text:
Synthesing tetrahydropyran from ethylene oxide

I thought that the first step must be using a Grignard reagent to add carbons, but I need a hydroxyl group on the carbon on the other side to make it a diol and produce the product given, but nothing seems to be working.
How do I proceed?

What if you used Grigniard reagent with chlorine present...

Think about how many carbons you need to introduce. In addition, you need a some kind of functional group handle so that you have something to work with.

it would solve it, so basically what can my R chain contain in gringard, i thought only hydrocarbons, it can contain halogens then?

Yes, halogen (fluorine in particular) may be present, but it may complicate preparation of the reagent.

Alternatively, you can incorporate some unsaturation into your Grignard reagent, or a different functional group that's protected.

Prepare the tetrahydropyran ether of commercially available 3-bromopropanol using dihydropyran, catalytic PTSA in DCM.

Make the Grignard of the THP-protected bromopropanol, react this with ethylene oxide. Work up and isolate to give the mono-THP-pentane-1,5-diol.

Make the tosylate or mesylate of the mono-THP-pentane-1,5-diol.

Remove the THP group using cat. PTSA in MeOH

Form the alkoxide using LiHMDS in Et2O and cyclisation will occur to give the pyran. Isolate by careful distillation.

For preparation of the Grignard reagent, use allylmagnesium bromide.
$$\ce{CH2=CH-CH2-Br ->[Mg][ether] CH2=CH-CH2-Mg+Br-}$$
Now we react the Grignard reagent formed with ethylene oxide, our starting compound.
$$\ce{C2H4O ->[CH2=CH-CH2-Mg+Br-][Et2O] HO-CH2-CH2-CH2-CH=CH2}$$
This gives us pent-4-en-1-ol. Now we protect the alcoholic group by esterification, using trifluoroacetic anhydride ($\ce{(CF3CO)2O }$).
$$\ce{HO-CH2-CH2-CH2-CH=CH2 ->[(CF3CO)2O][Pyridine] CF3-COO-CH2-CH2-CH2-CH=CH2}$$
Now, we use $\ce{HBr/R2O2}$ to form 5-bromopentyl 2,2,2-trifluoroacetate.
$$\ce{CF3COO-CH2-CH2-CH2-CH=CH2 ->[HBr/R2O2] CF3COO-CH2-CH2-CH2-CH2-CH2-Br}$$
Now we remove the trifluoroacetate protecting group to get 5-bromopentanol.
$$\ce{CF3COO-CH2-CH2-CH2-CH2-CH2-Br ->[K2CO3/MeOH] HO-CH2-CH2-CH2-CH2-CH2-Br}$$
Now, under basic conditions, the compound undergoes intramolecular SN2 to form the six member cyclic ether
$$\ce{HO-CH2-CH2-CH2-CH2-CH2-Br ->[OH-] C6H10O}$$

Let us continue this discussion in chat.

Actually, I'm not sure if you might actually just get spontaneous cyclisation during the deprotection. Would be a nice bonus.

That should work

Just summarizing the steps already written in the comments,

One equivalent magnesium will react with iodine preferentially due to it being a better leaving group than chlorine. The bond also has more covalent character.
Subsequent step of reaction will open the oxirane ring easily.
Then hydrolysis produces 5-chloropentan-1-ol.
This is followed by intramolecular SN2 attack by lone pair on oxygen of hydroxide at the chlorine carbon causing cyclized product formation.

I would suggest that instead of I, you might want to use Br. It is just a thought since Br would still satisfy the conditions and it seems to be more prevalently used.

How does the magnesium not react with chlorine?  Can we use $\ce{SCH3}$ at the far end instead?

Do you have a reference for the preparation and use of 3-halomagnesium halides?

@Waylander https://pubs.acs.org/doi/pdfplus/10.1021/acs.jchemed.5b00453?src=recsys Not exactly for the above compound, but preferential reaction is observed.

That is a 1,4 dihalo aryl ring.  So it doesn't exactly equate to what you have done..

@Safdar I couldn't find an article with exactly 3-halomagnesium reagents.

I think the route falls over at that point. AFAIK you cannot form a Grignard of 3-halo propyl halides and get a standard addition. There is scope for all kinds of alternative reactions - cyclopropane formation possibly.

@Waylander And wouldn't transmetalation of bromofluoropropane work to make such a Grignard reagent?

@Mithoron  Still full of problems plus alkyl -F is a very poor leaving group for the cyclisation.


text:
What is the mechanism for this intramolecular SN1 reaction?

What is the mechanism for the following intramolecular $\mathrm{S_N1}$ reaction?

For this question, my professor hinted at an intramolecular reaction, and he asked us to recall the bromonium ion as well. As such, I believe the $\ce{SMe}$ group attacks the open carbocation when $\ce{OTs}$ leaves. I think I understand how the left-side enantiomer is formed: $\ce{SMe}$ attacks the open carbocation, then water attacks from the back because the front-side is occupied by the hydrogen (not shown). However, what is the mechanism for the right-side enantiomer? How does the $\ce{SMe}$ end up in front (on a wedge) along with water after hydrolysis?

Do you know how mustard gas works?

I do not know how mustard gas works.

Take a look here: https://www.ncbi.nlm.nih.gov/books/NBK236079/

Right, what I gathered briefly from that reading was a cyclic intermediate is involved, but how does the SMe group in my diagram end up on a wedge (in front) in the right-side enantiomer?

Draw the cyclic intermediate - include the hydrogens that are not shown in the diagram in the question - then look at the possible points of attack of water.

Flip your second product 180 degrees about the horizontal axis and the offending solid wedges will become dashed wedges.

You are correct on your initial assumption. This hydrolysis is assisted by the neighboring thiomethoxy group ($\ce{SCH3}$) as depicted in the following scheme:

The cyclic methylsulfonium intermediate, which is achiral in this case, is a mimic of bromonium ion in bromination of alkenes (also see Chemistry of Sulfur Mustard on releasing chlorine through cyclic alkylsulfonium ion). Realistically, this intermediate is mimic of bromonium ion of of cis-2-butene (also achiral), and hence would give a racemic mixture. Similar to bromonium ion intermediate, the upcoming nucleophile ($\ce{H2O}$) can attack at two places: through path $a$ or through path $b$. Keep track on the (R,S)-configurations of two stereo centers. The original compound has (S,S)-configuration. The (S)-configuration on carbon attached to $\ce{SCH3}$ is unchanged but carbon attached to $\ce{OTs}$ has been changed to (S)-configuration in cyclic intermediate, because it is necessary to have an anti-periplanar conformation for attacking group ($\ce{SCH3}$) and leaving group ($\ce{OTs}$), similar to $\mathrm{E2}$ mechanism.
During the path $a$ mechanism, $\ce{H2O}$ attack the carbon where $\ce{OTs}$ originally was. Therefore, the carbon bearing $\ce{SCH3}$ group does not change its (S)-configuration. However, the $\ce{H2O}$ attack happens in the opposite face of $\ce{SCH3}$ group, the (S)-configuration retain in the carbon bearing $\ce{OH}$ (as in the original compound).
However, both stereochemistry would be reversed during the path $b$ mechanism (see the scheme). Accordingly, $\ce{H2O}$ attack the carbon where $\ce{SCH3}$ originally was, which has (S)-configuration. Since the $\ce{H2O}$ attack happens in the opposite face of $\ce{SCH3}$ group, the (S)-configuration would be changed to the (R)-configuration in this carbon, now bearing $\ce{OH}$. (as in the original compound). The second carbon, now bearing $\ce{SCH3}$ group, would not change its (R)-configuration (as in the cyclic intermediate).
As a result, you get racemic mixture.

I think this answer could be slightly improved if you noted somewhere that the cyclic intermediate is achiral and thus, the product must be either achiral or a racemic mixture.

@  Zhe: Very good point. I edited as suggested. I appreciate your careful reading.

No problem. I already upvoted before my comment, since it was a good answer.

@user55119: I can't see why you are not able to post your answer. I still see "Add Another Answer" button so it is still available. The question has received two close votes already, so you may try to post your answer before it get closed. Good luck.

It should be, because there are no other way to answer this question. :-)

Time to remove this conversation.

The key, as emphasized in Zhe'a comment to Mathew's answer, is that the cyclic  intermediate is achiral. This ensures a perfect racemic product mixture ("erasing" the stereochemical information, i.e. you could have started with the other enantiomer as a reactant and obtained the same product mixture).
Here is a 3D rendering of the stereochemistry of the carbon atoms in the intermediate (ignore the geometry around the sulfur):

In the orientation shown, oxygen would attack from the bottom while the methyl sulfide group would remain on the top of the cisoid conformation. In one case, you have a double displacement with retention of configuration, and in the other case single displacements on both carbon centers, inverting both:

Rotating around the central carbon-carbon bond would give the product conformations as shown.

source: Stack Exchange

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

sort: votes


text:
How do you know what elements will bond in a reaction?
I know all about the types of reactions, synthesis, decomp. etc., but when a bond is broken, how do you know that the free element will bond to another molecule?
Is it because that element has a higher electronegativity? Is it because the product is lower energy?
Why does the free element choose a specific element to bond and how do I know what element it'll bond with?

There is no pat answer to your very broad questions. As a chemist you either have the knowledge, or you often can often make an educated guess.

You know that by applying a few simple rules, which are not really simple and not really few. In fact, there are entire books devoted to this subject. They are typically titled "Chemistry" on the top cover.

Haha wish it was that easy :( .But then if it was then Chemistry would probably  not even exist.

Why does the free element choose a specific element to bond and how do
I know what element it'll bond with?

Science has more answers related to "how" rather than "why"? Can we answer why do you have five fingers and two eyes? Not as yet, but scientists can talk a little bit how the eye works.
Similarly, your open ended query depends on what is meant by "bond"? These are perhaps the mysterious forces of nature. Conventional textbooks teach that noble gases are nonreactive with a few exceptions, but someone who is doing microwave spectroscopy knows that Argon gas forms "clusters" with many many organic molecules in the gas phase i.e., this gas is too reactive for a spectroscopist who is interested in studying the behavior of single organic molecules.

To add an analogy: imagine you have a friend who married a jerk, and you wonder why. After all, there are so many people in the world, nearly $10^{10}$ at the latest count! How do people decide who to pair up with? Well, in that case it might have been due to a chance encounter at a mixer at which both got slightly tipsy (and believe it or not he happened to be the nicest guy there).
The analogy is that atoms don't choose. They usually just mix randomly, sometimes slowly, sometimes at extremely high speeds. There is sometimes no chance to evaluate all possible matches, it's the first atom you encounter you end up with. At times that one is a rare catch, other times it's one like any other. And sometimes an atom dumps a weakly-bound jerk when it later meets another more attractive atom.

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


text:
Parallel between polarisability of donor atom and CFS strength
Observing the strength of crystal field splitting by different ligands, can the following be considered?:

Most of the members in the series are ligands where the donor atom is nitrogen, oxygen or carbon.
If we consider only those can a parallel bases on the HSAB principle be drawn i.e. carbon which is softer and more polarisable than nitrogen generally is the donor atom in strong field ligands like CO or CN-. Similarly, it is followed by ligands having nitrogen, which is comparatively softer than oxygen but harder than carbon.
It may be argued that the halogens like iodine and bromine which are very soft are at the end but then they are down the group and also in different periods.
Although the series is dominated by C, N and O donor atoms, they are in the same period. Similarly, PPh3 where phosphorus is the donor atom is a stronger field ligand than the comparatively harder sulpur atom (in ligands like SCN- or S2-) in the same period.

Can such a parallel be drawn?

No, the spectrochemical series is far more complex than just HSAB theory.


text:
How to obtain optically pure mixture from racemic mixture
I recently studied stereo chemistry in school.All i know is that compound are optically active if they have no plane of symmetry and about the nomenclature we follow nomenclature.......
My questions

What is the general principle behind obtaining a optically pure mixture?

Theoretically, Sn1 reaction gives raecmic mixture, more invertion occurs depending upon the time when the carbocation is attacked by the strong base. So refluxing and HEATING a tert butyl alcohol with HI (to avoid elimination) may induce partial raecmization by Sn1. See A guidebook of Organic Mechanism by Peter Sykes. Nucleophilic Substitution reaction.

Welcome to Chemistry SE! It is preferred if you can post separate questions instead of combining your questions into one. That way, it helps the people answering your question and also others hunting for at least one of your questions. Thanks!

What is the general principle behind obtaining a optically pure mixture? (correct it as optically pure compound)

You have too many separate questions. In short, organic chemists would tell you that there is full fledged branch of synthetic chemistry in which only one enantiomer is produced rather than a racemate. Read more about enantioselective synthesis, also called asymmetric synthesis on Wikipedia.
Sometimes, it is not possible to produce a pure enantiomer in a reaction. Then chiral chromatography comes to rescue. You can separate enantiomers using a special type chromatography called chiral chromatography. Pharmaceutical companies routinely use this approach and more commonly, chiral supercritical fluid chromatography.

Don't discount selective crystallisation

Right, but in pharma research, prep chiral SFC/HPLC is heavily used-almost exclusively for obtaining single enantiomer.

yes, but in scale-up and production crystallisation is still very important

Thanks, scaling aspect is very important.


text:
Raman or FTIR spectroscopy to identify metals?
Metals are opaque, so doesn't most of the light reflect? And does the IR source even cause vibrations in the metal?
Same goes for Raman spectroscopy, is it useful for use of metal identification?
Is it better to use EDS to identify unknown metals?

Well, you obviously cannot measure in absorption, unless you have a very thin film. ATR will of course work. Spoiler: The spectra will look terribly boring, and you won´t be able to differentiate much.

I guess it is better to use XRD to identify metals, instead of raman or FTIR.


text:
Can you measure an IR spectrum of a sample inside a glass vial?
If I have a sample of some compounds dissolved in water inside a glass vial (let's say quartz). Would it be possible to get a usable measurement through this glass vial without too much absorption by the glass itself?

Maybe Raman: https://www.spiedigitallibrary.org/conference-proceedings-of-spie/9995/999506/New-capability-for-hazardous-materials-ID-within-sealed-containers-using/10.1117/12.2241540.short?SSO=1

Why don´t you measure the empty vial?

Thank you for your answer. Let me try to rephrase my question. I HAVE to analyze a glass vial containing an aqueous solution. I am also using FT-Raman, but I was curious if it would make sense or not to use IR or FT-IR to analyze the content in the vial without opening it. Or if the glass or water would overlap the signals from the sample. I would like to conclude wether or not it contains e.g. m-cresol.

You can use ft Raman for your purpose. It works well. If you want to use IR,use AgCl liquid cell set up .you can use aq.solution.

As EdV said, when you have aqueous samples, Raman spectroscopy is the way to go. The problem is not the container (which should not be glass etc),the problem is water. Its absorption will hide the absorption of everything.
If you ever open an FTIR instruments, you will see that there is nothing which is made of glass or quartz because they absorb too strongly in the useful IR range.

Thank you for your answer. Let me try to rephrase my question. I HAVE to analyze a glass vial containing an aqueous solution. I am also using FT-Raman, but I was curious if it would make sense or not to use IR or FT-IR to analyze the content in the vial without opening it. Or if the glass or water would overlap the signals from the sample. I would like to conclude wether or not it contains e.g. m-cresol.

If there is water and the container is made of glass, the absorption by these two components will be so strong that IR beam may not even cross the sample. You would have 0% transmittance and infinite absorbance.

In short, there is no point in doing FTIR.

Thank you very much for your replies. They are very appreciated.
Just a bonus question; Would FT-Raman be usable? I only have access to FT-Raman equipment and not regular Raman. Would it be sensible to measure the samples through the glass vial?

FT-Raman is a very fancy instrument, it will work, why not? What you have to do it is to run the same container with water filled in and subtract this from the sample signal.

Great, thank you. I will look forward to see the results from my FT-Raman measurements. Have a nice day.


text:
Why does the solubility of haloalkanes in water decreases as the length of the carbon chain increases?
Due to the polar halogen-carbon bond in haloalkanes, they are more soluble than hydrocarbon in water. But the solubility is still very low because the overall influence of the dipole-dipole interaction is small.
Cool.
But why does the influence of the halogen decreases as the length of the carbon chain increases?
Shouldn't this be increasing the solubility of haloalkanes in water instead? As an increase of carbon in the chain translate to more electrons and more movements of electrons increase the dispersion forces. This increase in inter-molecular forces means that its closer to of the strength of hydrogen bonding in water

The hydrophobic chain becomes larger so the hydrophobicity increases, whereas the hydrophilicity remains similar. So it becomes less soluble.

As Safdar said the increases size of the hydrocarbon in length or branching causes increase in hydrophobic character. Moreover, the C-X bond is not polar enough to form hydrogen bond with water. Only dipole forces exist between molecules of  haloalkanes, not hydrogen bonds.

Because the chain has hydrophobic character. More hydrophobic character decreases the solublity

Hydrogen bonds do exist between chlorine atom and hydrogen. But these bonds are rather weak, with respect to usual hydrogen bonds with N, O or F.

@Maurice Hydrogen bonds may exist between Cl and H, but for that to be efficient doesn't the chlorine also require a high partial negative charge on it. Like HF is associated with H bonds because of the high polarity of the H-F bond itself. Such a high charge seperation cannot from the relatively small polarity of the C-Cl bond. I doubt if the C-F bond will be polar enough, though for hydrogen bonding.

@ADITYAROYCHOWDHURY Hydrogen bonding mainly needs a free electron pair on one side and a sufficiently polarised covalent H-X bond on the other. Of course without a large negative partial charge on the first atom, this hydrogen bond is not really worth mentioning for most practical uses. The IUPAC definition is an unusually smart one, because it additionally demands that there must be "evidence of bond formation". When in doubt, you can always say "no H-bond". ;-) Whoever disagrees has to produce the evidence himself.


text:
Aspartic acid conjugate base acidity
Question

Which hydrogen among the four in the conjugate base of aspartic acid is most acidic?

Answer

 No answer given

My attempt
I am confused between Hb and Hc
I know Hd has $\mathrm pK_\mathrm a \approx  9$, so it will not be the most acidic.
$\ce{NH3+}$ has very high -I effect but no mesomeric effect, so Hc appears to be quite acidic.
But at the same time it is at the alpha position of the carboxylic acid and may not be effectively conjugated in the $\ce{C=O}$ bond.
Here can the electron pair if formed by loss of Hc be conjugated on the $\ce{C=O}$ bond? This will prevent the lone pair on oxygen from being delocalized, however, something that is not characteristic of carboxylic acids.
Hb may be acidic but no special effect may be attributed to it, similar to Ha.
I think the order will be Hb > Hc > Ha > Hd
Is my assumption correct?

A simple way to compare the $\mathrm pK_\mathrm a$ of $\ce{-NH3+}$ and $\ce{-COOH}$ in such a scenario would be to assume the situation when the amino acid becomes a zwitter-ion.
A zwitter-ion is a molecule that has both positive and negative charges on it. In an amino acid, the $\mathrm {pH}$ at which this is seen is known as the isoelectronic point.

Let's consider this molecule to have reached the isoelectronic point. We know that the value would be less than $7$ since there are two acidic groups (meaning that this is an acidic amino acid). When we reach that point, we see that the both the carboxyl groups are de-protonated and the amine group still exists as $\ce{-NH3+}$.
This means that the $\ce{-NH3+}$ group is less acidic than the the acid groups. Therefore, Hd is is less acidic than Ha or Hb.
As you further increase the $\mathrm{pH}$, we see that at a point, the $\ce{-NH3+}$ group becomes $\ce{-NH2}$, before Hc is removed. This means that Hd is more acidic than Hc.
Now, we need to compare the two acid groups. As you may have noticed, The $\ce{-NH3+}$ group is closer to Hb than Ha, since this group is a highly electron withdrawing group [-I effect], the negative charge formed by the de-protonation of Hb would be more effectively withdrawn making it more stable than the conjugate base formed by the de-protonation of Ha.
Therefore the order of stability would be:

Hb > Ha > Hd > Hc

I highlighted the some of your attempts:

$\ce{H_d}$ has $\mathrm pK_\mathrm{a} \approx  9$, so it will not be the most acidic.

This assumption is correct. For aspartic acid, $\mathrm pK_\mathrm{a}(\ce{H3N+}) = 9.6$.

$\ce{NH3+}$ has very high -I effect but no mesomeric effect, so $\ce{H_c}$ appears to be quite acidic. But at the same time it is at the alpha position of the carboxylic acid and may not be effectively conjugated in the $\ce{C=O}$ bond.

This assumptions are correct as well. The $\alpha$-hydrogens in acids and esters are not as acidic as those in aldehydes and ketones because carbonyl groups of them already involved woth attached $\ce{OH}$ or $\ce{OR}$ groups, respectively. However, $\alpha$-hydrogen in amino acids fairly acidic than those in aldehydes or ketones due to strong -I effect from $\ce{H3N+\!-}$ group.

Here can the electron pair if formed by loss of $\ce{H_c}$ be conjugated on the $\ce{C=O}$ bond?

Yes, that is the main reason it has being acidic (compared to $\beta$-hydrogen or other $\ce{C-H}$ in the same alkyl chain). However, it won't leave to make a carboanion until it has reached the correct $\mathrm{pH}$. The  $\mathrm pK_\mathrm{a}$ of $\alpha$-hydrogen in an $\alpha$-amino acid is in the range of $16$ to $17$ (Ref.1). The exception is been D-phenylglycine, $\ce{Ph-CH(NH2)-CO2H}$ $(\mathrm pK_\mathrm{a} =  14.9)$, which is not a natural essential amino acid. The reason for lower $(\mathrm pK_\mathrm{a}$ value is the additional resonance contribution from neighboring $\ce{Ph}$ group. The few examples are given below:
$$
\alpha\text{-Hydrogen $\mathrm pK_\mathrm{a}$ values (Ref.1)}\\
\begin{array}{c|ccc}
\hline
\text{Amino acid} & \mathrm pK_\mathrm{a} \\
\hline
\text{Alanine} & 16.5 \\
\text{Valine} & 17.0 \\
\text{Isoleucine} & 16.9 \\
\text{Leucine} & 16.7 \\
\text{Phenylalanine} & 16.2 \\
\text{D-Phynylglicine} & 14.9 \\
\hline
\end{array}
$$
Therefore, it is safe to assume that $\alpha$-hydrogen of acid is the least acidic among other acidic hydrogen in the molecule, even lower than that of $\ce{H3N+\!-}$ group.

$\ce{H_b}$ may be acidic but no special effect may be attributed to it, similar to $\ce{H_a}$.

This assumption is incorrect. As a matter of fact, $\ce{H_a}$ and $\ce{H_b}$ are the most acidic hydrogens among other acidic hydrogens in the molecule (this was already proved in the other answer, so that I'm not going to repeat it again). Between the two, $\ce{H_b}$ is more acidic than $\ce{H_a}$ since it is closer to $\ce{H3N+\!-}$ group and feel more -I effect (through less numbers of $\sigma$-bonds). The actual $\mathrm pK_\mathrm{a}$ of $\ce{H_a}$ and $\ce{H_b}$ are $3.65$ and $1.88$, respectively.
Thus, the correct order should be $\ce{H_b} \gt \ce{H_a} \gt \ce{H_d} \gt \ce{H_c}$.
Reference:

Eric D. Stroud, Dennis J. Fife, Grant Gill Smith, "A method for the determination of the $\mathrm pK_\mathrm{a}$ of the $\alpha$-hydrogen in amino acids using racemization and exchange studies," J. Org. Chem. 1983, 48(26), 5368–5369 (https://doi.org/10.1021/jo00174a045).


text:
A theoretical chemistry question about elements and their presence over time
I'm not a chemist myself, but I can't think of a better place to get some insight about elements and chemistry than from people who live and breathe it here on stackexchange. This is not any kind of homework problem just my personal curiosity, and I'd appreciate any input you might have on it.
Imagine we are travelling back in time towards the big bang. Universes being pulled back together, the density of energy is larger, atoms still have their neutral state but energy is colliding at a much higher frequency.
Is it possible that in these times, the amount of energy would permit atoms larger than we have observed in nature or been able to create in the lab?
If we then travelled to a far future, hundreds of millions of years from now, when the energy of the universe is much more spread out, and atoms have had more time to decay and lose electrons, wouldn't there be, at some point, a time where we no longer observe certain heavier elements in nature?
At either of these points in time wouldn't you expect to find more ionic elements in nature?

Of course, radioactive elements like uranium will slowly disappear. But you will have to wait much longer than the duration of the universe. Probably billions of billions of billions of years.

We don't call it exactly that, but well, basically yes. Any supernova would create a bunch of radioactive elements. Back in the past there were more of them, and in the future there will be fewer, then none at all, then all stars will fade and all radioactive elements will decay. Some say that even the protons will decay given enough time, but that's a question for Phys.SE.

" decay and lose electrons"? " find more ionic elements in nature"?  Man, I have a feeling you have too many weird ideas to fix in one post...

You have a lot of false belief in your question, and you base your question on these false beliefs. Some of them are close to be incomprehensible (like "universes being pulled back together", but there are around 5 of them in your question). Ask a much simpler question first, and then go step by step further, after you understood the answers.

The "primordial elements" were hydrogen, helium and a small amount of lithium. Heavy elements were only created in the first stars that went supernova.

You are a little confused about the nature of atoms and nucleosynthesis. The ionisation state of an atom has nothing to do with the creation of new atomic nuclei (or new elements). The relative abundance of elements over time is purely about nucleosynthesis and how it happens and nothing to do with ionisation.

Is it possible that in these times, the amount of energy would permit atoms larger than we have observed in nature or been able to create in the lab?

Quite possibly, but we do not need to go back in time. In fact, the early moments after the Big Bang are the least likely time to find very heavy elements, as the primary elements produced initially are believed to have been almost exclusively hydrogen and helium.
Since those early moments, the universe has certainly expanded, but there are still places where energy and mass are highly concentrated, such as in supernovas and black holes. Whether or not very heavy elements are transiently created in these places is an open question.
The problem is that elements heavier than uranium tend to be unstable, so they break apart (nuclear fission) very quickly into lighter elements. In order to make very heavy elements, you need to combine two less heavy elements (nuclear fusion), but it's hard to get two heavy atoms together to fuse if they keep falling apart themselves. A good discussion of the possibility of very heavy element creation in stars is
Goriely, S. and Martinez Pinedo, G. (2015) The Production of transuranium elements by the r-process nucleosynthesis. Nuclear Physics A 944:158. DOI:10.1016/j.nuclphysa.2015.07.020

If we then travelled to a far future, hundreds of millions of years from now, when the energy of the universe is much more spread out, and atoms have had more time to decay and lose electrons, wouldn't there be, at some point, a time where we no longer observe certain heavier elements in nature?

The answer to this question relates to the first answer. Since heavy atoms are constantly being synthesized within stars, the elements that are decaying get replaced. But as the number of stars decreases over time, the rate of synthesis will necessarily decrease as well, and elements that are susceptible to fission will slowly decrease in abundance. The time scale for this, however, is much longer than even hundreds of millions of years. Our Sun, for example, is expected to last about another 4.5 billion years, and the final dying out of all of the stars in the universe is expected to take somewhere between 1 and 100 trilliion years. I encourage you to read up on ideas about the "heat death" of the universe if this is a topic you are interested in.

At either of these points in time wouldn't you expect to find more ionic elements in nature?

Since it takes energy to separate opposite charges from one another, ionization happens with high energy, so places like stars where there is a lot of energy will have a lot of ionic matter. Likewise, when high energy emissions from those places interact with other matter (like cosmic rays hitting the Earth's atmosphere), ionization will take place. So you're right that when the universe's energy is more concentrated, there were probably more ions.
In the distant future, the opposite is true. Since the universe decays to a low energy state, and ionization takes energy, we expect that oppositely charged matter will combine to make neutral atoms and molecules as the universe gets older. Some have theorized, however, that the expansion of the universe will mean that matter is so spread out that oppositely charged ions get too far away from each other to interact, so it is conceivable that the final state of the universe is separated ions, but that is just speculation.
And as noted in the comments, you may get a much better answer over at physics.SE.

I'm not so sure OP would get this better answer - you did well for an answer for such a bad question. Another thing is cross-posting is frown upon.


text:
Name of this bond-line formula?
I don’t know much about chemistry but I recently found the below bond-line structural formula painted on a wall near my place. I really can’t let go of trying to find the name of the compound. If anyone would be so inclined as to relieve me of my suffering, I sure would be grateful!
Thank you and greetings from Sweden!

The full name of the compound is 3,4-methylenedioxymethamphetamine, which is typically abbreviated as "MDMA". It is commonly used recreationally, and is casually referred to as "ecstasy", "molly", "E" and many other names.


text:
Do reactions necessarily need to be redox in order to occur?
When l asked my teacher why the reaction between $\ce{S2O8^2-}$ and $\ce{Co^3+}$ did not occur, he said both were oxidising agents so their reaction would not be redox and thus the reaction would not occur. So am l right to say then that all reactions that are not redox will not occur ?

no, e.g. common neutralization reactions are not redox

No!! $\ce{}$

You say the reaction, but which one? You need to propose products and stochiometry, then one can say yes or no.

I think some of the responses are not getting to the crux of the problem. The issue is not about whether other reaction types can occur. There are several classes of reactions, of course, such as redox reactions (which involve changes in oxidation state), acid-base reactions (exact definition depends on how you define acid and base, but generally these don't involve oxidation state changes), and more.
I would also say that, as quoted, this explanation provided to you is exceedingly poorly phrased. So let's explain what's the matter.
The issue is that for any redox-type reaction to occur, you need a reducing agent and an oxidising agent. Changes in oxidation states are essentially additions or subtractions of electrons. If an atom gains electrons, its oxidation state is lowered, and vice versa. But in a chemical reaction, you cannot spontaneously create electrons out of thin air. So for atom X to gain electrons, those electrons have to come from somewhere else: in other words, there has to be a simultaneous loss of electrons from atom Y. In this situation one would say that X is reduced and Y is oxidised.
One cannot have a reaction in which both X and Y are gaining electrons at the same time, because electrons can't be magically pulled out of a hat. Likewise, one cannot have a reaction in which both X and Y lose electrons at the same time, because those electrons would have nowhere to go to, and you can't just vanish them away into thin air.
Now, both $\ce{S2O8^2-}$ as well as $\ce{Co^3+}$ are species which have pretty high oxidation states. Consequently, they don't have any electrons to give away, or at the very least you'd need some serious firepower to get them to give up the electrons that they do have. This means that there's very little chance of any electron transfer occurring between them, as neither is keen on giving up their electrons to the other.
This doesn't preclude other types of reaction which don't involve electron transfer from occurring, of course. Ruling these out would require a whole new explanation, so is beyond the scope of this answer. But I get the feeling that that isn't the point of your question.

We cannot say something to be a truly oxidising or truly reducing agent. Although, some compounds can be observed to display solely reducing activity or solely oxidising property in many reactions, there may be reagents which cause them to reverse their activity.
Eg. Xenon fluorides are considered oxidising agents in majority of reactions. It can oxidise bromate ion to perbromate, $\ce{Ce(III)}$ to $\ce{Ce(IV)}$, $\ce{Fe^2+}$ to $\ce{Fe^3+}$ etc. But it would be incorrect to name it as an absolute oxidising agent. In the reaction,
$\ce{O2F2 + XeF4 -> XeF6 + O2}$
The $\ce{XeF4}$ is a reducing agent, while $\ce{O2F2}$ is the oxidizing agent. Thus the attributes interchange.
And there are several reactions which are not redox. Neutralization, as in the comment is an example.

Then why does the reaction between S2O8 2− and Co 3+ not occur?

Nobody knows why a reaction does not happen. Why does charcoal not react with oxygen at room temperature ? Why does potassium permanganate not react with water at room temperature ? It is probably a question of activation energy. That's what we say when we don't know the answer.

@Rabbit Which reaction between $\ce{S2O8^2-}$ and $\ce{Co^3+}$ should happen ? It is like asking why a wooden log does not burn in petrol vapour, or why vater vapour mixed with oxygen does not ignite. Reactions are like business, there must be a seller and a buyer. If there are 2 sellers and no buyer, there is no business. And vice versa. Both above ions are buyers ( of electrons ), but there is no electron seller.


text:
Why cupric ion oxidises iodide to iodine
I have learnt in the qualitative analysis of cupric ions that upon treating its aqueous solution with excess $\ce{KI}$, triiodide ions are formed which can be further tested using starch.
The explanation provided is that $\ce{Cu^2+}$ ions oxidise iodide ions to iodine which combines with additional $\ce{KI}$ to form $\ce{KI3}$.
$$\ce{2Cu^2+ + 2I- -> 2Cu+ + I2}   \tag1$$
$$\ce{I2 + KI -> KI3 }   \tag2$$
I looked up the reduction potentials of cupric to cuprous and iodine to iodide, and noticed that the latter is greater than the former.

I figured that maybe the reaction is reversible and lies to the left but the little amount of iodine formed is sufficient to turn starch blue black as it is a sensitive test.
But my doubt stems from the fact that it is generally regarded that cupric ions can oxidise iodide to iodine and also that cupric iodide does not exist (at least not in aqueous solutions). These statements would imply that the reaction (1) would lie to the right hand side which according to me should not be true.
Is it true that the reaction (1) actually lies to the right possibly due to formation of cuprous iodide which is a precipitate and the triiodide complex.
If not, then wouldn't it be incorrect for us to say that cupric ions oxidise iodide ions?

Have you done this reaction in the lab before? $\ce{Cu+}$ doesn't just hang out on its own. Solid $\ce{CuI}$ is precipitated. https://en.wikipedia.org/wiki/Copper(I)_iodide You need to either look up the reduction potential for $$\ce{Cu^2+ (aq) + I-(aq) + e- -> CuI(s)},$$ or somehow take this precipitation into account when doing the thermodynamics.

Have you taken into consideration copper(I) iodide solubility and Cu2+/Cu+ concentration ratio ?

Related: https://chemistry.stackexchange.com/questions/53768/what-happens-when-potassium-iodide-is-added-to-a-copper-ii-salt

Chemistry of the reaction is also discussed here https://www.researchgate.net/post/Why_CuI2_is_not_stable_but_CuI1 .


text:
Order of rate of acid catalysed dehydration of alcohols

The relative rate of acid catalysed dehydration of following alcohols would be:

According to me, we have to arrange them in order of their carbocation stability.
Q will be the most stable as it is stabilized by Resonance and Hyperconjugation. P should come next. After that I think it should be S>R as in case of R there is no hyperconjugation and hyperconjugation is a “stronger” effect than the Inductive effect.
But in my Book R>S is given. Is Inductive effect stronger here? Or is there some other reason to it?

R is capable of rearrangement (methide shift) to give a stable tertiary carbocation. S needs two hydride shift to get there.

@Mathew In the dehydration of alcohols,the formation of the first carbocation is the rate determining step(that is,the one with the biggest activation energy), right? Subsequent rearrangements may lead to lower energy intermediates,but shouldn't the stability of the first carbocation(and thereby,the energy of the first transition state) be the governing factor for rate,and not what the cation can potentially rearrange into?

@Yusuf Hasan: True, but R and S have no choice but make primary carbocation. However, that would be assisted by concomitant migration of methide and hydride migration, respectively. Even though this assistance is significant, the activation energy of each is still higher than P and Q. The activation energy of R is comparatively lower than that of S, since methide migrarion is giving a tertiary carbocation. Meantime, S has to go through two activation barriers (ist one is higher). The first activation barrier decide the rate.

@MathewMahindaratne That poses another issue in my mind: If the migration is concomitant, then why is the activation energy of R significantly greater than P and Q? If such a high difference exists, then it must be because due to the formation of the first primary cation as opposed to P and Q straight off the bat forming secondary and tertiary cations respectively. If this is the case, then why isn't such an initial difference being observed between the energies of primary cation-like transition states of R and S, which should exist if there has to be a big difference b/w R & S and P &Q?

@Yusuf Hasan: Think this way: In transition state, $\ce{C-^+OH2}$ bond is partially broken. That need very high energy regardless getting assist or not. That bond needed to be partially broken to get this assistance at the first place.

Your basic assumptions are correct.
It can be observed as such:

In Q the carbocation is stablised by resonance, inductive and hyperconjugative effects so it is quite stable and will be formed fastest.
Comparing P and R: In P, a secondary carbocation is formed and there are 4+1 or 5 hyperconjugative structures possible. Moreover, the inductive effect is also greater in P than in R. Hence, it forms carbocation faster.
Both R and S form primary carbocations. Inductive effect is more prevelent in R than in S, while S has 3 hyperconjugating structures.
But R offers capacity to rearrange and stablise the carbocation, by methyl migration to produce highly stable tertiary carbocation, also stablised by resonance over the adjoining phenyl ring.
But, S also forms a carbocation which is primary, but offers no stablising effect other than 2 +1 or 3 hyperconjugating structures.

Similarly, this rearrangement may also be attributed to the comparison of P and R, where hydride shift and methyl shift occur. The migration amplitude of hydride, being lower, leads to rearrangement of positive charge quickly and hence is kinetically also favoured.
Thus finally the order may be:
Q>P>R>S

Won't both R and S rearrange to form the same carbocation?

R has to have only one rearrangement to form tertiary resonance stablised carbocation. But, carbocation in S is quite isolated and primary. It can be argued that in S the first hydride shift will form secondary carbocation comparatively more stable than the initial. But then a second rearrangement has to occur to get resonance stablised carbocation, similar to Q. Then also inductive effect and hyperconjugation will be higher in R than S in the final carbocation formed after multiple rearrangement steps jn S.

Then the whole process of multiple rearrangement will be somewhat tough, and although Hammond Postulate provides for the carbocation rearrangement to occur i.e. from primary, secondary to tertiary and transition states are effectively stablised, the process will be long, tedious and the whole process will make the rate determining step slow.

An acid catalysed dehydration reaction of alcohols will always proceed with the removal of -OH as H2O molecule and therefore, will be dependent on the formed carbocation's stability because the rate deternining step for this process is formation of carbocation. Carbocation Carbocation can be stabilized by inductive effect, field effect, mesomeric effect (or) resonance and hyperconjugation effect.
Q should be the fastest because it's corresponding carbocation is the most stable because it is favoured by all the effects.
P is the next fastest as it's carbocation gains maximum stability through a hydride shift as the secondary carbocation becomes tertiary carbocation which is stabilised by both resonance with the phenyl group and the hyperconjgation of H in the attached methyl and ethyl group.
R would be faster than S because it's carbocation becomes stable after it undergoes a methyl shift which is less favourable than hydride shift but results into similar stability.
S shall be the slowest because it has the most unstable carbocation after its formation.


text:
What is the meaning of pi selectivity?
I came across this term while going through a research paper for the first time of organic chemistry on diastereoselectivity of tricyclo[2.1.0.02,5]pentan-3-ones. I am not familiar with the term and on doing a google search I came across papers on pi facial selectivity.
This is my first time reading a research paper which is of certainly a different standard all together, so any kind of help in letting me know about the general definition of the term and how it is related to diastereoselectivity will be very much appreciated.
Thank you very much.

What paper is this? Context will help.

Is the paper this one? https://www.researchgate.net/publication/288797274_1-Substituted-tricyclo210025pentan-3-ones_Revisiting_the_diastereoselectivity

Pie Selectivity - Which pie do you like best? (a) apple, (b) blueberry or (c) cherry. .

Beside addition of the reference will help the discussion; assuming it is about the reaction on the carbonyl group with a nucleophile (e.g., a Grignard), check if it is not a typo.  A plausible alternative were re / si selectivity (https://en.wikipedia.org/wiki/Prochirality).

@Safdar yes that is the paper, I was working on.


text:
Acid dehydration to ether

Can anyone please help me come up for a curved arrow mechanism for the reaction pictured? I'm familiar with the mechanism for the general acid catalyzed dehydration of butanol, where the arrow comes off a hydrogen bonded O on the sulfuric acid to the hydrogen on butanol to the oxygen and the following steps for that reaction are pretty simple too but I can't for the life of me come up with a curved arrow mechanism for this side reaction.
Currently, I think the butanol takes a hydrogen from the sulfuric acid (arrow from butanol O to Sulfuric acid H to O), followed by the removal of the hydronium atom from the butanol chain (arrow from sulfuric acid O to carbon on butanol chain to the O on butanol chain), followed by maybe a combination of butanol and the synthesized butanol sulfate???? I'm really not sure any help would be greatly appreciated.
SOS
Edit:
This is the final mechanism I came up with. Is this right?

In the fourth step your curly arrows are not right. The proton transfer in the first step is correct, though, so you certainly do know how to draw it correctly.

Also, proton transfer from protonated ether could be to another alcohol molecule as well.

@orthocresol Bisulphate is not a very strong base, due to I- effect and reverse d-orbital hyperconjugation. So should it attack the proton for deprotonarion, or will some other stronger base like water deprotonate?

@ADITYAROYCHOWDHURY I don't know what "reverse d-orbital hyperconjugation" means, but I'm pretty sure that no such thing exists. You don't need, and shouldn't use, d-orbitals to describe the bonding in sulfate ion (or its protonated forms).

@orthocresol I meant that will back bonding of the electron pair over the d-orbital of sulphur not decrease basic strength of the ion? In such a situation will it still deprotonate?

@ADITYAROYCHOWDHURY I don't make any claim as to which species deprotonates which. Truthfully, I don't think it matters much; as long as the thing gets deprotonated, it doesn't really matter what molecule does it. But yes, water, and the alcohol itself as Mathew suggests, are stronger bases than HSO4-, so yes it is probably better to use water to deprotonate. I do take issue, though, with you talking about d-orbitals of sulfur. That is an outdated theory of bonding which is not really true at all.

I think this blog will help you: https://www.masterorganicchemistry.com/2014/11/14/ether-synthesis-via-alcohols-and-acid/

The best part is that the final ether spells 'WOW' :)


text:
ʟ-Glutamine solubility in hot water
What happens when you try to dissolve ʟ-glutamine in hot water (above 80 °C)? Does the heat destroy the ʟ-glutamine?

Thermophilic proteins are depleted in glutamine compared to their mesophilic counterparts: https://pubs.acs.org/doi/abs/10.1021/bi027435e. This is a hint that maybe they tend to hydrolyze at higher temperature.

Glutamine in aqueous solutions degrades slowly when stored in room temperature. Therefore, we can expect that rate of degradation is faster at higher temperatures. It is evident that the hydrolysis product of glutamine is 5-pyrrolidone-2-carboxylic acid (not glutaric acid as shown for enzymatic degradation) and ammonia (Ref.1):

The Ref.1 states that:

The degradation kinetics of ʟ-glutamine (Gln) in aqueous solution was studied as a function of buffer concentration, $\mathrm{pH}$ and temperature. Stability tests were performed using a stability-indicating high-performance liquid chromatographic assay. The degradation product of Gln was 5-pyrrolidone-2-carboxylic acid. The reaction order for Gln in aqueous solution followed pseudo-first-order kinetics under all experimental conditions. The maximum stability of Gln was observed in the $\mathrm{pH}$ range from 5.0 to 7.5. The $\mathrm{pH}$–rate profile described by specific acid–base catalysis and hydrolysis by water molecules agreed with the experimental results. Arrhenius plots showed the temperature dependence of Gln degradation, and the apparent activation energy at $\mathrm{pH}$ 6.41 was determined to be $\pu{9.87 \times 10^4 J mol−1}$.

Factors affecting the stability of ʟ-glutamine in solution has also been studied (Ref.2). They also found that glutamine degradation resulted in the equimolar formation of ammonia and no associated formation of glutamate. At $\pu{22–24 ^\circ C}$, the degradation rate of glutamine was variable depending on the type of solution used. It was 0.23% in water $\mathrm{pH}$ 6.5; 0.22% in dextrose/water (15% $w/v$); 0.8% in mixed total parenteral nutrition (TPN) solution; and on the $\mathrm{pH}$, molarity and type of buffer used. The degradation rate was essentially unaffected by light and $\ce{O2}$.
Nonetheless, the solubility of ʟ-glutamine in water increases with temperature (Ref.3).
References:

Kanji Arii, Hideyuki Kobayashi, Toshiya Kai, Yukifumi Kokuba, "Degradation kinetics of L-glutamine in aqueous solution," European Journal of Pharmaceutical Sciences 1999, 9(1), 75-78 (https://doi.org/10.1016/S0928-0987(99)00047-0).
K. Khan, M. Elia, "Factors affecting the stability of ʟ-glutamine in solution," Clinical Nutrition 1991, 10(4), 186-192 (https://doi.org/10.1016/0261-5614(91)90037-D).
Jan P. Amend, Harold C. Helgeson, “Solubilities of the common ʟ-$\alpha$-amino acids as a function of temperature and solution $\mathrm{pH}$,” Pure and Applied Chemistry 1997, 69(5), 935-942 (https://doi.org/10.1351/pac199769050935).

This paper here which describes the isolation of Glutamine from beets, contains a note that Glutamine is rapidly hydrolysed in water at high temperatures. The paper includes procedures in water at 60C.
edit:
This paper here describes the degradation kinetics of L-Glutamine in aqueous solution to 5-pyrrolidine-2-carboxylic acid.

Thank you for your answer! What does glutamine hydrolyse into?

Do you have an idea of the order of magnitude that could be. For instance, if left an aqueous solution at saturation point for 10 hours at 60, 80, 90 Celcius degrees, what respective order of degradation can I expect to have? 0.1% / 1% / 10% / 50%?

I have added a second reference that may answer that

Thank you, but I can't manage to access the full text, it's asking me for money :'(

I do not have access either.

Apart from normal hydrolysis by applying heat, this reaction is also enzyme catalyzed as it is discussed in this paper1. ʟ-Glutamine hydrolyze to glutamic acid and ammonia with the help of glutaminase enzyme. This reaction is also catalysed by PabA subunit of p-aminobenzoate synthase. Another enzyme called glutamin-(asparagin-)ase (EC 3.5.1.38) is also known to catalyse hydrolysis of both ᴅ- and ʟ-isomers glutamine and asparagine.

Reference

Kishore, N.; Tewari, Y. B.; Goldberg, R. N. A Thermodynamic Study of the Hydrolysis of ʟ-Glutamine to (ʟ-Glutamate + Ammonia) and of ʟ-Asparagine to (ʟ-Aspartate + Ammonia). The Journal of Chemical Thermodynamics 2000, 32 (9), 1077–1090. DOI: 10.1006/jcht.1999.0496.

Is this relevant to the OP's question? I interpreted the question to mean that they made a solution of Gln in water, so there is no reason to believe that degradation enzymes would be present.

@Andrew Yes, glutamine solutions can be hydrolysed by applying heat as the other answerers noted. My answer points out that hydrolysis can also be catalysed with the help of enzymes to give a different product, It might not to be relevant to you but I think it is still relevant to OP's question if not directly. Heat is not the only factor in hydrolysis reactions.


text:
If we take a adiabatic closed container and allow CaO and water to mix. Will reaction take place? will entropy of system change?
my attempt. If reaction takes place then it releases energy (exothermic) which will increase temperature inside the container this implies internal energy of container increases but because heat exchanged this zero and work done is also zero by first law of thermodynamics    change in internal energy is also zero. this contradicts that reaction takes place. So can we say that in any adiabatic process in closed container change in entropy is zero.

The internal energy may change when work or heat in exchanged with the environment. But it can also change if a chemical reaction happens, like here.

Internal energy remains constant in isolated systems.
Chemical energy released/consumed by rearranging of chemical bonds during chemical reactions transforms usually to/from thermal energy, leading to temperature changes.
Entropy of isolated systems during spontaneous processes increases.


text:
Vapour pressure and Raoult's Law
The following question from my textbook assignment reads

A mixture contains $1$ mole volatile liquid $A$ ($P_\mathrm A^\circ = \pu{100 mmHg}$) and $3$ moles of volatile liquid $B$ ($P_\mathrm B^\circ = \pu{80 mmHg}$). If the solution behaves ideally, the total vapour pressure of the distillate is?

Using Raoult's Law:
\begin{align}
  P_\text{total} &= P_\mathrm A^\circ \times \chi_\mathrm A + P_\mathrm B^\circ \times \chi_\mathrm B\\
  P_\text{total} &= \pu{100 mmHg} \times 0.25 + \pu{80 mmHg} \times 0.75\\
  P_\text{total} &= \pu{85 mmHg}
\end{align}
But the answer is $\pu{85.88 mmHg}$.
I tried searching online for an explanatory solution but all of them seem to use the mole fraction of solute in vapour phase without any explanation. I figured it was due to the word distillate. Can someone explain why do we need to use mole fraction of the vapour phase and not of the solute in solution?

The distillate is the condensed form of the vapor.

@Safdar You mean take out all the vapours of the solution separately and then condense them into a solution?

Exactly.. Then attempt to use Dalton's law of partial pressures to find the mole fractions in distillate/

@Safdar hey doesn't that mean the (condensed) distillate will form its own vapours?

It would but they have asked for the distillate of the given mixture and not the distillate of the distillate ...

@Safdar Could you possibly post a half solution? I'm confused whether it is the pressure due to the vapours separated out alone in another beaker (if so then it would be on the base of the beaker, isn't it?) Or the pressure due to the distillate's condensed as well as vapour existence?

@Safdar what is causing the pressure and on what is the pressure being applied. These answers might help me understand, thanks!

There is no such external pressure that is being applied(this is irrelevant), they are asking you to find the vapour pressure of the mixture that you get after you condense the vapours formed into a new liquid. Due to this, you would have a new solution whose vapour pressure you are asked to find

@Safdar Is it right if I put it this way? The vapours of the initial solution were condensed and separated in another container. The condensed solution developed vapours over itself. The vapour pressure to be calculated is of these vapours, on whatever the condensed solution is left.

Exactly. That should solve it..

Also do not use the homework tag anymore, it is deprecated(should not be used).

here "distillate" is to be considered as the vapour (of the orignal liquid) which is then condensed so you will have to find the composition of the liquid in vapour phase then use the composition in vapour phase as liquid phase as it was distilled.

During the process of distillation, we condense the vapours (B) of the mixture (A) to give a mixture (C) that has the same mole fractions as that of the vapour (B) of mixture (A).
In order to solve this question there would be three steps.

Firstly, finding the vapour pressure of the initial mixture.
Second, we find the mole fraction of components in the vapour phase of this mixture.
The third and final part would be to find the vapour pressure of the distillate leading to our answer.

However, finding the actual composition of the vapours here, would be the biggest challenge in this question.
Step $1$
In order to find the vapour pressure, we use Raoult's Law which mathematically states:
$$P_T = \sum{P^\circ_i\chi_i}$$
Since this is a two-component system, the total pressure would be expressed as a sum of two terms for each component A and B.
$$P_T = P^\circ_\mathrm a \chi_\mathrm a +P^\circ_\mathrm b \chi_b$$
Substituting the values given in the question, we get $P_T= \pu{85 mmHg}$
Step $2$
The next step is to find the composition in the vapour phase. For this, we use the  property that the partial pressure of a component in the vapour phase is equal to the contribution of the same component in the liquid phase. Mathematically, it is equivalent to stating:
$$P_Ty_\mathrm a= P^\circ_\mathrm a \chi_a$$
Here, $y_a$ is the mole fraction of component A in vapour phase. Solving for $y_\mathrm a$, we get that $y_\mathrm a = \frac{5}{17}$, which implies $y_\mathrm b = \frac{12}{17}$.
Step $3$
The final is exactly the same as the first step, wherein you use Raoult's law to find the final vapour pressure of the distillate (mole fraction of components equal to that mole fraction of vapour in initial mixture).
Solving for the vapour pressure, you get $P_T^{'} = \pu{85.88 mmHg}$, which is the answer provided.


text:
what is the graphic difference between ionic and covalent bonds?
Reading my chemistry 101 again I can not find a graphical difference between an ionic bond (gaining-losing and electron) and a covalent bond (sharing) an electron when I try to think of the orbits/cloud of the electron. Why do they say that one is gained/lost or shared?
I would love to see a graphic example (meaning a drawing, video or something) that represents if there is a difference between different orbits (of NaCl and H2 or something). Since the electron occupies a cloud instead of an orbit in both cases, where is the difference?
May be is not a great question, but I can not visualize it! Thanks for helping starters

It is hard to visualise, even the idealised cases. Furthermore, the bonds are in between these two concepts, which makes it even harder to grasp. In an ideal ionic bonds, electrons get transferred from one partner to the other, so in approximation the electrons only experience the field of one nuclei. The resulting ions then are attracted to reach other electrostatically. In a covalent bond, the binding electrons are shared, they experience the field of both bonding partners.

In an ionic bond, one atom has lost one electron, given to the second. Afterwards, the atoms are becoming ions, and they are separated. They can move independently from one another. They attract themselves, but they are not necessarily in contact. In a covalent bond, the two atoms are attached one another with their common electrons. They stay together. The distance between their nucleus is constant, or nearly, as the bond can be stretched.

Thank a lot to both of you. In NaCl, Maurice, how they are together when they are together? or there is Na with 10 and Cl with 18 in their own orbits (living happily, but close to each other?). Is that it? But what about HF? The hidrogen lives alone without electron, but it is still hidrogen? This might be to knaive, Sorry if it doesn't make sense.

@ Mike. The sodium atom Na is like a mother who must go to work without her baby. She gives him for a while to a nurse, thinking that later on she will pick him back. This is like an ionic bond. Na gives an electron to a foreign atom acting like the nurse. The baby is the image of the electron. A covalent bond is different. It is like a child sleeping in a bed between mother and father, or walking between his parents who are holding him by the hand. The child belongs to the two parents simultaneously. Not the ionic bond.

You may map the electron density around an atom will find regions where this probability is relatively high.  Similar to a hiking map where lines mark the same elevation in a terrain, contour maps offer a representation in 2D, where the lines mark regions of same electron density. You find these used in quantum mechanics (like the ones below), or in crystallography (example).
These contour maps may be drawn for atoms in vicinity of each other, as shown below.  The comparison of multiple electron density maps may reveal you the differences between a non-polar molecular bond, a polar molecular, and the one in the ionic state.

(reference)
For nitrogen ($\ce{N#N}$) shown on the left, the map depicts a symmetrical electron density distribution which is symmetrical along the line you may imagine between the two nitrogen atoms.
Again, similar to a hiking map, these maps are a projection of a terrain as seen from a plane; seen perpendicular to this direction yields a profile like the following for the hydrogen atom (in this figure, read «charge density» as synonym for «electron density»):

(reference)
For a polarized bonding, like in the $\ce{H-Cl}$ molecule, the minimum electron density between the two binding atoms no longer is at half distance between the two atoms.  Contrasting to a polarized bond, in an ionic state, like in NaCl, the electron density between the two atoms not only passes a minimum, but may become practically negligible:

(reference, slide #48)
The discern between a «polarized bond» in molecules and the «ionic state» is somewhat arbitrary.
Addition:
@Mike suggested to compare the situations above with the one in a typical metal like sodium, or copper.  In the solid state, the spatial arrangement of metal atoms is regular; for example in a bcc lattice (sodium), or a fcc lattice (copper).  Similar as in the case of NaCl, the metal atoms again donate the outmost electrons, formally yielding a charged cation.  Contrasting to NaCl, however, these donated electrons are shared among all sodium / copper atoms; colloquially, this distribution is called electron gas because these electrons may move freely and contribute for electric conductance.
If you compare the electron density map of solid copper with the one of solid NaCl, passing from one atom to the next one, only the later contains extended volumes with practical zero electron density (marked by the blue arrow) insulating the ions from each other:

(figure slightly modified from this reference)

Thanks @Buttonwood you really understood what I could hardly explain. I'm not sure about making another question (nonsense?): When I think about H2 (g-deuterium) I think of two atoms of H together and then space, but when I think about Na2 in solid I imagine that the two atoms are together and next to each other (Na2 in a solid form or another element like H2 but solid). Does Na2 form a crystal or solid structure like ice water does? Do they "chain" like Polyethylene? If it makes sense to ask that, I will create a question. Thanks again

Thanks @Buttonwood it is really helping. I'm reading Chemistry for dummies and refresing my chemistry from highschool and some of the stuff in your answer is way too advance for me, but the graphs are really helping me to see it. Is there like an hydrogen bonding for NaCl molecules (or they are just solids and they are close to each other)? I also have this question: https://chemistry.stackexchange.com/questions/138549/what-is-exposed-in-the-surface-tension-of-water. Thanks again

@Mike I'm not aware about NaCl molecules.  For normal molecular bonds (e.g. H2), or hydrogen bonds (e.g., the dimers of carboxylic acids), there is significant electron density all along the line between the two atoms, there is direction.  However, NaCl is kept together as a crystal with 1) Coulomb attraction between opposite Ions ($\ce{Na+}$ and $\ce{Cl-}$) and 2) Coulomb repulsion between Ions of same charge (either $\ce{Na+}$ to $\ce{Na+}$, or $\ce{Cl-}$ to $\ce{Cl-}$). In an analogy: like with the little kitchen magnets, you seek an (energetically speaking) optimal arrangement, but in 3D.


text:
Illustrating moles with sand grains, Are there more moles of SiO2 in a grain than moles of sand on the earth
I have been trying to understand the scale of atoms, thinking about sand grains as an example.
With approximately $7.5\times10^{18}$ grains of sand on the earth that is $1.25\times10^{-5}$ moles of sand grains. Source
I've found that a beach sand has approximate density $2.5\times10^{3}$ grams/litre or 2.5g/cm$^3$. Assuming a quartz grain of diameter 0.05cm, so volume of approx $5\times10^{-4}cm^3$ and mass of approx 0.001g. Source
With moelcular mass 60, then there are $2.2\times10^{-5}$ moles of $\ce{SiO2}$ per grain.
This surprised me. Are there really more moles of $\ce{SiO2}$ in a grain of sand than moles of sand on the Earth? And is there any verfication of this by a more serious study?

Actually, I do mean moles I think here.

This site reports heavier sand grains, averaging 0.0044 grams.  So by that measure the SiO2 entities win by a wider margin.  Lots of atoms fit into a little space.

Your question should be: Are there really more moles of $\ce{SiO2}$ in a grain of sand than amount of sand grains on the Earth (in mols)?  Seemingly, yes. Similar to number of molecules in 10 drops of water, which is even bigger than that.

https://what-if.xkcd.com/83/

What "more serious" study do you want? The math is correct, the assumption for the sand grain weight makes sense, and if you doubt that number for the amount of sand on earth, I fear you are in the wrong SE. ;-)

@Karl Awfully anthropocentric of xkcd to normalize a fine grain of sand to our sun... what did Rigel ever do wrong?


text:
Remembering Chemical formula
How do you all remember the name of chenical formula of Glucose, Lime, Vinegar ... etc. becasue these substances have big big chemical formula. So, How do I remember their chemical formula?

Why do you think lime and vinegar have big formulas ? BTW, vinegar is not a substance, it is basically diluted water solution of acetic acid. It is remembered by the same way as you remember things you cannot derive from the general principles. Or in case of substances, from their systematic names. You just know simple things from your heart over time, if you are a chemist. You look up for complex things if you need them.

Practice.$\mathstrut$

It is like a language. A logical language.

How can I know What Is the Chemical formula of Glucose? How do you derive the formula?

You don't. This is not a thing that can be deduced by logic. Just read the formula in a book and memorize it.

@Khan_Jamal Chemistry is in many aspects as learning other language. There  is many things you cannot derive, you just remember them or look them up.


text:
rate of reaction and RDS
An article on Khan-Academy, states the following:

The overall reaction rate is determined by the rates of the steps up to (and including) the rate-determining step

Here, it says that rate is determined by steps before and including that of RDS. But from what I know the rate is determined by one single elementary reaction, which is the slowest, that is, the RDS. So why does it include steps previous to RDS?
(I would appreciate an example of this)

Could you share the link where you saw this statement?

https://www.khanacademy.org/science/chemistry/chem-kinetics/arrhenius-equation/a/reaction-mechanisms

It states the overall rate equation is dependant on the steps, that means the overall reaction equation can have compounds that are not seen in the RDS. This link explains this better.

https://courses.lumenlearning.com/boundless-chemistry/chapter/reaction-mechanisms/ This was the link this statement was referenced from.

The interpretation of the statement depends on exactly what is meant by "overall rate."

The rate-determining step corresponds to the largest free energy difference from a transition state to any preceding stable state (reactant or intermediate).
In the example above, the biggest forward barrier is from the reactant to the second transition state, so the second step is rate-determining. However, the barrier, as identified above, is comprised of both the first and second step, so they both figure into the overall rate of the reaction, as determined by the slow step.
EDIT:
Based on comments, I think a clarification is in order. Note that the above conclusion is not true in the general case. You could make the indicated claim for the depicted reaction specifically, but not, for example, in the case below:

The second step here is still rate-determining because the transition state is involved in the largest barrier. However, the biggest barrier is from the intermediate to the second transition state. In this example, I would not say that the overall rate depends on the first step.

"they both figure into the overall rate of the reaction" - do they? How?

They must because the rate-determining step has a corresponding $\Delta G^{‡}$ that spans the reactant and the second transition state. The barrier is made up of both the first and second step. In this example, if you were to write the rate law, it would have to include the rate of the first step, but rate constant for step 1 will  be absorbed into the overall rate constant, via the equilibrium from the reactant and the first intermediate. @user6376297

I see. Yes, OK, if you consider equilibrium constants as ratios of rate constants then they 'figure' in the overall rate constant. But I think this confuses the point. The OP, with whom I agree, was puzzled by a statement seeming to indicate that the rate (constant) of the overall reaction is influenced by the rate (constants) of the RDS and of the steps that precede it. Instead, under the RDS approximation the preceding steps can only influence the overall rate constant by their thermodynamics, not by their kinetics. It does not matter how fast they are, just that they are 'much faster'.

To be more explicit. In the example I posted below, the equation for $C(t)$ when the first step is fast and the second step is the RDS only contains rate constant $k_B$, not $k_A$. The numerical value of $k_A$ has no bearing at all on it. Even if the first step were a fast reversible reaction, it would still only influence the overall $C(t)$ by its equilibrium constant, indeed as you say. It will not be necessary to know the rate constants of the direct and inverse reactions. This IMO is no minor difference, and it does make a big difference (operationally) when one models such systems.

@user6376297 It's not so much that the rate constant matters. It's that the first step matters. The overall rate includes contributions from both steps, and therefore, you cannot ignore the first step. Is the rate constant explicit? No, it is not. But if you remove the first step, you've changed the kinetic properties of the system.

@user6376297 I added a second example. Hopefully, that at least partially addresses your concerns in the general case.

It is perhaps worth noting that the 'rate determining step' is an approximation.
It is technically possible to model any reacting system without it, by using numerical methods.
It is even possible to model certain systems fully analytically.
Take for instance a sequence of 2 consecutive irreversible reactions:
$\ce{A ->[k_A] B ->[k_B] C}$
which is described by:
$\frac{dA}{dt}=-k_A A$
$\frac{dB}{dt}=+k_A A-k_B B$
$\frac{dC}{dt}=+k_B B$
This system of differential equations has an analytical solution. Assuming that only $\ce A$ is present at $t=0$, at a concentration $\ce{A(0)}$:
$A(t) = A(0) e^{-k_A t}$
$B(t) = \frac {k_A A(0)}{k_B-k_A} [e^{-k_A t}-e^{-k_B t}]$
$C(t) = \frac {A(0)}{k_B-k_A} [k_B (1-e^{-k_A t}) -k_A (1-e^{-k_B t})]$
Suppose that the first reaction is 100 times faster than the second ($k_A = 1, k_B = 0.01$).
If you plot the 3 concentrations for $t \in [0,100]$, you will see $A$ disappear quite fast and $B$ reach its maximum when $C$ is still quite low.
In other words, you can assume that the second step is the 'rate determining' one, i.e. assume that all $A$ almost immediately turns into $B$, so $B(0) \approx A(0)$, and model only the second and third reaction:
$B(t) \approx A(0) e^{-k_B t}$
$C(t) \approx A(0) [1-e^{-k_B t}]$
The same result can be reached directly from the exact analytical solution, by setting $k_A >> k_B$:
$B(t) \approx \frac {k_A A(0)}{-k_A} [-e^{-k_B t}] = A(0) e^{-k_B t}$
$C(t) = \frac {A(0)}{k_B-k_A} [k_B (1-e^{-k_A t}) -k_A (1-e^{-k_B t})] \approx \frac {A(0)}{-k_A} [-k_A (1-e^{-k_B t})] = A(0) [1-e^{-k_B t}]$
Suppose instead that the first reaction is 100 times slower than the second ($k_A = 0.01, k_B = 1$), and plot the 3 concentrations for $t \in [0,1000]$.
You will see $A$ disappear according to a rather typical first order decay, $C$ form equally typically, whereas $B$ remains barely visible at the bottom of the plot (in some contexts $B$ would be called a 'stationary intermediate').
You can assume that the first step is the 'rate determining' one, i.e. assume that $\frac {dB}{dt} \approx 0$, and consequently $\frac {dC}{dt} = k_B B \approx k_A A$, and model only the first and third reaction:
$A(t) = A(0) e^{-k_A t}$
$C(t) \approx A(0) [1-e^{-k_A t}]$
Here, too, the same result can be reached directly from the exact analytical solution, by setting this time $k_A << k_B$:
$A(t) = A(0) e^{-k_A t}$
$C(t) = \frac {A(0)}{k_B-k_A} [k_B (1-e^{-k_A t}) -k_A (1-e^{-k_B t})] \approx \frac {A(0)}{k_B} [k_B (1-e^{-k_A t})] = A(0) [1-e^{-k_A t}]$
In both these 'extreme' cases, the approximate curves described by the 'rate determining step' approximation fit quite closely the exact curves, but there is nothing better than the latter, for a more accurate account of the system.
Also considering that one does not know a priori if a given approximation holds true or not for a particular system.
It is sufficient to set $k_A = 1, k_B = 1.01$, i.e. have two consecutive reactions of comparable rate, and plot the exact concentrations for $t \in [0,10]$, to see that $C(t)$ is not really well described by either approximate rds solution.
Luckily these days there are plenty of excellent software solutions allowing to model systems of any complexity via very accurate numerical methods.

It might be worth mentioning that the RDS approximation may break down completely in catalytic cycles (or yields the right observation by accident).


text:
Conductivity of molten or aqueous salts

If there is a 1M aqueous solution of 1 liter NaCl (aq), and the molar conductivities of the solution and the chloride ions were found to be $x$ and $y$ respectively, is it accurate to say that the molar conductivity of sodium ions is $x-y$? If not, how else can I calculate the molar conductivity of sodium ions from the given data? (I'm just using NaCl as an example)

Is there any way to compute the molar conductivity of an ion without any experiments?

it would be x+y since they are ions and that is what is needed. Check Kohlrausch's law (feels related).

You cannot determine individual ion's molar conductivity at 1 M. The ions are not independent enough. Kohlrausch's law of independent migration of ions does not apply, it only applies at a fictitious concentration of 0 M NaCl! This value is found mathematically by extrapolation to zero concentration

Is there any way to compute the molar conductivity of an ion without
any experiments?

Yes, if you look up the tables of molar conductivities at infinite dilution, then you can determine molar conductivities of individual ions.


text:
Is melting aluminium in the presence of carbon dioxide desirable?
So aluminium cans melting is prone to formation of aluminium oxide due to presence of oxgygen and thin sheet of aluminium (large surface area). To mitigate this I can design a cruicible which is filled with $\ce{CO2}$ from bottom so it displaces air before the start of melting process as well as a check valve to eliminate fumes and built up of pressure while heating.
Question: Will this process be successful in mitigating formation of aluminium oxide? I know in argon welding, they keep purging argon on the welding area (dynamic seal) to prevent formation of oxide, but in case of aluminium, will $\ce{CO2}$ suffice? Also is there any cheaper alternative?

This idea seems reasonable

Press the cans into a solid block and you don´t have to worry about air. But why do you want to do this? There are large aluminium recycling factories, and they do this far more energy-efficient than you, simply because of their size advantage.

@Karl Makes sense, so maybe use hydraulic press and press all the cans into a blob right? Also to answer your question, hobby and DIY stuff.

Al can react with CO2 to displace carbon, but at the comparatively low MP of Al, I don't know how rapidly. Just prevent rapid air exchange in the crucible, and the small amount of O2 in air will be quickly removed, leaving an Ar/N2 atmosphere.

https://en.wikipedia.org/wiki/Aluminium_recycling just in case ... ;)

@DrMoishePippik, makes sense. I guess formation of small alumina is inevitable in kitchen chemistry, its just a matter of knowing how much is called significant to render the process useless. The crucible will have a lid and a check valve so no air would enter and potentially all the gases produced from the coatings on the can as well as added $\ce{CO_2}$ will be purged out through the valve when temperature is gradually increased. I dont know if this is a recipe for disaster so any valuable input is greatly appreciated.


text:
Why is PLGA so expensive in comparison to PLA?
A Google search has lead me to believe that PLGA goes for anywhere between 8 USD and 100 USD per gram. Meanwhile, for PLA (3d printer grade), prices float around 50-100 USD per kilogram.
I understand there are differences in the two, but why is PLGA so much more expensive? Is it solely because of PLGA's application in medical-grade uses, while PLA has widespread consumer-grade application?

Pretty much anything becomes a great deal more expensive when purified to medical grade.

https://en.wikipedia.org/wiki/PLGA  poly(lactic-co-glycolic acid)

Mass production?

@IvanNeretin - and it isn't just the purification, but all the overhead of quality control and certification to spec to get it approved as medical grade.

Poly(lactic-co-glycolic acid) or PLGA is a biodegradable copolymer of lactic acid ($\ce{CH3CH(OH)CO2H}$) and glycolic acid ($\ce{HOCH2CO2H}$). Poly(lactic acid) or PLA is, on the other hand, a polymer of lactic acid, which is also biodegradable. Since both polymers are linked together by ester linkages of their monomers, the resulting polymers can be categorized as linear, aliphatic polyesters:

PLGA is one of the most successfully biodegradable polymers, and hence it attracts considerable attention in the development of drug delivery systems due to its desirable properties and to its Food and Drug Administration (FDA) and European Medicine Agency (EMA) approval for parenteral administration. PLGA encapsulated drug delivery enjoys drug protection from degradation. In addition, PLGA also provides prolonged controlled release of encapsulated drugs. Therefore, as you have suspected, these medical uses (current and developing applications) needed GMP manufacturing of the product for the purpose of the use in Humans. This GMP manufacturing of drugs is always costly. The price of the compound also depend on the other factors such as the size of the polymer (molecular weight), percent enantiomeric ratio (Lactic acid has two enanthiomers), end groups, etc., which are critical in human use. Also remember, synthesis of copolymer with desired ratio is not a easy task.
PLA has become a popular material due to it being economically produced from renewable resources. In 2010, PLA had the second highest consumption volume of any bioplastic of the world. For example, PLA is the most widely used plastic filament material in 3D printing. This commercial success of PLA makes its bulk production (no need of GMP manufacturing processes). Thus, it is much cheaper than demanding PLGA. Yet, PLA can be expensive as well, depending on its use, which require careful synthetic processing. For example, D- and L-lactide copolymer is used in bone engineering (Ref.1).
Sigma-Aldrich listed series of PLGA and PLA with some different properties:
$$
\begin{array}{c|ccc}
\hline
\text{EXPANSORB$^®$} & \text{PLGA or PLA} & \text{L/DL Ratio} & \text{End group} & \text{Molecular weight} & \text{Price}/\pu{10 g}\\
\hline
\text{DL 100-2A} & \text{PDLLA} & 0/100 & \ce{COOH} &  \pu{10-25 kD} & \pu{501.00 USD}\\
\text{DLG 50-2A} & \text{PLGA} & 50/50^a & \ce{COOH} &  \pu{5-20 kD} & \pu{446.00 USD} \\
\text{DLG 75-9E} & \text{PLGA} & 75/25^b & \text{Ester} &  \pu{100-150 kD} & \pu{458.00 USD} \\
\text{DLG 75-2A} & \text{PLGA} & 75/25^b & \ce{COOH} &  \pu{5-20 kD} & \pu{446.00 USD} \\
\hline
\end{array}\\
^a: \text{D/L-Lactic : Glycolic} = 50:50; \ ^b: \text{D/L-Lactic : Glycolic} = 75:25
$$
The list was posted to show when synthesized for biological usage, either polymer can be expensive. For example, the entry 1 of the above table is essentially PLA (made by 100% D/L-lactic acid), but it is the most expensive one on the list.
References:

C. X. F. Lam, R. Olkowski, W. Swieszkowski, K. C. Tan, I. Gibson, D. W. Hutmacher, "Mechanical and in vitro evaluations of composite PLDLLA/TCP scaffolds for bone engineering," Virtual and Physical Prototyping 2008, 2(4), 193-197 (https://doi.org/10.1080/17452750802551298).


text:
Thermal equilibrium in reversible processes
In whatever resources I have consulted to study reversible thermodynamic processes, a common statement seems to be :

The system and surroundings (if the boundary allows) are in mechanical ($P_{sys}=P_{surr})$ and thermal $(T_{sys}=T_{surr})$ equilibrium at all instants.

However, in various answers on the Chemistry stack exchange, and in physical chemistry by peter atkins, it is claimed that the "surrounding" is an infinite reservoir: so the temperature can be assumed to be constant.
1.Is this not contradictory to the "thermal equilibrium" I mentioned in the beginning?
2.Thermal equilibrium of a system with its surroundings means that the entropy change of the surroundings , should be given by $\int\dfrac{-dq_{sys}}{T_{surr}}$. However, everywhere, this equation is simply given as $\dfrac{-Q_{sys}}{T_{surr}}$, which seems to be explained by the claim that the Temperature of the surroundings doesnt change. This equation, according to me should be applicabe only to Irreversible processes happening quickly. (or to isothermal processes).
so what exactly should we consider: $T(sys)=T(surr)$ at each instant, or $T(surr)$=constant?

You might want to elaborate on point #1. What exactly is the contradiction?

If the system changes its temperature from T1 to T2,(T2 $\neq T1$) then the surroundings also change from T1 to T2, in a reversible process. Doesnt this clearly contradict the statement :"surrounding temperature remains the same"?

In a cases where where the temperature is changing during the reversible process, instead of just using a single reservoir at a constant temperature, what you do is use a continuous sequence of reservoirs, all at slightly different constant temperatures (over the range from the beginning to the end of the process), with a tiny (differential) amount of heat exchanged with each.  So, for each reservoir, you have $$dS_{surr}=\frac{dq_{syst}}{T_{surr}}$$ and, if you add all these differential changes in entropy up, you end up with $$\Delta S_{surr}=-\int{\frac{dq_{syst}}{T_{surr}}}$$

With regard to question 2, the correct equation to describe the entropy change during a finite reversible process is
$$\Delta S_{\text{surr}} = \int\dfrac{-dq_{sys}}{T_{surr}} \tag{1}$$
which follows from the statement $$dS_{\text{surr}} = \dfrac{-dq_{sys}}{T_{surr}} \tag{2}$$
Only for an isothermal process does Eq. (1) become
$$\Delta S_{\text{surr}}=\dfrac{-Q_{sys}}{T_{surr}} \tag{3}$$
It is possible to alter the temperature of the surroundings during a process (in practice and conceptually), so Eq. (3) is not a general statement.
The series of pseudo-stationary states traversed during a reversible process are equilibrium states, attainable under the limiting condition that the process is carried out infinitely slowly. At each step during a reversible process we can write $T_{sys}=T_{surr}$, but this does not require the temperature to be equal for each state$^\dagger$. The use of the word infinity is meant to convey an ideal process that can only be approached approximately in reality. An infinitely large heat reservoir means that it has an infinitely large heat capacity, so that changes in the system do not alter the temperature of the surroundings:
$$dT = \frac{dq}{C_{\text{surr}}}=\frac{dq}{\infty} =0 \tag{4}$$
A textbook example of a process carried out reversibly while altering the temperature is the cycle of heating and cooling a pure substance before and after bringing it to a phase transition temperature, in order to compute the entropy change for the spontaneous process under supercooled conditions.
$^\dagger$ If this appears to be like one of Zeno's paradoxes, well, it is.


text:
Analysing reversible processes

A reversible process is a process in which the system is in mechanical and Thermal equilibrium with its surroundings.

Consider a simple thermodynamic system: a gas in a vessel, with a piston (free to move).A large no. of pebbles are kept on the piston, each pebble having a very small mass: as we remove pebbles, we decrease pressure by an infinitesimal amount, and thus the pressure is varied slowly.
We wish to carry out a reversible expansion.Initially, The pressures and temperatures of the system and the surroundings are equal.

My (guess) for what happens in this process: Suppose we remove a single pebble: the pressure decreases infinitesimally, and the gas expands: This causes the Temperature and volume of the system to change slightly.(suppose the temperature decreases from $T$ to $T-\Delta(T)$).Since the surrounding Temperature is now higher, some heat flows into the system.

I had initially thought the some heat simply flows and decreases the temperature of the surroundings, (and increases the temperature of the system) to some value $T^{'}$, and then the situation becomes static. We then remove the subsequent pebbles,in a similar fashion.
however, this picture leaves me confused with how exactly can we carry out an isothermal reversible expansion: There doesn't seem to be any mechanism that enforces $T_{sys}=T_{surr}=T$ throughout the process.

is my picture correct?
If my picture is correct: How exactly can we carry out an isothermal, reversible expansion?

Please correct this: " some heat simply flows and decreases (and increases) the temperature" - do you mean decrease or increase? Also, hopefully my answer to your other question provides an answer here as well. If you assume the surroundings has infinitely large heat capacity, then it can transfer to or absorb heat from the system without a change in its temperature.

@BuckThorn corrected

After you remove a pebble, the same thing will start out happening as in the adiabatic process, and the gas temperature will decrease slightly.  This would cause heat to flow from the reservoir to the gas, and ithe gas would re-equilibrate to the reservoir temperature again.  So the overall result of removing just the tiny pebble would be that the pressure would decrease a little and the temperature would end up the same.

As already in the answer by above, the process occurs infinitesimally, slowly and is a reversible process. As it occurs slowing the conditions such as net temperature of the system remains constant i.e. at each step of the reaction the equilibrium is restored by taking heat from the surroundings, but an overall isothermal process will occur. Similarly, we can consider it to be an action of two forces, where the driving force which is the force exerted by the gas due to removing the pebbles one by one, and the weight of the pebbles which is opposing. In such a situation if the we can consider the whole system to be at equilibrium. And if the weight of the pebbles is increased even slighly by dx then the system reverts back to the initial equilibrium, just before the pebble having mass dx was extracted.


text:
What happens when isopropyl alcohol expires?
My isopropyl alcohol has expired. What happens when it expires? Does it break down into something else? Can I still use it to clean or sanitize?

Nothing, unless the container has  been opened. If opened, it could slowly absorb water from air and/or evaporate isopropanol, but would still be useful.

Isopropyl alcohol is a fairly stable compound, so it probably won't break down.  Instead, past the expiration, even if you've not opened the container, some of the alcohol might escape (thus lowering the percentage of alcohol*) and, in addition, some plastic might leach from the container into the alcohol.  *You can tell whether this has happened significantly by comparing the level of liquid in your container with that in new containers; note that there is typically variation in the levels even in new samples, so this will only tell you if there's been significant loss.

Expiration of stable compounds like isopropyl alcohol has nothing to do with their stability. It is rather commercial requirement.

@Karl PE containers aren't typically 100% pure PE.The plastic often contains additives as well as impurities. And these compounds can leach into the isopropanol. Here's just one example (it discusses extraction with EtOH, which is close to isopropanol):  https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3222987/  And here's a genl. discussion about leaching from HDPE pipe. It's discussing leaching of organics into water, but most water-soluble organics will be as soluble, or more so, in isopropanol:  https://pprc.org/2015/p2-rapid/is-high-density-polyethylene-hdpe-a-good-choice-for-potable-water/

That latter article states: "a 2003 study by Skjevrak, identified a wide variety of leachates from HDPE pipe. Odors associated with these leachates were above acceptable levels set by USEPA’s non-mandatory quality standards. Skjevrak found that the dominant source of contamination was likely due to breakdown-products of common polymer antioxidants. While these contaminants are not of remarkable toxicity, a host of other minor contaminants were identified, including benzene and xylene."  [I'll add that, when I look at the bottoms of my isopropanol containers, they say they are made with HDPE.]

@theorist I take back the "zero", of course I knew that PE contains processing aids (mineral oil usually) and stabilisers. One shouldn´t drink acohol that was stored in a PE bottle.

A mixture of all comment above here. Mostly I came to say that the question seems legit and should not have been down voted.

Chemicals don't (usually) suddenly expire. But impurities from slow reaction products or atmospheric contamination slowly build up over time. The "expiration date" is something the manufacturer calculates based on their knowledge of those processes specifying when they can no longer guarantee the specified purity.

For pure and stable chemicals like isopropanol, it is rather the expiration of the package, after which it should not be sold, not expiration of its content.


text:
Pyridine synthesis by tin(II) chloride reduction of 5-nitronorbornene
What is the reaction mechanism here?
I guess the $\ce{SnCl2}$ reduces $\ce{NO2}$ to $\ce{NH2}$ but then what happens?

I'm thinking there is a retro-Diels-Alder in there somewhere

Can you please show me how exactly, I tried didn’t work

If I knew exactly I would post an answer. I do not have database access to find the paper where this is reported.

@Waylander A side-note about the speculated occurrence of a retro-DA, I'm sceptical; because in the starting material, I count 12, but in the product only 11 carbon atoms.

@Buttonwood Count again! There are 13 in the SM

@Waylander Agreed, I forgot the bridge.  A difference of two C like for ethene, maybe.

This reaction was conducted by Ho and Liao1. Nitro compound 1 is reduced to the nitroso compound 2 which undergoes a [3+3] electrocyclization to form 3 (4). Tautomerization of 4 to 5 is followed by a retro-"Diels-Alder" reaction to open to 6. Conjugate addition of the nitrogen to the unsaturated aldehyde moiety of 6 to afford dihydro pyridine 7. Tautomerization of 7 to 8 is followed by the loss of acetaldehyde. [Structures are from the paper; my numbering.]

ADDENDUM: The mechanism provided in the paper does not address a number of stereochemical issues. Nitro compound 1 is obviously formed as an endo-Diels-Alder product between (E)-(2-nitrovinyl)benzene and 1,3-cyclopentadiene. The bicyclic compound 3 (4) must have a cis-fusion given the mechanism of the [3+3]-electrocyclization (2 → 3) (vide infra). However, the stereochemistry arising from the retro-Diels-Alder reaction would not be expected to kinetically form the (Z,Z)-diene 6 but rather the (E,Z)-isomer 6a. Clearly, 6a is incapable of cyclization as prescribed. But 6a is a masked, vinylogous malonaldehyde that is susceptible to enolization and isomerization to 6b or 6c. While 6c can cyclize by the reported procedure of conjugate addition, there is also the probability that ring closure occurs by electrocyclization of 6b.

Tse-Lok Ho and Po-Yau Liao, Tetrahedron Lett., 1994, 35, 2211. DOI:10.1016/S0040-4039(00)76799-2

I already upvoted your answer. But I'm curious about the elimination of acetaldehyde. What is the driving force? Is that mean $\beta$-amino acids are prone to leave $\ce{CO2}$?

@Mathew:
Wouldn't be surprised if the zwitterion of beta-aminopropionic acid gives ethylene under some set of conditions. In the present case, becoming aromatic is certainly worth something. If you mix 9 and 10 under some set of conditions, I'd be surprised to see any 8.

@Mathew: I've thinking about your question again. There are stannic salts generated in the reduction. Perhaps they facilitate loss of acetaldehyde. So the question is, will Sn(IV) catalyze the addition of acetaldehyde to 3-phenylpyridine (9) to form 8?

Might not. Otherwise, the last elimination step would have been reversible. I think gaining aromatically is a very good point for the elimination. In my experience, I didn't find any decomposition of $\beta$-amino acids during their synthesis.

Mostly a quibble, but I was wondering if the last step could be conveniently drawn as such. A bit reminiscent of beta-ketoacid decarboxylation. https://i.stack.imgur.com/Byn29.jpg (cc @MathewMahindaratne)

@orthocresol: Good point. Done

@orthocresol: Good point. Why I didn't think thay before? Now, I don't have any doubts.


text:
Is the width of a Soxhlet Condenser Inherent to its Function?
Soxhlet extractors are often coupled to a specifically designed condenser with a very large lower joint.
Is this inherent to it's function as a condenser in the apparatus or only because the Soxhlet Extractor part underneath it needs a big upper joint to get stuff in and out? In other words, is there any drawback to using an adaptor and replacing the condenser by a longer (but thinner) graham condenser?
Preventing flooding might be another reason (for high boiling solvents mainly?).

The condensate needs to drop into the carboard vessel with whatever you try to extract from! Also, using a Graham condenser for refluxing is a gross piece of lab malpractice. Never use anything but a regular reflux condenser for refluxing! https://chemistry.stackexchange.com/questions/608/choosing-the-right-condenser/41460#41460

Hi Karl, thanks for answering! What about a graham atop a dimroth/allihn? My concern is, can I use a standard width condenser above it, or does it really need to be as large as the soxhlet part? Is there any big drawback to an adaptor?

No! Put a Dimroth reflux condensor on top of your refluxing setup, and nothing else. If it is too small, use a bigger one. If you think you need to put another one on top as a safety measure, you have already strayed from the path of GLP, don´t do it!

I'm a stray dog! :D What are the disadvantages of using condensers in series of each other (other than structural/weight-related)? Can I use a long conventional Dimroth vs the unusually "fat" Soxhlet condensers (long vs fat)?

Of course you can reduce the NS 45 or larger of the Soxleth down to NS 29 of a regular Dimroth. The joint of the soxleth just needs to be so oversized to get the cardboard tube with your material in.

Yes, excess weight, you need more clamps, which means danger of tension. And stacking condensors is useless.  The first one must be able to do its job. If it floods, then you need one with bigger diameter, or cooler cooling water.

Give the TA in your lab a kick in the butt from me. Or tell him to send me a check. ;)