text:
Extraction of spilled methylamine from mud
I was reading this question about Breaking Bad on the Movies & TV Stack Exchange site and I'm wondering:
If you spill dirt in methylamine, can you extract the methylamine back out?
My very basic level of understanding leads me to believe that if you heat the resulting mud, you should be able to dry out the dirt and return it to its normal state, and likely create some type of vapor.  But, I'm probably completely wrong.
So, is this possible?

Yes, it's possible, and quite easy at that.

But the more pure/clean the starting material, the better the extraction yield will be. If you mix a few grams of methylamine into a pound of soil, you will not be able to recover all of the methylamine. You might be able to get some back, but you will lose most of it. whereas if you mix a few grams of soil with a few pounds of methylamine, you should be able to get most if it back.

Yeah, but is anyone aware that methylamine is a gas at room temperature?

@vapid Methylamine is used as crystalline hydrochloride.

@Mithoron In the movie it was liquid.

The wording of the question (and confirmed in a comment) suggest a major problem with the scene in question. If mixing dirt with methylamine made mud, then it was either some solution of dissolved methylamine, or liquid methylamine, but not a solid methylamine salt like methylamine hydrochloride. The first scenario wouldn't make sense for the synthesis of methamphetamine, and the second case is the real problem: methylamine boils at -6oC.
Assuming they had a solution containing methylamine that had become contaminated with dirt, the recovery would actually be pretty easy. Because methylamine has such a low boiling point, simply basifying the solution during distillation would readily separate the methylamine from anything else that could be present in the contamination.

I know little about science (hence my need to ask this question), but in the TV show in question, what was referred to as methylamine was a blue-ish liquid that came in large plastic drums.  It was used by the main characters as an ingredient in Meth.  At one point they either spilled dirt into a drum of it or a drum fell over, I can't remember which, and they were still able to extract the methylamine and use it.  I was just curious if that's a believable scenario.


text:
How to visualize the rearrangement of a pentacyclic triterpene alcohol?
I was looking into the exam paper of last year when I found this:

Propose arrow-pushing mechanism for the reaction below.

I suppose it's a Wagner–Meerwein rearrangement. (Sorry for the dirty mechanism graph.)

Is the answer correct? How would I explain the stereochemical configuration of the reaction? And why doesn't the elimination (the leaving of proton) happen at carbon A, as the cation intermediate is just as stable as the others?
I tried drawing chair conformation graph of the reaction. I marked the possibly important groups in red for convenience.

Have you tried drawing the 5 rings in their chair conformations?

@NotWoodward Thank you for the reply! chair conformation graph added. The structure of reactant did inspire me the difference of conformation in the fifth ring, but further explanation didn't come to me yet.

You are correct about the beautiful chain of eight Wagner-Meerwein rearrangements. The stereochemistry of each of these is given by the fact that this arrangement is suprafacial and proceeds under retention, thus a group on the top of the molecule will stay above the ring plane and a group below will stay below. During the rearrangement process, since each intermediate cation is planar, the rings A to D (from where the hydroxy group was to where the double bond will be) will undergo a ring flip each, preserving the general trans-decalin structure for each pair of rings.
Once the final rearrangement has occurred, we arrive at that position in the molecule which features a cis-decalin system: note how in your drawing of the starting material the final ring E is not aligned with all the others but pointing downward. Assuming that final proton, which is eliminated, would actually migrate, we would generate the final carbocation at the cis-decalin junction. The methyl group between the eliminating proton and carbon A is in the wrong orientation to migrate; the carbon belonging to ring E has a much greater tendency because it is better aligned (anti to the proton). That migration would lead to a 6,5-spiro compound and a trisubstituted double bond. That isn’t bad in itself, but the pathway asked for in the question leads to a trans-decalin system by elimination, which features much less strain than a spiro compound. Additionally, a tetrasubstituted double bond is also more stable.
To arrive at the all-trans-polcyclic system, the proton on the junction of rings D and E must eliminate which it is happy to do. Ring D flips, which now turns the cis-decalin system of DE into a trans-decalin system, ring strain is released and the reaction terminates.

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text:
How to derive these general formulae for number of stereoisomers of a compound with a possible plane of symmetry?
These formulae are used if the molecule has a possible plane of symmetry. One such example would be:

Here the carbons marked with an asterisk are stereogenic centres (the asterisk is not used to mark isotopes). We can clearly see that if carbon number 2 (in the entire longest chain) and the carbon number 4 have opposite stereogenic configuration, then the molecule will be achiral because it will have a plane of symmetry. For example, (2R,3S,4S)-pentanetriol will be a meso compound with a plane of symmetry. This compound clearly satisfies the criteria we have set for the type of molecules being discussed.
Let us take another example.

Again, we see that if the molecule has the same geometrical configuration at the first and the third double bond, then the molecule has a plane of symmetry. For example, (2,7)-diphenylocta-(2Z,4E,6Z)-triene clearly has a plane of symmetry.
Our goal is to find the total number of stereoisomers such compounds can have in total. We can assume that a molecule does not have both, a double bond and a chiral centre.
In our class notes, we have written these formulae:

For geometrical isomers (i.e. in case of polyenes),

If 'n' is even (here n is the number of double bonds):$$\text{Number of stereoisomers} = 2^{n-1}+2^{n/2-1}$$
If 'n' is odd, then: $$\text{Number of stereoisomers} = 2^{n-1}+2^{(n-1)/2}$$

And for optical isomers (molecules with chiral centres):

If 'n' is even (here n is the number of chiral centres): $$\text{Number of enantiomers} = 2^{n-1}$$ $$\text{Number of meso compounds} = 2^{n/2-1}$$ $$\text{Total number of optical isomers} = 2^{n-1}+2^{n/2-1}$$
If 'n' is odd: $$\text{Number of enantiomers} = 2^{n-1}-2^{(n-1)/2}$$ $$\text{Number of meso compounds} = 2^{(n-1)/2}$$ $$\text{Total number of optical isomers} = 2^{n-1}$$

How to derive these formulae?

related http://chemistry.stackexchange.com/questions/46674/find-the-number-of-stereoisomers http://chemistry.stackexchange.com/questions/10620/is-there-a-mathematical-generalization-for-the-number-of-isomers-structural-and

How many strings of length $n$ using only the characters "A" and "B" are palindromes?

@f'' I tried using the same method! But I'm getting a different formula for optical isomers when 'n' is odd (I get one that is identical to n=odd for geometrical isomers).

Oh, I see. This case is trickier because of the central stereocenter. If the left and the right sides have identical configurations, then it is no longer a stereocenter because it has two identical substituents, so you actually only have half as many isomers of this kind as you would expect. This is why the total number of isomers is lower than you calculated by the amount $2^{(n-1)/2}$.

Related: Is there a formula to determine the total number of constitutional isomers?

Will a mathematical approach be more appropriate? In that case, this question is worth sharing with the Math StackExchange community...

I remember seeing them with a note that they are valid only if structure is symmetric without considering stereoisomerism as in the examples you have mentioned above.

I want to know whether this question still requires an answer, or has it already been answered somewhere else (that is, in some other question on chemistry or maths SE)? I've got an answer but I want to make sure whether it's still required before investing my time into it.

I have explained the proof of these formulae in this video: https://youtu.be/fmkFZ7Gqa0s

I managed to crack the formula for optical isomers with odd chiral centers, so I'll share my attempt here. Hopefully others may innovate on it and post solutions for other formulae.

Pseudo-chiral carbon atoms - an introduction
The Gold Book defines pseudo-chiral/pseudo-asymmetric carbon atom as:

a tetrahedrally coordinated carbon atom bonded to four different entities, two and only two of which have the same constitution but opposite chirality sense.

This implies that, in your case:

If chiral carbons 2 and 4 both have configuration R (or both S), then the central carbon 3 will be achiral/symmetric, because now "two and only two of its groups which have the same constitution" will have the same chirality sense instead. (Your approach by "plane of symmetry" is wrong. Find more details on this question)
Hence, there can be two stereoisomers (r and s) possible on the 3rd carbon due to its pesudochirality. But, there will be only one if both substituents on left and right have the same optical configurations.

Building up an intuition by manual counting
For optical isomers with odd number of chiral centers and similar ends, you can guess that, if there are $n$ chiral centers, then the middle ($\frac{n+1}2$-th) carbon atom will be pseudo-chiral. To build up an intuition, we'll manually count optical isomers for $n=3$ and $n=5$:
Case $n=3$
Take the example of pentane-2,3,4-triol itself. We find four (=$2^{n-1}$) isomers:
$$
\begin{array}{|c|c|c|}\hline
\text{C2}&\text{C3}&\text{C4}\\\hline
R&-&R\\\hline
S&-&S\\\hline
R&S&R\\\hline
R&S&S\\\hline
\end{array}
$$
As expected from the relevant formula, we find that the first two ($=2^\frac{n-1}2$) are meso compounds, and the remaining two ($=2^{n-1}-2^\frac{n-1}2$) are enantiomers.
Case $n=5$
Take the example of heptane-2,3,4,5,6-pentol:

We expect $16~(=2^{n-1})$ isomers, with the C4 carbon being pseudo-chiral. To avoid a really large table, we observe that the number of meso isomers is easily countable (<< number of enantiomers). Here is a table of those four (=$2^\frac{n-1}2$) meso isomers:
$$
\begin{array}{|c|c|c|c|c|c|}\hline
\text{C2}&\text{C3}&\text{C4}&\text{C5}&\text{C6}\\\hline
R&R&-&R&R\\\hline
R&S&-&S&R\\\hline
S&R&-&R&S\\\hline
S&S&-&S&S\\\hline
\end{array}
$$
Note that the total optical isomers are given by $2^{n-1}$ isomers (more on that below). Hence, the number of enantiomers is easily $12(=2^{n-1}-2^\frac{n-1}2)$.

A formula for the number of meso isomers
As you must have observed from the table, the sequence of optical configurations, when read from the fourth carbon atom, is exactly the same towards both left and right. In other words, if we fix an arbitrary permutation for the optical configurations of the carbon atoms on the left (say RSS), then we will get only one unique permutation of the optical configurations on the right (SSR).
We know that each carbon on the left has two choices (R or S), and there are $\frac{n-1}{2}$ carbon atoms on the left. Hence, the total number of permutations will be $2\times2\times2\cdots\frac{n-1}{2}\text{ times}=2^\frac{n-1}{2}$.
Since, our description ("the sequence of optical configurations, when read from the fourth carbon atom, is exactly the same on both left and right") describes meso isomers, we have hence counted the number of meso isomers, which is $2^\frac{n-1}{2}$.

A formula for the number of total isomers
We note that there are $n$ chiral carbons (including that pseudo chiral carbon). Again, each chiral carbon has $2$ choices. Hence, the maximum possible number of optical isomers is $2\times2\times2\cdots n\text{ times}=2^n$. This is the maximum possible, not the actual total number of isomers, which is much lower.
The reduction in number of isomers is because the string of optical configurations reads exactly the same from either terminal carbon. Example: RSsRS is the same as SRsSR. This happens because the compound has "similar ends"
Hence, each permutation has been over counted exactly twice. Thus, the actual total number of isomers is half of the maximum possible, and is  $=\frac{2^n}2=2^{n-1}$.

Conclusion
Hence, we have derived that, if 'n' (number of chiral centers) is odd for a compound with similar ends, then:

$\text{Number of meso isomers} = 2^{(n-1)/2}$
$\text{Total number of optical isomers} = 2^{n-1}$
$\text{Number of enantiomers} = 2^{n-1}-2^{(n-1)/2}$

Regarding your statement under the introduction part: "If chiral carbons 2 and 3 both have configuration R (or both S), then the central carbon 3 will be achiral/symmetric,..." : I think you concluded that such a molecule would be achiral due to the absence of a pseudo-asymmetric carbon. Your linked question deals with the opposite case where the molecule has a pseudo-asymmetric carbon but is achiral. Similarly, even if the molecule lacks a pseudo-asymmetric centre, it can be chiral. It doesn't matter for our counting purpose but I wanted to make a note of it.

"If chiral carbons 2 and 3 ...then the central carbon 3 will be achiral/symmetric" - here, did you mean "if chiral carbons 2 and 4"? Similarly, isn't it the last two entries in the table for n=3 which are meso compounds? Also, in that same case, aren't the total isomers RSR, RRR, SSS, SRS(all four of these meso), and then RSS and RRS(both enantiomers)?

@HarryHolmes Every pair of meso compounds can be considered to be one and same. RSR and RRR are effectively the same compound due to them being meso

@SafdarFaisal: I saw the edit, thanks. About the meso compounds; when you pass plane polarized light through RSR and RRR separately, yes, you wouldn't be able to differentiate them, but how does that change the fact that the relative arrangement of atoms on the central carbon are different? Wouldn't they be technically different compounds, though you couldn't practically tell them apart?

@SafdarFaisal: also, what does it mean when you left the configurations for $C_3$ blank?

Cut a mirror across carbon 3 vertically across, and then tell me if they look the same or not. there is no difference in relative arrangement.

$\ce{C - 3}$ can be R or S doesn't matter..

Let us continue this discussion in chat.

I have explained the proof in this video: https://youtu.be/fmkFZ7Gqa0s

We can calculate no. of optical isomers for straight chain molecules with even and odd no. of Chiral centres very easily by observing symmetries and using simple mathemetical logics. I will briefly elaborate on even no. of centres. Odd no. cases can also be calculated by similar idea.
Molecule with even no. of  Chiral Carbon Centres where molecule can be divided into two equal halves
Calculation of total no. of Optical isomers include calculation of Meso forms and Enantiomeric pairs. In that case, (and also for Odd no. of chiral carbon centres) calculation of meso forms are very easy.
For calculating no. of Meso isomers, consider a general straight chain molecule with even no. of chiral centres (say, $n = 2k $) as below,

The groups ($A_i$, $\forall i = 1,2,...,2k+1$) are arranged so as to represent a meso compound in general. Now observe that between $k$th and $k+1$th Carbon centre, by an imaginary plane the molecule can be divided into two equal halves. That's why, as a whole the above molecule is a Meso compound. Now, to become a meso compound, an isomer of this compound also has to be symmetric w.r.t this imaginary plane.
So,we can only change configuration the upper $k$ or lower $k$ Carbon atoms to form a new Meso compound, because the configuration of the symmetric Carbon atom will automatically change to maintain symmetricity.
Now, we can change $0$ atom (i.e. our starting compound) $1$ Carbon atom or $2$ atoms or $3$ atoms and so on upto $k$, and thus we are actually calculating each isomers twice.Because , inversion of $k-1$ atoms is nothing but the mirror image of inversion of $1$ atom ($k$ th atom) which is already considered before.  Thus,total no of Meso compounds = $$\frac{\binom{k}{0} + \binom{k}{1} + .... + \binom{k}{k}}{2} = 2^{k-1}  = 2^{\frac{n}{2} -1}$$ which gives us the result.  Now for Enantiomeric pairs,fix the first Chiral Carbon centre ($\ce{C_1}$), because if we don't fix it, we will end up counting mirror images also, which we don't want for counting pairs. Now,fix its symmetrically opposite carbon atom ($\ce{C_{2k}}$) as the relative configuration opposite to $\ce{C_1}$ i.e. $A_3$ on the left and $A_2$ on the right.Thus, remaining $n-2$ centres can be oriented in $2^{n-2}$ ways and all will be different optically active isomers and there are no mirror images also. Thus, Enantiomeric pairs will be  = $2^{n-2}$.   So, total no. of Optical isomers = $2 \times 2^{n-2} + 2^{\frac{n}{2} -1}$ = $ 2^{n-1} + 2^{\frac{n}{2} -1}$
For, odd no. of carbon atoms also similar idea can be applied. If you take $n = 2k+1$, no. of meso isomers will be, $$\binom{k}{0} + \binom{k}{1} + .... + \binom{k}{k} = 2^k = 2^{\frac{n-1}{2}} $$
The logic behind it and the remaining part is left for the reader to think about...

while calculating enantiomers, why did you fix relative configuration of C1 opposite to C2k?

@shreya The fixing of relative configuration of C1 has to be done because we actually want to count distinct Enantiomeric pairs to be precise, and after calculating distinct pairs we can multiply it by 2 to get total enantiomers. Now this fixing concept is basically similar to the permutations of people around a round table if you think carefully. If you can understand permutations around a round table, this problem is actually very similar. And also the molecule overall should be dividable into two equal halves and hence the fixing should be done for C2k as well.

Aren't there also optically active structures in which C$1$ and C${2k}$ are not oppositely oriented?

I have explained the proof in this video: https://youtu.be/fmkFZ7Gqa0s

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text:
Acid-Base determination of organic compounds
I'll start this question with the problem that led me to ask this question...
USNCO 2009 Problem 8 part c:

Part c. states that all four of the compounds above are more basic than ammonia, and the solution explains that it is because the "carbon-containing groups" that are attached to the nitrogen are "better electron donors" than just hydrogen (in ammonia). This "inductive effect" causes the lone pair of the nitrogen to attract the proton ($\ce{H^+}$) better, or as the solution said, "donate the electron pair" to the proton better.
This got me thinking.
My thought process was that the polarity of the hydrocarbon group and the nitrogen overall makes the atom extremely polar towards nitrogen. What I mean here is that the nitrogen has a large negative dipole. This also, in corollary, means that nitrogen has more electrons. In turn, it would attract the proton more strongly.
However, I do not know if my thought process is correct.
This is the official ACS solution to the problem in question:

What does it mean they are "better electron donors?" What is this inductive effect? If you can provide it, are there any other rules that are important pertaining to this and is there a source where I could learn these rules?
Thank you in advance.
Note: this is NOT homework, and my thought process for part c HAS been specified.
Source: https://www.acs.org/content/dam/acsorg/education/students/highschool/olympiad/pastexams/2009-usnco-exam-part-ii.pdf

It's just the thing you say - your reasoning is equivalent, simply you somehow don't know the standard wording.

Would you tell me an example of what the standard wording would be so I can keep it as a reference for the future?

I meant things you're citing...

There is a slight error in your sentence. The nitrogen does not have more electrons, but it has a higher electron density because of it's neighbours. But I think this is how you meant it. Often it is described in terms of donating electrons, whilst these electrons are not donated in a cation/anion sense, so I think electron density is a better to grasp description.
Anyway, with being "better electron donors", it is meant that they are more readily able to donate part of their electron density, just like you describe with your description of polarization.
The inductive effect is the ability of an atom or functional group to donate part of it's electron density through $\sigma$-bonds. So in this we are talking about the electronegativity of a group. It is something you have already encountered in chemistry, but this time it has received the proper name 'inductive effect'.
What you would want to be reading, is about Hammett plots. Hammett plots are a tool to picture the ability of substituents on benzoic acid to influence the acidity. Originally Hammett used hydrogen as the reference substituent. The cool thing is that you can compare the effect of substituents on a different molecule, with the effect they have on benzoic acid. I can advise you to get to the library and find 'Modern physical organic chemistry' by Anslyn and Dougherty. Especially chapter 8.2 about Substituent Effects and chapter 8.3 about Hammett Plot's as a tool will be useful.


text:
Reason for prescence of oil bubbles while frying
When frying with oil, it's generally assumed that heavy bubbling indicates frying, and little to no bubbling indicates a lack thereof. What's the connection between bubbles and frying? If you create bubbles in some other way, baking soda and vinegar in oil perhaps, will you increase frying?

I am extremely skeptical that a typical cooking oil and water form a significant azeotrope. Such a claim demands an authoritative reference. Most known azeotropes have boiling points only fractions of a degree below the lower boiler, making little difference in the boiling point. Likewise, most known azeotropes contain over 99 wt% of the lower boiler again making most azeotropes insignificant.
It is simply my opinion, but I believe the oil transported from the bubbling liquid is in the form of an aerosol or perhaps even larger droplets.
Those issues aside I agree the bubbling is boiling of lower bp chemicals, almost certainly mostly water. As cooking occurs, decomposition into lower bp compounds is possible (and if significant amounts form, they may form azeotropes with the oil (but see comments above)).
I am also not aware of the solubility of the various atmospheric gasses (N2, O2, CO2) in oil. Gasses are in general less soluble as temperature increases (eg carbonated beverages), so they also would boil out if present in significant amounts (I'd guess they aren't very soluble at 1 atm pressure).
I should also add that while most azeotropes aren't "significant", most liquids do form them.

This is not an answer to the OP's question but rather a comment on another user's answer.

@ToddMinehardt No, it is. While also addressing previous answer OP gives many other informations. I feel it's atually better answer, it mainly lacked formatting.

I agree this answer is better than mine. I simply repeated what my teachers told me, even though I know I should be more critical than that. I could not find any source for my claim and therefore enthusiastically rescind my answer.

@Mithoron - I stand edified. Retracting down-vote.

Aw what a shame, oil mist is what I thought was responsible in the first place, but the azeotrope idea seemed so cool! I figure both can actually happen simultaneously, but the oil droplet mist is likely the major contributor by far.

EDIT: This is very probably wrong, and to be considered so unless someone can find a reliable source for these claims. See @Li Zhi's answer to know why.
The bubbles are not the cause of the frying, but its consequence.
The main purpose of frying an aliment (apart from cooking it, of course) is to remove its moisture. It does so by boiling the water away. Hence the bubbles.
To delve a bit on the technical side, the water/oil mixture probably forms an heteroazeotrope. This means it is not water which boils, but a mixture of (mostly) water and (a bit of) oil, with a boiling point below either. This is why frying things is so messy: you actually "boil" oil which recondenses into droplets that cover everything around, like some disgusting, greasy morning dew.

Oh wow, I never realized the oil and water formed an azeotrope and that's why everything got greasy, amazing! Now I wonder if there's a market for a Dean-Stark apparatus lid accessory for pots and pans...

I'm afraid Li Zhi is right.

I am going to come down on the water is boiling because the temperature of the oil is higher than $100\ ^\circ \mathrm{C}$


text:
Why so much color change in synthesis of triphenylmethanol?
In lab today, I prepared a solution of Grignard reagent (phenylmagnesiumbromide) in anhydrous ether. I then added a solution of diphenylmethanone in anhydrous ether (AKA benzophenone) to the Grignard reagent and observed a brilliant array of color change.
Immedietely when the first drops of the benzophenone were added to the cloudy grayish grignard mixture (grey color due to impurities in starting magnesium), the reaction turned a dark purple before quickly changing to a crimson red, before once again changing to a bright pink. As more benzophenone was added, the reaction mixture changed, once more, to a milky white color and promptly hardened.
I understand the white final color and hardness is due to formation of my intended product: triphenylmethoxide magnesium salt...
But WHAT in the world was happening in that reaction with all those fantastic color changes?

The various shades of blue-purple are a trademark color of benzophenone dianion that could be created by reduction of benzophenone by  remaining excess magnesium. Benzophenone and sodium are commonly used in distillation of solvents to produce especially pure and oxygen-free solvents. At lower concentration the anion might appear to have brighter shades.

@permeakra, magnesium should be relatively fresh, but yes, that's also a possibility, good catch :)

It sounds like you might have generated a species with this structural moiety:

This can happen if your solution was concentrated enough or the boiling of your ether was vigorous, so the local overheating caused the formation of $\ce{MgO}$, the guy above, and $\ce{Br-}$, which gives a very distinct bright red-pink color, which is VERY bright and only requires a bit of the side product to form. Though this carbocation is kinda stable, it will find the $\ce{Br-}$ in the solution and form triphenylmethylbromide as a minute impurity.

This is a pretty common lab experiment in the undergraduate organic chemistry lab, including in courses I teach. I would love to see a reference for this cation so that I can share it with my students.

The tri  para amino derivative is called crystal  violet a well known dye.

the cation you've drawn is bright yellow, see wiki for the corresponding alcohol

@BenNorris exactly, every other student in my lab observed similar color array and its described in other protocols for synthesis of triphenylmethanol I found online. Neither the TA's nor professor could pinpoint the exact cause of all the short-lived color change.

@permeakra, yes, you're right, I didn't mean the cation itself as is, it would need an EDG to give more of a blue hue in low concentrations, or shift towards blue when "stacked" with, say, benzophenone or other aromatix.

The colour is because of electron delocalisation over the three rings rather than just one because the cation is planar. (see structure in answer from @ChemistryHelpCenter). Think of particle on a ring, bigger ring , lower energy, thus the absorption of the compound changes from being in the uv (only benzene rings delocalised ) to all three rings, and absorption in the visible. The final product will have a more or less tetrahedral structure so this delocalisation is lost and the colour is again in the uv and outside out visual perception.

Well, maybe it's again nitpicking, but trityl isn't planar, or one may say only as planar as it can get before steric repulsion of ortho hydrogens blocks further planarisation.

@Mithoron , yes not quite planar as you say, so I should have been more exact in my description. The rings are twisted rather like  propellor blades, but nevertheless the delocalisation produces the colour as it were. In fact, the photophysics of these compounds is very interesting showing very non-exponential fluorescence decays and solvent dependent dynamics.


text:
Fischer to Haworth projection: which carbon connects to the oxygen?
For D-erythose, carbon 4, (the last carbon), connects to the O which connects to Carbon 1.

However, for D-glucose, carbon 5 (the 2nd last carbon) attaches to the O which attaches to carbon 1.

Why is it the last carbon for D-erythose but the second last carbon for D-glucose that is connected to the oxygen in the Haworth projection? How do I decide which one gets connected when trying to convert Fischer to  Haworth projections?

Related: Why is H bonded to C5 drawn below the plane in Haworth formula for D-glucose?

It just is. There are pyranose and furanose forms for all aldoses and hexoses (given they are long enough), including glucose:
$\hspace{3cm}$
You will just have to remember which connections correspond to which sugars. You can read more about anomers ($\alpha$ vs. $\beta$) here, and more about carbohydrate chirality (D vs. L) here.

I don't think the pyranose and furanose, the  α and β or the D / L isomers are of any relevance here

All sugars attach to carbon n-1 when making pyranose rings, except for the ones with 4 carbon atoms. The reason being, a ring with four atoms wouldn't be stable (too much strain).

I don't get your point and it looks like comment not answer.

It supposed to mean, that it only attaches there, because it can't attach anywhere there

Oh, you mean that tetrose it's the only option... Yes that's a good answer, but you should elaborate and be more precise.

It’s not important whether the last, penultimate, fifth-to-last or any other carbon’s oxygen reacts with the carbonyl group to form a cyclic structure. The only thing that matters is the ring size you generate.
A concept called Baeyer strain exists which was introduced by Adolf von Baeyer in the 19th century. It bases on the fact that the tetrahedral angle is approximately $109.5^\circ$ but a the bonds in e.g. a three-membered ring must feature bond angles of $60^\circ$ due to the laws of geometry. The difference in bond angles induces strain.
This Baeyer or angle strain is particularly prevalent in small rings (three- and four-membered) and semi-large rings (eight- to twelve-membered). However, five- and six-membered rings are notable exceptions; both are almost unstrained. For five-membered rings this is immediately obvious since a pentagon has an inner angle of $108^\circ \approx 109.5^\circ$. For six-membered rings, one needs to account for the chair conformation, which allows perfect tetrahedral angles.
Since five- and six-membered rings are rather stable compared to other ring sizes, monosaccharines will primarily form these ring sizes. If both are possible, a five-membered ring is often favoured kinetically (i.e. it forms more rapidly than a six-membered one) but the six-membered ring is favoured thermodynamically (i.e. it has a lower overall energy). For many sugars, six-membered rings overwhelmingly dominate the species found in solution: in glucose solutions, the five-membered (furanose) forms are only present in less than $1~\%$ of all possible forms.
However, if you take erythrose or threose, six-membered rings are impossible since only five atoms can from the chain. Thus, these sugars will exclusively form furanose structures, never pyranoses (six-membered rings).
When determining which oxygen attacks, start counting on the carbonyl carbon and try counting to 6, landing on an oxygen. If that is possible, a pyranose form (six-membered ring) is very likely. If not, try again attempting to reach an oxygen on 5. If that is possible, furanose forms will form. Other ring sizes can typically be disregarded as they only account for minute quantities of the solute distribution.


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treatment of alkyl chloride with aqueous KOH and alcoholic KOH
The treatment of alkyl chloride with aqueous KOH leads to the formation of alcohol but in the presence of alcoholic KOH alkenes are major product. Why?

I have added an image to answer your question, always look at the acid base reactions which are possible to take place, that helps in understanding the reactions and their respective mechanisms.

Please cite your source, and consider typing what's included in the image.

As I couldn't type the reactions as answer I wrote them on MS-word and uploaded the screenshot.


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Chromene and cumarin nomenclature
We have these nomenclatures of the same family :
4H-chromene
2H-chromene
Why do we use 2H and 4H to name them? And how ?

There are two isomers of chromene (1-benzopyran): 2H-chromene and 4H-chromene.

According to Subsection P-14.7 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the name can be made specific by indicating the position of the extra hydrogen atom in the structure, i.e. the position where no multiple bond is attached. This is accomplished by adding to the front of the name an italic capital ‘H’ preceded by the appropriate numerical locant for this atom.
In 2H-chromene, the ‘indicated hydrogen’ locates one extra hydrogen atom in position ‘2’ of the chromene structure; and in 4H-chromene, the ‘indicated hydrogen’ indicates an extra hydrogen atom at position ‘4’, i.e. one hydrogen atom more than the number present if there were a double bond in the ring at that position.


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What is the state of stationery phase in paper chromatography?
As per my textbook:

Partition Chromatography: Partition chromatography is based on continuous differential partiotioning of components of a mixture between stationary and mobile phases. Paper chromatography is a type of partition chromatography. In paper chromatography, a special quality paper known as chromatography paper is used. Chromatography paper contains water trapped in it, which acts as the stationary phase. (highlight mine)

According to the book, paper chromatography is a kind of partition chromatography where the stationary phase is water, as I think it should be, because in partition chromatography a solid is not the stationary phase. For instance, the Wikipedia entry on partition chromatography states:

Chromatographic separation was considered to occur by an adsorption process whereby compounds adhered to a solid media and were washed off the column with a solvent, mixture of solvents, or solvent gradient. In contrast, Martin and Synge developed and described a chromatographic separation process whereby compounds were partitioned between two liquid phases similar to the separatory funnel liquid-liquid separation dynamic. This was an important departure, both in theory and in practice, from adsorption chromatography

But according to the Wikipedia and this site, the stationary phase in paper chromatography is paper (a solid).
So which one is correct? Is paper chromatography really a kind of partition chromatography or adsorption chromatography?
The picture is from my textbook: NCERT class 11, pg-353.

It's a bit of both, actually, and also ion-exchange chromatography.
The cellulose and other components of paper do exhibit adsorption and ion exchange properties with some analytes, but as paper chromatography is usually performed, partitioning between the flowing solvent and trapped water is the major mode of separation.
Paper chromatography typically uses water-saturated solvents and enclosed chambers where water vapour can keep water trapped in the paper (the paper is equilibrated with this vapour before development).
Since cellulose is very hydrophilic, immiscible or poorly miscible solvents are usually used, which keeps the water stuck to the cellulose as it has much higher affinity for water than the solvent.
This is an early paper showing the apparatus used for paper chromatography of amino acids and this book details the theory involved starting on page 4.

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


text:
Computational methods to obtain relative energies of electronic configurations of atoms
When one learns the Aufbau principle to “predict” electronic configurations, and the $(n + \ell)$ ordering rule (or Madelung rule), one also learns of the exceptions to the rules… some of which (Cr, Cu, Mo, Ag, Au) can be half-explained by adding another “rule”, some of which (Nb, Ru, Rh, …) cannot. We have on this site our fair share of electronic-configuration questions dealing with these: Nb, Pt, Pd, Cu, Th, etc.
My question here is not about rules or theories to rationalize and understand these complicated electronic configurations, but rather about computational tools to predict them. What computational methods (and software) would one use to compute the different energies of several electronic configurations of an atom?
For example, take niobium. How would I answer, with computational quantum chemistry methods, the following question:

what is the energy difference between the 5s1 4d4 and 5s2 4d3 electronic configurations of niobium?

And would that method also give some interpretation in terms of what these energy differences arise from?

For atoms, Full CI should do the job. Otherwise DMC.

The different occupations you suggest correspond to different spin states. Therefore, they are quite simple to distinguish.
On the other hand, the ground states are often degenerate (in the terms of 5 equivalent d orbitals available). Therefore the lowest (and sufficient) level of theory is the CAS-SCF with inclusion of the s- and d- orbitals.
I tried several calculations using ORCA in def2-svp basis and the results correspond to what one would expect, the time to perform the calculations is in order of seconds.
Zr (should be $4d^2$ $5s^2$, ie. MULT=3)
    CAS-SCF STATES FOR BLOCK  1 MULT= 5 NROOTS= 1
    ROOT   0:  E=     -46.4247189198 Eh
    0.62280 [     0]: 211110
    0.37690 [     2]: 211011

    CAS-SCF STATES FOR BLOCK  2 MULT= 3 NROOTS= 1
    ROOT   0:  E=     -46.4308999710 Eh
    0.49082 [     0]: 221100
    0.47988 [     2]: 221001
    0.01125 [    18]: 210111
    0.00276 [    27]: 201120

Cr (should be $3d^5$ $4s^1$, ie. MULT=7)
    CAS-SCF STATES FOR BLOCK  1 MULT= 7 NROOTS= 1
    ROOT   0:  E=   -1043.1949498785 Eh
    1.00000 [     0]: 111111

    CAS-SCF STATES FOR BLOCK  2 MULT= 5 NROOTS= 1
    ROOT   0:  E=   -1043.1559672632 Eh
    0.40000 [    15]: 111111
    0.30000 [     0]: 211110
    0.30000 [    30]: 011112

Nb (should be $4d^4$ $5s^1$, ie. MULT=6)
    CAS-SCF STATES FOR BLOCK  1 MULT= 6 NROOTS= 1
    ROOT   0:  E=     -56.3066169626 Eh
    0.96220 [     0]: 111110
    0.03771 [     2]: 111011

    CAS-SCF STATES FOR BLOCK  2 MULT= 4 NROOTS= 1
    ROOT   0:  E=     -56.2843641219 Eh
    0.89474 [     0]: 211100
    0.04255 [     2]: 211001

The lower the energy the better.
From the numbers I'd conclude that it is possible to obtain the ground state configuration by reasonably simple quantum chemical calculations. Nevertheless, I'd expect also less elaborate theories to predict such a fundamental properties.
Edit: I added the multiplicity to the expected ground state configuration

How do you then distinguish electronic configurations that have the same spin state? Excited states, for example?

@F'x: As this was not a part of the original question, I took the simplest approach. Nevertheless, it is possible to calculate the excited states as well.
But to check whether the ground state has one or two electrons in the s-orbital, the described procedure is enough, as they have to differ in total spin.

In addition to Riccardo's answer, I would like to mention software like Gaussian and GAMESS which can be used for electronic configuration studies (computational approach).

How would you use typically quantum chem software to compute atomic electron configurations?

You can input different spin multiplicity values (Molecule specification) and check which one is giving you minimum energy. This is helpful in finding stable electronic state (using high level of theory).

See the @ssavec's answer below. Here, different multiplicity values [MULT = 5, 3 etc] are giving different energy values and hence helpful in predicting electronic state.

To have such detailed information you need to use a highly correlated method (CI) adding spin-orbit coupling. Dirac should do the job.
Of course you cannot "force" the electrons on orbitals, but check the different spin states.


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is there a database of known chemical reactions?
I am looking for a complete directory of chemical reactions. It should be searchable by element or substance and also the various factors (pressure, temperature, catalyst, etc) should be mentioned. Is there something like this? Thanks!


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Ultrasound Vegetable Glycerin homogenization
We are trying to homogenize VG (USP) with propylene-glycol using ultrasonic homogenizer with titanium probe, but we are experiencing very strange - bad odor from the emulsion after sonication.
We tried different amplitudes, different time settings, different power settings. We can control amplitude for 2 modes only, I believe they are 50% and 100%.
Also if I am sonicating pure vegetable oil for just 5-10 sec I can smell that strange odor, it is very similar to that, I am getting with VG.
I just want to know, what this smell can come from? From our probe? No it is clean, we can sonicate water well.

Depending on intensity, ultrasound can produce intense heat at some locations. In fact, cavitation can induce temperatures far exceeding 5,000 K. Even a small ultrasonic room humidifier can burn through a plastic comb held in the upwelling stream. The sound intensity is decomposing the oil and/or propylene glycol.
Try lower frequencies and lower sound intensity. You might experiment with different shape containers, as well, to avoid focusing the sound into a high-intensity spot.

Can you recommend device that is perfect to work with vegetable glycerin? I am accepting the answer, but wanted more information. Can't find any special info about VG and sonication, in my case sonication is used for homogenization process only.

Sorry, I don't have actual experience with using ultrasound for homogenization. You might compare your design with a mayonnaise emulsifier, though: http://www.hyfoma.com/en/content/food-branches-processing-manufacturing/oil-margarines-sauces/mayonnaise/. These rely on mechanical stirring and use of a lecithin (egg-yolk for mayonnaise) to form the emulsion.


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How does adsorption work?
I would like to know how adsorption works particularly between a gas and a solid. I know that there is an attraction of intermolecular forces between the gas molecules and the solid surface molecules. But how can this attraction suck all the gas molecules onto the solid surface? I thought the intermolecular forces cannot act at such long distances. Any idea?

The pressure exerted on the wall by the gas is macroscopic representation of collisions of gas molecules with surface. Every such collision is an opportunity for chemical or physical interaction resulting in adsorption.

Adsorption (with a d) is caused by intermolecular Van der Waals Forces allowing molecules to stick to a solid surface.  The solid surface can be described using a Langmiur model.  In this model a surface can bind with a number of gas molecules at any number of active sites. These active sites may be packed much closer together than molecules dispersed in a gaseous state. When dispersed gaseous molecules in a closed container interact with a surface packed with active sites, the surface may appear to "suck the gas molecules" as the gas molecules bind to the densely packed active sites.
This suction is particularly true of several activated carbons because the porous surface causes a large number of active sites to pack together.
The Langmuir model is only one model. In this model once all of the active sites are filled, the gas around the surface forms a monolayer. The monolayer is expanded upon in more sophisticated models such as the BET model.  The BET model is used to analytically determine BET surface area of a material like activated carbon.
Sometimes the behavior of the surface is described empirically such as in the Freundlich adsorption isotherm.  In this case, as with all empirical models, there is no explanation for the attraction sucking molecules to the surface.  It is simply accepted the model works.

Intermolecular - between different molecules would be more appropriate IMO

@Mithoron - you are right ..  trying to write too fast..  changed it

On colliding with a surface the molecules do not necessarily bounce off because, depending on the collision energy, Van-der-Waals forces can act to hold the gas molecule close to the surface. These are typically London dispersion forces but depending on type of surface and molecule, any type of VDW force could occur. This process is called physisorption and the molecule may be slightly distorted by the surface interaction, but they are always held at a greater distance than it would be if a chemical bond were formed, typically  > 4 typical bond lengths. (The energy released is comparable to the energy of condensation.)
The surface to which the gas is attracted will have atoms with unsatisfied valence. Thus it is possible that a reaction analogous to chemical bond formation can occur between the surface's atoms and the molecule. By 'analogous' is meant that there is significant wavefunction overlap from surface states to molecule. This can only happen if the molecule can escape from physisorption and so approach more closely to the surface. The sketch shows the potential energy vs. distance from surface. The enthalpy of chemisorption (250$\pm$100kJ/mol) is far larger than for physisorption (50$\pm$20kJ/mol).

(the dotted lines show continuation of the potentials if no interaction between them occurs)
As the molecule approaches the surface the VDW forces can trap it provided that the energy released can be carried away by vibrations/phonons or electrons in the surface faster than the molecule could separate from the surface by rebounding off the repulsive part of the potential. The molecule is now somewhat trapped, the time scale for this this will depend on the size of the attractive well compared to thermal energy. To reach the chemisorbed state, a larger barrier than separating from the surface may have to be breached, so this will be an unlikely process, but not impossible, in the way the sketch is drawn. However, it is quite possible for the barrier to chemisorption to be below zero, in this case the rate constant towards chemisorption will be faster than separation from the surface. As before, the energy released on chemisorption is taken up by the surface states. The potential forming chemisorption and that for physisorption must interact to make the potential barrier between them, this means that there has to be some symmetry elements common to both, or, put another way, there has to be some type of interaction, say spin-orbit coupling or electronic/vibrational anharmonicity to allow this to happen.
The kinetics of the on-off process is given in its simplest form by the Langmuir isotherm. See my answer to the question Is the rate of desorption dependent on pressure?  for a description of this isotherm.


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Nomenclature of aryl halides
How do you know when to name a benzene ring substituted by a simple haloalkane as a derivative of benzene or as a derivative of the alkyl halide.
For example, is it 1-chloro-4-isopropylbenzene or 2-(4-chlorophenyl)propane?

See answer in http://chemistry.stackexchange.com/questions/57245/naming-of-a-hydrocarbon-with-a-benzene-functional-group

So to sum up, both are okay but the first is preferred?

In the usual substitutive nomenclature, simple halogen compounds are always expressed by prefixes (‘bromo’, ‘chloro’, etc.). Hence, the senior parent structure of such compounds is not determined by a principal characteristic group (a characteristic group chosen for citation at the end of a name by means of a suffix or a class name, or implied by a trivial name).
According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the ring or the chain can be the senior parent structure; however, for the preferred IUPAC name, the ring is always selected as the senior parent structure.

P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)
Two methods are recognized to name systems composed of rings and chains (exclusive of linear phanes).
(1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.
(2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.
(…) For selection of a preferred IUPAC name, see P-52.2.8.

P-52.2.8 Selection between a ring and a chain as parent hydride
Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

Therefore, the benzene ring is selected as the senior parent stucture for the preferred IUPAC name of the compound given in the question, which leads to ‘1-chloro-4-isopropylbenzene’. However, according to current recommendations, the preferred prefix for the isopropyl substituent is ‘propan-2-yl’; the prefix ‘isopropyl’ is retained for use in general nomenclature. Thus, the complete preferred IUPAC name is ‘1-chloro-4-(propan-2-yl)benzene’.


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Why is copper (II) oxide used in quantitative analysis of carbon?
What is the significance of using cupric oxide in the quantitative analysis of carbon?
As far as I can tell from my research, only $\ce{CuO}$ is used to oxidize carbon to $\ce{CO2}$. Why can't we use other oxidants like $\ce{HNO3}$ or other metal oxides such as hematite? Is the use of $\ce{CuO}$ prevalent because of low cost or does it has some other special property?

"The quantitative analysis of carbon" is done using many different procedures. Exactly what type of sample and what procedure are you interested in?

@MaxW          But many sites and my textbook mention only one compound CuO for estimating the presence of carbon in an organic compound, so out of curiosity I asked this question . Could you please refer any book or website where oxidant is other than CuO? A particular sample is not mentioned and it is given in general.

For example it is common to analyze for carbon in steels and rocks. You won't use the same procedure for analyzing rocks that you use for steels.

You got the crux of it: low cost.  And it works, regardless of what the starting compound was.  And it's not just low cost to purchase, it is easy to regenerate the used cupric oxide multiple times by heating it under oxygen.


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Why does an anion not only bind to a single cation?
When $\ce{Na}$ binds with $\ce{Cl}$ to form $\ce{NaCl}$, why does the Na bind with multiple $\ce{Cl}$'s and not only one? A Na-ion is positively charged due to donating an electron and a $\ce{Cl^-}$ion is negatively charged due to accepting an electron. That's why they bind. But salts form a grid where each Na-ion binds with 6 $\ce{Cl^-}$ions and vice versa.
Shouldn't be the Na-ion already be "used" after binding with a $\ce{Cl^-}$ion? Or does it stay charged and attracts all ions that are negatively charged?
I hope someone can help me to understand.

In the vapour phase where NaCl can exist (obviously at high temperature) then the molecule is a diatomic, with a covalent bond, albeit with a lot of ionic character.
In the crystal, to which I assume you refer, there is exactly one $\ce{Na^+}$ to each $\ce{Cl^-}$. You can see this if you look at the structure of the crystal, however to achieve this in the cubic crystal each ion has to be surrounded by six of the opposite charge as you correctly state. However, the charges are not 'used up' with one neighbour but shared equally between them, so a whole perfect crystal is electrically neutral. (If you want you can think of the charges as electric fields spreading out radially from each ion, they cancel each other out at some point between ions making a complicated shaped surface of zero potential). The bigger chloride ions form a simple cubic array with the sodium in the gaps. The crystal has this structure, as opposed to another (say face centred cubic), because this is the one than minimises the total energy, by keeping (large) chloride ions far enough away form one another so that their repulsion does not destroy the crystal.
In aqueous solution the water, which is very polar, means that it is more energetically favourable to dissociate into ions and solvate them rather than keep NaCl as a diatomic molecule, i.e. form $\ce{Na^+(aq) + Cl^-(aq)}$ where each ion is surrounded by several water molecules.


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Reacting sodium metal with aqueous sodium hydroxide, what would happen?
I know sodium metal, when reacted with water, produces flammable hydrogen gas but I was wondering if that would also be the case if I had aqueous 1M sodium hydroxide?
Would this reaction be safe or will it also produce a harmful compound like hydrogen gas?

Yes, the sodium metal will react with the water to produce more sodium hydroxide and hydrogen (gas)... until there is no water left to react with.

ok, so will the sodium metal be consumed? And how fast would that be?

1M aqueous sodium hydroxide is no different from water in this regard.

Follow up question that (to me at least) is slightly less easy to answer: what if you react molten sodium metal with molten "anhydrous" sodium hydroxide?  Do you get sodium oxide and hydrogen?

Consider first the solvation of sodium metal in water to give a sodium ion and a hydrated electron.
$$\ce{Na(s) \xrightarrow{\ce{H_2O}} Na^+(aq) + e^-(aq) }$$
The hydrated electron will then act as the reactive species to reduce water to hydrogen atoms and hydroxide anions.
$$\ce{e^-(aq) + H_2O(l) \rightarrow H(aq) + OH^-(aq)}$$
Note that two of these hydrogens come together to form the diatomic hydrogen gas.
$$\ce{2H(aq) \rightarrow H_2(g)}$$
When we multiply the coefficients in the first two equations by two, we are able to consider the net ionic equation through the elimination of the reactive intermediates.
$$\ce{2Na(s) + 2H_2O(l) \rightarrow 2Na^+(aq) + 2OH^-(aq) + H_2(g)}$$
We see that the dissolution of sodium metal in aqueous sodium hydroxide would be affected by Le Chatelier's principle due to the presence of the sodium hydroxide. We would expect there to be a reduction of solubility and rate of dissolution. The greater the initial concentration of sodium hydroxide, the more appreciable the effect.
Given that a solution of sodium hydroxide at $\ce{25 ^{\circ}C}$ is saturated at $\ce{25 M}$, a $\ce{1 M}$ solution would not be expected to have an appreciable effect. Hydrogen gas would remain a product, evidenced by the net equation.
As far as the reaction of molten sodium and molten sodium hydroxide, I expect that is what they were referring to on the Wiki page for sodium oxide when they gave this equation.
$$\ce{2NaOH + 2Na \rightarrow 2Na_2O + H_2}$$
The source given is Greenwood, Norman N.; Earnshaw, Alan (1997). Chemistry of the Elements (2nd ed.). Butterworth-Heinemann. ISBN 978-0-08-037941-8.
The solubility of sodium hydroxide was resourced from the Wiki page for sodium hydroxide. The source given appears to be Haynes, CRC Handbook of Chemistry and Physics.
The first set of equations were sourced from Walker, David C. Production of Hydrated Electrons. Canadian Journal of Chemistry. Volume 44. 1966.

The reaction starts Na(s) ->Na+(aq) + e-(aq). Hydrated electrons are much more reactive than in liquid ammonia, but still catchable by their colour by 10000 fps fast cameras. It was also backup up by quantum chemistry simulation of behaviour of cluster of sodium atoms. Hydrated electrons than reduce water to H2 and OH-.

Thank you, @Poutnik. I believe I have correctly updated the equations to reflect this.

Sodium metal reacting with aqueous sodium hydroxide will continue to react, producing hydrogen ($\ce{H2}$) and sodium hydroxide ($\ce{NaOH}$). Eventually, however, the solution will become saturated with sodium hydroxide and the $\ce{NaOH}$ produced will form a coating on the outside of the sodium, preventing further reaction from occurring.

Eventually, however, the solution will become saturated with sodium hydroxide and the NaOHNaOH produced will form a coating on the outside of the sodium, preventing further reaction from occurring. || sodium gladly reacts with sodium hydroxide at elevated temperature. Furthermore, 'saturated' aqueous solution for sodium hydroxide is 1:1 molar mixture.

@permeakra I didn't know this before. What product will the reaction make, as a matter of interest?

Sodium hydride and sodium oxide.

what about room temperature, then it will follow what you said of coating or will there be a totally different reaction?


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What does the "total energy" of a molecule mean for the Chem3D MM2 minimized energy calculation?
I was fiddling around with Chem3D and the MM2 calculation for finding a molecules minimized energy.
Benzene was assigned a total energy of -0.5741 kcal/mol while biphenyl was assigned a total energy of 4.5397 kcal/mol.
What is the "total energy" relative to? In other words, why does benzene have a negative total energy while biphenyl has a positive total energy? What is the 0 energy state that they are being compared to?

Typically the force field energies are parameterized to attempt to reproduce heats of formation. Otherwise, the answers below are pretty good.

General definition of energy
Usually, the total energy of a system is subdivided into kinetic energies of all particles and potential energies of the interactions between them, where the later terms are most often limited to the strongest Coulomb interactions only. In atomic units we get then:
$$
    E
    =
    - \sum\limits_{\alpha=1}^{ν} \frac{1}{2 m_{\alpha}} \nabla_{\alpha}^{2}
    - \sum\limits_{i=1}^{n} \frac{1}{2} \nabla_{i}^{2}
    - \sum\limits_{\alpha=1}^{ν} \sum\limits_{i=1}^{n} \frac{Z_{\alpha}}{r_{\alpha i}}
    + \sum\limits_{\alpha=1}^{ν} \sum\limits_{\beta > \alpha} \frac{Z_{\alpha} Z_{\beta}}{r_{\alpha \beta}}
    + \sum\limits_{i=1}^{n} \sum\limits_{j > i}^{n} \frac{1}{r_{ij}} \, ,
$$
where $m_{\alpha}$ is the rest mass of nucleus $\alpha$ and $Z_{\alpha}$ is its atomic number, $r_{\alpha i} = |\vec{r}_{\alpha} - \vec{r}_{i}|$ is the distance between nucleus $\alpha$ and electron $i$, $r_{\alpha \beta} = |\vec{r}_{\alpha} - \vec{r}_{\beta}|$ is the distance between two nuclei $\alpha$ and $β$, and $r_{ij} = |\vec{r}_{i} - \vec{r}_{j}|$ is the distance between two electrons $i$ and $j$.
Now, the zero of energy corresponds to the case when all the terms are zero. Kinetic energy is zero when velocity is zero, while the Coulomb interaction energy is zero when particles are infinitely separated from each other. Thus, the zero of energy corresponds to the case when all the particles are infinitely far away from each other and not moving. In accordance with that choice of the zero of energy, the energy of any stable molecular system in the modeled environment (usually, an isolated molecule in the gas phase) has to be negative.
Molecular mechanics definition of energy
Molecular mechanics (MM) operates with a different kind of energy. To start from, usually, it is just the potential energy. Secondly, in MM we think about a molecule in terms of atoms and bonds (rather than nuclei and electrons) and the potential energy is divided into contributions from bond stretching, angle bending, torsional rotations around single bonds, non-bonded interactions, etc.
$$
    E = E_{\mathrm{stretch}} + E_{\mathrm{bend}} + E_{\mathrm{torsion}} + E_{\mathrm{nb}} + \dotsc \, ,
$$
where the first term, for instance, is typically proportional to the sum of the squares of extensions from the equilibrium length for all the bonds:
$$
    E_{\mathrm{stretch}} = \sum\limits_{\mathrm{bonds}} k (r - r_{\mathrm{e}})^2 \, .
$$
Now, we could take the energy corresponding to the equilibrium length $r_{\mathrm{e}}$ as the zero of energy and proceed similarly for the other bonded terms, which would mean that (ignoring usually small non-bonded interactions) the MM potential energy is zero when the molecular system is completely "unstrained" (i.e. when all geometrical parameters have their "natural" equilibrium values). However, different molecules are bonded differently, and thus, the zero of energy would, in general, be defined differently.  Each potential energy will thus be computed relative to a different zero of energy which makes the comparison of energies for any two different molecules meaningless.

Perhaps the steric effects from the methyl groups as well as the repulsive effects from a higher electron density conjugated system contribute to the positive total energy? I'm sure nothing is wrong with the set up and even a simple MM2 model should be capable of approximating a positive or negative energy.

The Chem3D manual confirms your definition of total energy as sum of kinetic energy and potential energy. However, the “Total Energy” reported by MM2 in Chem3D is actually the total steric energy, which is the sum of the individual terms for bond stretching, bond angle bending, torsion, van der Waals interactions, and electrostatic interactions. (I can reproduce the results given in the question with default values for MM2.)

The assumption of zeros at infinite separation are false for bonded potentials in force fields (which MM2 is). Take a closer look at the parametrization of force fields. Usually bonded potentials are zero at equilibrium. Hence you will get positive energies when these potentials are coupled while adding electrostatics can go either way.

The total energy is indeed the sum of kinetic and potential energies. In force fields such as MM2, the potentials are split into bonded and non-bonded terms. The bonded terms are generally such that they are zero at their respective global minimum. For instance, a harmonic bond potential has the form $k_{ij}(x_{ij} - x_{ij}^{(0)})^2$. When you combine enough bonds, it is no longer generally possible to find a geometry that all bonds are at respective $x_{ij}^{(0)}$. Since the potential is non-negative, this means the energy has to be non-zero and positive.
The nonbonded terms generally do not follow this definition, but rather tend to zero at infinity, e.g., the Coulomb term or a Lennard-Jones vdW potential. These may furthermore have any sign so you may witness negative or positive energies.
In your specific case of benzene and biphenyl, you need to look at the individual contributions to discern why biphenyl is positive in energy.

Why do you think that "it is no longer generally possible .. that all bonds are at [the minimum]"? While I agree that in complicated systems, angle, dihedral, non-bonding terms, etc. might cause problems - it's certainly possible to have large systems where all bonds are at the equilibrium bond length.

I am using "general" in the same sense as two lines in general position aren't parallel and I am talking about the minima of the constituting terms, so while a large system may be at the bond minima of the constituting harmonic terms, it generally won't be. Consider three bodies with 3 bonds and one of those bonds' minimum energy separation exceeds the sum of the other two bonds' minimum energy separations. Add angles and non-bonded terms and the "problem" is exacerbated.


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How to determine the order of nucleophilicity for given chemical species?
How to determine the order of nucleophilicity for given chemical species?
Like I came across this question, to rearrange $\ce{RCOO-, OR-, OH-, H2O}$ (alkyl acetate, alkoxide, hydroxide, and water) in decreasing order of nucleophilicity and it ranked $\ce{OR-}$ first.
Shouldn't it be $\ce{OH-}$ since $\ce{OR-}$ would be more sterically crowded?

It might be but if it's simple alkyl group then inductive effect is usually more important.

I guess if this question attempts to become a canonical question on how to determine the order of nucleophilicity for a given chemical species, then a better answer with more elaboration on steric effects/electronic effects/etc. would be necessary, at least more than the currently accepted answer.

Within a group of nucleophiles with the same atom, the nucleophilicity decrease with decreasing basicity of nucleophile.
Decreasing basicity means the decreasing affinity of electron pair for a proton. The decreasing order of nucleophilicity is shown in the figure.

But the relationship between nucleophilicity and basicity can be reversed by steric effect. Less basic but steric unhindered nucleophile therefore have higher nucleophilicity than strong basic but strically hindered nucleophile.

In this case you can look at the pKa values of their conjugate acids. Higher pKa of conjugate acid means a weaker acid hence a stronger base. The pKa of water is 15.7, while most alcohols have pKa of 16-18.
I guess this question is asking for a general trend. Although alkoxide could be sterically hindered, for which an extreme case would be a tertiary alkoxide, but if the electrophile is not hindered at all (e.g. primary alkyl halide), then OR⁻ is still stronger than OH⁻.

In addition to looking at pKa's, you must also consider the stabilizing effects of the surrounding groups.  RO- has an alkyl group attached, allowing a greater amount of polarizability.  This means Oxygen's loan pairs will be more readily available to react (ie they will not be held in as close and as tight) in RO- than in HO-.  Consider Fluoride vs Iodide, the same concept applies.  Fluorine has a smaller atomic radius and thus less polarizability (less area to distribute the charge over) than Iodide.  Fluroide holds onto it's loan pairs much tighter than Iodide, and therefore the loan pairs are much less available to react.
Another great example of this, consider sulfides.  They are generally better nucleophiles than similar compounds that contain oxygen instead of sulfur.  This is due to the larger atomic radius of sulfur and therefore greater polarizability/charge distribution.
Also, negatively charged species are almost always more nucleophilic than neutral species (when considering the charged form vs the neutral form of the same molecule, such as water vs hydroxide or ammonia vs amide).

OR¯ is NOT RO¯ !!!
Clearly, R¯ is an unknown species and nothing else can be said about it.
OR¯ would be an unusual structure, given oxygen's electronegativity compared to carbon's. In the case of RO¯, the R "group" would have to be very large and bulky to dramatically sterically affect (hide) the O¯ from other molecules. So for simple organic compounds, the effect will be quite small. As already said pKa of the conjugate acid would be an excellent guide to nucleophilicity (in the relevant solvent).

sigh OH- = HO- doesn't it? So where did you get this idea?


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Would sodium metal react differently with aqueous solutions than with pure water?
I know sodium metal reacts violently with water, but what about aqueous solutions (like sodium bicarbonate). Also, what would the reaction kinetics be?
I have watched YouTube videos were a small piece of elementary sodium reacts completely with water in a matter of seconds. Would that also be the case of elementary sodium in aqueous solutions (for example sodium bicarbonate)?

Yes, it would be pretty much the same as with water.

what its the solution is very concentrated (like 6M sodium bicarbonate) would it still behave like it will with water?

You can't make sodium bicarbonate 6M. Water simply wouldn't dissolve that much.

This video shows what happened in 1947 when the U.S. Army dumped 20 thousand pounds of sodium (in 3500 pound containers) into frozen alkaline Lake Lenore, WA, USA: https://www.youtube.com/watch?v=HY7mTCMvpEM . Not sure how alkaline the lake was prior to the sodium addition.

It helps to know what the reaction actually is. When sodium metal reacts with water, it is oxidised and the hydrogen in water is reduced according to the following equation:
$$\ce{2Na (s) + 2 H2O (l) -> 2 Na+ (aq) + 2 OH- (aq) + H2 ^ (g)}\tag{1}$$
Note that no other components play a role; the only thing that does is the $\mathrm{pH}$ value of the water (lower $\mathrm{pH}$ would mean a more violent reaction) — this is a spoiler and later shown in equation $(8)$. Since this is a redox reaction, we can use standard half-cell theory to analyse it. Most importantly, thanks to equation $(2)$, only the effective cell potential influences $\Delta G$ ($z$ is the number of electrons transferred while $F$ is Faraday’s constant) which in turn can be separated into two half-cells according to equation $(3)$ which can then be analysed by the Nernst equation $(4)$.
$$\Delta G_\text{cell} = - n_{\ce{e-}}FE_\text{cell}\tag{2}$$
$$E_\text{cell} = E_\text{cathode} - E_\text{anode}\tag{3}$$
$$E = E^0 + \frac{RT}{zF} \ln \frac{a_\text{Ox}}{a_\text{red}} = E^0 + \frac{0.059~\mathrm{V}}{z} \lg \frac{a_\text{Ox}}{a_\text{Red}}\tag{4}$$
The half-cell reactions are shown in equations $(5)$ (oxidation) and $(6)$ (reduction) respectively, along with their standard potentials $E^0$ (under standard conditions, i.e. most importantly $\mathrm{pH\ 0}$).
$$\ce{Na+ + e- <=> Na (s)} \qquad E^0 = -2.71~\mathrm{V}\tag{5}$$
$$\ce{2 H+ + 2 e- <=> H2 (g)} \qquad E^0 = 0.000~\mathrm{V}\tag{6}$$
Let’s take a look at the anodic oxidation first. The initial concentration of sodium ions could be, say, $3~\mathrm{M}$. Remember that the activity is defined as $1$ for solids. With the assumption that $a \approx c$, that would mean our oxidation potential is $-2.68~\mathrm{V}$ as shown in equation $(7)$.
$$E_\mathrm{anode} = -2.71~\mathrm{V} + \frac{0.059~\mathrm{V}}{1} \lg \frac{3}{1} = -2.71~\mathrm{V} + 2.82 \cdot 10^{-2}~\mathrm{V} = -2.68~\mathrm{V}\tag{7}$$
The corresponding calculation for the reductive side, equation $(8)$, is a little harder since we have no real starting value for the partial pressure (and thus activity) of hydrogen gas.
$$E_\text{cathode} = \frac{0.059~\mathrm{V}}{2} \lg \frac{a(\ce{H3O+})}{p(\ce{H2})} = -0.0295~\mathrm{V} \times \mathrm{pH} - 0.0295~\mathrm{V} \times \lg p(\ce{H2})\tag{8}$$
But we do know that a reaction will happen if $E_\text{cell} > 0~\mathrm{V}$ and we have a good idea of the anode’s potential. Thus, if we want no reaction we must set the $E_\text{cell} < 0$ (equation $(9)$):
$$\begin{align}0 &> E_\text{cathode} - E_\text{anode}\\
- 2.68~\mathrm{V} &> E_\text{cathode}\\
- 2.68~\mathrm{V} &> -0.0295~\mathrm{V} \times \mathrm{pH} - 0.0295~\mathrm{V} \times \lg p(\ce{H2})\end{align}\tag{9}$$
The second summand is negative if $p(\ce{H2}) > 1~\mathrm{bar}$. (The equation itself requires a dimensionless pressure which is typically defined to have a the same value as if it were in $\mathrm{bar}$). However, that basically requires performing the reaction under a hydrogen atmosphere. This setup is rather hard and not part of the question. Instead, let’s take the natural concentration of hydrogen which is $1~\mathrm{ppm}$ according to Wikipedia. That results in $0.177~\mathrm{V}$ for the second summand (equation $(10)$).
$$-0.0295~\mathrm{V} \times \lg 10^{-6} = (-6) \times -0.0295~\mathrm{V} = 0.177~\mathrm{V}\tag{10}$$
At least we didn’t lose too much. But there is still the first summand, which is also negative by sign. We just need a high enough $\mathrm{pH}$ value to compensate. The calculation is in equation $(11)$.
$$\begin{align}-2.68~\mathrm{V} &> -0.0295~\mathrm{V} \times \mathrm{pH} + 0.177~\mathrm{V}\\
-2.86~\mathrm{V} &> -0.0295~\mathrm{V} \times \mathrm{pH}\\
96.9 &< \mathrm{pH}\end{align}\tag{11}$$
So we would need a $\mathrm{pH}$ value of almost $97$ to prevent the reaction from happening due to too basic conditions. Of course, this value is unobtainable in water or practically any solvent known to us. Even tert-butyllithium, one of the most basic species known has a $\mathrm{p}K_\mathrm{a}$ value of only $\approx 50$.
You could now try to argue that you would just need to increase the hydrogen gas partial pressure appropriately. So how does the value change if the reaction is performed under hydrogen atmosphere? See equation $(12)$ for the answer.
$$\begin{align}-2.86~\mathrm{V} &> -0.0295~\mathrm{V} \mathrm{pH}\\
90.8 &< \mathrm{pH}\end{align}\tag{12}$$
As you may or may not have guessed, a logarithmic difference of $6$ for the partial pressure results in a lowering of the $\mathrm{pH}$ by $6$ units. So to arrive at achieveable $\mathrm{pH}$ values, your reaction would need to happen at a hydrogen pressure of $10^{75}~\mathrm{bar}$ — go figure.

All things considered, the reaction is so violently endergonic that there is basically no way to inhibit it whatsoever. If metallic sodium sees a sufficient amount of water, it will start reacting, end of discussion.


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Nomenclature of catenanes and rotaxanes
Looking at the IUPAC Blue Book, I noticed that the words “catenane”, “rotaxane” (or “interlocked”, “mechanical bond”, …) do not appear within the rules of nomenclature. Given that a catenane or rotaxane is typically considered a single molecule, I was wondering how these compounds are named. I have read on Wikipedia that catenanes with n bonds are referred to as a [n]-catenanes… but how are specific compounds named?

See also Nomenclature for rotaxanes and pseudorotaxanes (IUPAC Recommendations 2008)


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How does fat affect the precipitation of proteins?
I am doing an experiment using milk, heavy cream, and vinegar. We are trying to precipitate proteins from milk and heavy cream.
When I did the experiment with milk, it was very quick reaction and easily to isolate the casein polymers with vinegar added to the heated milk. This does not occur when the same process is used with heavy cream. I have tried to find out how this process fails with the heavy cream online and in my book there is a decent portion of the chapter dedicated to fats, but it doesn't mention why polymer creation can not occur in this experiment.
Can some one please help me with the reason the heavy cream doesn't curdle?

Casein is the main protein in milk. When you perform the polymerization experiment, the result is an "aggregation" of said protein, which gives the product a solid consistence.
Products such as butter and cream are made with the fats from milk (not the proteins!), hence they are almost casein-free. Of course, being them almost without proteins, the polymerization doesn't occurr.
If you look up the nutrition facts of milk and heavy cream, you see that proteins constitute the 20% of calories of milk, and only the 2% of heavy cream.


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Gases have no fixed volume does that mean that one mole of any gas can occupy whole universe?
First of all , how can one prove that gases have no fixed volume ? And if (though i know that it have) lets take a case . Take small amount of gas and fill it into a large box , about size of earth , consider that box has vacuum in it . The gas should fill all the earth but if does that  , then the intermolecular attraction force should be inconsiderable or negligible at such large distances . What happen then ? There should be a limit to the volume ...
Another question is why molecule let H2 not attract another gas particle let He as there is same van der Waals force ........

Even vacuum can occupy the whole universe...

Of course this isn't entirely true. The "normal" gas laws neglect gravity. Fortunately for us a sufficiently large amount of hydrogen gas in outer space can collapse into a star.

Gases have no fixed volume, if you put it in an infinite vacuum, a gas will at equilibrium have an infinite volume. You'll be waiting a while for it to reach equilibrium though.
H2 will have some interaction with He, but it's very weak, so for most purposes can be neglected.

Why b/w he and h2 is weak force  there but not b/w h2 and h2

@dr.honey Yes there is also an attractive force between individual H2 molecules.

Ya i know .....i want to know why only He-He has attraction but not in  H2-He?

@dr.honey All gas molecules interact, but He-He interaction is weakest known. See comments here


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Dividing different units of measurement?
1) $$\frac{35 \; \text{cm}^2}{0.62 \; \text{cm}} = 56.4516$$
For the units in the answer, do you put $56.4516$ cm$^2$ or just $56.4516$ cm?
1) $$\frac{0.075 \; \text{g}}{0.0003 \; \text{cm}^3} = 250$$
Same here, do you put $250$ g, $250$ cm$^3,$ or $250$ g/cm$^3$?

You can do 'math' with units in the same way you do with numbers... if the units are the same then the indices can add/subtract when multipled/divided. If not, then you just string them together (or convert the units so they match)

Related: May I treat units (e.g. joules, grams, etc.) in equations as variables?

as well as treating units as variables you need to consider the number of decimal places you quote in your answers relative to the number of significant digits in the numbers.

What does $g/cm^3$ mean? It means you have divided a mass in grams by a volume in $cm^3$ (the slash character"/" is really a division). So yes, your second value is 250 $g/cm^3$.
In the first case, you can simplify $cm^2/cm$, just like in any equation, to $cm$. So your value is 56.4516 cm.


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Is it correct to say that H2SO4 is an acid in this reaction?
Is it correct to say that $\ce{H2SO_4}$ acts as an acid in the following reaction?
$\ce{2Ag + H2SO_4} \rightarrow \ce{Ag_2SO_4 + 2H_2O + SO_2}$
I know it acts as an oxidizing agent, but is it correct to say it shows acidic properties?

Do you get the same reaction with sodium sulfate?
I do not think so, at least under "normal" conditions.
When the sulfuric acid acts as an oxidizing agent giving off $\ce{SO_2}$, the excess oxygen comes off as oxide ions which must be somehow put into more stable form.  The protons from the sulphuric acid do that by turning them to water.
Getting the reaction to go requires the sulfuric acid to act as both an acid and an oxidizing agent.


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Why is ice sticky between −20 to −8 degrees Celsius but not below −30 °C?
I was watching an Air crash investigation video where the cause of the plane crash was determined to be due to blockage of fuel pipes by ice crystals (from the water in the jet fuel) which would have formed while the plane was cruising over Siberia where temperatures are as low as −74 °C. Now the engines were working fine until the fuel temp. was −30 °C. But as soon as it started getting warmer near London, Heathrow and the fuel temp hit −20 °C the ice crystals coalesced to form a mesh that completely blocked the fuel pipes and caused the engines to lose power and end up crashing.
My Question is: What caused the ice to not stick at a colder temp (−30 °C) when I believe the molecules would be vibrating slower than those at −20 °C? It seemed from the investigation's experiment that below −20 °C big ice crystals don't form at all, if they do they're tiny. So why does the increase in temperature help ice sticking?

The official accident report can be found here: https://www.gov.uk/aaib-reports/1-2010-boeing-777-236er-g-ymmm-17-january-2008

I think I remember that episode. Basically, crystals of water in kerosene were not a problem during the whole flight (at minus thirty something) but crystals were too sticky when the temperature raised to -20°C and prevented the fuel to go to the engines just a few millimeters before entering the "warming cell" which was supposed to prevent any such blockage.
Crystals at low temperature (let's say -35°C) will be quite hard and quite big, even if some molecules of the fuel will be trapped in the crystals of ice (because there is much more fuel than water). If the temperature goes up to a certain level (let's say -20°C), crystals will break where there was molecules from the fuel. As a consequence, the big crystals will break into very small crystals. Ice crystals do not like their environment (i.e. the fuel) because they cannot find any stabilizing effect with it, but they love other ice crystals. Moreover, the overall surface greatly rises when the size of the crystals diminishes. As a consequence, they will try to stay together and form a "slush".
In that accident, it was not the only factor. The other two problems were that acceleration increased the flux of kerosene (so more small crystals of ice were available to create a slush), and there was a misconception in the heat exchanger which was aimed at avoiding water crystals.

so basically your explanation is that the increase in surface area allowed for better stickiness? Wouldn't the bigger ice crystals at (-35 C) be more efficient in blocking the fuel since it's bigger? But according to the video I think those crystals aren't big enough below -30C

Big crystals could allow some kerosene to go through but small crystals (especially if they are sticky) would not. Imagine water going through sand. The quite large and solid particles leave some space between them and water can go through rather easily. With flour however, the tiny particles leave much less space (it is more compact) and water hardly runs though.

But is surface area the only reason to account for the extra stickiness?

No but it makes stickiness more problematic in that case. Maybe my answer was a bit misleading.


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reaction between copper metal and aqueous sodium bicarbonate
I wanted to know what reaction would occur between copper metal and sodium bicarbonate?
I believe that once in sodium bicarbonate solution, the copper metal will go into solution as copper ions and replace the sodium ions. Is this wrong?
Also, if it is indeed true, what would the reaction between copper ions and bicarbonate ions be?

There would be no reaction.

why? is it because copper cannot replace sodium

You may put it this way, if you like. In my opinion, non-existence of something does not require a reason; existence does.

Aqueous sodium bicarbonate indeed reacts with $\ce{Cu^2+}$ but not copper metal. see this:- http://chemiday.com/en/reaction/3-1-0-5035

okay, but will there at least be copper ions in solution from the copper metal?

First, reduction potentials.
Theory page
https://en.wikipedia.org/wiki/Reduction_potential?wprov=sfla1
Table of standard electrode potentials
https://en.wikipedia.org/wiki/Standard_electrode_potential_%28data_page%29?wprov=sfla1
Standard electrode potentials of sodium is -2.70V while copper (I) is 5.20V, single electron reduction of copper (II) has 1.59V, full reduction potential is 3.37V
Higher the reduction potential, more the tendency for the species itself to reduce. Reduction potential of Copper is much higher than Sodium.
So conclusion is no.
Lets say if copper can replace sodium. By how much? Under room conditions, RT/F*lne is approximately 0.06V, so the amount of copper ions that will replace sodium will be orders of magnitude 71.5 to 115 times dilute. In other words, non existant.

The standards electrode  potential go copper 2 is 0.337V and 0.159 for copper 2 copper ion.

It can react, but at oxygen presence to oxidize copper, slowly forming famous greenish basic copper carbonate. In a mineral form, it is known as malachite.
One of possible reactions:
$$\ce{2 Cu + O2 +  HCO3- + H2O -> CuCO3 . Cu(OH)2 + OH-}$$


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Boron enolates double bond geometry
The usage of dialkylboron triflates produces Z enolates from ketones predominately, while the usage of dialkylboron chlorides changes the selectivity favoring the E enolate:

How is this trend explained?

here is Evans' paper on it: JACS 1981, 103, 3099

The image below (Taken from the Evans' lectures, which you might like to consult as there is a whole lot more detail in them) sums up the situation.
Essentially you're interested in how good a leaving group is attached to the boron. The triflate is excellent and as such fully dissociates, whilst the chloride remains (at least partially) in association.
Depending on whether the group on boron dissociates, the 'boron' is significantly more/less bulky, changing the conformation from which enolisation is preferred, which in turn leads to E/Z enolates.
The choice of base is also important. With boron triflates, DIPEA is often used as it is just that bit bulkier and so biases even more towards the Z enolate. In the boron chloride aldols, the bulky DIPEA is a bit too much, and so Et3N is commonly used.
If you study the TS's below, it might be apparent why the boron chloride aldols are typically a little less selective than the boron trifoliate aldols.

Source: Dave Evans' Harvard CHEM207 Lecture notes.

Yes, his lectures are very enlightening. So, to sum up, regarding the triflate, BL2OTf-Me steric repulsion is much less becase the boron is not tetrahedral and the R-Me repulsion determines the reactive conformation. In the case of the chloride, boron is tetrahedral and the repulsion BL2Cl-Me is significant, so the reactive conformation is the one suffering Me-R repulsion which is why the lower stereoselectivity in this case. Did I get it right?


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Boiling Point Reversal
If substance A boils at a lower Temperature than substance B under atm. pressure, can it be that under some other condition, say, elevated Pressure or under vacuum a reversal of the trend occurs?
i.e. substance B boiling lower than substance A?
Is this phenomenon known? What are some examples?
In other words, If the vap. pr. of A is higher than B at one given T there's no fundamental reason for that inequality to persist at all temperatures, correct?

Correct.  The temperature dependence of the vapor pressure depends thermodynamically on the heat of vaporization.  So, depending on the heats of vaporization of the two substances, the vapor pressure curves can cross.

Thanks Chester! But do they, in practice? I'm just curious since I've never heard qualification of the sort "X is low boiling than Y so long as we are above 500 torr" etc. I guess I could try various Antoine equations & simulate things........

$$\frac{d\ln P}{d(1/T)}=-\frac{\Delta H}{R}$$  This is the Clausius-Clapeyron equation.  Look in Perry's Chemical Engineer's Handbook or a Thermo book for a graph of vapor pressures as a function of temperature.  Notice all the crossovers.


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Lassaigne's test for sulfur

The sodium fusion extract is acidified with acetic acid and lead acetate is added to it. A black precipitate of lead sulphide indicates the presence of sulphur.

What is the need for acidification of extract before adding lead acetate? Wouldn't Lead sulfide be formed if we add directly lead acetate? Can we use any other acid and its salt of lead?

The silver test for halides (using silver nitrate) requires the mixture to be acidified to remove carbonate because silver carbonate could precipitate and be confused for a silver halide. Lead carbonate is also insoluble in water, so maybe this is the reason for using acid here.

@f" But lead carbonate is not black whereas leadsulfide is black

Still, it is better not to have any unnecessary side reactions with their precipitates that may obscure your vision.


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Acid strengths for HBrO4 vs HIO4, and HBr vs HI
So I was looking at factors that control the relative strengths of acids and bases.
We know that $\ce{HI}$ is a stronger acid that $\ce{HBr}$ because $\ce{I}$ is a much larger atom.
But then, I read somewhere that $\ce{HBrO4}$ is stronger than $\ce{HIO4}$, because in this case it's electronegativity that plays the important role.
How do you know whether to use size or EN as a guideline for comparing strengths of acids/bases?

In HI and HBr, the hydrogen is directly bonded to the halogen, but in the perhalic acids it isn't.

"We know that HI is a stronger acid that HBr because I is a much larger atom"

That is only half the story. It is about the strength of the H-Br versus H-I bond. Since I is larger than Br, the H-I bond is longer, weaker, and therefore more prone to dissociation. See here for more in-depth discussion.

$\ce{HBrO4}$ and $\ce{HIO4}$

both have the acidic proton bound to oxygen:

The H-O bond is approximately of equal strength in both acids and therefore you turn to electronegativity arguments.


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What factors affect the chemical stability of a compound?
My question is essentially about the factors that make certain atoms in certain compounds leave the compound they are currently in in order to bond with another atom/compound.
For example, I've been taught that if you react a halo-alkane with a strong base, like KOH, a substitution occurs in which OH and the halogen change places. Why does this happen (if it does)?
Would an electronegative atom rather want to bond with another extremely electronegative atom, or one that has extremely little electronegativity? Basically, would fluorine want another fluorine, or rather something like caesium?
$$\ce{Cs + F \to CsF}$$
$$\ce{F + F \to F2}$$
My thought is that fluorine 'loves' electrons so much, and there is nothing else that loves electrons more than this guy, so this has to be the most stable compound ever that is not dissociable by any other compound/atom.

There is a nice video about the reaction between cesium and fluorine: https://youtu.be/TLOFaWdPxB0

Any change which is spontaneous, takes place to lower the usable energy present in that system (This principle is so important, that even universe obeys it), which depends upon the conditions in which change in taking place. To demonstrate this, lets take an example of a stone at height h, one can do a lot of work with that configuration, it can do work by hitting someone on their head. But in contrast, when it is at ground, it does not do anything much on its own, the reason being it has used up some of its energy stored in the form of gravitational potential energy. Now of course, if conditions change, the stone can again do some work, in fact it is always loosing some of its usable energy energy, in the form of increasing entropy since it is breaking down by the natural forces of wind.
Coming back to your question, why does substitution takes place in case of halo alkane with hydroxide anion. Now, we can see that haloalkane has transformed into an alcohol. If one calculates the usable energy (Gibbs free energy, this encompasses several other factors  which include enthalpy, entropy and thermal energy) stored in haloalkane and corresponding alcohol, it will be clear that usable energy has decreased going from haloalkane to alcohol. To understand more about that, you can look up gibbs energy which quantifies that usable form of energy. Will fluorine bond with fluorine or Cesium- The answer, which I think is obvious to you, is surely cesium. why? Looking at their atoms, imagine cesium has no or very little pull on its valence electrons, while fluorine nucleus has so much pull, that electrostatic force permeates farther than atom. Hence, when two fluorine atoms come close to each other, there is some attraction between both atoms due to their individual pulls, but because both pulls are of same strength, it results in a covalent bond, but in case of cesium, the electrostatic force of fluorine nuclei pulls the valence electrons of cesium taking them away from the cesium nucleus, and thus fluorine's thirst for electrons is satisfied (the rock is at surface now!), while cesium is totally cool about it, since it had very little attraction for those electrons. After this process fluorine has negative and cesium has a positive charge, which attracts both of them and also it decreases the energy stored in fluorine-cesium systemIs CsF so stable that it cannot be dissociated into its components? No its not, put in water and water will dissociate the compound easily and energy is supplied by hydration and increasing entropy.

The first few lines of your answer are unfortunately either unrealistic explanations/interpretations that are sometimes purposely given to stupid kids (note that not all are thus) who will not be able to grasp the reality of the matter, OR very inaccurate examples. Universe does not obey rules, rules are our observations of how the universe works. Stone does not fall in order to reach a lower energy state, since it has no brain whatsoever to know that it will end up that way, rather, the reason it falls is because there is a reason making it do so (earth applying force on it), not, a purpose.

2 questions:
1- Can I then infer that, the fluorine atoms in F2 would want to bond with atoms that are less electronegative than Fluorine? For example, if I had some fluorine gas and some pure carbon that doesn't have any crystal structure whatsoever (if possible), or, let's just say I have some phosphorus, will that fluorine gas react with those substances?
2- I believe, just as with the falling of the stone, when it comes to CsF dissociating in water, it doesn't do so out of a purpose, but rather, because of electrostatic forces (H2O is dipolar). So, would CsF dissolve in smt like C6H6

In reply to questions, yes you can infer that fluorine atoms will choose to bond with less electronegative atoms in preference to fluorine atom since ionic bonds have lower energy state than covalently bonded compounds, for example if you mix chlorine atoms with Fluorine fluorine will bond with chlorine since bond formed will have some ionic character due to electronegativity differences , for reference, here is the link, where you can watch video of fluorine reaction with carbon (Charcoal).https://www.youtube.com/watch?v=TLOFaWdPxB0.

In response to your second question, no CsF will not dissociate in CsF, since there will be very little interaction with benzene, which will not overcome the lattice energy associated with CsF.

Regarding very first comment about so called unrealistic explanations. I am not sure, whether kids are provided with these explanations. This statement that every change takes place in the direction of decreasing free energy, can be thought of as a law, which helps to make sense, but in no way answers the why question. Just as newton's law of gravitation does make sense but in no why explains why.

Firstly, I should thank you for the link to that video, it made me learn and kind of witness 2 things I've been wondering, and also thanks for the answer to my second question.
As for so fore-called unrealistic explanations, I called them unrealistic because they seemed like attempts to explain 'why' changes occur, and they seemed to suggest that matter can have 'purpose' like a conscious being. The better explanation to 'why' is most likely 'we'll never know why', or, 'even if we do find out, there will be inner things that we won't be able to know'. Some things are just not comprehensible


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Can we freeze water at −1 °C?
Is it possible to freeze water if it is directly exposed to $-1\ \mathrm{^\circ C}$ and not $0\ \mathrm{^\circ C}$? Is freezing point the minimum temperature at which a substance can be froze?

Yes, it will start cooling down to -1, going though 0 and then freezing at 0. You can in theory supercool water, but that's another story. The freezing point is the temperature that a substance is frozen below it.

Yes it is possible. Even if you expose water to an object at −50 °C, it will freeze. This is due to the fact that when you place two objects, heat transfer occurs. It is not a spontaneous process, it will take time for water to release heat and cool and eventually freeze.
Freezing point is merely the degree temperature at which a substance changes from liquid to solid phase. It is the temperature at which the substance will freeze.
Hope this helps.

I down voted this. To me you missed key points. "Heat" transfer takes place from hot to cold. Liquid water would decrease in temperature to 0 C at which point the liquid water would start to freeze. Once all the water has frozen to ice, then the ice would cool to  -1 C.

Yes, I've missed all that...thanks @MaxW, for telling me what's wrong before downvoting, which for me is seldom seen.

At atmospheric pressure it is possible to supercool water down to almost -50°C before it freezes. The water has to be very pure and not in contact with surfaces which will act as nucleation sites but supercooled water droplets are common in clouds and pose a danger to aircraft (ice formation on the wings).
0°C is the "freezing point" because it is the temperature that most fairly pure water will freeze at and the point at which ice will melt. If you slowly apply heat to ice (Ice 1h, hexagonal ice, is the stable form of ice at ~0°C and 1 atm) starting below 0°C, temperature rises until 0° when temperature will stop rising until the ice is melted (phase change from solid to liquid) after the ice is melted, temperature will begin to rise again (the rate of rise per energy input is known as its specific heat capacity, Cp, and that rate is different for liquid and solid water.)

It is possible by pressure too! Above $10~\mathrm{MPa}$ melting point of water decreases. At $209.9~\mathrm{MPa}$ melting point become $-21.9°\mathrm{C}$. For more information, see this page.