to 6.17.
Unbuffered, the range of pH change from addition of 0.1 M strong acid or strong base is ~4 pH units around the starting pH of 3; buffered, the range is only 0.35 pH units around the starting pH of 6. Adding less of the strong acid or base produces an even smaller change in the pH of the buffered solution.


text:
Why is lead (IV) chloride covalent while lead (II) chloride ionic?
I would like to find out more about what makes a compound ionic and what makes one covalent. However, I felt this was too general a question so I thought that asking about the chlorides of lead would be more appropriate. I have already come up with an explanation of my own, which I have provided below. However, I would also like to hear the opinions of others.
My explanation is as follows: I consider the interaction between the atomic orbitals of the lead (II) and lead (IV) cations with the chloride ions. Firstly, we must note that their valence electronic configurations are different: lead(II) has its valence electrons in 6s orbital while the valence electrons of lead (IV) reside in the 5d orbital. On the other hand, the valence electrons of the chloride ions reside in the 3p orbital. Based on what I have read from Clayden's Organic Chemistry, I understand that atomic orbitals of similar energy interact strongly to produce strong covalent bonds. I was thinking that since the interacting orbitals of the lead (IV) cation and the chloride ions are closer in energy, a favourable covalent interaction would occur. However, in the case of lead (II), covalent bond formation is not favoured since the orbital energies are considered to be too far apart. I am not confident that this is correct. Please feel free to point out any inaccuracies.

At its simplest it is just more positive charge -> more polarising -> more covalent bond. Pb(IV) can't form bonds with a fully filled 5d subshell, btw, unless you invoke some kind of d-s hybridisation. It's probably best represented by 4x dative bonds from chloride into 6s + 6p orbitals of Pb, which have covalent character because of strong polarisation. Same applies to Pb(II) and its fully filled 6s orbital.

@orthocresol Ok. That sounds simple enough. On a side note, how would we explain that titanium (IV) iodide is covalent?

https://chemistry.stackexchange.com/questions/19763/why-is-pbcl%e2%82%84-covalent https://chemistry.stackexchange.com/questions/42784/would-tin-ii-sulfide-be-considered-a-covalent-network-solid?s=1|0.1778 https://chemistry.stackexchange.com/questions/9038/how-would-one-compare-the-magnitude-of-covalent-character-between-sncl4-and-snf2

@orthocresol Actually, if you think about it, the argument using positive charge seems to fail when we consider neutral atoms in the first step. This argument only works if you first assume that the interacting species are all ions, not neutral atoms. Then only can we say that it is the high polarising power of lead (IV) cation etc.

Is it possible to explain this in a "less ionic" manner, meaning that we do not invoke the concept of charges and polarisation. I mean the electronegativity of lead is invariant in whatever compound it forms. So why would believe that the lead atom in lead (IV) compounds is more polarising than in lead (II) compounds.

Does this answer your question? Why is PbCl₄ covalent?


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Calculation of irreversible surrounding entropy
I have a specific query. In my undergraduate book Physical chemistry by Engel and Reid, sample problem 5.8 calculates the change in entropy of the surroundings and system for an irreversible isothermal compression.
My question has to do with the manner in which they calculate the entropy of the surroundings.
Entropy is first operationally defined as a result of a new cyclic integral satisfying the state function criteria. This integral includes the heat transfer of all reversible parts of the carnot cycle: $$\oint\frac{\require{cancel}\cancel{d} q_{rev}}{T} = 0$$
hence the definition of entropy becomes: $\Delta{S}=\int\frac{\cancel{d} q_{rev}}{T}$     and it is from here that we derive our requirement that all subsequent calculations of entropy be done along reversible paths.
This is relevant because it is my understanding (albeit one I expect to be incorrect) that sample problem 5.8 completely disregards this without explanation. I've since moved on in Chemistry, but my confusion on this point has persisted. Let me explain:
The problem has 1 mole of ideal gas at 300 K be isothermally compressed via a constant external pressure from $25.0 L$ to $10.0 L$.
The constant pressure is what causes the change to be irreversible. However, we are required to calculate a change in entropy along a reversible path, so we proceed as follows: We know the process occurs isothermally, so $\Delta U_{syst} = 0$, and $q_{reversible_{syst}} = -w = nRT\int_{V_i}^{V_f} \frac{dV}{V} = nRTln\frac{V_f}{V_i} $
After some algebra, this gives us $q_{rev_{syst}} = -2.285* 10^3 J$ and $\Delta{S}=\int\frac{\cancel{d} q_{rev_{syst}}}{T} = \Delta{S}=\frac{q_{rev_{syst}}}{T} = \frac{-2.285* 10^3 J}{3
00K} = -7.62 J*K^{-1}$
And all is well with the world. HOWEVER, this is where I run into trouble. We now turn to the surroundings, and again, there is a constant pressure, which means this process is irreversible. We are duty-bound by the definition of entropy just discussed to pursue a reversible path, considering the surroundings as our new system.
In my mind, this requires that we consider the change reversibly, ie: occurring in infinitesimal increments, which would look equal and opposite to the calculation we had just performed for the system. But when finished, this would result in us having $\Delta S_{total}=\Delta S_{syst}+\Delta S_{surr} =0$, which is not true for an irreversible process. So instead, the writers proceed as follows:
They find $P_{i}$ and $P_{external}$ using $P = \frac{nRT}{V}$
"$P_{external} = \frac{1mol*8.314Jmol^{-1}K^{-1}*300 K}{10.0L*\frac{1m^3}{10^3L}} = 2.494*10^5 Pa$
$P_{i} = \frac{1mol*8.314Jmol^{-1}K^{-1}*300 K}{25.0L*\frac{1m^3}{10^3L}} = 9.977*10^4 Pa$"
They then apply the same reasoning as in the system's case:
"Because $\Delta U = 0$, $$q = -w = P_{external}(V_f-V_f)= 2.494*10^5 Pa*(10.0*10^{-3}m^3 - 25.0*10^{-3}m^3)$$
$ = -3.741*10^{3}J $
The entropy change of the surroundings is given by:
$\Delta{S_{surr}}=\frac{q_{surr}}{T} =\frac{-q}{T}=\frac{3.741* 10^3 J}{3
00K} = 12.47 J*K^{-1}$
...$\Delta S_{total}=\Delta S_{syst}+\Delta S_{surr} = -7.62 J*K^{-1}+12.47 J*K^{-1} = 4.85 J*K^{-1}$"
I've enclosed segments borrowed directly from the book in quotes. My problem, is that although we are told time and time again that only a reversible path may be used to calculate entropy, in the case of the surroundings, $P_{external}$ is used to directly calculate heat, and subsequently that heat is used to calculate entropy. That value for heat was NOT calculated along a reversible route. How then, can we justify using it in the calculation? And if we can't, how are we to ever find anything other than $\Delta S_{total}=0$?
I know that my confusion is caused by some underlying misunderstanding, but I have not been able to pinpoint it. I'm hoping one of you will. Much of my research done on this topic, including the reading of Atkins' book, and MIT lecture notes, as well as other forums, focus on residual entropy, or worry themselves over whether the surroundings can be considered infinite or not, but none of these have seemed to answer my problem. I italicize that, because I am not confident that the answer I seek was NOT in those explanations, but I haven't seen it.
Thanks again!

What they are doing is modeling the surroundings as something like a weight sitting on top of the massless frictionless piston plus an ideal infinite reservoir at 300 K.  Lifting a weight is a mechanically reversible process that does nothing more than increase the (conservative) potential energy of the weight.  So this part doesn't figure in the entropy change.
Now for the reservoir.  An ideal infinite reservoir is one that can absorb or reject any amount of heat without its temperature changing because of its infinite thermal capacity.  Plus, its change in entropy is always $\frac{Q_{res}}{T}$, where $Q_{res}$ is the heat absorbed by the reservoir.
So, in this problem, the surroundings are regarded as an ideal reversible entity.
What they are doing in the calculation in parenthesis is determining the actual amount of heat transferred to the ideal infinite reservoir in the process.  This is equal to minus the heat transferred to the system.  So the entropy change of the reservoir is $\frac{-Q_{syst}}{T}$

Thank you so much for responding Chester! Maybe I can clarify my confusion even further. When you say "What they are doing in the calculation in parenthesis..." I assume you're referring to step with "P_external(Vf−Vf)."
If the corresponding entropy were indeed equal to -(that of the system), then when considered together, \D S_{syst} + \D S_{surr} would equal zero no? The argument I see being made in the book is that the entropy changes of the syst and surr are different, via that calculation with P_ext, and I am unsure why we are allowed to use P_ext in a reversible calculation for entropy.

I am unsure about this because P_{ext} is being used in the calculation of irreversible heat transfer. By the logic of the text, this produces q_{irrev}, which I would have thought could not be used to calculate entropy change on the grounds that entropy is defined explicitly using reversible heat.
Furthermore, if the q for the surroundings were to be calculated using a reversible path that traverses the same change, it would result in dS_surr = -dS_syst, and total S would then be zero, which cannot be true due to the process being irreversible.
Thanks again, I only hope to clarify!

In looking at reversible paths to get entropy change, the reversible path for the surroundings does not have to be the same as the reversible path for the system.  Each entity can be treated separately in determining its entropy change (which is really a material property which depends only on the initial and final states of that entity).  However, in the present irreversible process, even though the system experiences an irreversible path, the actual path for the surroundings is reversible.  Here is a simple primer:  https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

Here are the initial and final states of the surroundings:  Initial:  A pair or weights totaling  (W+w) at elevation z = 0 and  an ideal reservoir at temperature T, with no heat added yet.  Final: A weight W at elevation $z=z_f >0$, a weight w at elevation z = 0 (this weight w was removed from the piston at time zero), and an ideal reservoir at temperature T that has received Q heat added.  Irrespective of what happened to the system, what is the entropy change for the surroundings?

In the above example, $P_{ext}=W/A$, where A is the cross sectional area of the piston and cylinder and $\Delta V=Az_f$.

The initial system of the surrounding is P1, V1, and T1. This is an isothermal compression, so weight is added and the final state is P2, V2, but T1, due to heat outflow which balances the work done on the system.
The reversible path is one in which P and V are allowed to change infinitesimally, which requires the inclusion of an integral does it not? As in $\Delta U = 0 -> q=-w = \int{PdV}$ While I see your point of $P_{ext} = \frac{W}{A}$, I also would be tempted to substitute $W = nRTln\frac{V_f}{V_i}$ in order to retain the reversible path.

My confusion is although $\Delta S$ is a state function, $q$ and $w$ are not, and are path dependent, so using $W = P_{ext}(V_f-V_i)$ calculates the work along an irreversible path, which when equated to $q$ cannot be said to be $q_{rev}$? Or so I thought. We would require that integration over the entire change occur, ensuring $P_{ext} = P$ at each stage. This is what leads to the $W = nRTln\frac{V_f}{V_i}$ term no? Thank you again for replying so thoroughly!

The surroundings are not P1, V1, T1.  We don't know anything about the volume and temperature of the surroundings, except that there is also a reservoir at 300 K.  The surroundings path is reversible for our process because it just involves raising a weight against gravity.  The work $W=p_{ext(V_f-V_i)}$ is irreversible for the system, but reversible for the surroundings.  Your determination of the reversible work for the system (which differs from the actual work done by the system) is correct.    And the entropy change you determined for the system is correct too.

In this process, as I said, even though the path executed by the system is irreversible, the path executed by the surroundings is reversible.  It just involves a transfer of heat to an ideal (reversible) reservoir and raising a weight (which is also reversible).  And the amount of heat transferred to the reservoir is equal to the actual work done by the system.

I have a question for you.  If there were an actual physical weight being raised rather than an alluded-to $P_{ext}$, would you be able to more easily accept the idea that, even though the system path is irreversible, the surroundings path is reversible?  Another question:  Did you read that primer I referenced?

I did read your primer, and I think I might understand. This sentence: "The work $W=p_{ext}(V_f−V_i)$ is irreversible for the system, but reversible for the surroundings." May have helped. Are you saying that because the surroundings are modeled as an infinite reservoir, $P_{ext}$ is inexhaustible much in the same way that the $T_{surr}$ is? And therefore the reversible path calculation in the surroundings can take the form of what would be considered an irreversible path for the system? I mean to say inexhaustible in the sense that $P_i = P_f$ for the infinite surroundings?

The reversible model you are proposing for the surroundings is indeed valid, but is just one of many that would deliver a constant external force per unit area to the system while, at the same time, experiencing a reversible path itself.  It would involve suddenly removing a constraint on the piston at time zero.  An equally valid model would be the one I indicated, involving sudden removal of part of the weight from the top of the piston.  There are other models as well.

Yes, I believe my confusion is that sudden to me implies non-reversible, via the infinitesimal description of reversible processes. I think I am reading you say that you can circumvent this issue by treating the surroundings as a reservoir of infinite temperature and pressure? So my initial problems with both my proposed method and the one you have just proposed (2 rev paths between same i and f states) is centered on the use of both sudden and reversible at the same time.
$$ $$
If that is in any way correct, then I believe you have helped a great deal. Thanks Chester!

You're welcome.  Note also that for some types of energy (such as potential energy and kinetic energy), changes to not have to occur quasi-statically for reversibility to apply.

I think the key learning here is that a system may experience an irreversible change while its surroundings experience a reversible change (without specifying in detail how the reversible change in the surroundings is brought about, other than it involves an ideal reservoir).  In most cases, it is implicitly assumed in thermodynamics problems that the surroundings undergo changes that are reversible.

The gas pressure equals to the external pressure during isothermal irreversible expansion. To maintain P(gas)=P(ext), heat is supplied reversibly. By this reason delta S(surr) must be calculated above. Am I clear?


text:
Why have citric acid only a denticity of 3 and tartaric acid of only 2?
Why can't the hydroxyl-groups in citric acid (denticity 3)

or in tartaric acid (denticity 2)

contribute to the denticity, as there are doing it in aminoethylethanolamine (denticity 3)

Why do the hydroxyl-groups can't contribute in the first two compounds?

Their denticity can get up to 4, maybe even higher, just not with only one cation, because of steric reasons.

So, the molecule itself gets in the way?

This sort of thing is where you need a good 3D model kit to play around. It really helps.


text:
Grey "skin" on gallium
I have a sample of gallium (a few grams). I've noticed every time I melt the gallium and refreeze it, it forms this grey skin on the outside. When I remelt the gallium, this portion doesn't melt and remains a dull grey color. I have skimmed it off every time and collected it. I looked it up and gallium doesn't seem to create an oxide or other compound unless under extreme conditions. I have noticed this when melting it in a plastic bag, a glass beaker, and a plastic container. I have been using hot water to melt it, but the water doesn't contact the gallium, only the container. It seems I am slowly losing gallium from my sample and have a small amount of this grey material saved in case it can be recovered. Does anyone know what this is and how it can be reverted to elemental gallium?

It is gallium (III) oxide. You can't reclaim it in any practical manner.

@MaxW I read it doesn't react to air a room temperature, only when red hot. Do you have a source?

I have no real knowledge of it, and I've found almost no literature on its handling and storage. From Scientific American 1878:"Unlike lead, however, it acquires only a very slight tarnish on exposure to moist air..." and "In aerated water it tarnishes slightly." My very very old copy of Cotton and Wilkinson (Adv. Inorg. Chem) states that it is a moderately reactive metal and will react with sulfur. And of course we know it readily alloys (I'd say "amalgamates" except that term is specific for mercury alloys) with a large variety of metals. The (III) oxide is white, so unless your scum is white, it ain't that. The (I, sub) oxide is brown-black so that's a possibility. As are mixed oxide /hydroxides. Since the original investigation says it does tarnish in air, I'd take that at face value. There is the obvious possibility that your sample is either contaminated or was sold adulterated. If it's been in contact with any other metal (especially Aluminum) then it is almost certainly contaminated. There's no practical way you'll be able to recover the metal - the expense will far exceed the price to simply replace it. The best way to check for any large contamination is to determine its melting point, it should be very very close to its stated MP, shouldn't even be off by 1°.


text:
Major product of addition reaction of substituted alkene when both carbons have same number of hydrogens and different substituents
What will be the major product in an addition reaction if a substituted alkene has same number of hydrogens in both carbons of the double bond and different groups attached to them?
For example, What will be the major product of the addition  reaction between 1-chloroprop-1-ene and hydrogen bromide?
$$\ce{CHCl=CH-CH3 + HBr -> ?}$$
And what will be the major product of the addition reaction between 1-bromo-2-chloroethen and hydrogen bromide?
$$\ce{CHBr=CHCl + HBr -> ?}$$
Can Markovnikov rule help us in this regard?

The Bromine just attacks the carbon on which the carbocation formed would be the most stable (as the intermediate is a carbocation and hence the rate of reaction would depend on it).
This is the basis of the "Markovnikov Rule".
So, in the first case, bromine would attack the chloro-substituted carbon, as the +ve charge  on it would be stabilised by resonance by chlorine.
And, in the second case, it will attack the bromo-substituted carbon as bromine would be a better lone-pair donor than chlorine.

I don't think it's resonance that is important here for the chlorine atom. It should be the inductive effect. The methyl group can stabilise the carbocation by hyperconjugation while the chlorine atom destabilises it by drawing electron density away. Thus, in the first case, the carbocation should form on the methyl-substituted carbon and the bromide ion would attack that

@TanYongBoon   But, resonance effect dominates over the inductive effects for halogens in the case of aliphatic compounds. (mostly, the reverse is true only for aromatic ones)

//Resonance effect dominates over the inductive effect for halogen// But in the first case, there's a fight between methyl's inductive effect and chlorine's resonance effect. Which would win?

@Hisab         The general order of stabilisation is resonance> hyperconjugation> H-Bonding>Inductive effect.

@Ayushman Can you suggest me a book in which this topic is given in details? couldn't find this topic in my textbook.

@Ayushman Plz, help me

@Ayushman Why does resonance effect dominate over inductive effect for halogens in aliphatic compound and the reverse in aromatic compound?

In the first case bromine will attack the chlorine substituted carbon because the carbocation formed would be stabilised by the $\ce{-I}$ effect of chlorine and resonance of lone pairs of chlorine.
In the second case bromine will attack the bromine substituted carbon because of $\ce{-I}$ effect of chlorine and also bromine is a better lone pair donor. Also the carbocation formed will be in resonance.
These criterias fullfil the Markovnikov rule as carbocation formation is the most important step in the process.


text:
Physical state of the carbon denoted with "gr"
I would like to know what does a gr mean when used to represent the physical state of the carbon, for example, in the synthesis of the glucose
$$\ce{6C_{(gr)} + 6H2_{(g)} + 3O2_{(g)}-> C6H12O6_{(\ell)}} \quad,$$
what does the $\mathrm{gr}$ mean?

The letters in brackets indicate state of aggregation (includes state of matter, crystal structure (solids) and atomicity (gases)), therefore $\ce{C_{(gr)}}$ most likely refers to graphite. Often one can also see $\ce{C(s, gr)}$ or $\ce{C(s, graphite)}$ There is a nice demonstration in [1, p. 248]:

$$\ce{C(gr) + O2(g) <=> CO2(g)}$$
  Here $\ce{(g)}$ indicates that the compound is gas, [...], and $\ce{(gr)}$ that carbon is graphite (not many people will burn diamonds, even if they are chemists).

P. S. According to IUPAC recommendations, subscript notation is deprecated.

Metiu, H. Physical Chemistry: Thermodynamics; Garland Science, 2006. ISBN 978-1-315-31491-4.


text:
Listing errors in experimental section
I am fairly new to report writing, and am currently writing an experimental section.
Obviously we are required to list measurements, but I was wondering whether or not we are supposed to list the corresponding errors with these?
E.g. 5 mL $\pm$ 0.01 mL sodium hydroxide was added to...
Or if not, why this isn't included?

5 mL ±0.01 mL  doesn't make much sense. I guess you meant $\pu{(5.00 \pm 0.01) mL}$ or $\pu{5.00(1) mL}$. If you speak German and have access to DINs, I would suggest to have a look at DIN 1319 series and DIN 1319-3 + 1319-4 in particular. Otherwise any lab manual explains the measurements handling more or less OK.

Ask your supervisors?

It depends a lot on the context.
For example if we are talking about synthetic chemistry the exact amount often isn't relevant, for example in case of solvents for the reaction. It doesn't really matter if you do it in 200 mL or 198 or 201 mL. Therefore no one cares to measure it exactly in the first place and no one really cares about those errors, especially since you would need to measure temperature as well.
In other cases, like many analytical methods, you need to know the exact amount which was used and in those cases you should definitely include errors.
In the end you need to include it if it's relevant information, it's common to omit it if it's not relevant.


text:
Why don't alkynes undergo preferential anti bromination to the degree that alkenes do?
Alkenes, for the most part (unless there is a phenyl group or highly ionising solvent) undergo almost exclusively anti addition of bromine. However, under similar conditions alkynes, although undergo preferential anti-addition, it is not as exclusive as alkenes. Even with electron withdrawing substituents like two $\ce{COOH}$, only 70% is anti. (Peter Sykes, page 182)
Why?

I've checked two of my books but I didn't find anything. If I could make an assumption I would say that maybe the mechanism followed by alkynes differentiates from the one followed by alkenes? In alkenes' mechanism a three-membered ring is formed among one Br atom and the two double bonded carbons thus leading the other Br to bond with the carbons from the other side of the bond to create an anti-bromination

From what I know of the mechanism, Addition of Br2 takes place in the following mechanism
For an alkyne, the intermediate would not be stable due to the ring strain on 3 member ring with a double bond in it, thus reducing the rate of reaction.
Mechanism referred from: https://en.wikipedia.org/wiki/Halogen_addition_reaction

One factor that can be responsible for lesser extent of anti brominationot of alkynes is may be to do with intermediate formed during the reaction - cyclic bromonium ion.
In case of alkyne, the intermediate consists of 3 membered ring including the halogen. The ring has a double bond thereby increasing the angle strain in the ring. This makes alkynes less reactive.

In alkenes,  the cyclic intermediate does not have a double bond in the ring and so there is far less angle strain as compared to the one in alkynes. Hence they are more likely to undergo bromination i.e. to a far greater extent as compared to alkynes.

In alkynes, the bonding arrangement is linear (180 degrees) and as an advantage of this fact, they do not suffer the effect of steric hindrance as do alkene or more specifically bonding arrangements that contain only 1 set of Pi electrons. Also, electrophilic addition reaction mechanisms tend to initiate by forming an intermediate-complex with the Pi electrons being shared in the substrate, and as a result, the nucleophile has 2 sides of electrophilic attack in alkenes compared to alkenes in which there is only one. That being said, anti-arrangement will dominate in the reaction regardless due to the steric repulsion between the two Bromine atoms, however there are mechanisms and methods of achieving a greater percentage of syn products such as using a inorganic catalyst like Palladium.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

authors: Αντώνιος Κελεσίδης


text:
How is this Steampunk patina being applied?
Can anyone explain what is being deposited in this video @ 11:40?

It's Steampunk patina technique to make things look old and decrepit.

"The mixture consists of: 50 g of 30% vinegar, 20 g of salt and 50 g of water. I used a 5 volt and 2 amp power supply. Connect the positive terminal to the part, and the negative terminal is lowered into the water.  Time of blackening of the metal somewhere in 10 minutes"

And the cathode is stainless steel. It wipes off with a scourer. So specifically,

What is being deposited?
Is the deposit related directly to the cathode material?
What happens if I change the cathode to say normal steel /copper wire?

I'm not a chemist, so I won't understand anything clever or those molecular spider diagrams...

Looks like copper oxide to me. No, the cathode material is irrelevant.

I haven't found much in literature about this process, we can try to find the solution reasoning, but some experiments should helpful.
What is being deposited?
If he connects the positive terminal to the object it means that the object is the anode. So something is reduced on the anode.  Copper(II) oxide quite common oxide used for balckening. How can we obtain this by reduction? I would say that it could be the reduction of copper acetate to copper(II) oxide.
So where the copper acetate come from?
In fact, to the cathode there is also a copper wire wrapped around the steel, it could be that the copper ions are produced from that wire on the cathode:
$$\ce{Cu + 2H^{+} -> Cu++ + H2 }$$
Vinegard contains acetic acid and reacts with copper:
$$\ce{2CH3COO- +Cu++ -> Cu(CH3COO- )2}$$
Or directly from the acid attack of acetic acid:
$$\ce{2CH3COO- + 2H+ + Cu ->Cu(CH3COO- )2 +H2}$$
What could happen is the following (decribed here):
$$\ce{Cu(CH3COO- )2 +H2 -> (CH3COO- )2CuH2}$$
$$\ce{(CH3COO- )2CuH2 + Cu(CH3COO- )2 +H2O -> Cu2O +4CH3COOH}$$
Is the deposit related directly to the cathode material?
If this is true there is no relationship with the steel used at the cathode.
What happens if I change the cathode to say normal steel /copper wire?
The cathode may oxidize. If the copper wrapped around the steel has a part on the process, the process could not work if you don't use copper.
Do we not understand the role vinegar and some salt plays?
Salt is probably used to increase the conductivity of the solution.
The vinegard could be used for different reasons, but as  I said before it is probably used for producing copper acetate.
Why is this simple process open to debate and conjecture?
This is an excerpt from an article written at the New Zeland Institute of chemistry:

Electroplating is both an art and a science. Although based on several
  technologies and sciences, including chemistry, physics, chemical and
  electrical engineering, metallurgy, and perhaps others, it retains in
  some ways the aspects of an art, in which experience is the only
  teacher. In fact, of course, all the sciences have elements of art
  which can be learned only by experience; all the reading of textbooks
  on chemistry will not produce a chemist. No text on electroplating
  will produce an expert electroplater; there is no substitute for
  experience and what is somewhat inelegantly termed know-how.

Is not easy to predict the role of each component if the experiment is poorly described, there are many variables.

I've had a bounty expiration warning, but (there's always a but).  Why is this simple process open to debate and conjecture?  Do we not understand the role vinegar and some salt plays?  This guy isn't demonstrating cold fusion...


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Hydrogen evolution during thermal spray
Firstly, let me state that I'm no chemist. My specialty is in thermally applied coatings such as electric-arc metalizing.
During the course of my career I've always been presented with vague warnings to use "proper" ventilation to prevent hydrogen buildup within a contained area when spraying Al alloys.
Could someone please explain the process that creates hydrogen gas when arcing between two Al electrodes?
Also, what would the "proper" ventilation be? Would dry dust collection/ventilation suffice or would wet collection methods be required? Would this same reaction occur when using other alloys commonly used in thermal deposition processes?

The culprit in the undesired production of hydrogen gas from electric-arc metalizing is water vapor, as it is the only significant source of hydrogen atoms in the atmosphere. The concentration of water vapor is then of course a major factor in how much hydrogen gas can be produced. In very humid conditions the air can contain as much as a few percent water vapor. The chemical reaction for the conversion of water vapor to hydrogen gas under the conditions of this process is as follows:
$$\ce{H2O <=> H2 + 1/2O2}$$
A few percent water vapor may not seem like enough to produce explosive levels of hydrogen gas, but hydrogen gas becomes explosive in air at levels as low as $\pu{4\%}$.
A recirculating dust filtration system, dry or wet, is not an appropriate system for removing hydrogen gas. The only reliable way to prevent the possibility of hydrogen buildup in a closed area is with an appropriate fume hood (or similar approved device) that vents the gases out of the building.
I can't give an exhaustive list of the metals/alloys that could result in hydrogen production via this process, although base metals like copper, lead, nickel and zinc have the potential to do so.

Hmm. While I also like to err on the side of safety - hydrogen spreads out into a room extremely rapidly and mixes well with air and would be vented out. It would take quite a lot to make a significant (25% LEL) buildup in a normal room with normal ventilation. However, in an unventilated room - things will go boom.

@StianYttervik Yea, since the OP mentioned warnings about proper ventilation in his procedure, I figured I should address the issue anyway. In any kind of open space it shouldn't be a problem, but as you say, there is the potential for going boom in small unventilated areas ;)

Typically when I am working in a spray containment, I size my negative pressure collection systems to cycle the air 1-2 times per minute. Some of these systems can handle up to 100,000 CFM and for larger contained areas we would use multiple units to ventilate different zones. I'm not familiar with flow rates of fume hoods. How many full evacuations per minute are achieved by a basic fume hood?

There are a couple different basic kinds of fume hoods: those that cycle air through some kind of filter and thus don't actually move any air out of the area at all, and those that vent the air out of the area, typically up a stack to the outside. The 100,000 CFM hood sounds like the later, which is what you would need to remove hydrogen gas. The number of evacuations per minute is a function of the flow rate and the size of the room, but if you are working inside the hood that isn't a big issue with the small amount of hydrogen you will produce.


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Freezing point of oxygen-18 water
The freezing points of heavy water (3.8 °C) and tritiated water (4.49 °C) both seem to be well-known. I can't find anywhere that gives the freezing points of $\ce{H2^18O}$, $\ce{D2^18O}$, or $\ce{T2^18O}$, though. Are any of these values known? If not, is it at least known whether they'd be higher than with ordinary oxygen?

related https://chemistry.stackexchange.com/questions/79189/what-kind-of-water-would-oxygen-15-and-tritium-create

Probably slightly higher.

I also agree with @Mithoron. A decrease in the intermolecular distance due to a decrease in the amplitude of molecular vibrations when the light isotope is replaced by a heavy one very likely results in increased freezing temperature.

Increasing the mass of the O doesn't have near the effect on physical properties of water like MP as increasing the mass of H does.  I think those values would all be within a degree of their oxygen-16 analogs, and a quick look through Sigma-Aldrich seems to confirm this for your first 2 compounds anyway.

In this paper [1] from 1963, various properties of heavy-oxygen water are measured including the melting point of both $\ce{H2^{18}O}$ and $\ce{D2^{18}O}$.
The melting temperatures were measured at:
$$\ce{H2^{18}O}:\ T_\mathrm{mp}=\pu{0.28\pm0.02^\circ C}$$
$$\ce{D2^{18}O}:\ T_\mathrm{mp}=\pu{4.02\pm0.02^\circ C}$$
So, as some commenters mentioned, the melting (or freezing) point of the heavy-oxygen water is slightly larger when it is protium on the heavy-oxygen, and quite a bit larger when deuterium is present. This basically means that the liquid has become more structured on average, which is to say that the average lifetime of a hydrogen bond is slightly longer mostly due to an increase in the average strength of a hydrogen bond.
I would be very surprised if there has been a measurement of $\ce{T2^{18}O}$ because tritium is very rare and not very easy to isolate. Additionally, heavy-oxygen is also quite uncommon, so purifying the $\ce{T2^{18}O}$ to make any reliable measurements would likely be very difficult. I would be very willing to guess, however, that the melting temperature is $\ce{T2^{18}O}:\ T_\mathrm{mp}\approx\pu{4.7^\circ C}$ by just noting that the isotopic substitiution effects seem to be roughly additive.

Extra Credit: Trying to Interpret the Data:
As a note which is really not what you're asking but I think is really interesting, is why there is such a sharp increase in freezing temperature for deuterium (and hence why heavy-oxygen has these effects only on a smaller scale). Indeed, it is true that the $\ce{O-D}$ vibrational frequency in $\ce{D2O}$ is quite a bit smaller than that mode in $\ce{H2O}$, but it really isn't obvious that lowering this frequency should make $\ce{D2O}$ freeze at a higher temperature. For instance, all else being equal, a larger frequency in $\ce{H2O}$ would predict a shorter hydrogen-bond distance because the vibrationally averaged bond length will be longer (because the long bond lengths contribute more than the short ones because the potential is Morse-like). But, what I just said about the vibrationally averaged bond length did not allow the $\ce{O-O}$ distance to relax in $\ce{D2O}$, so if the $\ce{O-O}$ distance shrunk in $\ce{D2O}$, then we could say it's just the lower zero-point energy because then the shorter $\ce{O-O}$ distance would explain the stronger hydrogen-bonding we seem to be observing in $\ce{D2O}$. Well, very interestingly, the $\ce{O-O}$ distance in liquid $\ce{D2O}$ has been both theoretically and experimentally observed to increase slightly or stay exactly the same. See ref. [2] and experimental references therein. So both geometric parameters point towards water actually having a higher melting point even though this obviously isn't true.
What's the deal then? Well most of the answer seems to be that ordinarily we think of nuclei as being classical, i.e. the atom can be located precisely, but this type of reasoning is quite bad for hydrogen. The important point then is still that deuterium is heavier than hydrogen, though not because of the explicit contribution which the frequency lowering would have on the geometric parameters if everything else were the same. Rather, a hydrogen atom is much more spread out than a deuterium atom due to the fact that lighter atoms will essentially behave more quantum mechanically. In the sense of being more wavelike.
The fact that this delocalization effect for hydrogen is most important was demonstrated in ref. [3], where the authors find that the hydrogen-bond length cannot be directly related to the vibrational frequency in a one-to-one sort of way. This is due to a large sensitivity to the $\ce{OHO}$ angle. This then explains what we were missing up above because even though the average bond length in $\ce{H2O}$ is larger than the in $\ce{D2O}$, this sensitivity to the hydrogen-bonding angle greatly decreases the hydrogen-bonding strength in liquid $\ce{H2O}$ compared to $\ce{D2O}$ which does not delocalize as much and hence does not suffer from this angle problem.
As a final fun point, if one does a simulation of liquid $\ce{H2O}$ using classical nuclei, and then the same simulation is done using path integral methods to include nuclear quantum effects, the change in freezing point is a decrease of about $\pu{4^\circ C}$, which can only be due to this delocalization effect I describe.
I basically included all this to say that interpreting what this heavy-oxygen change really means can actually be pretty complicated. Whatever it is, there must be some structural change, and one would always expect isotopic substitution for a heavier isotope to make liquids more structured both from the frequency effects and the nuclear quantum delocalization.
Sorry if that was overkill :)

[1] Steckel, F., & Szapiro, S. (1963). Physical properties of heavy oxygen water. Part 1.—Density and thermal expansion. Transactions of the Faraday Society, 59, 331-343.
[2] Chen, B., Ivanov, I., Klein, M. L., & Parrinello, M. (2003). Hydrogen bonding in water. Physical Review Letters, 91(21), 215503.
[3] Rey, R., Møller, K. B., & Hynes, J. T. (2002). Hydrogen bond dynamics in water and ultrafast infrared spectroscopy. The Journal of Physical Chemistry A, 106(50), 11993-11996.

One could readily get T2-18O by reacting tritium gas with isotopically pure 18O. No isolation required. It is mainly a question of price vs desire...


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What is structure of lactose?

Please look at stereochemistry of anomeric carbon or carbon no.  1 of glucose unit.
Which among two is correct structure of lactose?
If, both are correct, then how?
Note : Image 1 is from Harper biochemistry

Related  How do we assign alpha/beta status in polysaccharides?

Both of the images are lactose. The first image is $\alpha$-lactose and the second is $\beta$-lactose. The difference between $\alpha$ and $\beta$ lactose is the orientation of the most right hydroxy (OH) group in the pictures.

Sorry, can you please focus on OH group at carbon 1..

When you say carbon 1, I assume you are refering to the carbon next the red "1". There is no OH group at this particular carbon. The four bonds to this carbons is  -O-, -O-, -C- and -H

Sorry again, anomeric carbon please.

@AnubhavGoel I edited my answer to include this difference

But both show beta glycosidic bonds?

@AnubhavGoel If I remember correctly what determines the difference between alpha and beta lactose is the form of the glucose. I.e. I think the galactose is always in beta form in lactose

Harper says lactose is O-α-D-galactopyranosyl-(1→4)-β-D-glucopyranose.


text:
Difference between cleaning products
I have a few cleaning products and after I looked on the ingredients, they feel pretty redundant:

Bath cleaner: Less than 5% anionic tensides, anionic tensides, scent.
Shower cleaner: Less than 5% anionic tensides, anionic tensides, scent.
Glass cleaner: Less than 5% anionic tensides, color components, helper components, scent.
Kitchen cleaner: Less than 5%: anionic tensides, scent, lime, linalool.

Others seem to have significantly different chemicals in them:

Bref Power Universal Cleaner: Less than 5% soap, non-ionic tensides, anionic tensides. Additionally: Scent (dutylphenyl methylpropional, citronellol, lime), preservative (sodiumhydroxymethylicinat)
Frosch pH-neutral cleaner: 5–15% anionic tensides, less than 5% amphoteric tensides, preservatives (sodium benzonate, benzyl alcohol), scent, color.

The “kitchen cleaner” has a distinct scent and cleans much better. When I get it on my hands, it feels really slippery, so perhaps it is alkaline? The “bath” and “shower” cleaner seem to be virtually the same except for the printing on the bottle.
I know that in order so solve grease with water, one needs to have a solvent that has both polar and Van der Vaals bonds. Ethanol would do, for instance. For limestone, one needs to have some acid, like citric acid (my chemistry teacher one said that he uses diluted hydrochloric acid, but that's another story).
Are those cleaning products really all that different? Is “anionic tensides” broad enough that there could be a significant difference between them such that using one or the other actually makes more sense?

I always wondered how we are easily forced to be redundant. I am not neglecting that there can be a lot of chemistry and fine effects in the detergents fields. But still, surely there is redundancy. A very similar case is that of cosmetic: cream for day, cream for night, cream for men, young men, woman, and so on. I think they invest a lot in scents and textures to drive the same chemicals as specifically oriented for a purpose. Think about dishes: lemon like smells are fine, but lavender might be not. The latter will be better suited for a bath gel or a bathroom cleaner....


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How can I detect contamination with acetic acid
how can I detect the contamination with acetic acid in a home brewed spirit? I fear the batch might be spoiled.

If you feel acetic acid smell, most likely it is in there. Detection of species based on smell is common, at least for species with a very typical odour. I do not say that one should go around smelling chemicals, but if s(he) feel things like ammonia or acetic acid, this is almost a positive test. You might further confirm acetate, in your case a drop of conc. Sulphuric acid should be fine. https://en.wikibooks.org/wiki/Inorganic_Chemistry/Qualitative_Analysis/Tests_for_anions#The_acetate_ion

@Alchimista would you mind changing your comment into an answer? Apparently you sufficiently answered the OP.

A preliminary consideration first: it is not a good lab practice to smell unknown substances and this shouldn't be done. With this in mind, is however possible to detect or get a hint of substances on the base of their physical characteristics as well as they particular colour or odour - at least sometimes this is unavoidable.
Obviously a sort of knowledge of the sample in handling is required. For instance, a smell of garlic would naturally indicate Allicin, however the same smell while burning a (in)organic mixture would rather point to As.
To answer your question directly: if you smell acetic acid out of a homemade brew, acetic acid is there. The first test for acetate ion is its displacement from the matrix by adding, say, hydrochloric acid. In your case acetic acid is already free and due to its volatility can be smelled and detected.
Given that you beverage is alcoholic, a nice confirmation test to try would be adding a drop (test tube scale) of conc. Sulphuric acid. Esterification should occur, and the smell should get a sweet note, like in nail enamel removers.
For this and other acetate tests look at
https://en.wikibooks.org/wiki/Inorganic_Chemistry/Qualitative_Analysis/Tests_for_anions#The_acetate_ion
By the way, you will find that smell is atop the list.


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Determining the acid dissociation constant change with temperature in sparkling water
I already have the different values of pH at the different temperatures, but what I'm trying to figure out is obtaining the acid dissociation constant using these pH values. Is there a way I can do that without having the initial and final concentrations of CO2 in the sparkling water or would I have to do an extra step?

Check out pKa table?


text:
Is bamboo based plastic less likely to leak chemicals?
Some plastics have been known to leak plasticizers (bisphenol and phthalates) into the food they contain, especially when heated.
Because such compounds are suspected of being endocrine disruptors it seems best to avoid eating hot food from a plastic container.
Recently bamboo bowls, cups and plates have become available. As far as I can tell these are made from some kind of composite material which contains bamboo fibers and something that glues these together.
The only information  is that it's: renewable, dishwasher safe, bio-degradable (the last two claims seem contradictory), food safe, but not suited to be microwaved.
Is it possible that these kitchen-wares are safer than other plastic bowls, plates and so on? Is it possible to even assess the risks?

While there has been plenty of noise about the risks of bisphenol-A (a monomer of the plastic not a plasticiser) and phthalates, there is little or no hard evidence they cause measurable harm. sure we could go back to glass (but that is not energy efficient and we are worried about global warming) or plant-based containers (but they will leak natural materials into your food which are just as likely to be harmful as man-made materials as possibly more so as they don't get tested as much).

Inspired by the other answer and the comment to my question, i have dug a little deeper. The tableware in the picture is sold by a company called Zuperzozial and made from bamboo and corn reinforced with melamine resin.
Many similar products are sold, almost always without a brand name and always without any specification of the actual material. So I will answer my question for this specific set of cups, there is no way to tell if the answer holds for 'bamboo' tableware in general.
Melamine from tableware may leak into the food, especially when heated, according to the FDA. It is ok to serve cold drinks in a melamine resin cup. The claim that these cups are micro-wave safe is not true.
As for the amount of BS of the rest of the marketing claims: bamboo and corn are renewable sources. But i'm not sure how degradable a resin is.

It is probably not "bamboo based". Most likely it is bamboo (saw dust) filled. That totally doesn't answer the question about what the plastic is that it's made from. The only restriction I can think of in using bamboo fiber is that if virtually all of the moisture isn't removed before its added to the mix, then the water will boil and you'll have real problems with bubbles in the plastic. So, you have to dry the bamboo. Degradation, rats, bugs, fungus, bacteria - where is it being dried? how long? under what conditions? No way for us to know. As far as I can see, unless you work for the manufacturer (or have access to their factory), the only thing you can rely on is brand name. Is it made by/for a reputable brandname? "food safe" has no "official" definition. It is marketing hype. My suspicions are raised when things like "renewable" are slapped on a label. That's 99% marketing BS. And "biodegradable" is a great idea but. A newspaper (remember those?) or a plastic bag rapidly degrades when exposed to the sun and rain. Put them underground and it will take centuries. That's just another meaningless (deceptive) marketing term. Like "natural" or "organic" or "pesticide free". And then there's "local"...we never seem to learn. My best advice is if it's a fad, avoid it. Of course, that's what I've been saying about lasik...

This answers my second question: safety of 'bamboo' kitchenware cannot be accurately assessed without more information. And nicely words the suspicions that claims about food-safety and bio-degadability are just "meaningless (deceptive) marketing" terms. +1


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Why can't you achieve the equilibrium exactly in the middle (K=1) when you add enough heat/energy?
My reasoning would be that given the high amount of energy added, both the activation energies of the reaction from left to right and right to left are excessively surpassed thus both directions are as easily achieved. Please correct my reasoning if this is wrong. Thanks!

If I borrow 1000$ from you, are the sums like 100$ excessively surpassed so that they have no meaning at all?

Both directions are easily achieved but one direction might still be much faster than the other...

If you'd ask about temperature, it'd be OK, but it not equals "heat/energy".

There are indeed reactions where the equilibrium constant passes through unity as a function of temperature.  The water-gas shift reaction
$\ce{CO + H2O <=> CO2 + H2}$
has an equilibrium constant that gradually decreases with increasing temperature (it's slightly exothermic as written), and the constant is close to $1$ at $\pu{800°C}$.


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Why mass spectra of 2-methylbutane do not show big peak at 71 coming from tertiary carbocation?
Mass spectra of 2-methylbutane show strong peak at 57 by loss of methyl radical to give secondary carbocation while a loss of hydrogen radical from 2-position would form tertiary carbocation with m/z = 71 peak.
Since rules suggest that it is the stability of carbocation that determines the fragmentation pattern.So this process must be more favorable process still we do not get base peak at 71 instead we get a small peak at 71.
Same pattern is observed in case of 2-methylpentane where M-1 peak is almost absent. Other compounds also do not favor loss of hydrogen radicals to give M-1 peak.
I had read about this somewhere that it had something to do with energy of overall process but not able to recall. Kindly help me to explain the reason behind this phenomenon.

Possible duplicate of Why is the molecular peak not observed in the spectrum of 2,2-dimethylbutane?

So tertiary carbocation gets fragmented further..plz do explain it further. My apology for not able to understand from linked question.

@MelanieShebel Not dupey enough, I think, nice from you though :)

Basically, you have 2 pathways - 1) losing H atom 2) losing CH3 radical. Both give you comparable carbocations.

I don't see how the linked thread answers the regioselectivity question raised here.

How do you know that losing a CH3 radical gives you a secondary carbocation?
Maybe it will perform a hydride shift and you land at the tertiary carbocation.
Always take this into account, because you can shoot yourself in the foot if you assume carbocations to be too stable. In EI, they rearrange all the time (e.g. benzylium <-> tropylium).
However, seeing that the ion with $m/z~57$ subsequently looses $\mathrm{\text{:}CH_2}$, it seems reasonable to assume that the secondary carbocation is formed here.
The important point is the following: You propose that instead of loosing the methylradical, yielding a secondary carbocation (or a tertiary, considering the argument from above), there should be a loss of a hydrogen radical.
However, the hydrogen radical is far less stable than the methyl radical, and I assume this is the reason for the regioselectivity, here.
Also keep in mind that you supply a lot of energy (70 eV). Hence, the induced fragmentation processes are highly endothermic, and should be under kinetic control.
If you really want to find out what is going on, you should probably have a look at McLafferty's/Turecek's classic textbook "Interpretation of Mass Spectra".

Come think of it, both homolytic bond breaks are effectively a $sp^3 \mapsto sp^2$ transition. For the secondary carbocation, only two hydrogens have to move in the plane to yield the $sp^2$ center. For the tertiary carbocation, all three neighbouring carbon atoms have to move into the plane of the new $sp^2$ center. This makes it very plausible that the carbocation is kinetically favored.

Thanks a lot. I think unstability of hydrogen radical is behind this apart from planarity..


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Why is methyl α-D-glucopyranoside preferentially formed from glucopyranose in acidic methanol?
According to my textbook Organic Chemistry 2e by David Klein, the formation of a glycoside from glucopyranose favors the α form (see picture below).
Why does this happen? Shouldn't the equilibrium favor the β form, where the methoxy group is in an equatorial position and has less steric interactions?

In general, if there is overlap between a filled orbital and an unfilled orbital, a stabilising interaction will take place between the two. A simple example is the resonance present in esters, where the ester oxygen lone pair (a filled orbital) overlaps with the C=O π* orbital (an unfilled orbital):

In this context, anomer refers to a diastereomer differing only in the configuration at C-1. So, the two diastereomeric products you have drawn above are often called the α and β anomers.
In the α anomer, there is one lone pair on the ring oxygen (a filled orbital) which can overlap with the C–OMe σ* orbital (an unfilled orbital):

This leads to a stabilisation of roughly 6 to 7 kJ/mol, and is called the anomeric effect.
In the β anomer, neither of the two lone pairs on the ring oxygen overlap well with the C–OMe σ*.

Of course, steric effects would disfavour the α anomer, which possesses an axial methoxy group. However, this factor only leads to a destabilisation of roughly 3 to 4 kJ/mol, and is therefore outweighed by the stereoelectronic stabilisation provided by the anomeric effect.
The rough numbers are taken from Kirby, A. J. Stereoelectronic Effects (OUP), p 16.

While the lone pair -> σ* explanation is the most commonly invoked, there are also likely to be other factors that are also present: for example, in the α anomer, the dipole moments of the ring oxygen lone pair and the exocyclic C–O bond oppose each other, which is a favourable electrostatic interaction:

Good explanation of the anomeric effect. However, the question should also be ‘… and why is the methyl glucopyranoside different from simple D-glucose (which prefers beta in solution)?’

why is the anomeric effect not operative in case of glucose?

@Infinite That's a different question, ask it as a new question please; and I don't know the answer anyway.


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How can a Hydrogen make such a difference?
Looking at familiar bodily fluids for a learning session. At first glance they both look the same. Then I noticed the N=C in biliverdin and HN-C in bilirubin. How can a N=C (imine group) and HN-C (secondary amine) make such a difference in colour? They look like structural isomers of each other to me. Or is there any stereoisomeric differences as well to this colour functional group change?
(Jaundice is directly related to bilirubin, urine is a product of bilirubin, I reckon there must be a real good reason and help aid from the colour change representation.)
e.g.
image 1: bilirubin is a brown pigmented substance while biliverdin is green pigment

image2

That's actually two hydrogens. The compound on the right is an oxidized form of the compound on the left.

@Zhe are you referring to first image? I don't see two hydrogens :@@ Bilirubin (brown) oxidised to create urobilinogen, then to urobilin for what we see in urine (yellow).

There are indeed two hydrogen atoms. There's one on the nitrogen and one that's implicitly associated with the missing double bond.

Actually there is little crucial on H yes or not. The entire pi system is changed in between.

Check how the conjugation is altered.

@Zhe so does this conjugation (resonance) result in the two molecules? Does it mean the interchange from -verdin to -rubin is easy and quick? And why would it takes long to reverse the process of someone having more bilirubin the body?

Not sure about that. The ease of oxidation/reduction is dependent on a bunch of factors that I can't claim to understand just by looking at the pair of the molecules.

The difference in structure of these two compounds is  perhaps the presence of a more conjugated structure in biliverdin compared to bilirubin. These products (degradation products of haeme1), just as most pigments, make use of the conjugated electron systems to absorb visible light, giving rise to strong colors. The long chain conjugated system in bilverdin leads to its strong green color, the reduced biliribin (methenyl bridge) has a less conjugated system with a different absorption consenquently different color.
This is a case of effects of conjugation in pigments than merely a difference in functional groups as I see it.
1Note The degradation of haeme occurs at  the end of the red cells’  lifetime, where they are removed from the circulation and their components degraded, thus depends of many factors most importantly regulations carried out in the body, enzymes etc.
References

Voet and Voet Biochemistry Section 26-4. Amino Acids as Biosynthetic Precursors.
Conjugated system

I am terrible at replying, I took a look at your reply then, but usually take a long time to digest it, when not sure.


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Was Wöhler’s urea synthesis carried out with or without oxygen?
I learnt and was very impressed by the urea synthesis by Wöhler in 1828. The reaction was the heating of ammonium cyanate. One of my textbooks in the library said that ammonium cyanate was heated ‘in the absence of oxygen’.
But when I tried reading other organic chemistry textbooks, they also did mention the synthesis but never talk about whether it was anaerobic. Could anyone suggest a reliable source which clearly state about this matter?

Wöhler’s synthesis of urea is not from 1828 but from 1825. We’ll get to that in a second.
In cases like these, one should always rely on the original publications most. Unfortunately, publications of that era did not follow the somewhat stricter standards of experimental sections that nowadays’ publications do. Still, one can make very educated guesses based on the presence (or absence) of statements. The relevant quotation from the 1828 publication, in which Wöhler notes that the substance he created is indeed urea reads:[1]

Ich habe schon früher angegeben, daſs man die oben erwähnte krystallisirte weiſse Substanz am besten erhält, wenn man cyansaures Silberoxyd durch Salmiak-Auflösung, oder cyansaures Bleioxyd durch flüssiges Ammoniak zersetzt.

Roughly translated:

I have previously disclosed that the best method of obtaining the above-mentioned white crystalline substance is by decomposing cyanacidic silver oxide[2] with salmiak or cyanacidic lead oxide with liquid ammonia.

It took a while to locate the previous reference which — in accordance with common custom at the time — was given as ‘In einer früheren kleinen Notiz, die in dem III. Bande dieser Annalen abgedruckt ist, …’ (‘In a previous little note that was printed in the IIIrd volume of these annals …’). In the end, I located the most probable candidate to be the artice Ueber Cyan-Verbindungen (On cyan compounds) also published in the Annalen der Physik.[3] Therein, Wöhler reports two syntheses of a novel crystalline compound; the second one being the interesting one for us:

Rein erhält man [die eigenthümliche kryſtalliſierte Materie], wenn cyanſaures Blei durch kauſtiſches Ammoniak, oder cyanſaures Silber durch Salmiak zerſetzt wird. Sie kryſtalliſirt in weiſsen durchſichtigen, ſtrahligen Kryſtallen, iſt leicht in Waſſer und Alkohol auflöslich.

Translated roughly:

[The weird krystalline material] can be acquired in pure form if cyanacid lead is decomposed with caustic ammonia or cyanacidic silver with salmiak. It crystallises in white transparent spear-like crystals and is readily dissolved by water and ammonia.

Wöhler then goes on to describe how different substances interact (or rather: don’t) with what he newly synthesised as was common in the days.
Here, we cannot really make any educated guesses on the reaction procedure by what is written because the fragments are extremely short. However, we can make educated guesses by what is not written. Wöhler made no reference to using sealed glassware or any specific atmosphere. He does not not any specific purification of the reactants. And he does not indicate any special care that may be required. Therefore, it is reasonable to assume that all reactions were performed at ambient pressure in the presence of oxygen and water (although water content will have been neglectable in the case of liquid ammonia).

References and Note:
[1]: F. Wöhler, Ann. Phys. 1828, 88, 253–256. DOI: 10.1002/andp.18280880206.
[2]: My note: Wöhler (incorrectly) assumed the acid anion to be $\ce{(CN)2O}$ which meant that the other constituent must have been silver(I) oxide giving $\ce{[Ag2O][(CN)2O]}$ — which gives the correct sum of $\ce{Ag(OCN)}$.
[3]: F. Wöhler, Ann. Phys. 1825, 79, 177–182. DOI: 10.1002/andp.18250790208.