How are the electrons of benzene actually configurated? I found this diagram:

According to it, it is not true that the six electrons of carbons in the pi-orbitals above the ring are all identical.

But, how can it be the case, seeing that this molecule should have an order-6 rotational symmetry?

So, as I told you in chat what you have here is the orbitals of something which looks like, but which is not, benzene. It is the orbitals of one the all its two mesomeric form, it is so a static view.

Now let me show you the orbitals which are in the benzene. I only draw with the p-orbitals, I won't do the hybridation with the hydrogen it will be too difficult for me two draw and it will not be more effective.

How to draw a molecular diagram when the molecule has a lot of atoms?

First draw the diagram like if in the molecule there is just its squeleton (atoms which interact to create the shape of the molecule, for example with benzene only for the carbon atoms) and do like if these atoms were hydrogen atoms.

Second draw the correct orbitals with the same symetry of what you got with hydrogen just before.

You are obviously not obliged to do it but it is easier to see which orbitals in the fragments have the same symetry or not.

First step
Here for the benzene I will draw the MO's diagram of benzene with $\ce{H_6}$ first, with a molecule which is a rectangle of hydrogen atoms and the other fragment is a longer dihydrogen molecule than the original one. After seeing which of them have the good symetries I get this

Second step

Here we want the same but with the p-orbital of the carbons so keeping the same symetries or antisymetries you get this

And then here you have the correct orbitals.

Be careful

The color on the orbitals represent the sign of the wave function which has no meaning in the reality. So you can make the black and the white anywhere you want if you obviously respect the molecule symetries.
In this example I draw exactly the same orbital.

I say that because the orbitals I drawn for benzene are not exactly in the same color logic of the MO's diagram of $\ce{H_6}$ but as you can see each one have the same symetries or antisymetries.

And then the answer of your question

On the MO's diagram you show they show the interaction of the p-orbitals in one of the mesomeric structure, like this

And then the symetries are not the same as in the benzene but at one of its mesomeric form which is a static view like this

In fact you need to imagine the same but with an interaction of all p-orbitals and the get a ring orbital for $\psi_1$, and the it explain the real shape of the benzene like this

Remember that orbital diagrams are done to think about what you observe in the nature and not to predict it. Well you can use quantum theory to predict things but it is an other subject.

I hope it can help !


"Alkanes must be vaporised before they will burn" - what does this mean? What does this sentence mean - I assumed that burning vaporised something? If something is vaporised, isn't that turning it into a gas? How can you burn a gas? This is a sentence in my AS Chemistry book and I am obviously missing something.

"Alkanes must be vaporised before they will burn"

Burning an alkane means reacting it with oxygen (in air). The larger the contact surface between the alkane phase and the air, the better the reaction. That's why cars have carburators: there's no larger "contact surface" than between the molecules in a gas mixture.


Markovnikov Rule of Alkenes $$\ce{CH3CH(CH3)CHCHCH2CH3 + HBr ->}$$
According to Markovnikov Rule, "the rich get richer". But, as we can see the 2 carbons have the same amount of $\ce{H}$. So, what's probably the answer?

My suggestion is to stop trying to predict the product by rules like “rich becomes richer” or “proton goes to carbon having greater number of hydrogens attached to it”, etc. for Markovnikov addition and opposite for anti-Markovnikov addition.
These are subject to many exceptions.
Instead, draw the direction of dipole in the molecule at the site of the double bond, (order of precedence: +R effect, +H effect, +I effect for considering direction). After doing this, $\ce{H+}$ will attach the carbon with δ− charge (as electrophile will attack first, and carbocation formed on δ+ carbon which is more stable), and nucleophile going to other carbon atom in case of Markovnikov addition.
The opposite holds for anti-Markovnikov addition, i.e. $\ce{H+}$ will attack carbon with δ+ charge.
Now coming to your question. After you figure out on which carbon $\ce{H+}$ will attack, the carbocation can further be stabilized by Hydride shift (always check for rearrangement possibilities like hydride, methyl, ethyl shift, ring expansion, etc.). After rearrangement only does the nucleophile $(\ce{Br-}$ here) attack.


Does a frying pan use radiation? I have a project to prove that the colour black absorbs more light than the colour white in a creative way. Our first idea was to time a white frying pan and black pan and see which would melt the ice cube first, but is this proving radiation? Because I looked at saw pieces of metal above where the flame would go, making me guess pans use conduction. Will a flame ever give off enough radiation to be noticeable?

You may estimate the contribution of thermal radiation by using the Stefan–Boltzmann law, which describes the power $P$ radiated from a black body in terms of its temperature $T$:

$$P=\sigma\cdot A\cdot T^4$$

where $\sigma=5.670\,367(13)\times10^{-8}\ \mathrm{W\ m^{-2}\ K^{-4}}$ is the Stefan–Boltzmann constant[reference] and $A$ is the surface area.

The formula can be corrected for real materials by introducing the emissivity $\varepsilon$:

$$P=\varepsilon\cdot\sigma\cdot A\cdot T^4$$

For an ideal black body, the emissivity is $\varepsilon=1$. For polished metal surfaces, the values are significantly lower.

For example, for a hot black frying pan with $\varepsilon\approx1$, $A\approx600\ \mathrm{cm^2}=0.06\ \mathrm{m^2}$, and $T\approx230\ \mathrm{^\circ C}\approx500\ \mathrm K$, the power may be estimated as

$$\begin{align}
P&=\varepsilon\cdot\sigma\cdot A\cdot T^4\\
&=1\times5.670\,367\times10^{-8}\ \mathrm{W\ m^{-2}\ K^{-4}}\times0.06\ \mathrm{m^2}\times\left(500\ \mathrm K\right)^4\\
&\approx200\ \mathrm W
\end{align}$$

However, in addition to thermal radiation, heat transfer also occurs via thermal convection and thermal conduction. You could exclude these modes by placing the frying pan in a vacuum.

You could also estimate the total heat transfer due to all three modes from the electrical power of the stove that is required to keep the frying pan at a constant temperature.


Is the proper formula for phosphorus oxide(V) P4O10 or P2O5? I would like to ask about formula of phosphorus oxide(V). Should it be $\ce{P4O10}$ or $\ce{P2O5}$? Which is right when I have to write the molecular formula?

The simplest types of chemical formulas are called empirical formulas, which use only letters and numbers indicating atomic proportional ratios (the numerical proportions of atoms of one type to those of other types). Molecular formulas indicate the simple numbers of each type of atom in a molecule of a molecular substance, and are thus sometimes the same as empirical formulas (for molecules that only have one atom of a particular type), and at other times require larger numbers than do empirical formulas. (Wikipedia)

For phosphorus pentoxide the molecule seems to be an adamantane-like compound with 4 phosphorus atoms.

So the correct molecular formula of the compound would be $\ce{P4O10}$.


Does dilution of a buffer affect pH? The Henderson-Hasselbalch equation for the $\mathrm{pH}$ of a buffer solution of the monoprotic acid $\ce{HA}$ is given by
$$\mathrm{pH}=\mathrm pK_\mathrm a+\log{\frac{[\ce{A-}]}{[\ce{HA}]}}$$
Since concentration appears in both the numerator and denominator of the fraction $\frac{[\ce{A-}]}{[\ce{HA}]}$ and $\mathrm pK_\mathrm a$ is constant (at a fixed temperature), it appears that dilution of the solution with pure $\ce{H2O}$ would not change the $\mathrm{pH}$. However, since
$$\mathrm{pH}=-\log{[\ce{H+}]}$$
 the amount of substance of $\ce{H+}$ must increase in order for $\mathrm{pH}$ to stay constant upon dilution.

Where is this additional $\ce{H+}$ coming from? I know that diluting an acid causes it to dissociate to a greater extent. But at the same time, you would be diluting its conjugate base and causing it to associate more, cancelling the dissociation of the acid.

In the Henderson-Hasselbalch equation, $K_\mathrm{a}$ is a product of concentrations and considered a constant.

In reality, $K_\mathrm{a}$, when defined as a product of concentrations, is not a constant:

Upon dilution (decrease in ionic strength) the $\mathrm{p}K_\mathrm{a}$ will change, and therefore the pH of the solution will change.

In addition to the above reason, pH will always approach 7 at extreme dilution as it approaches being pure water.


Assigning 1H NMR spectrum and predicting peak coupling patterns for 2H-1-benzopyran-2-one I have a the proton NMR of 2H-1-benzopyran-2-one (2H-chromen-2-one):

There are 6 individual hydrogens and none of them are equivalent. In the spectrum I have been given there are only four peaks.

I have been able to assign two double peaks at chemical shifts 6.43 ppm and 7.73 ppm to the two hydrogens on the alkene in the ring on the left. This leaves the four hydrogens on the benzene ring which should all be non-equivalent, but this spectrum shows them as two 2H peaks. Does this mean they have overlapped, and if so please help me figure out how. Should the multiplicities have been two ddd for the two H's in the middle and then the two H's on the ends should be a dd or ddd?

This is the expansion of those two 2H peaks in my spectra shown below.

Don't you have an integral for this spectrum? This would show you clearly that you have two sets of overlapping signals, just as you have surmised.

Looking at the numbering for your molecule described:

You are correct with the assignment for the vinyl protons. Proton E will be strongly downfield shifted, and F only moderately shifted from the base double bond chemical shift of 5.25. These values could be determined using additive tables found in lots of NMR textbooks.

The proton shifts for the aromatic ring can also be roughly calculated using additive tables, but what you have are indeed overlapping signals; Protons A and C are at ~7.3 and protons B and D are at ~7.55 ppm.

To predict the splitting patterns you anticipate for the aromatic system, consider first the general couplings you'd expect:

ortho coupling ~7-8 Hz
meta coupling ~2-3 Hz
para coupling ~1Hz or less

Let's ignore the para couplings for simplicity - they are not often seen.

So,

Proton A will have one ortho coupling and one meta coupling, so a will be a dd
Proton B will have two ortho couplings and one meta coupling, so will be a ddd
Proton C will have two ortho couplings and one meta coupling, so will be a ddd
Proton D will have one ortho couplings and one meta coupling, so will be a dd

Let's now build in double bond :

Protons E and F have a cis coupling (~10 Hz).
Proton F will appear as a doublet, but proton E is likely to have a homoallylic type coupling to proton D (4J), splitting both of these signals to an additional doublet with a coupling of about 1-2 Hz. This means that proton D may be a ddd, and proton E a dd if this coupling is observable.

So that should be the expected splitting pattern from this molecule. Now to justify the appearance of the peaks. Overlapping signals can often make things difficult to interpret. Without knowing what frequency this spectrum was recorded at, you also need to consider whether there are second order effects starting to manifest, certainly with the meta couplings. There will definitely be tenting of the peaks giving rise to distorted peak intensities from what would normally be expected.

So, I wouldn't be devoting too much time to the interpretation of overlapping signals like these. One thing I will say is that if you label the peaks as (say) a ddd, then you should be able to report all couplings for each d. Otherwise, if you cannot report couplings (and this means couplings, not splittings) report them as a multiplet (or in this case overlapping multiplets).

Complex splitting patterns of second order and overlapping signals can be analysed and simulated, but it takes a lot of work.You can try drawing out a splitting diagram for each signal using the approximate couplings for ortho and meta couplings, and then start overlapping through constructive addition to see what you get. It gets very complicated very quickly.


What are positive and negative voltage? I read about electrode potential today, the book says that:

  Metals more reactive than hydrogen will have negative electrode potential value and metals less reactive than hydrogen will have positive electrode potential.

I'm a bit confused by the sign of the voltage, can voltage have negative value? What does negative voltage mean?

Another question is why metals more reactive than H have negative electrode potential?

Negative voltage means the same as positive, except the electric current caused by it would flow in the opposite direction.

As for your another question, that's just what it is: having negative potential means being more reactive than H. See, that electrode potential is not an absolute value, it can be measured only relative to something. And the general agreement is to take the standard hydrogen electrode for that "something".

Suppose you'd make an electrode out of your metal immersed in a solution of some salt of itself, couple it to standard hydrogen electrode and see what happens. If a metal is more reactive, it will dissolve, the released electrons will run to the hydrogen electrode and reduce hydrogen there. With less reactive metals, it is the other way around: hydrogen will dissolve and send the electrons running in the opposite direction, to eventually reduce the metal.


Additive for water to increase the viscosity to a value between 5 and 20 cP Is there an additive that I can add to water that will increase the viscosity to a value between $5$ and $\pu{20 cP}$? Ideally, I would like the  mixture to be newtonian and since I will be working with gallons of mixture, it would be nice if I didn't have to add too much of the stuff.

You wasn't clear about your purpose to increase viscosity of water (water has viscosity of $\pu{1 mPa\cdot s}$ or $\pu{1 cP}$). However, whatever your reason, xanthan gum would be the materiel you are looking for. Kyoto Electronics has published a note on measurement of viscosity of xanthan gum solution. According to their results (average of 5 tests), 0.1% xanthan gum solution has shown viscosity of $\pu{19.1 cP}$. Five sample averages of 0.2% and 0.3% solutions have given viscosity readings of $\pu{57.54 cP}$ and $\pu{134.6 cP}$.
Accordingly, you can use range of percentages (0.01-0.1%)of xanthan gum to get water viscosity of $\pu{5-20 cP}$. The amount of material to add to water is minimal as you wanted. In addition, Casson plot (shear rate versus shear stress) of 0.1% xanthan gum solution is a straight line. Hence this solution is believed to be a Newtonian solution.


Systematic Approach to Finding Number of Isomers I have this compound

I want to know how many isomers (including stereoisomers) may have this constitution. Is there a systematic approach towards solving such types of problems related to the total number of isomers?

Note: The above compound has 64 stereoisomers (from textbook).

To my knowledge, there is no direct method or formula to find out how many isomers for given framework of an organic compound. Interestingly, this could be a good question for a mathematician. The only formula of this kind is the one to find theoretical optically active stereoisomers of a compound if the number of stereocenters (chiral centers) are given.

For instance, the given framework has 4 stereo centers, hence theoretically, you have $2^4 = 16$ stereoisomers for that compound alone. If your book says it has 64 stereoisomers, it probably has counted geometric isomers of conjugated double bonds (E/Z isomers) as well. All these 64 isomers are illustrated as compound 1 in the scheme below where each red asterisk represent a chiral center. Note that, since the ring has three different substitution groups, none would lead to symmetry in the molecule. Thus, all theoretical amount is equal to the actual number of isomers. Now, question is whether you need to count the stereoisomers for this framework only or not. If it is only for this framework, then the number is 64. But, if you want to change the framework by moving the substitution around, that would be a different story.

Now, for example, if you just move the carbonyl group around three other available carbons (say, move to the carbon between two hydroxy groups as shown in the compound 2 in the diagram), then you have additional 64 stereoisomers by that single move (compound 2). There are other 4 movable groups in the ring and all of them create multiple combinations such as 2,3,4,5-substituted cycloocta-1-one. Also note that chloro group and one of the hydroxyl groups can switch their positions to give another option. Or, the chloro group can just occupy the ring as well, creating another chiral center (e.g., 6-alkenyl-3-chloro-1,4-dihydroxy-8-methylcycloocta-1-one) or move to side chain. Upon all, there are changeable positions for the two double bond on heptyl chain as well (see compound 3). Double bond(s) in the side chain and an alkyl radical(s) can also move to ring (as methy, ethyl, etc.) are also possibilities. Thus, you are going to have tons of stereo isomer as any moving to the ring creating more stereo centers. Good luck on that, if you are not a mathematician.


How does partial reduction of esters to aldehydes occur? Wikipedia says that esters can be reduced to aldehydes using DIBAl-H. Can anyone please explain how this reaction occurs?

DIBAL exists as a bridged dimer, and it becomes a reducing agent only after it has formed a Lewis acid–base complex, so it reduces electron-rich carbonyl groups most rapidly.

DIBAL will reduce esters even at –70 °C, and at this temperature the tetrahedral intermediate that forms during the reaction may be stable. Only in the aqueous acidic work-up does it collapse to the aldehyde (by expelling EtOH in the example reaction below) when excess DIBAL has been destroyed so that no further reduction to the alcohol is possible.

A stable tetrahedral intermediate is more likely in the reduction of lactones, and DIBAL is most reliable in the reduction of lactones to lactols.

Edit:

As requested by the OP here are some more details about the collapse of the tetrahedral intermediate: During the aqueous acidic work-up the alcohol group is turned into a good leaving group via protonation. Thus, it is easily expelled from the molecule, when a water molecule removes the $\ce{Al^{$i$}Bu_2}$ group and the $\ce{C=O}$ double bond is reformed.

A different route, where the former carbonyl-oxygen is protonated twice - thus being turned into a good leaving group - and afterwards expelled from the molecule to give an oxonium ion is possible too. But such an oxonium ion is rather unstable and highly reactive, so that the reverse reaction is very fast and the oxonium ion is not produced in any appreciable amount.


Is there any rule that says that oxygen should have at least one free pair of electrons in Lewis dot structure? For example, Ozone has a double and single bond with at least one free pair of electrons in each oxygen atom. My doubt is whether ozone can have a triple bond and single bond with one oxygen atom without a free pair of electrons.

$\ce{H4O^{2+}}$ is a counter example to such a rule.

http://pubs.acs.org/doi/abs/10.1021/ja00265a031


What is a Sodium wire? I read in a paper:

  [...] we prepared the $\ce{PhTeNa}$ in situ from $\ce{(PhTe)2}$ and sodium wire in THF.

Can anyone explain me this procedure (the mechanism) focusing on the definition of "sodium wire"?

Sodium is normally stored as cubes under petrol or similar solvent. As a metal it is very soft, you can easily cut it with knife. For the reaction you need fresh reactive surface of the metal, so the wire is one of best choices. For lab-scale preparation see e.g. https://www.youtube.com/watch?v=btvQdW4fXJE


UV Absorption of iodine with some solvent $I_2$ in solution is typically absorbing around $525 \; nm$ this is why it appear to be magenta, I recently learn that with some solvent :

Benzene
Toluene
$o$-Xylene
$p$-Xylene

The coordination complex between the solvent and $I_2$ is also absorbing around $300 \; nm$ I guess there is some explanation with the MO diagram ?

Iodine is highly polarisable and will form a (Mulliken) charge-transfer complex with many aromatics. Very many pairs of donor - acceptor pairs do this and have been studied for approx $100$ years, see for example tetracyanoethylene–pyrene (J. Chem. Phys. 105,1996 p2287) for an ultrafast time resolved study.  In  absorbing a photon the ground state complex DA transfers  an electron directly to a charge transfer state $DA \overset{h\nu} \to \mathrm D^+A^-$. The interaction is thus that of all the orbitals in one molecule with all those in another.

Books

R. S. Mulliken and W. B. Person, 'Molecular Complexes' Wiley, New
York, 1969. and G. Briegleb, 'Elektronen Donator Acceptor Komplexe' Springer, Berlin, 1961.


Is there anything we lose by using Google Colab for DeepChem? DeepChem does not run on windows, so I am considering to try Google Colab. Is there any key feature I would be missing by using Google Colab remotely from Windows rather than the original DeepChem on Linux?

DeepChem's project web page indicates further documentation and access to the source files used.  Just click the «fork me on github» sign in the top right hand corner leads you to this page.  From the notes there, it becomes obvious that many tasks the program suite handles are managed with Python which you should install in first place (e.g., Python's Python, or Anaconda's Python) then offering a platform independent use of deepchem in Linux, Mac, or Windows (see this report).

Once you have this installed, you just have to solve the «must have» dependencies», in Python's Python with python package installer pip3 on the CLI

pip3 install joblib pandas sklearn tensorflow pillow deepchem

and maybe the ones listed as «soft requirements» such as RDKit.

A nice thing is that a package manager like pip will check if additional requirements have to be adjusted, and if so, will offer to adjust your system accordingly «on the fly».  Just be aware that you might need a considerable chunk free memory on permanent memory for the modules deepchem relies.  Don't forget the inofficial package (wheel) index for Python modules in Windows by Christopf Gohlke which may assist you in solving the dependencies, too.


Why would the hydrogen sulfide ion be a better nucleophile than a bromide ion? Why would the hydrogen sulfide ion be a better nucleophile than a bromide ion?

My professor says that hydrogen sulfide ion is a better nucleophile.

I thought I could determine the better nucleophile was based on available electron density and polarizability. The bromine ion has more available electron density and is more polarizable than the $\ce{HS-}$. Why would it be less nucleophilic?

There are a number of factors you have to take care of while deciding nucleophilicity. I will try to cover all the important ones.

First we need to clear something, polarizability is the most important factor only if the nucleophile is electrically neutral (uncharged)$^{[1]}$. Of course, this is not hard and fast, but in most cases we generally deal with, this is true. For example, triphenylphosphine is a better nucleophile than triphenylamine, because the former's central atom has a more polarizable lone pair.

In our examples, the ions are charged, so we will put aside polarizability and look at the other factors. For charged species, size of the anion should be small in order for it to be a better nucleophile. Why? Two reasons:

One, the same charge packed into a small volume is attracted more strongly by the electron-deficient area, and it can reach quite close to the positive region. This means that potential energy of the system will fall more, thus increasing stability of the product.
Two, the central atom being small, its overlapping orbital can overlap more effectively with the empty orbital of the electrophile, since the orbital will be smaller in size, and hence less diffuse.

The last factor that remains is the solvent. In fact, the order is same for both the nucleophiles in both the kinds of solvents (protic and aprotic), so I will not go into the analysis of solvent effects. Just a comment, here the solvent effect seems unimportant, seeing that bromide ion is quite behind $\ce{SH^-}$ in the order of nucleophilicity in protic solvents (my source is Solomons and Fryhle and a website mentioned later).

Note: Electronegativity is not important here either, even though it would produce the same order. We generally use electronegativity only to compare nucleophiles with central atoms from the same period.

So ultimately we see that $\ce{SH^-}$ is more nucleophilic than $\ce{Br^-}$.

Most of what I said can be found in the textbook Solomons and Fryhle's Organic Chemistry, and the following link:

$^{[1]}$http://chem.libretexts.org/Core/Organic_Chemistry/Reactions/Substitution_Reactions/SN2/Nucleophile

Here are some more useful links:

- http://chem.libretexts.org/Core/Organic_Chemistry/Alkyl_Halides/Properties_of_Alkyl_Halides/Introduction_to_Alkyl_Halides/Leaving_groups
- http://chem.libretexts.org/Core/Organic_Chemistry/Alkyl_Halides/Reactivity_of_Alkyl_Halides/Alkyl_Halide_Reactions/Substitution_and_Elimination_Reactions_of_Alkyl_Halides/SN1_Substitution_Reactions/Nucleophilicity_and_Solvent_Effects
- http://chem.libretexts.org/Core/Organic_Chemistry/Reactions/Substitution_Reactions/SN2/Substrate


Can =NH2+ in Arginine's side chain form hydrogen bonding at physiological pH? I know that at physiological pH, Arginine will have its $\ce{=NH}$ protonated to $\ce{=NH2+}$. I was wondering if this $\ce{=NH2+}$ can still form a hydrogen bond by being a donor? And does the positive charge on Nitrogen affect the hydrogen bonding's strength? My confusion stems from the definition of hydrogen bonding, which says that hydrogen's electron cloud needs to be withdrawn by a more electronegative atom like $\ce{F, O, N}$ to be partially positive enough to participate. Therefore, I believe that the positively charged N will withdraw electrons from hydrogens more strongly, leading to more partially positively charged hydrogen and eventually stronger H-bonding with another molecule. Could someone please verify this for me?

Moreover, I have seen $\ce{=O}$ and $\ce{=N}$ forming hydrogen bonding by being an acceptor. However, I tried search but did not find any example of $\ce{=NH3+}$ or $\ce{=NH2R+}$ or NHR2+ forming hydrogen bonding as a donor. So can a Nitrogen with a double bond still participate in hydrogen bonding as a donor?

Yes, the arginine side chain is an excellent hydrogen bond donor. Charged N-H groups are even better hydrogen bond donors than the corresponding neutral N-H groups. Here is an article that gives examples of arginine side chains in hydrogen bonds. It showcases bidendate hydrogen bonds (i.e. two hydrogen bonds from the arginine side chain) with carboxylates and with nitrogenous bases of nucleic acids.

Specifically for proteins, if the hydrogen bond donor is positively charged (e.g. Arg+) and the hydrogen bond acceptor is negatively charged (e.g. Glu-), you have a combination of ionic and hydrogen bond interaction, which is particularly strong. It is often referred to as a salt bridge, such as in this example. If you take a close look at the arginine side chain, you will see that it is protonated (i.e. positively charged) and interacts with two negatively charged carboxylate groups.


Flavor components of high end vodka? I always thought vodka was just alcohol and water, and that the ideal vodka was one lacking all flavor components characteristic of the grain or potato mash that it was distilled from. I once saw a vodka process diagram showing multiple distillation and charcoal filtration steps--suggesting an effort to remove all flavor components. But looking at the prices being charged for high end vodkas, I am wondering what the draw is. Are there flavor components in these high end vodkas that make them that much more desirable than pure alcohol that people pay more than $50 a bottle for them?

High end vodka is primarily good vodka+branding. You pay extra to get that designed bottle. Perhaps the manufacturer also does something extra that they claim makes the vodka special. Whether it actually does depends on what they have done.

Good vodka has most flavours removed. Some flavours - specifically those that both have a boiling point close to that of alcohol and the ability to pass through an active carbon filter will remain.


Why is H in OH group is more shielded than in H in CH2 group (ethanol NMR spectroscopy)? I am confused with this $\mathrm{^1H}$-$\mathrm{NMR}$ spectroscopy result:

Shouldn't $\ce{H}$ in $\ce{OH}$ group the singlet be more towards the left of the $\mathrm{NMR}$ spectrum?
It is near the highly electronegative atom $\ce{O}$ rather than the two $\ce{H}$ in $\ce{CH2}$ group they have a weak eletronegative atom $\ce{C}$ so it shouldn't be that desheilded.

It is a myth that electronegativity is the only things that effect chemical shifts. However, electronegativity is only one of many things that effect chemical shift of a proton. Other things effecting chemical shift are magnetic anisotropy (e.g., large deshielding effects on aromatics and alkenes), steric compression (e.g., measurable $\Delta \delta$ values of alkene protons on cis- and trans-isomers), anisotropy of double bonds (e.g., measurable positive $\Delta \delta$ values of methyl group of methylcyclohexene and  methylcyclohexane, $\delta_{\mathrm{sp^2}(\ce{CH3})}-\delta_{\mathrm{sp^3}(\ce{CH3})} \approx 1.60-0.92= 0.68$), solvent effects, etc.
All of above mentioned factors effect chemical shifts of proton attached to electronegative atom such as $\ce{OH}$ and $\ce{NH}$. In addition, solvent used and solute concentration are also fators changing chemical shifts of these protons. For example, let's consider alcohol $\ce{OH}$ protons. In dilute solution of alcohols in non-hydrogen-bonding solvents such as $\ce{CCl4, CDCl3,}$ and $\ce{C6D6,}$ the $\ce{OH}$ signal generally appears at $\delta \ 1\!- \!2$. For instance, residual peak in $\ce{CDCl3}$ is always appeared at around $\delta \ 1.6$. At higher concentrations in non-polar solvents,  the $\ce{OH}$ signal moves downfield ($+\delta$) because when concentration increase, the $\ce{H}$-bonding increases as a result. For instance, the $\ce{OH}$ signal of ethanol comes at $\delta \ 1.0$ in a 0.5% solution in $\ce{CCl4}$, and it appears at $\delta \ 5.13$  in the pure ethanol (5-HMR-2 Chemical Shift):

The solvent used to take the given spectrum is unknown. Based on the chemical shift of $\ce{OH}$ signal, one can assume solvent might be $\ce{CDCl3}$, which is the most common among NMR solvents, which does not promote $\ce{H}$-bonding. As you can judge by the image above, it might have taken in the concentration of 1-5% range.


The best chromatographic technique to analyse a solution of benzene, toluene, and ethylbenzene in n-hexane What is the best chromatographic technique to analyse a solution containing benzene, toluene, and ethylbenzene in n-hexane?

Is gas chromatography (GC) the best way to analyse this solution? Or is using liquid chromatography with a non polar eluent and column better than GC in this case?

The choice of analytical technique is based on three simple factors: what is available, and how low concentration do you want to go, and how much time do you have? Do you have a gas chromatograph with flame ionization detector or a UV high performance liquid chromatograph or both? This type of mixture is readily amenable to both GC and reversed phase HPLC. The only problem is hexane in reversed phase HPLC, but if less than 1 uL is injected it should be fine.


Why do bonding and anti-bonding states split in energy (mathematically)? I am a physicist, and when I hear about the splitting of energy levels for MO antibonding and bonding states, I have a rough intuition that tells me this is reasonable to expect, but I can't think of (nor find) a good formal explanation.

My best shot at understanding this is borrowing from Avoided Crossing and saying that if I have a 2 atom system with a total Hamiltonian $ H = H_1 + H_2 $ that is the sum of each individual atom Hamiltonians, then I can take the Eigenstates for each individual Hamiltonian as a basis.

$$\begin{align}
H_1 | \psi_1 \rangle &= E_1 | \psi_1 \rangle \\
H_2 | \psi_2 \rangle &= E_2 | \psi_2 \rangle
\end{align}$$

Then the total hamiltonian in this basis becomes,
$$\begin{align}
H &= \left(\begin{matrix}
E_1 & \langle \psi_2 | H_1 | \psi_2 \rangle \\
\langle \psi_1 | H_2 | \psi_1 \rangle & E_2
\end{matrix}\right)
\end{align}$$

In the limit of the atoms being really far apart, I would assume the off diagonal terms would vanish, and then the eigenstates are just the atomic orbitals. But if the atoms come closer together, then we get off diagonal terms which causes the splitting as described in avoided crossing.

Is this correct?

You are (essentially) correct and the off-diagonal terms are related to so-called "overlap" between atomic orbitals (AOs). It is common to talk of the AOs "overlapping" to form molecular orbitals (MOs), and as you rightly said, the off-diagonal terms (or "overlap") goes to zero in the limit of infinite bond length. The term "avoided crossing" is not commonly seen in introductory MO theory, but I think it is fair to describe it as such.

Of course there are some slight subtleties. Let's stick to a one-electron system such as $\ce{H2+}$ for now. $\psi_i$ refers to the 1s orbital on hydrogen atom $i$ and $E$ is the energy of a 1s orbital on hydrogen. The minor issue is that $\hat{H} \neq \hat{H}_1 + \hat{H}_2$ (following your notation where $\hat{H}$ is the molecular Hamiltonian and $\hat{H}_i$ is the Hamiltonian for hydrogen atom $i$), because:

$$\begin{align}
\hat{H} &= -\frac{\nabla^2}{2} - \frac{1}{r_1} - \frac{1}{r_2} + \frac{1}{R} \\
\hat{H}_1 &= -\frac{\nabla^2}{2} - \frac{1}{r_1} \\
\hat{H}_2 &= -\frac{\nabla^2}{2} - \frac{1}{r_2} \\
\end{align}$$

(I used atomic units. $r_i$ is the distance between the electron and nucleus $i$, and $R$ is the distance between the two nuclei). So actually, it is more like

$$\hat{H} = \hat{H}_1 - \frac{1}{r_2} + \frac{1}{R}$$

and therefore the first term in the matrix

$$\begin{align}
\mathbf{H}_{11} &= \left<\psi_1\middle|\hat{H}_1 - \frac{1}{r_2} + \frac{1}{R}\middle|\psi_1\right> \\
&= E + \left<\psi_1\middle|-\frac{1}{r_2}\middle|\psi_1\right> + \frac{1}{R} \neq E
\end{align}$$

(since $\langle\psi_1|\hat{H}_1|\psi_1\rangle = E$, and also $\langle\psi_1|(1/R)|\psi_1\rangle = (1/R)\langle\psi_1|\psi_1\rangle = 1/R$).

For more information I refer you to Atkins' Molecular Quantum Mechanics 5th ed., pp 262–266. The terminology can be a bit confusing and also Atkins does not use atomic units (in his text he uses $j_0 = e^2/4\pi\varepsilon_0$ which is simply equal to $1$ in atomic units). Adjusting for this, the final quoted formulae for the matrix elements are:

$$\alpha = \mathbf{H}_{11} = \mathbf{H}_{22} = E - j' + \frac{1}{R}; \qquad j' = \left<\psi_1\middle|\frac{1}{r_2}\middle|\psi_1\right>$$

(as found earlier), and

$$\beta = \mathbf{H}_{12} = \mathbf{H}_{12} = \left(E + \frac{1}{R}\right)S - k'; \qquad S = \langle\psi_1|\psi_2\rangle;\quad k' = \left<\psi_1\middle|\frac{1}{r_2}\middle|\psi_2\right>$$

If you now solve the generalised eigenvalue equation $\mathbf{Hc} = E\mathbf{Sc}$ (the simple $\mathbf{Hc} = E\mathbf{c}$ only works when your basis set is orthonormal, but in this case it is not) you obtain the two eigenvalues

$$E_\pm = E + \frac{1}{R} - \frac{j' \pm k'}{1 \pm S}$$

Note that when the atoms are infinitely far apart both $S$ and $k'$ vanish, so (1) the off-diagonal element $\beta$ vanishes and (2) the "bonding" and "antibonding" orbitals are of the same energy, i.e. $E_+ = E_-$.


How to apply the Principal of Atom Conservation (POAC) 27.6 g $\ce{K2CO3}$ was treated by a series of reagents so as to convert all of its carbon to $\ce{K2Zn3(Fe(CN)6)2}$. Calculate the weight of the product.

I am asking here is that, why we are  applying POAC only on Carbon atom not on Potassium Atom.

Because in the question it is specifically written
"All of carbon of K2CO3 has been converted ".
Whereas it has not mentioned anything about potassium.
You have to understand that K2CO3 is conberted to K2Zn3(Fe(CN)6)2 by a series of reactions therefore there might be other potassium compounds which had been formed by the series of reactions.
But as it is given all of carbon converted therefore out of the series of reactions there was only one carbon containing compound K2Zn3(Fe(CN)6)2 .
Hope this helps..


Composition of Portland Cement before Hydration I recently came across a question-

Which of following chemicals is NOT found in Portland Cement?
A. Di-calcium silicate
B. Tri-calcium silicate
C. $\ce{CaSiO3}$
D. $\ce{Ca3(PO4)2}$

I answered D. $\ce{Ca3(PO4)2}$. But according to the examiner the answer is C. $\ce{CaSiO3}$
I strongly doubt that. Any pointers to credible sources/books settling this please!!

According to Ullmann’s Encyclopedia of Industrial Chemistry (sixth edition), Portland cement consists mainly of tricalcium silicate, dicalcium silicate, tricalcium aluminate, and calcium aluminoferrite.

$$
\textbf{Phase composition of Portland cement clinker} \\
\begin{array}{llll}
\hline
\text{Constituent} & \text{Formula} & \text{Content in wt %} \\
& & \text{Range} & \text{Average}\\
\hline
\text{Tricalcium silicate} & \ce{3CaO.SiO2} & 46{-}79 & 61 \\
\text{Dicalcium silicate} & \ce{2CaO.SiO2} & 5{-}30 & 15 \\
\text{Calcium aluminoferrite} & \ce{2CaO.(Al2O3,Fe2O3)} & 4{-}16 & \phantom{0}8 \\
\text{Tricalcium aluminate} & \ce{3CaO.Al2O3} & 6{-}18 & 12 \\
\text{Free calcium oxide} & \ce{CaO} & 0.1{-}4 \\
\text{Free magnesium oxide} & \ce{MgO} & 0.7{-}1.5 & \phantom{0}1.5 \\
\hline
\end{array}$$

Büchel, Moretto, Woditsch Industrial Inorganic Chemistry (second edition) lists the same six constituents as important clinker phases. This reference also includes a diagram for the three component system $\ce{CaO}$–$\ce{SiO2}$–$\ce{Al2O3}$ with composition ranges for various cement types, which shows that calcium monosilicate (CS) is not necessarily within the typical range for Portland cement.

Furthermore, this reference includes a table with a typical chemical analysis of a Portland cement, which shows that the $\ce{CaO}$–$\ce{SiO2}$ ratio is too high for calcium monosilicate and also indicates that small amounts of phosphate could be present:

$$
\textbf{Typical chemical composition of a Portland cement clinker} \\
\begin{array}{ll}
\hline
\ce{CaO} & 58{-}67\ \% \\
\ce{SiO2} & 16{-}26\ \% \\
\ce{Al2O3} & \phantom{0}4{-}8\ \% \\
\ce{Fe2O3} & \phantom{0}2{-}5\ \% \\
\ce{MgO} & \phantom{0}1{-}5\ \% \\
\ce{SO3} & \phantom{0}0.1{-}2.5\ \% \\
\ce{Mn2O3} & \phantom{0}0{-}3\ \% \\
\ce{P2O5} & \phantom{0}0{-}1.5\ \% \\
\ce{TiO2} & \phantom{0}0{-}0.5\ \% \\
\hline
\end{array}$$


Can entropy in a closed system remain constant when you heat a liquid? This is in regards to an exam question that I did not understand and got wrong.

The question was: When you heat acetone from 200 K to 400 K, what happens to the entropy?
Answers: Increases, decreases, more info needed

I put more info needed, but this was incorrect. My logic was that if you have a sealed container completely filled with acetone (there is no gas or empty space in the container) and you heat it, there are no new states that the liquid particles can occupy so there is no change in entropy. Is this reasoning incorrect?

The quantitative differential classical thermodynamic definition of entropy is $$\mathrm{d}S=\mathrm{d}Q/T$$

Therefore providing heat $Q$ to the acetone increases its entropy as
$$\Delta S=\int_{T_\mathrm{1}}^{T_\mathrm{2}} \frac{C(T)}{T}\cdot \mathrm{d}T$$

where $C$ is the heat capacity of the system.

The statistical thermodynamics definition has an alternative definition to $$S=k_\mathrm{b} \cdot \ln \Omega$$ as $$S=-k_\mathrm{b} \cdot \sum_\mathrm{i} p_\mathrm{i} \cdot \ln p_\mathrm{i}$$ as the sum over all states.

The higher energy states of electrons, of bond vibrations and molecule rotations are not occupied at low temperatures, or are occupied unevenly.
In such a case, even for the same number of states, more evenly distributed probabilities at higher temperatures lead to higher entropy.


How adding one more electron increases the ionization energy? Elements of group 6A, compared to 5A, require less ionization energy due to the paired electrons of 6A.

But a question arises: why does group 7A, compared to group 6A, require more ionization energy in spite of the paired electrons in group 7a too.

Regardless of the general trend that the ionization energy increases as we go from left to right in periods, why didn't we apply the same procedure as the first comparison?

It's not just adding one more electron. Another proton is also added to the nucleus. The overall trend across a row is driven by shielding effects.

In general, as you go across a row, the shielding effect is fairly small compared to the effect of increasing nuclear charge.

As you note, there is a slight deviation in this trend between 5A (i.e., $\ce{N}$) and 6A (i.e., $\ce{O}$) because the extra electron in oxygen is paired. This increases the electron-electron repulsion, so the ionization energy decreases slightly.

As you note, $\ce{F}$ (7A) resumes the overall trend of increasing ionization energy driven by the attractive effects of the higher effective nuclear charge $Z_{eff}$.


What can be a harmless electrolytic additive for water for creating an oxy-hydrogen flame? I was thinking of preparing a tiny oxy-hydrogen flame for soldering. I am going to use 6V DC and aluminum sheet as the electrode for this. Now, I have to add some electrolyte to water, that can be a salt, base or acid. But, if I add $\ce{NaCl}$ or $\ce{HCl}$, they will produce chlorine. If I add $\ce{NaOH}$, it will eat electrodes. Adding $\ce{H2SO4}$ may produce poisonous gases like $\ce{SO2}$ and $\ce{H2S}$. So, what can be a harmless/least harmful electrolyte for such purpose? Is $\ce{Na2CO3}$ a good choice?

Electrolysis of water has become an increasingly common means of creating torches for soldering and welding, and there are commercial devices available as well as plenty of internet do-it-yourself sites. According to this site regarding the SafeFlame consortium:

  "[...]the European Union-funded SafeFlame consortium has developed a
  torch system that generates a flame using nothing but H2O and
  electricity.   SafeFlame utilizes an electrical current to electrolyze
  ordinary water, separating it into hydrogen and oxygen gas. Those
  gases are then mixed and ignited as they exit the torch's nozzle. By
  fine-tuning the proportions of the two gases, different types of
  flames can be produced for different applications."

A commonly used electrolyte / electrode combination is $\ce{KOH}$ or $\ce{Na2CO3}$ with stainless steel electrodes. I believe this would be pretty corrosive toward aluminum, but stainless steel for such a setup should be cheap and easy to come by.


(How) Does ICP-OES determine the quantity of detected elements, or just their occurance in the sample? ICP-OES is a very common analytical method (Wikipedia page here), and I have heard a number of times (from chemists, too) that it is used for quantitative analysis.

However, from the description of the process, I'm not certain if it really does give quantitative information, how it does so, and whether it can do so on its own, or whether there are additional methods run in combination (like with gas chromatography and mass spectrometry) to achieve that result.

Although fp's answer is absolutely correct and I see that you have something more than high school chemistry, I'd like to give a more elaborated, yet simplistic answer, if you or anyone else reading this has any interest on the basic principles of an ICP-OES - you only need high school chemistry to understand it.

I'm not certain if it really does give quantitative information

It does, and very "accurately". Detection limits can be in ppb, that is, it can detect 0.0000001g (0.1 µg) of something in a 1L sample. That's really, really impressive if you stop to think about it.

how it does so

Wizardry! Well, almost. Read on and you'll see.

First of all the sample must be prepared, you can't simply toss something in it. Depending on what your sample is, it must be mixed with a known amount of this or that chemical (you'll see later why). This answers your next question:

and whether it can do so on its own.

Although the actual analysis and preparation can indeed be done purely by the machine in some cases (here's a video), and there are methods such as laser ablation which make it possible to almost just toss something in the ICP, the know-how of a trained analyst is still require to tell the machine what to do, to interpret the results it outputs, and to tell what chemicals to mix in the preparation process, if there is one at all. Equipments such as gas chromatographs can (and might need to) be coupled to assist in the sample preparation.

Back to how it does so:

When you introduce the prepared sample into the machine (it basically sucks it) some processing is done to it, and at some point it reaches the atomizer - the plasma torch. The atomizer does what it's name suggests: It turns stuff into separate atoms (by heating them to absurd temperatures). Here's a picture I took from an ICP atomizer:

Note: Not 100% of the sample is atomized. In fact, only a small percentage is.

When it's turned on it looks like the picture you saw at Wikipedia. You can see by the size of the parts that only a few milliliters of sample - if not less - is atomized at a time. Now that it got our sample atomized, it heats the sample up just a bit more, so the electrons get exited and go from their current level of energy to a higher one. While the sample passes through the equipment it gets cooler, so the electrons get back to the lower energy level, and by doing so, the atom emits light, hence the name optical EMISSION spectroscopy.

(source: wikimedia.org)

But what elements are emitting light, and at what wavelength? The principle of every spectroscopy analysis is that each element emits and absorbs specific, known, predictable wavelengths ranges, and only when heated to specific temperatures (in the case of emission). Because the temperature was set to a specific one for the analyte and a specific set of chemicals were mixed with the sample, only the atoms of the element we want to quantify will emit light at a certain element-specific wavelength range. We set the equipment to detect only that wavelength range and ignore others and that's how we know how much of that element there is in the sample.

But if you kept track of the proportions you might be thinking: "if it detects ppb amounts in only a few ml of sample where only a few % of it is atomized and only certain elements will emit only a certain wavelength, how the hell can it detect that little light?". That's the photomultiplier's job. It's a fantastic piece of equipment and you should definitively read about it.

After it detects the light emitted by the analyte in the sample, it compares it to a blank, does some calculations and displays the results in the form of a graph. Here's a screenshot of an ICP-OES software:

As I mentioned, this is a simplistic explanation, most of the process isn't that simple in reality. So as you can see, it's basically witchcraft.


Stereoisomers of metaldehyde and paraldehyde (1) In case of paraldehyde I see 3 similar chiral carbons. So I divided it into 2 subcases
-two methyls above the plane and one below the plane
-all three above the plane
so, that makes it 2. When I looked this up on Wikipedia I found this,

can,(1) and (4) not be obtained by simply ring flipping, doesn't that make them conformers? Do I need to count them separately in my count of stereoisomers?

this according to me should have 4 stereosiomer(not including conformers) can someone confirm this answer? Are there any more im missing out?

Your analysis for metaldehyde is correct. There are four achiral stereoisomers. The example cited here is a bit different. To understand the use of R, S, r and s, see this discussion on the inositols.


Bridgehead carbocation: Violation of Bredt's Rule This is the solution of a university's problem set (nucleophilic substitutions)
It says that the given compound undergoes SN1 type substitution.

What confuses me is that even though iodide is a good leaving group, bridgehead carbocations (sp² bridgeheads) are extremely unstable. According to me, this reaction shouldn't occur.

Could anyone explain what's going on here? If the explanation is just thermodynamic (hopefully not), could some data be provided? Thanks!

The reaction in question is known and has been studied fairly extensively, both experimentally and computationally, and is accepted to go via the bridged carbocation that you propose.

To give some data, one experiment  measured the relative reaction rate of a series of such structures towards solvolysis in 80% aqueous ethanol (basically the reaction you describe above):

The intermediate carbocation is known as a non-classical carbocation- a carbocation that does not conform to the standard organic chemistry definition of a planar sp2 intermediate (as we often see with SN1 reactions).

In the case of the norbornyl cation, the carbocation can be observed spectroscopically by treating the parent halide with a superacid such as $\ce{SbF5.SO2ClF}$ at very low temperatures:

One issue with non-classical carbocations is their tendency to rearrange, such that you wouldn't get a single product as predicted, but instead end up with a mixture of products resulting from other classical carbocations formed through hydride shifts on the non-classical carbocation (the exact mix of products would depend upon the conditions of the reaction).

References

All figures taken from Carey, F. Advanced Organic Chemistry - Part A


Reaction of Diols with HX hydrohalic acids, Competing SN2 Reactions Suppose 1,5-pentanediol were exposed to something like hydroiodic acid. Would the alkoxonium end of the diol be exposed to a nucleophilic attack by the remaining hydroxide at the other end for a ring formation? Or would the newly-free iodide ion be more competitive for substitution? Thanks!

edit: It was pointed out to me "newly-free iodide" was quite confusing. I meant it as though HI dissociated, the Hydrogen ion attacked one end of the diol, and the iodide from the HI being "newly-free". Now that one end of the diol has a good leaving group (H2O), is it more susceptible to be attacked by the iodide, or by the other end (OH) of the diol.

Bob:
Here is the likely process. The alcohol 1 is protonated to provide intermediate 2 which undergoes SN2 displacement. The closure of 3 to  tetrahydropyran 6 is unlikely under the reaction conditions. Rather a second displacement via 4 would lead to diiodide 5. If the tetrahydropyran did form, ethers are cleaved by HI. The conversion of 3 to 6 is normally accomplished in the presence of base (NaH/THF).


Does 95.8% silver tarnish as fast 92.5% silver? I've been trying to find out whether 95.8% (Britannia silver) would tarnish as fast as 92.5% (Sterling silver)? I haven't been able to find much on tarnish for Britannia silver.

Given that both silvers are kept in the exact same environment, would the Britannia silver tarnish at the same rate as Sterling silver? If not which of the two tarnishes quicker?

If you look here: https://blog.centimegift.com/does-sterling-silver-tarnish/, it says that pure silver is very tarnish resistant. Sterling silver on the other hand being mixed with some Cu to enchance the metal's mechanical properties tarnishes as a result of the Cu in it. Therefore, considering that the amount of Cu in Britannia silver is smaller, it should tarnish more slowly than sterling silver. In practice i dont know if there will be a significant difference as the increse in silver purity is not that much.


Determining which hydrate of calcium chloride I have? I have calcium chloride in the form of small (1-2 mm) round white pellets, originally bought at a home-brew shop. The packaging gives no indication what hydrate it is. Since each water molecule adds ~16% of the weight of anhydrous CaCl2 this introduces a big error when using it to treat water to a specified amount (in ppm).

Naturally I figured I could throw it in the toaster-oven for a while at 350°F and weigh it every 15 minutes, to figure out how much water was driven off. Results:

0 min.  -  1.15 g.
15 min. -  0.92 g.
30 min. -  0.90 g.
45 min. -  0.90 g.

I originally assumed this would be anhydrous, but after reading that calcium chloride dihydrate decomposes at 347°F, and calcium chloride monohydrate decomposes at 500°F, I'm not so sure.

I couldn't find any other information on figuring out the hydrate online  or indicating what state calcium chloride is typically in (at least that I could understand). Am I correct in thinking I'm left with the monohydrate? Or is it actually anhydrous?

The mass of your sample decreased by 22%.

If you originally had $\ce{CaCl2.4H2O}$ and quantitatively dehydrated that to $\ce{CaCl2.H2O}$, then you should have seen a decrease in mass of 27%.

If you originally had $\ce{CaCl2.4H2O}$ and quantitatively dehydrated that to $\ce{CaCl2.2H2O}$, then you should have seen a decrease in mass of 18%.

Although it's tempting to say that you just have some mix of mono-and-dihydrates, I think that we just can't ignore the stability of your mass measurements and the temperature of your sample, which due to the radiative heating from the toaster heating coils was probably even greater than $\mathrm{350^oF}$. The most reasonable option remaining is that you originally had primarily $\ce{CaCl2.4H2O}$, but also some $\ce{CaCl2.6H2O}$ mixed in, and now you have just the monohydrate remaining.


Attachment of disulfide (protected) thiolated oligonucleotide on gold surface What will happen if I incubate a clean gold surface in un-deprotected
($\ce{oligo-C6-S-SH}$) thiolated oligonucleotide solution instead of deprotected ($\ce{oligo-C6-SH}$) solution. I have read clean gold surface can cleave disulfide bond.

What actually happens in sulfur-gold bonding is not very well known. There have been many proposals and studies done, well-summarized here. There is some evidence, including computational studies, that suggest that dissociation into two thiolates is energetically favourable. Bridging across protruding bits of gold (we often draw gold surfaces as straight lines, but they're seldom that flat) may also be involved. This case, where you only have a chain on one of the two sulfurs, may be less interesting and I haven't found many studies of terminal disulfides so the exact process might be different.

To answer the practical question, Can I stick a disulfide-functionalized molecule onto gold?, yes you can, and they will stick very well.


How to justify this contradiction in acid-base equilibrium of methanol and water? Let's talk about the following compounds - water and methanol.

We know that methanol is slightly more acidic than water, because water's ability to donate a proton as an acid is reduced due to extensive hydrogen bonding.
The $\mathrm{p}K_\mathrm{a}$ values are approximately $15.5$ and $15.7$, for methanol and water respectively.
Now consider their conjugate bases, i.e. methoxide ion and hydroxide ion. It is very clear that methoxide is a stronger base compared to hydroxide, due to the +I (inductive) effect of the methyl group (electron releasing).

How is this possible? Isn't it contradictory?
It is a well known fact that a strong acid has a weak conjugate base, and a weak acid has a strong conjugate base.

This is a consequence of  $\mathrm{p}K_\mathrm{a}$ + $\mathrm{p}K_\mathrm{b}$ = $\mathrm{p}K_\mathrm{w}$, where Kw is the ionic product of water, a constant at a given temperature.

We know that methanol is slightly more acidic than water, because water's ability to donate a proton as an acid is reduced due to extensive hydrogen bonding. [...]

[...] It is very clear that methoxide is a stronger base compared to hydroxide, due to the +I (inductive) effect of the methyl group (electron releasing).

You are pointing out competing factors that, in abstract, should lead to either methanol or water being more acidic than the other (that is. lead to either hydroxide or methoxide to be more basic than the other, respectively). Once we move from considering properties in abstract to an actual system in which an acid-base equilibrium is present, the competing factors are superimposed, with a non-contradictory result: either methanol will be more acidic (and hydroxide more basic) or water will be more acidic (and methoxide more basic), depending on which factors dominate. (In particular, as pointed out in Why is methanol more acidic than water? and Is methanol really more acidic than water?, the relative acidities depend on the solvent.)


In rovibrational spectra, when exactly is the Q branch forbidden? I'm aware that the Q branch ($\Delta j = 0$) is usually not observed in rotational-vibrational spectra of diatomic molecules in the gas state, however, I've heard different things about when exactly this transition is forbidden.

I've sometimes been told that the transition is not allowed in general, for any diatomic molecule, but I've sometimes heard that this is only true if the electronic state has a $\Lambda \neq0$, so which is the correct selection rule here?

I'd love to have this clarified for me. If it is possible I would also like to know when is a Q branch observed in vibronic spectra of heteronuclear and homonuclear diatomics.

I would like tell you what I know about the answer to this question.

So, it's known that the selection rules for a fine vibrational spectrum (rotational-vibrational spectrum) for diatomic molecules make $\Delta J = 0$ or $\pm 1$ transitions allowed between the rotational levels of two vibrational states (usually $V''=0$ to $V'=1$, where the symbols have their usual meaning). But one must be careful in realizing that the $\Delta J = 0$ transition is forbidden for all parallel vibrations in the oscillator and only allowed for perpendicular vibrations.

Now, let's see what are parallel and perpendicular vibrations!

Any vibrational mode in infrared active only when the change in the dipole moment vector is sinusoidal w.r.t. time and thus can interact with the electric field of the electromagnetic radiation. If this change in the dipole moment vector is parallel to the major axis of the molecule, then the vibrational is labelled as a parallel mode of vibration. If the change in the dipole moment is perpendicular w.r.t. the major molecular axis, it is a perpendicular mode of vibration.

Let's look at the case of diatomic molecule, as the question is focused on that.

The major molecular axis for it is the bond axis (containing both the atoms). Since a hetero-nuclear diatomic molecule has only one vibrational mode, viz., stretching, it can only occur parallel to the bond axis. Therefore, $\Delta J = 0$ transition is forbidden in the case of parallel vibrations. Thus, on observing the fine infrared spectrum of hetero-nuclear diatomic molecules, we only observe $P$ and $R$ branches. $Q$ branches are not observed.

References:

Fundamentals of Molecular Spectroscopy - Banwell and McCash
Introduction to Molecular Spectroscopy - Gordon M. Barrow
Atomic and Molecular Spectroscopy - Rita Kakkar


The difference between the different types of rhodamine What is the difference between different type of rhodamine, for example rhodamine red and rhodamine green?
are these two type of rhodamine (red & green) have a good Raman signal?
We would like examples of Raman signals of these two types of rhodamine.

Rhodamine green
CAS: 189200-71-3

Rhodamine Red
CAS: --


Why is Oxygen compulsory for free radical reactions? Why is presence of oxygen compulsory for free radical mechanism? Can it take place in presence of any other element?

There is no such condition that oxygen must be present to generate free radicals. Chlorine forms free radicals, and so do many organic molecules. In fact the presence of oxygen is bad for free radical reactions.


Why is H bonded to C5 drawn below the plane in Haworth formula for D-glucose? I have always seen the hydrogen atom bonded to C5 drawn below the plane as in the following figure. Shouldn't it be drawn above the plane since it is drawn on left side in Fischer projection?

The Haworth projection depicts the pyranose (or furanose, as the case may be) as a planar ring. The Fischer projection depicts the open chain carbohydrate with the carbon backbone in a single plane. Importantly, the groups on the horizontal bonds are coming out of the plane and the vertical bonds are going into the plane. The result of this is that the carbon backbone is curving toward itself, and the compound is partially poised to close a ring. However, the ring of the pyranose is not all carbon. In glucose, it's the oxygen at C5 that adds to the aldehyde to close the ring. We can manipulate the Fischer projection so that this oxygen at C5 is coplanar with the rest of the atoms that will make up the ring. Rotating around the C4-C5 bond places this oxygen on the Fischer projection's vertical axis. The Haworth projection is essentially a closed version of the Fischer projection tipped on its side with C1 at the right. Forming the hemi-acetal gives the pyranose form of glucose.

You imply in the question that groups that appear on the right side of the Fischer projection should appear on the bottom of the Haworth projection and groups that appear on the left side of the Fischer projection should appear on the top of the Haworth projection. That idea is true. However in the process above the -CH2OH is rotated to the the left side of the Fischer projection. Consequently, it appears on the top of the Haworth projection.


Modern alternatives to Gaussian Specifically, I'm curious if there are any programs which come close to Gaussian in breadth but also take advantage of things such as GPU processing.

There's a page on wikipedia that goes through several options but I'd like to get feedback from actual users.

"Specifically, I'm curious if there are any programs which come close to Gaussian in breadth"

Not really, as far as I can tell. Gaussian is a massive work by many, many people over many, many years, and the reason it's still one of the most popular packages, as far as I can tell, is that it does most things (notably excluding non-GTOs), even if it's not necessarily the fastest or the best at the things it does in every case.

"but also take advantage of things such as GPU processing."

Double no. As LordStryker says above, GPU programming is pretty new, and most of the resources I've seen for it are C-based, while a significant chunk of scientific development still takes place in Fortran. I suspect library-based encapsulation of GPU-running numerical routines may improve this over time (I mean, by introducing people to the idea of getting speedup by using the GPUs more easily), but I wouldn't hold my breath. It'll be interesting to see what hybrid chips with multiple types of core, or the Xeon Phis will be able to do, though.


flory ( or theta) temperature I know the flory temperature is the unique temperature at which the attractions and repulsions of a polymer in a solution cancel each other.

but what does this mean in practice? does it mean that this is the temperature at which the polymer and the solution mix best? and if you move away from this temperature it mixes less well?

I'm going to use Theta-Solvent and Theta-temperature since that is what Wikipedia uses, and Wikipedia usually has the preferred names.

In order to get a polymer to the Theta-Condition two factors must be considered.

(1) Solvent

A Theta-Solvent is a poor solvent for a polymer is one such that the interactions between the polymer segments and the solvent are minimized. This results in polymer segment interactions to be maximized.

A good solvent for a polymer is one such that the interactions between the polymer segments and the solvent are maximized. This results in polymer segment interactions being minimized.

(2) Temperature

The Theta-temperature is the temperature for the particular theta-solvent
which minimizes interactions between the polymer segments and the solvent.

Point

The combination of using a theta-solvent and being at the theta-temperature satisfies the theta-condition.

This makes experimental determination of important quantities such as the root mean square end-to-end distance or the radius of gyration much simpler.

Note

This assumes that polymer will self-coil into its lowest energy configuration. For polymers with same unit repeating this is not unlikely.

For proteins in biological systems this self-arrangement is unlikely. Proteins do not have a "repeating unit." Therefore biological systems have a mechanism to fold a protein in a certain way.


Aromatic Elimination Addition Chlorobenzene will react with hydroxide to give phenol. But will the reaction on Chlorobenzene also work with methoxide, to give anisole?

Yes, It will.
It's a Nucleophilic Substitution reaction,

Chlorobenzene + Sodium Methoxide --> (Under extreme conditions ) Anisole

Extreme conditions because Aryl halides don't go Nucleophilic Substitution easily.


How are phosphoglycerides different from phospholipids? (Currently studying high-school Biochemistry)

I'm unable to distinguish between phospholipids and phosphoglycerides from their "definitions" (Courtesy: Wikipedia)

On phospholipids:

Which suggests that phospholipids necessarily contain a glycerol "bridge" that connects the fatty acid tail to the phosphate head. To me, this looks like,

On phosphoglycerides:

"Glycerol-based phospholipids"?! But aren't phospholipids already 'glycerol-based'?

If both phospholipids are glycerol-based, and if phosphoproteins are "glycerol-based phospholipids"...how are they any different?

What am I missing here?

Phospholipids of a lipid bilayer can be divided in to two groups:

Those derived from glycerol -  phosphoglycerides.
Those derived from sphingosine -  sphingomyelin.

Phosphoglycerides are indeed a type of phospholipid.


Why do so many biochemical reactions require enzymes? Studying biochemistry as part of the Great Courses, I am struck that all 10 steps in glycolysis require an enzyme.

I’d have thought that evolution would have selected for a “simpler” pathway with fewer enzyme-catalyzed steps. If a reaction A + B requires an enzyme, three molecules have to be in the same compartment and collide instead of two.

My hunch is that having enzyme-catalyzed reactions instead of more reactive molecules reacting without catalyst gives cells the ability to control their metabolism, making homeostasis and regulation possible.

all 10 steps in glycolysis require an enzyme

We count it as 10 steps because there are 10 enzymes. There might be other steps that are fast (like a product acting as an acid or base) that we don't count because they don't need an enzyme.

  I’d have thought that evolution would have selected for “simpler” pathways.

What is simple is that everything is happening in the same reaction vessel, the cytosol (well, there are more compartments, but just a handful). This is different from the organic chemist, who works with very reactive substances in pure form, carefully mixed in separate reaction vessels with distinct solvents refluxed at distinct temperatures for hours.

  Is the very fact that enzymes are required make life more probable since the cells ability to control their quantities make homeostasis and regulation possible?

Yes, that is true. It also allows for a web of reactions rather than a single pathway (you can funnel fructose or galactose into glycolysis, and you can turn pyruvate into lactate instead of decarboxylating it to carbon dioxide and acetic acid).

Stable molecules require catalysts

One requirement for life as we understand it is a way to store, replicate and access genetic information. In our living world, that requirement is met by having nucleic acids. Under physiological conditions, RNA is fairly stable (good enough for a virus) and DNA is very stable (to the degree that sometimes DNA sample from extinct organisms preserve genetic information to this day). If these molecules do not form or decay on their own, there must be enzymes to act on them.

Granted, there could be other pockets of metabolism that work without catalysis. The other requirement, though, is to capture free energy to drive reactions away from equilibrium (e.g. to make the carriers of genetic information mentioned above). This is done through coupled reactions, where e.g. the free energy from food being oxidized (with oxygen as terminal electron acceptor) is captured in proton gradients across membranes and in the phosphorylation of ADP to obtain ATP. If any of the reactions going toward equilibrium would occur without catalysis, the free energy would be lost instead of captured by a coupled enzyme reaction.

General thoughts on complexity

In any complex system (government, computer chip, industrial process etc.), as the complexity grows, the number of steps that seem inefficient or over-complicated will also grow. Many of these steps might not make sense from the present view, but were put in place (or evolved) earlier to meet a historic need. Once in place in a system that has to run continuously, there is often no path to swap it out for a more efficient way of doing things.

One example from biology is the human eye, where the nerve cells run through the optical path, giving rise to the blind spot. The octopus eye "lucked out" and evolved in a way that makes more sense, but human eye sight is just fine (there is a "software patch" in the brain to fill in the blind spot):

Source: https://external-preview.redd.it/xWn0rxCSyu9TrnXozPrsF3aq2_uILbUEJwzBv87TZo4.jpg?auto=webp&s=7a1bccf554ec19a813f9da4466c114ac84c426b7


Goldschmidt alumino-thermic process I recently came across this question in a book.

  Among the metals $\ce{Cr}$, $\ce{Fe}$, $\ce{Mn}$, $\ce{Ti}$, $\ce{Ba}$ and $\ce{Mg}$, the one that cannot be obtained by reduction of its metal oxide by aluminum is:-
  (a) $\ce{Cr}$
  (b) $\ce{Fe}$
  (c) $\ce{Mn}$
  (d) $\ce{Ba}$
  (e) $\ce{Mg}$

So, one method to answer this question is by Ellingham Diagram and to remember the values of $\Delta G^0$(Gibbs free energy change) for oxidation of each metal into its oxide and it's variation with temperature.

So, we can see that $\Delta G^0$ for oxidation of $\ce{Al}$ is much lower than other metals except $\ce{Mg}$ at lower temperatures. Thus, $\ce{Al}$ can reduce other metals except $\ce{Mg}$ at lower temperatures.
Thus, the answer is (e) $\ce{Mg}$.
So, this was the answer based on the experimental data for metals.
Couldn't we determine this answer theoretically as we can't remember all these data?

Actually, activity series works well in aqueous medium, while these reactions happening at high temperature will leave all waters long gone in vapours.

Well, here we go by (delta)G for the coupled reactions.
We know that a negative (delta)G amounts to SPONTANEOUS process.

The 2 reactions which need to be coupled must be such that sum of (delta)G1 for the reaction-1 and (delta)G2 for the reaction-2 is NEGATIVE.

Seeing the curve, we can conclude, Al can reduce all the metals listed except Mg, which of course could be above ~1600 degree celcius.

The conclusion: upto 1600 degrees Mg can reduce Al2O3, while above 1600 degrees Al can reduce MgO [Unlike what happens in aq. medium!]


Why are radicals unstable? Why are unpaired electrons especially reactive? Does pairing electrons decrease the reactivity of the electrons? But then doesn't forcing two electrons into the same orbital cost energy?

Free radicals are usually very reactive.  If we bring two free radicals or a free radical and a molecule with electrons available for bonding together a bond will usually form.  Yes, it does cost energy to bring those electrons close together.  But look at the payoff, a new chemical bond is formed and this is usually quite exothermic (stabilizing), usually more than enough energy to offset the energy cost of bringing those electrons close enough to form a new bond.  Because of the exothermicity of bond formation the reaction profile usually has a small activation energy.  Translation: a small activation energy means a  fast reaction.  A caveat, at the outset I said that "free radicals are usually reactive".  There are many known free radicals that are quite stable.  This is usually due to steric factors (the reactive free radical center is surrounded by bulky groups and just not sterically accessible for reaction) or electronic factors (some free radicals exist in very large delocalized systems, hence the spin density at any one atom in the system is so small that reaction is unlikely).  One further point, most of what I've said relates to free radicals, which was the title of your question, however you did mention "unpaired electrons" in the beginning of your post.  Molecules with unpaired electrons (note the plural, electrons) are different than free radicals with (usually) just one unpaired electron.  Some molecules like $\ce{O2}$ are quite stable even though they have two unpaired electrons.


2-Deoxy-D-arabinohexose, 2-Deoxyglucose Aldrich says that these two names are synonyms of one another.  Does anyone know the etymology of arabinohexose used in this context?

Actually arabinohexose got its name from the family of six carbon membered mono saccharide(i.e. aldohexose) and the first name arabinose is taken from the another popular name i.e. gum arabica because it was first extracted from it!


Pepto-bismol molar bismuth concentration I want to measure the Bismuth molar concentration in the OTC drug "pepto bismol".

525 mg bismuth subsalicylate (C7H6BiO4) / 30 mL (17.5 mg/mL)
molecular weight 363.10054 g/mol

If I convert to mM (17.5 mg/mL / 363.10054 g/mol * 1000), I get 48.2 mM. However, this value seems very high. Am I missing something obvious?

I am sorry if this question seems basic to you. I can't find the answer anywhere else though. Thank you so much in advance!

Your calculation is correct. You are not missing anything obvious. Your set up, which converts mL to L and then mg to mmol is correct and you get the correct units.

$$\require{cancel}
\mathrm{\dfrac{17.5\ \cancel{mg}}{\cancel{mL}}\times \dfrac{1000\ \cancel{mL}}{L}\times \dfrac{mmol}{364.10054\ \cancel{mg}} =48.2\ mM}$$

This concentration is not "too high". It is what it is. 48.2 mM is 0.0482 M, which is a pretty low concentration.

For comparison, your 17.5 mg per mL is at most 2% by mass (low-balling the density of the suspension to be around 1). 1 M hydrochloric acid (which perhaps seems like a high concentration) is around 3% by mass. Concentrated HCl is 36% by mass and about 12 M.


Flame test for copper salts I have read from sources here, here and even in my textbook that Copper (specifically $\ce{CuCl2}$ ) burns with a bluish green flame. All the sources mention that this is due to electron excitation and de-excitation in the orbitals of Cu. However, Why is copper the only transition element to give a coloured flame?

Most other flame tests are given by alkali metals and alkaline earth metals, which is understandable and is a common trend across the group. Copper stands out as an exception. I would like to know if there are any other transition elements that can give flame tests and the colours of the corresponding flames. I would also appreciate if someone could explain the electron excitation, either via Valence bond theory or by a Molecular Orbital treatment of a copper compound.

There is a famous saying that paper never refused an ink, which means anyone can write anything on the paper and the paper won't say anything back to the writer. Internet has made this situation worse. One should develop a healthy level of skepticism when reading website and forums.

  Why is copper the only transition element to give a coloured flame? Most other flame tests are given by alkali metals and alkaline earth metals, which is understandable and is a common trend across the group. Copper stands out as an exception.

This is not true. Boron compounds, when burned in alcohol, burn with a similar green color. Now here is the fundamental point to understand. When we see colors in the flames, usually it is due to very simple molecules of those metallic compounds in the flame. It is a molecular emission. If you can buy a pocket spectroscope, you will just a broad bands of color. So a flame spectrum, specially a low temperature flame, is not equal to an atomic emission spectrum. If I introduce copper into a very hot flame (>2000 Kelvin), you will not see anything, because copper atoms emit light mainly in the UV range.

Therefore all the common flame tests with calcium, barium, strontium show the flame emission spectrum of simple diatomic molecules of Ca/Ba/Sr in the flame. Copper is no exception. You may also wonder why a hydrocarbon flame is blue, there is no transition element there?

Vanadium is yellowish green

Cerium makes the flame yellow

Lead is blue white

Yttrium is beautiful crimson, which often appears on textbook covers on inductively coupled plasma spectroscopy (very high temperature atomic emission). Wikipedia has a big list.

Atomic spectroscopists use Grotrian diagrams to assign the transition lines the emission spectrum. Nobody uses VBT to explain the spectrum. Predicting the correct emission spectrum even of a diatomic molecule in the flame, will perhaps require the knowledge level of a PhD in physics.


Why shouldn't lithium batteries be discharged below 50%? I know both from online websites and from experience that discharging a li-ion battery below 50% greatly decreases its life and capacity.

But why is it so?

There are two main effects that cause damage to a Li-ion battery due to "deep cycling":

During the charge/discharge cycle, lithium ions are intercalated within the electrodes. This causes physical stress to the electrode material and spalling of their coatings. Deeper discharge causes increased stress. This effect is made worse with the temperature range from heavy current use as in a laptop.
Lithium metal may accumulate on the anode, taking it out of play. This effect may be due more to over-charging than to deep-discharge, though.


Amine group - electron donating or withdrawing group? Is the amine group a donating or withdrawing group? I think it depends on resonance. If the amine group can participate in resonance, the amine group is inductively donating. Resonance depictions below.

When the atome of nitrogen participates through its free electron pair in resonance, the amine group plays the role of a donating group by conjugation. When it cannot participate in conjugation through resonance, the amine groupe plays the role of inductive electron-withdrawer group, due to the higher electronegativity of nitrogen atom compared to carbon atom.

When the amine group plays the role of electron donating group through conjugation, its inductive withdrawing effect (still existing) is weaker and masked by the first effect.


What is the precise definition of maximum work in the context of electrochemistry and thermodynamics? I'd like clarification of the meaning of $W_\text{max}$ in the following equations (and thus in the combined result in the third equation):
\begin{align}
W_\text{max} &= \Delta G\\
W_\text{max} &= -nF\varepsilon\\
\Delta G &= -nF\varepsilon\\
\end{align}

In the first equation, we somehow come to the conclusion that the maximum useful work available from a process is that process' change in Gibbs Free Energy. I ask now what work that means - does it mean work done on the surroundings by release of energy from the system? Does it mean the maximum work done on the system via a chemical process (this makes more sense because it doesn't make sense that an increase in Gibbs Free Energy of a system would do work on anything but the system)?

Secondly, in the electrochemical definition of $W_\text{max}$, we have the point of view of the system; that is, positive work is done on the system and negative by the system. Therefore, we add the negative sign because when cell potential is positive the work done on the system is negative (it does work on the surroundings).

The only logical conclusion I can come to is that the two $W_\text{max}$ definitions must be both with respect to the system's point of view, which means that $W_\text{max} = \Delta G$ refers to the maximum amount of work done on a system during a chemical process (the intuitive interpretation that I posed). However, it still makes me uncomfortable to define "useful work" as work done on the system. How can we harness work from putting energy into something?

"Maximum Additional Work that can be extracted at constant pressure and temperature" to be precise, which may include electrical work, etc.  It means the work besides the expansion work.

The negative sign signifies the sign of the charge of the electron, since electrical work is proportional to the charge.


Solubility of epinephrine in water Do you know how can I dissolve epinephrine in water? because it is insoluble in ordinary situation. When I decreased the pH by $\ce{HCl}$ it really worked but I need pH=7 to continue my experiment. Can I add $\ce{HCl}$ and dissolve the EPI and later using buffer to reach pH=7 and in this way the EPI wont be precipitate?

What the context of aiming for a highly concentrated aqueous solution of epinephrine / adrenaline?  What would be a sufficient concentration?  Would -- assuming a chemical reaction is intended -- in sight of a low reaction rate a steady of epinephrine in low concentration be sufficient?

While this lists a typical concentration of saturation in water of about 180 mg/L, other conditions are provided, too.  The list equally includes a statement like "soluble in acetic acid" -- so perhaps you do not need to decrease the pH value down to 1 (by HCl), and maybe you were able to obtain a sufficient concentration of epinephrine in a mixture of water / acetic acid, too?

Are there restrictions, imposed by the intended application / compatibility  toward (biological?) materials to stay near neutral conditions?  One way to reply to this and to tailor solubility of pharmaceutical active ingredients is to refrain from their form as free acids or bases, and to use them as salt.  In this sight, would be epinephrine tartrate, or an acetate, an alternative option?


Is hybridization of the cyclopropyl anion sp5? I have read some answers about the $\mathrm{sp^5}$ hybridisation of carbon atoms in the $\ce{C-C}$ bond in cyclopropane as well as the $\sigma$-bond resonance in methylcyclopropyl cation.
Overall, the answers say that the $\ce{H-C}$ bond is due to overlap of $\mathrm{sp^2}$ orbital of carbon and $\mathrm{1s}$ orbital of hydrogen, while the bent bonds of the ring are formed by overlap of $\mathrm{sp^5}$ hybrid orbitals.
But, overall true separation of orbitals in such cases is not defined i.e. we cannot say that some orbital is completely $\mathrm{sp^2}$ hybridised.
Similarly, if the loss of a proton occurs from cyclopropane then the following:

In such an ion, will the hybridization change and $\mathrm{s}$ and $\mathrm{p}$ characters of the orbitals be redistributed?
My assumptions for this are:

In the $\mathrm{sp^2}$ bond the lone pair due to loss of proton will pose increased steric hinderance as it occupies greater volume. This will cause increased repulsion and decreases stability. Moreover, shouldn't the $\mathrm{s}$-character increase to accommodate the increased negative charge?

Accordingly, wouldn't the $\mathrm{sp^5}$ bonds will now have greater $\mathrm{p}$-character to ease even greater strain produced?

Then, what will be the change in the hybridisation in the overall molecule on formation of this ion?
References:

Sp5 hybridization in cyclopropane?
Are the triply-bonded carbons in pyridyne (and benzyne) sp-hybridised?
Non-integer hybridization
https://www.ch.imperial.ac.uk/rzepa/blog/?p=14548

Background
Charles Coulson was the originator of Coulson's Theorem, a useful tool for the chemist.  It allows you to make "back of the envelope" estimations of hybridization if bond angles are known. Conversely, if one knows the hybridization from, say, $\mathrm{p}K_\text{a}$ or $J_{C^{13}-H}$ data, then the bond angle can be estimated.
The key equation is $$\ce{1+\lambda_{i} \lambda_{j} cos(\theta_{ij})=0}$$
where $\ce{\lambda_{i}}$ represents the hybridization index of the $\ce{C-i}$ bond (the hybridization index is the square root of the bond hybridization) and $\ce{\theta_{ij}}$ represents the $\ce{i-C-j}$ bond angle.

See these earlier answers for interesting examples where the theorem is applied:

estimating the hybidization in cyclopropane
angle of attack of a nucleophile at a carbonyl
the lone pairs in water are non-equivalent - the resultant hybridization of water

The Question
The $\ce{H-C-H}$ angle in cyclopropane has been measured to be 114°.  From this, and using Coulson's theorem
$$1 + \lambda^2 \cos(114^\circ) = 0$$
where $\ce{\lambda^2}$ represents the hybridization index of the bond, the $\ce{C-H}$ bonds in cyclopropane can be deduced to be $\mathrm{sp^{2.46}}$ hybridized. Using the equation
$$\frac{2}{1 + \lambda_{\ce{C-H}}^2} + \frac{2}{1 + \lambda_{\ce{C-C}}^2} = 1$$
(which says that summing the "s" character in all bonds at a given carbon must total to 1), we find that $\lambda_{\ce{C-C}}^2 = 3.74$, or the C–C bond is $\mathrm{sp^{3.74}}$ hybridized.
Now, if we remove a proton from cyclopropane and generate the cyclopropyl anion, we move from a situation where we had a pair of electrons shared between carbon and hydrogen in a $\ce{C-H}$ bond to a situation (the anion) where we have a pair of electrons residing entirely in an orbital on the carbon atom.  In other words, we have increased the electron density in this carbon orbital.  Bent's Rule tells us that the molecular geometry will change so as to lower the energy of this pair of electrons and that it will lower the energy of these electrons by increasing the s-character of the orbital they are in.  In order to increase the s-character in this orbital, we will take some s-character away from the 2 $\ce{C-C}$ bonds and 1 remaining $\ce{C-H}$ bond.  Hence, $$\lambda^2_{\ce{C-C}}>3.74$$ $$\lambda^2_{\ce{C-H}}>2.46$$ and $$\lambda^2_{\ce{C-electron pair}}<2.46$$
Said differently, the interorbital (not internuclear) $\ce{C-C-C}$ angle at the anionic carbon will decrease and the $\ce{C-C-H}$ angle will decrease making the substituents about the anionic carbon appear more puckered.


Why does electron configuration in the periodic table is different when you follow the aufbau principle? . In periodic table for example Ag has an electron configuration of [Kr] 4d10,5s1 but when you do it yourself using Aufbau principle it should be [Kr] 5s2 4d9. Why is that?

When ask for Ag + valence electron configuration answers may now vary because of the existence of two electron configurations from the periodic table and using aufbau principle...

As a general rule, full electron shells are more stable than unfilled shells. In silver, promoting an electron from the 5s to the 4d shell completes the d orbital, which increases the stability of the 4d shell more than it destabilizes the 5s shell.
This is also explained here:

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=21&t=2807


Product of Friedel-Crafts cyclisation of (3-methylpent-4-en-1-yl)benzene My proposed mechanism and final product is:

But the book says the answer is:

I know that a 6 membered ring is more stable than a 5 membered ring so I made the product with the 6 membered ring (however, the carbocation was a secondary carbocation).

However, the book claims the answer to be the one with the 5 membered ring (the carbocation here was a tertiary carbocation).

Is the answer provided in the book wrong?

Though the question is old and pretty much answered in the comments, I ran into this question and draw the mechanism, so I thought about summarizing all that was brought up in an answer.

The carbocation is first formed on the secondary carbon. However, molecules are more stable when the charges are delocalized (this page talks about the stability of carbocations). When the positive charge is in the secondary carbon, it can only "share the burden" of having this charge with 2 neighbor carbons. So the tertiary carbon donates an hydrogen through a sigma to atomic orbital interaction (the sp3 hybdrization allows a π-type interaction) resulting in the more stable tertiary carbocation seen in the mechanism below:

Then the reaction follows analogously to the pathway you deduced for the 6 membered ring: First a π to atomic orbital interaction between the aromatic ring and the tertiary carbocation, which I believe to be the Rate Determining Step since it breaks the ring aromacity (bigger energy gap). Then finally the sigma to atomic interaction between the hydrogen and the ring carbocation, that restores aromaticity and releases the hydrogen.


A chemistry approach to environmentally friendly yoga mats… NBR, NR, TPE, PER I'm currently seeking your expertise while creating a line of environmentally friendly yoga mats, and I apologize in advance if any chemistry knowledge of mine has been absent since college and I lack the technical capability of this forum. I typically have a scientific approach to problem solving, and I'm hoping I can apply the same methodology to finalizing the material composition of my product.

Instead of blindly accepting the industry suggestions of natural rubber, NBR, TPE, and PER, I wanted to start with the physical characteristics required and work backwards to determine the ideal (perhaps innovative) material that would work best. My requirements are: 1) Eco-Friendly, 2) high fidelity printing compatibility via heat transfer

1) Environmentally friendly: PVC is not an option for the reasons I'm sure you would know. One would assume natural rubber would be the obvious choice, however if it degrades and has to be replaced every 2 years or so, I'd think the long term production effects might actually be more harmful then something more sustainable. From reading an article on PER it seems like it is essentially phthalate free PVC, and although processed different to be phthalate free the vinyl chloride monomer still exists and could be viewed as toxic, although to a lesser degree then PVC. Nitrile Rubber (NBR) is another option but it sounds like although it might be more heat resistant then natural rubber, it degrades similar to natural rubber. Various types of TPEs are also an option, but I've found minimum answers towards their actual eco-friendliness.

2) Heat resistant printing - Sublimation printing has typical been used on polyester, but from what I've researched should be compatible with any polymer. Unfortunately it requires a heat transfer of about 400 deg F, which I believe would scorch/melt natural rubber.

I know its a stretch, but I would love to get your input on the above materials or any other polymer that might be an option. I don't want to blindly commit to an "eco-friendly" material without fully investigating their chemical properties.

Thank you in advance for your help, and I'm definitely open to all of your respected critiques and suggestions.
Regards,
Shane

You will need a natural woven fiber. Any film will not help the skin in contact with it.


Distillation of methanol and water - Temperature Change We tried the synthesis of methyl benzoate in the lab via the Fischer Esterification reaction with excess starting methanol. After 1 hour of refluxing the reactants we immediately set up a simple distillation procedure to remove the excess methanol and here lies my question. By closely monitoring the temperature of the thermometer the mixture started boiling at 70+ oC and continued to rise slowly as expected. At one point the temperature started dropping slowly and after asking the professor she said that the distillation was over. I fail to understand why the temperature dropped. Can someone provide insight as to why it dropped and furthermore what should the thermometer show from start to finish and why? To my knowledge, the solution before the distillation contains methanol, some benzoic acid, methyl benzoate, sulfuric acid and water. Thank you in advance.

The thermometer that records the temperature is in the air high above the solution (usually above a Vigreux column). Your distillation was under ambient conditions, so there was air in contact with everything.

The thermometer can only records a rising temperature if the temperature around it rises. During the distillation, this is due to methanol vapour (around $\pu{70 °C}$) which can also condense on the thermometer to transfer heat.

Once all methanol has been distilled away, there is nothing coming from below that could transfer any additional heat: water, sulphuric acid and benzoic acid all have far too high boiling points and low vapour pressures. Thus, the thermometer is again in contact with the surrounding air and cools down.

While I have described this with air, basically the same thing will happen in a vacuum distillation, too.