text:
Do Metal Carbonyl complexes exhibit multiple back donation at the same time?
I came across a question which renewed one of my doubts that I had regarding bonding in metal Carbonyl complexes. According to me, metal should exhibit multiple back donation at the same time.
Everywhere on the web, I have only found metal carbonyl’s synergistic bonding in a single picture. But my doubt arises from the consideration from looking at the complete picture. According to me, In Metal Carbonyl complexes, the metal should exhibit multiple back donation to $\ce{ \pi^*}$ orbitals of Multiple Carbonyl Ligands according to the number of $\ce{d\pi}$ electron pairs present. I would like to explain this further through an example:
Bonding in $\ce{Cr(CO)_6}$
Now I do know that the orbitals taking part in complex bonding are a blend of d,s and p but I only know CFT but will still be explaining my doubt while keeping that in my mind.
Here the metal, Chromium has $\ce{3d\pi}$ electron pairs which according to CFT are present in $\ce{d_{xy},d_{xz},d_{yz}}$ orbitals, I don’t know how it will be according to LFT but the blended orbitals still(maybe) should be oriented in that direction and that is what I simply wish to use. Since, $\ce{Cr(CO)_6}$ is octahedral, thus the back donation for only 3 Carbonyl Ligands is possible at the same time, And this is where my question arises, that are there 3 back donations from $Cr$ at the same time or just a single back donation?
At first, I thought that I could simply confirm multiple back donation thingy by comparing bond lengths of $Cr-C$ in $\ce{Cr(CO)_6}$ and some other compound say $\ce{Cr(CO)_3(NH_3)_3}$, if I was correct the bond length of $Cr-C$ would be shorter in the latter case, BUT I didn’t find any data regarding the latter compound, I tried to search here and there but no luck but I found this post on Chemistry SE, if I assume it is right then crudely,I can say that my idea is correct about the bonding in $\ce{Cr(CO)_6}$, but I would like to have a further and deeper insight into this.
TL;DR Do Metal Carbonyl complexes exhibit multiple back donation at the same time or just a single back donation?

I find the question confusing. In BF3, the existence of the resonance hybrid means that all three pi interactions exist simultaneously, but they mutually weaken each other so the formal bond order on B is one total pi bond. So there's no difference between "multiple back donation at the same time" and "resonance hybrid"

@Andrew I mean to ask whether there exists 3 back donations as $Cr$ has $\ce{3d\pi}$ electron pairs or just 1. And Yes, I do think comparison with $\ce{BF_3}$ might be pointless.(I will be editing that out)

There are actually 12 "back donations" happening simultaneously, as each of the three off-axis d orbital interacts with all 4 ligands in its plane. Each ligand interacts via both of its orthogonal pi systems. For example, the CO located on the positive z axis interacts with dxz and dyz.

@Andrew As in $BF_3$ If one had to count the number of $B-F$ bonds, it would be 4(three sigma bonds and one back bond in resonance) That is what I wish to ask, In $Cr(CO)_6$ How many $Cr-C$ bonds are there. Also “For example, the CO located on the positive z axis interacts with dxz and dyz”, If that was the case then I don’t see how the correct representation would be M=C=O for such complexes.

With 3 filled d orbitals, you have three bonds worth of electrons, which divided by 6 ligands means a pi bond order of 0.5 (which is actually two 0.25 bonds, one in each of the two orthogonal pi systems on CO). But half bonds are a pain to draw, so M=C=O is a good way to represent that there are both sigma and pi interactions. No need to get too hung up on the exact bond order.


text:
Beckmann Rearrangement in cyclic compounds
Consider 1-indanone:

I am supposed to find the rearrangement product on treating this with hydroxylamine (which gives an oxime) and then $\ce{H+}$ (with heat). I am aware of the mechanism of Beckmann Rearrangement, and also know that the migration of the group anti to the hydroxyl group predominates. However, I find it difficult to predict the rearranged product as the structure is cyclic. I might be gravely wrong but If the group anti to the hydroxyl has to migrate, wouldn't the ring have to cleave? I cannot visualise any other way for this to happen. Any help would be appreciated.

Surprise! It doesn't. There's just an additional NH group adjacent to the carbonyl in place of the carbon atom in the cyclopentane ring. (Amide) formed without cleavage.

Cyclic has nothing to do with the rearrangement. Draw each of the oximes and do the reaction as you would for an acyclic oxime. Because the rearrangement is stereospecific, the answer depends one which oxime you are rearranging.

Without cyclic oximes, expanding to lactames, access to $\epsilon{}$-caprolactam and eventually nylon would not this significant today.

source: Stack Exchange

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text:
Calculate half-life from decay
I need to calculate the half-life in hours and I got the decay constant in min, how do I do? Which formula do I use? Is it just ln2/decay constant in minutes?

Don't you think converting from minutes to hours is not "chemistry"?

Ať what age are children taught how many minutes is 1 hour ?

Thank you for the kind answers :)))) my question was more about which formula to use.

The formula is always the same, just units of halftime and time constant differ. Or, if you insist to use mixed units, you have to involve explicit unit conversion factors. It is similar if you use mixed units for relation of distance, speed and time.

If you know the decay constant ($\lambda$), you use:
$$t_{1/2} = \frac{\ln 2}{\lambda}$$
The unit of $\lambda$ is $\pu{s-1}, \pu{min-1}, \pu{h-1}, \pu{d-1},$ or $\pu{a-1}$, while that of $\ln 2$ is unitless. Consequently, $t_{1/2}$ can be calculated as $\pu{s}, \pu{min}, \pu{h}, \pu{d},$ or $\pu{a}$, respectively. Since you have $\lambda$ in $\pu{min-1}$, what you have to do is converted it to $\pu{h-1}$ and calculate $t_{1/2}$ in $\pu{h}$ or calculate $t_{1/2}$ in $\pu{min}$ and converted the answer to $\pu{h}$.
For example, suppose your $\lambda = \pu{0.005 min-1}$.
$$t_{1/2} = \frac{\ln 2}{\lambda} =  \frac{\ln 2}{\pu{0.005 min-1}} =  \frac{0.693}{\pu{0.005 min-1}}= \pu{139 min} = \pu{2.31 h}$$
Or,
$$\lambda = \pu{0.005 min} = \pu{0.005 min-1} \times \frac{\pu{60 min}}{\pu{1 h}} = \pu{0.30 h-1}$$
Thus, $$t_{1/2} = \frac{\ln 2}{\lambda} =  \frac{\ln 2}{\pu{0.30 h-1}} =  \frac{0.693}{\pu{0.30 h-1}}=  \pu{2.31 h}$$

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text:
Is there an enolate equivalent for enamines?
I'm learning that enolates are stronger than enamines are stronger than enols in terms of general nucleophilicity. Makes sense. But what I can't find any explanation for online is why the trend shouldn't keep on going past enolates. Why isn't there a name for the species of enamine deprotonated on its heteroatom the same way there is for enols? (enols -> enolates, enamines -> ?)

I would imagine this species does not exist for some reason having to do with stability. Any explanation
or reasoning would be greatly appreciated and would help me get into the subject more. Thanks.

That would be rather called imine α‐anion https://onlinelibrary.wiley.com/doi/abs/10.1002/poc.357

Enamines are formed from secondary amines.

https://en.wikipedia.org/wiki/Enamine#Metalloenamines ; https://pubs.acs.org/doi/abs/10.1021/ja00262a025


text:
Source for UV cut-off values of buffers
When I want to check a $\mathrm{p}K_\mathrm{a}$ value I look it up in CRC Handbook. Is there such source for UV cut-off values of various buffers?
Usually books in chromatography contain tables with that data, but often no source is given.

Unless you use quartz cells the cut of will be limited by the glass(pyrex ,BK7 etc)/plastic at about 300nm. If there are any aromatics in your sample (and it is not coloured)  the cut off will be in the 250 -400nm region,  every other solution will absorb below about 210nm.

Snyder cites an article by Li [1] as the primary source for UV cutoff values ("Ref 7" below) for various buffers [2, p. 299]:

TABLE 7.1 Buffers for Use in HPLC Separation
$$\begin{array}{lrcr}
\hline
\text{Buffer} & \mathrm{p}K_\mathrm{a} & \text{Buffer Range}^\textit{a} & \text{UV Cutoff}^\textit{b} \\
\hline
\text{Trifluoracetic acid} & \gg 2 & 1.5{–}2.5 & \pu{210 nm}~(0.1\,\%) \\
\text{Phosphoric acid/mono- or di-K} & 2.1 & <3.1 & <\pu{200 nm}~(0.1\,\%) \\
\quad\text{phosphate} & 7.2 & 6.2{–}8.2 & \\
  & 12.3 & 11.3{–}13.3 & <\pu{200 nm}~(\pu{10 mM}) \\
\text{Citric acid/tri-K citrate} & 3.1 & & \\
  & 4.7 & 2.1{–}6.4 & \pu{230 nm}~(\pu{10 mM}) \\
  & 5.4 & & \\
\text{Formic acid/K-formate} & 3.8 & 2.8{–}4.8 & \pu{210 nm}~(\pu{10 mM}) \\
\text{Acetic acid/K-acetate} & 4.8 & 3.8{–}5.8 & \pu{210 nm}~(\pu{10 mM}) \\
\text{Mono-/di-K carbonate} & 6.4 & 5.4{–}7.4^\textit{c} & <\pu{200 nm}~(\pu{10 mM}) \\
  & 10.3 & 9.3{–}11.3 & <\pu{200 nm}~(\pu{10 mM}) \\
\text{Bis-tris propane}^\textit{e}\cdot\ce{HCl}/\text{Bis-tris} & 6.8 & 5.8{–}7.8 & \pu{215 nm}~(\pu{10 mM}) \\
\quad\text{propane} & 9.0 & 8.0{–}10.0 & \pu{225 nm}~(\pu{10 mM}) \\
\text{Tris}^\textit{d}\cdot\ce{HCl}/\text{tris} & 8.3 & 7.3{–}9.3 & \pu{205 nm}~(\pu{10 mM}) \\
\text{Ammonium chloride/ammonia} & 9.2 & 8.2{–}10.2 & \pu{200 nm}~(\pu{10 mM}) \\
\text{1-Methylpiperidine}\cdot\ce{HCl}/\text{1-} & & & \\
\quad\text{Methylpiperidine} & 10.1 & 9.1{–}11.1 & \pu{215 nm}~(\pu{10 mM}) \\
\text{Triethylamine}\cdot\ce{HCl}/\text{triethylamine} & 11.0 & 10.0{–}12.0 & <\pu{200 nm}~(\pu{10 mM}) \\
\hline
\end{array}
$$
$^\textit{a}$ $\mathrm{pH}$ range allowed with this buffer (conservative estimate).
$^\textit{b}$ Absorbance $<0.5\mathrm{A};$ from Ref. 7.
$^\textit{c}$ Requires addition of an acid (e.g., acetic or phosphoric).
$^\textit{d}$ tris(hydroxymethyl)aminomethane.
$^\textit{e}$ 1,3-bis[Tris(hydroxymethyl)methylamino]propane.

References

Li, J. B. Signal-to-Noise Optimization in HPLC UV Detection. LC GC 1992, 10 (11), 856–864.
Snyder, L. R.; Kirkland, J. J.; Glajch, J. L. Practical HPLC Method Development, 2nd ed.; Wiley: New York, 1997. ISBN 978-0-471-00703-6.


text:
Oxidation number of oxygen atoms in oxygen difluoride
My textbook says:

Oxygen is assigned an oxidation number of $+1$ in oxygen difluoride. The number assigned to each Oxygen atom will depend upon the bonding state of oxygen but this number would now be a positive figure only.

Can there be multiple bonding states for oxygen in oxygen difluoride?
What can be the oxidation state of each oxygen atom in oxygen difluoride?

Oxygen difluoride $\ce{OF2}$ has a single oxygen atom. You must mean dioxygen difluoride $\ce{O2F2}$.
The rules are the same as for hydrogen peroxide.
Both oxygen atoms in $\ce{O2F2}$  are equivalent, so their oxidation number is +1.


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Birch reduction in diethyl ether
I wanna ask, why I can't find any experiment using amonia dissolved in diethyl ether. You could bypass need of cooling the amonia under it boiling point by gassing the amonia to diethyl ether. Why it wouldn't work? Any experiments using dissolved amonia in solvent at normal temperature?? Thanks.

The reduction of enones with Li bronze in ether has been  accomplished. https://pubs.acs.org/doi/abs/10.1021/jo00418a024

If you can find another solvent in which to run the reaction, you don't need the ammonia as a solute.  The Birch reduction is a reaction with solvated electrons, not with ammonia, and the key is identifying a solvent that can hold the solvated electrons for reaction with the substrate.
Many solvents that are liquid at room temperature and atmospheric pressure can harbor these solvated electrons.  In fact even water does so, but (in addition to being a relatively poor solvent for many organic compounds) its highly polarized hydrogen atoms bonded to oxygen react so quickly with the electrons that the latter are gone before they can get to the intended substrate.  Solvents such as THF or ethylenediamine, in which the hydrogen atoms are bonded to carbon and nitrogen atoms and are less apt to be transferred as protons, are the ones that work.  From Wikipedia:

Since liquid ammonia has to be condensed into the flask and has to evaporate overnight after the reaction is complete, the whole procedure can be quite troublesome and time-consuming. However, alternative solvents have been employed, such as THF[1][2] as well as a mixture of n-propylamine and ethylenediamine,[3] both with comparable results.

Cited references
1.
Ecsery, Zoltan & Muller, Miklos (1961). "Reduction vitamin D2 with alkaly metals". Magyar Kémiai Folyóirat. 67: 330–332.
2.
Donohoe, Timothy J. & House, David (2002). "Ammonia Free Partial Reduction of Aromatic Compounds Using Lithium Di-tert-butylbiphenyl (LiDBB)". Journal of Organic Chemistry. 67 (14): 5015–5018. doi:10.1021/jo0257593. PMID 12098328.
3.
Garst, Michael E.; Lloyd J.; Shervin; N. Andrew; Natalie C.; Alfred A.; et al. (2000). "Reductions with Lithium in Low Molecular Weight Amines and Ethylenediamine". Journal of Organic Chemistry. 65 (21): 7098–7104. doi:10.1021/jo0008136. PMID 11031034


text:
In an closed adiabatic container, if we carry out an exothermic process, what will happen to the temperature of the system
I know that a closed adiabatic container acts as an isolated system, so I expect the internal energy of the system to be a constant as in an isolated system. But since it is an exothermic reaction, temperature will increase, and since internal energy is a function of temperature it should increase. This is a direct contradiction. Can anyone please help?

Have you heard about energy conversion ?

Temperature will increase. If system is adiabatic, it doesn't neccesarily mean that temperature of the system can't change because even though system can't exchange energy with surroundings, processes inside can generate heat. In your example, exothermic reaction. Note that, internal energy in your case stays the same even though temperature increased. This is because internal energy is the sum of kinetic energy of all molecules and atoms and all potential enegy of interaction between atoms chemically bonded and between all molecules (intermolecular bonds). When exothermic reaction happens, potential energy of interaction between atoms is partially converted into heat because if reaction is exothermic, chemical bonds formed (of products)are stronger than of reactants. Potential energy is only converted to heat which is kinetic energy of atomic and molecular motion (energy conservation - 1st law of thermodynamics). Since internal energy is sum of potential and kinetic energy of atoms and molecules it stays the same, only potential enegy is smaller and kinetic is bigger after exothermic reaction. If you have any misunderstandings, fire them to me.

If a reaction happens in an isolated system will it reach equilibrium? We know that entropy must be maximized. So what is the equilibrium state that maximizes entropy when a reaction takes place inside an isolated system?

Yes. You would need to solve for what extent of reaction entropy reaches its maximum, dS/dX = 0 (X is extent of reaction) in the same way as you needed to solve dG/dX = 0 for closed systems at constant p and T, to find equilibrium X.


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What products are formed when KMnO4 and KI react with each other in "weakly basic" medium?
While solving JEE Advanced 2020 (Paper 2) questions, I came across this particular question:

According to the final answer key, its answer is 6.
In my chemistry book (NCERT Chemistry), the following equation is given:
In neutral or faintly alkaline solutions:
$\ce{2MnO4- + H2O + I- -> 2MnO2 + 2OH- + IO3- ~~~~(1)}$
If I go by this equation, the answer should have been 0.
According to the same book,
In basic medium:  $\ce{MnO4- + I- -> MnO2 + I2 ~~~~~(2)}$
This gives the correct answer.
Is there a difference between "weakly basic" and "faintly alkaline"? What I can conclude from here is that the reaction in weakly basic medium is equivalent to that in basic medium, and not in faintly alkaline medium.
Could anyone please explain why the answer is 6 and confirm which equation is valid for the given case?

Side question:
I had learned that in basic medium, n-factor of $\ce{KMnO4}$ is 1 due to the following reaction:
$$\ce{MnO4- -> MnO4^2-}$$
This transition is not seen in the reaction (2). Could someone explain this?

**UPDATE:
After doing some analysis, I have been able to conclude the following:

Acidic medium:
$\ce{MnO4^- + I^- -> Mn^2+ + I2}$
Neutral / Faintly Alkaline medium:
$\ce{MnO4^- + I^- -> MnO2 + IO3^-}$
Basic (not alkaline - so that I2 is not oxidised to $\ce{IO3^-}$ due to presence of $\ce{OH^-}$ ions) / Weakly Basic medium:
$\ce{MnO4^- + I^- -> MnO2 + I2}$
Alkaline medium / Strongly Alkaline medium:
$\ce{MnO4^- + I^- -> MnO4^2- + IO3^-}$

Could anyone please confirm if the above equations are correct?**

See: https://chemistry.stackexchange.com/questions/67636/why-does-kmno4-oxidize-iodide-to-iodine-in-acidic-medium-whereas-iodide-to-ioda

With increasing pH, preference of MnO2 kicks in before I2 disproportionation of I2 to I- and IO3-.

@Poutnik so does that mean that even in the alkaline medium, MnO2 will be formed instead of MnO4(2-)

MnO4^2- is preferred in strongly alkaline medium. If not alkaline enough, it disproportionates to MnO2 and MnO4-.

Ok, thank you for adding that, so in the case of weakly basic medium, MnO4^(2-) did form but it got dissociated into MnO2 and MnO4(-) Hence, the ‘MnO2’ in the equation 3 (in the update section).

I still have some doubts. Coming to the equations 2 and 3 (in the update section), I had deduced solely from the answer key of the question that faintly alkaline and weakly basic mediums have a difference, but do they really differ? If yes, then in what ways?

Rather in mild alkalic solution, MnO4^2- does not form at all, unless as a not detectable intermediate product. // Those 2 ways of wording is hard to quantify.

Oh alright; So are all the four chemical equations correct? // Yeah, even I think that, thank you for confirming // I have considered that in basic (but not alkaline) and alkaline medium, different reactions occur due to the fact that in alkaline medium, OH^- ions are released. Could you please confirm if I am correct here


text:
Does lead(II) chloride dissociate into Pb²⁺ and Cl⁻ ions in HCl?
While solving JEE Advanced 2020 (Paper 2) questions, I came across this particular question:

Q.10 Choose the correct statement(s) among the following.
(A) $\ce{SnCl2 · 2 H2O}$ is a reducing agent.
(B) $\ce{SnO2}$ reacts with $\ce{KOH}$ to form $\ce{K2[Sn(OH)6]}.$
(C) A solution of $\ce{PbCl2}$ in $\ce{HCl}$ contains $\ce{Pb^2+}$ and $\ce{Cl-}$ ions.
(D) The reaction of $\ce{Pb3O4}$ with hot dilute nitric acid to give $\ce{PbO2}$ is a redox reaction.

According to the final answer key, its answer is options (A), (B).
Coming to the option (C), after doing some research, I have been able to conclude the following:

$\ce{PbCl2}$ is sparingly soluble in water.
$\ce{PbCl2}$ + little amount of dil. $\ce{HCl}$ decreases its solubility due to common ion effect.
Upon further addition of dil. $\ce{HCl}$ or $\ce{PbCl2}$ + conc. $\ce{HCl}$ leads to formation of soluble lead complexes: $\ce{[PbCl3]-}$ and $\ce{[PbCl4]^2-}.$
Hence, option (C) is incorrect as in one case, $\ce{PbCl2}$ doesn't dissociate at all and even in the other, $\ce{Pb^2+}$ and $\ce{Cl-}$ ions are not formed (so the dilemma of the $\ce{HCl}$ doesn't exist anymore as this option is incorrect regardless of the $\ce{HCl}$ being dilute or concentrated).

Are the above four points accurate? It'd be appreciated if someone could add more details and correct the above points if needed.

Your analysis about $\ce{PbCl2}$ in four points is correct.
Then ($\ce{A}$ and ($\ce{B}$) are correct, if they are in aqueous solution. The following equation shows that $\ce{SnCl2}$ is a reducing agent that is easily oxidized by air : $$\ce{6 SnCl2 + O2 + H2O  -> SnCl4 + 4 Sn(OH)Cl}$$ and, in ($\ce{B}$),  $\ce{SnO2}$ is soluble in $\ce{KOH}$ solutions according to $$\ce{SnO2 + 2 KOH + 2 H2O -> K2[Sn(OH)6]}$$ (D) is wrong, because the reaction is the following $$\ce{Pb3O4 + 4 HNO3 -> PbO2 + 2 Pb(NO3)2 + 2 H2O}$$ This reaction can be deduced by dealing with $\ce{Pb3O4}$ as if it was a mixture of $\ce{PbO2 + 2 PbO}$ so that only the two $\ce{PbO}$ react with $\ce{HNO3}$, without changing its oxidation number $$\ce{2 PbO + 4 HNO3 -> 2 Pb(NO3)2 + 2 H2O}$$


text:
What is the difference between radioactive equilibrium and transient equilibrium?
Consider the following radioactive decay:
$\ce{A\rightarrow B\rightarrow C}$
In my textbook, it is written that radioactive equilibrium is when rate of formation of $\ce{B}$ is equal to rate of disappearence of $\ce{B}$.  In my textbook, it is also written that when the net formation of $\ce{B}$ is zero, we say that the system is in transient equilibrium i.e. $\frac{dB_{N}}{dt}=0$ where $B_{N}$ is the number of atoms of $\ce{B}$ at time $t$. Now, $\frac{dB_{N}}{dt}=0$ means that the rate of formation of $\ce{B}$ is equal to the rate of disappearance of
$\ce{B}$ which is exactly what radioactive equilibrium is. Then, what is the difference between radioactive equilibrium and transient equilibrium? Is there even any difference between them? Pleas explain. I am confused.

I guess you have been caught in a semantic trap. Adjectives radioactive and transient describe  two different aspects of the same thing.
It could be as well written as "radioactive transient equilibrium". Or, even better as "radioactive steady state", as it is not true equilibrium.
Radioactive describes its nature . It may not be radioactive,by can be thermal, chemical, radiative, mechanical etc .
Transient describes its dynamics. It need not to be transient ( temporary, also called  a steady state ), but it can be a permanent, timeless, true  equilibrium.

Are steady state and transient state the same thing @Poutnik? @Poutnik please answer.

As it is already said in 2 different ways in the answer, answering it for the third time can be considered excessive.


text:
Lindemann–Hinshelwood mechanism for a gas reaction

The effective rate constant for a gaseous reaction which proceeds by a Lindemann–Hinshelwood mechanism is  $\pu{1.01E-5 s^-1}$ at a concentration of $\pu{4.6E-6 M}$ and $\pu{2.21E-5 s^-1}$ at a concentration of $\pu{8.71E-5 M}.$ Calculate the rate constant for the activation step in the mechanism.

From what I've understood the equation that I should be using is:
$$ \frac{1}{k_\mathrm{eff}} = \frac{k^{'}_\mathrm{a}}{k_\mathrm{a}k_\mathrm{b}} + \frac{1}{k_\mathrm{a} [\ce{A}]}\label{eqn:1}\tag{1}$$
However I don't quite understand how this equation is derived from this:
$$\frac{\mathrm d[\ce{P}]}{\mathrm dt} = k_\mathrm{eff}[\ce{A}],\tag{2}$$
where
$$k_\mathrm{eff} = \frac{k_\mathrm{a}k_\mathrm{b}[\ce{A}]}{k_\mathrm{b} + k^{'}_\mathrm{a}[\ce{A}]}\tag{3}$$
If we assume that  $k^{'}_\mathrm{a}[\ce{A}] \gg k_\mathrm{b},$ shouldn't $\displaystyle k_\mathrm{eff} = \frac{k_\mathrm{a}k_\mathrm{b}}{k^{'}_\mathrm{a}}?$
I have only seen a similar problem using the pressure and they get a new equation from \eqref{eqn:1}, which looks like:
$$\frac{1}{k}-\frac{1}{k^{'}} = \frac{1}{k_\mathrm{a}}\left(\frac{1}{p} - \frac{1}{p^{'}}\right)\tag{4}$$
which they use to solve for $k_\mathrm{a}$ and that is the rate constant for the activation step. I don't quite understand how they get that equation either.
How is \eqref{eqn:1} derived and how can it be rewritten as the last equation to solve for $k_\mathrm{a}$?

Write down the equations for $\ce{A + M<=> A^*  + M;; A^\to products}$ and apply steady state to $A^$.

@porphyrin Thank you! I think I understand how the equations are derived. I modified eq. (4) from pressure to concentration and got that $k_a = 0.191 M^{-1} s^{-1}$. Hopefully this is correct :)


text:
Why alpha-decay only occurs in elements with atomic number greater than 83?
In my textbook, it is written that $α$-decay only occurs in heavier nuclei. But why? Why is that so? There is literally no explanation given in my textbook as to why this is true. Please explain. Would greatly appreciate it!

I have growing feeling you may confuse CH SE site with an online interactive course of chemistry. At doubts over your textbook, I advice you to think, search online and other sources well, read them, think again, and then think more.  If not enough, ask with the explicit summary of your intermediate results of your prior effort. Without it, your questions may get frequently closed.

Define "heavier".

Heavier means "with an atomic number greater than $83$" .

@Pranita In practice, this mode of decay has only been observed in nuclides considerably heavier than nickel, with the lightest known alpha emitters being the lightest isotopes (mass numbers 104–109) of tellurium (element 52). Exceptionally, however, beryllium-8 decays to two alpha particles (Wikipedia)

First, this is a better fit for Physics.SE. Second, why fixate on number 83? Is this a number you came up with, or is there a reference, since α-decay is also possible for lighter elements? Third, a nitpick: particle symbols are always upright and it makes no sense to use MathJax to denote them in text; there is no upgreek macro yet and in math mode all Greek letters are slanted, which is a wrong notation.

@Poutnik actually I am an Indian student studying in Class 12 and preparing for a pre-medical exam called National Eligibility cum Entrance Test (NEET). This year is very crucial for me as I am preparing for NEET and willing to be a doctor. So, I don't get much time for researching stuff (I have only 1 year). So, I ask my doubts (not homework questions) in this site as this is an amazing website with great people who clear my doubts instantly! Thank you so much Poutnik! You have helped me a lot in chemistry! I don't have a good chemistry teacher. I wish I could meet you someday in real life.

@Poutnik, also I don't get time to write all the stuff I have come up with so far with my answer. You can google up NEET and see how hard the exam is.

@Pranita The exam toughness dosen't decide the eligibility of a question. Although I believe this was a valid question about a concept and not a homework question.

The premise is bluntly false as, for examples, ⁸Be decays to two alpha particles and ⁸B to two alphas and β⁺ (see Poutnik’s answer for more details).

A general equation for the alpha decay is
$$\ce{A -> B + ^4_2He + $\Delta E$}$$
You can check whether the decay is possible by calculating $\Delta E$ as
$$\Delta E=\left(m_{\ce{A}}-m_{\ce{B}}-m_{\ce{He}}\right)c^2$$
The alpha particle has a relatively high binding energy of $28.296\ \mathrm{MeV}$ ($7.074\ \mathrm{MeV}$ per nucleon). However, when going down from $\ce{A}$ to $\ce{B}$, some binding energy is lost. The alpha decay is possible if this lost binding energy is less than $7.074\ \mathrm{MeV}$ per nucleon.
You should already know that the graph of the binding energy per nucleon has a maximum at around iron. For heavier nuclides, the binding energy per nucleon is decreasing. So the increase of total binding energy is not linear but levels out a bit. If this increase per nucleon falls below $7.074\ \mathrm{MeV}$, alpha decay becomes possible.
From the graph of total binding energy, we can see that alpha decay should be possible starting at a mass number of $A=142$. In fact, alpha decay was observed for $\ce{^142_58Ce}$.

Generally, lighter nuclei have too high nuclear bonding energy to allow alpha particles to escape.
For heavy elements, coulombic repulsive forces weakens the overall bonding, that leads to alpha radioactivity.
Exception are all nuclei with nucleon number 5 or 8 that quickly decay, forming an alpha plus a nucleon,  resp. 2 alpha particles(eventually after beta decay )
Aside of the above exceptions, the lightest alpha emitter is $\ce{^{104}_{52}Te}$, probably because the target $\ce{^{100}_{50}Sn}$ nucleons have double magic numbers  50 + 50. Isotopes with proton or neutron count equal to magic numbers are often especially stable.

Yes. But why is this limit fixed at this number 83 ? Why 83 ?

If it was 81 or 87, would you ask why 81 or 87? :-) At some proton number, the nuclear/EM force balance must have its threshold.

If it was 81, 87, 52, or 2000, I would ask the same question. Why is there a limit here ? Why should it have a threshold indeed ? Even Beryllium-8 emits two alphas !

Rather, 8Be cracks into 2 halves. Alpha has exceptionally low energy. 83 is the proton number of the first unstable element (not counting Tc and Pm) 209Bi has the halftime approx 10^19 years.

I knew all what you say. The main problem is why. Why a threshold for alpha radioactivity ? There is no threshold for beta radioactivity. Why for alpha ?

Beta radioactivity need not to overcome strong nuclear force.

@Maurice I asked once long time ago, why all radioactive series terminate at lead? Why exactly lead? Why not other metals? In fact, I got 2 very beautiful answers. Turns out, there is no threshold. You can theoretically reach hydrogen from radioactive decay but it will take an immense amount of time for the decays to happen (more than the age of universe).

@NilayGhosh. Wonderful answer. Thank you !

@Maurice Very most of nuclei would not have energy to do so. Even $\ce{^{209}_{83}Bi}$ which does have that energy would take  billion times the universe age. All the process of big stars life, leading from hydrogen to iron, would not be spontaneosly reverted back.


text:
Ramachandran plot Phi(ϕ) Psi(ψ) dihedral angle Convention for Zero, Positive and Negative value- old and new
Update: possible duplicate: What is the precise definition of Ramachandran angles?. Question modified.
G.N. Ramachandran et al, in their own work (PMC) (DOI), did not used phi(ϕ) and psi(ψ) as we use today. They used phi(ϕ) and phi-dash(ϕ') defined as follows:

The conventions we have adopted for the standard configuration ϕ = ϕ' = 0°) and for the positive sense of rotation for the two angular parameters ϕ and ϕ' are indicated in Fig. 1.

Where Fig. 1 is:

Also they had adapted a very different values, 0 to 360 degrees for ϕ and ϕ'. They provided the Ramachandran plot like this:

This looks very different from the Ramachandran plot we see nowadays because:

This (Ramachandran's original) graph looks flipped from todays.

The zero degrees value of present day is perhaps different from Ramachandran's Zero degrees. Because the current day zero degree correspond to Ramachandran's 180 degree (of original graph).

So my question is basically:
On todays convention; how we define the zero value, positive value and negative values of phi and psi dihedral angle? and how does the new and old convention relate?
Prior research:
I did not find anything on intensive internet search. I got one graphic without any reference

I am doubtful about this diagram because its Zero degrees seem to be the zero degree of Ramachandran's original plot.
Many thanks in advance
Update: As the user @Buck Thom said

The meaning of angles with value of zero is the same.

But sill I looked both old and new graph and found what is labelled as 0 degrees in the new graph, has been labelled as 180 degrees in the old graph as follows

@BuckThorn Thank you. I did not mean 360 degrees I mean 180 degrees. Actually I tried to manually reconstruct the Left handed alpha helix, by considering the Web clip (3rd picture). Turned out I made a right handed helix. The confusion gone when I followed Ramachandran's original paper. Seemingly the 2 conventions are very different.

Related: https://chemistry.stackexchange.com/questions/114794/what-is-the-precise-definition-of-ramachandran-angles

Evidently the old $\phi$ is measured from an angle oriented $180^\circ$ relative to the new convention. This implies that the amide H ($H_N$) rather than carbonyl carbon was used as the reference fourth atom used to define the dihedral, since these atoms are related by a $180^\circ$ rotation. The old choice of atom is opposite on the chain to the modern convention of using the heavy backbone atom (carbonyl carbon). You can obtain angles in the new convention by subtracting $180^\circ$ ie $\phi \rightarrow \phi - 180^\circ$.

The meaning of angles with value of zero for $\phi$' aka $\psi$ is the same. However the values between $180^\circ$ and $360^\circ$ in the original Ramachandran plot can be shifted below zero, as $180^\circ\rightarrow -180^\circ$ and $360^\circ\rightarrow 0^\circ$, or more generally $\psi \rightarrow \psi - 360^\circ$.
By the way inspection shows that this is the correct interpretation. See for instance the following diagram. The extended (beta/turn/pII) region is usually larger than the alpha region. The beta region sits at large values (below  $180^\circ$) of $\psi$, whereas the narrower alpha region is close  to  $0^\circ$ (at negative angles):

In Ref. 1 Ramakrishnan and Ramachandran explain the differences in the  conventions:

At a recent conference of some of the representative workers in this field held in Bethesda, it was decided to denote the two dihedral angles about the bonds N-Ca and C.-C' by $\phi$ and $\psi$ respectively, the sense of rotation being the same as that adopted here. The fully extended chain, with N-H and C'=O trans with respect to one another, is to be taken as the standard conformation with $\phi$=$\psi$=0.
It is readily seen that the new $\phi$ is the same as the old $\phi$, but that $\psi$=$180^\circ$+$\phi'$ (and $\phi'$=$180^\circ$+$\psi$). All the data reported here are thus readily converted into the ($\phi$,$\psi$) coordinates. The diagrams in Figs. 2, 3, and 6 have to be shifted up by $180^\circ$, or half the total length, along the vertical direction. In particular, the right-and left-handed alpha helices will have ($\phi$,$\psi$) equal to ($133^\circ$,$123^\circ$) and ($227^\circ$,$123^\circ$). As before, a helix with ($-\phi$,$-\psi$) will be inverse to one with ($\phi$,$\psi$); i.e., it will  be of opposite sense, but having the same number of turns per unit.
This paper was finalized well before this meeting, and so the older conventions are adopted here. However, it is proposed to use the new notation in the following papers in this series.

References
1.Ramakrishnan C, Ramachandran GN. Stereochemical criteria for polypeptide and protein chain conformations II. Allowed conformations for a pair of peptide units.  Biophysical Journal, 01 Nov 1965, 5(6):909-933. DOI: 10.1016/s0006-3495(65)86759-5.

"The meaning of angles with value of zero is the same" Then why 0 of new graph is corresponding with 180 degrees of old graph? I am not getting this.

It's because in a circle zero degrees is the same as 360 degrees. Or, alternatively, you can move clockwise say and call the angles positive, or counterclockwise, and call the angles negative.

Not 360 its 180 i think I failed to express what I wanted to say. let me update the question

This is the clearest answer yet. However looking for some atomic diagram showing the frame of reference, and if possible, reference, since it is a convention question.

The Ramakrishnan et al convention also seems different because it says " It is readily seen that the new ϕ is the same as the old ϕ" where as we can see the newest phi is 180 degrees different from old phi

I agree, that is confusing!  My current assessment is that they confused the symbols somehow (seems unlikely though). In any case the explanation in my answer shows the difference between the different coordinates, even if that quote from the article is not easy to decipher.

The paper compares two historic definitions. With the current definition main chain torsion angles are 180 degrees for an extended conformation, not zero like in the historic reference.

@KarstenTheis Thanks, I understand that there are differences, but I manage to still not understand the difference between $\phi'$ and $\psi$. The bottom figure in my answer shows a configuration with $\psi=0$. Would you agree? The caption calls this $\phi'=0$ The text says that at $\phi'=0$ the conformation is extended, but it clearly is not by modern standards, or?

I should have written in my answer "Ramakrishnan and Ramachandran attempt to explain the differences in the conventions". I am a little surprised that I am still confused by the text. In the text they say the meaning for $\phi$ is unchanged, while $\phi'=\psi+180^\circ$. OTOH I think I understand the standard definition of the dihedrals and that this matches what I wrote in my answer.

I tried to find out what the Bethesda conference decided on the the 1960s, and whether it is different from the current definition, but did not succeed. So maybe the footnote in the Ramakrishnan and Ramachandran paper is inaccurate, or the community decided to switch the atoms defining phi and psi later on. With the sources at hand, it is impossible to tell.

@KarstenTheis I would say the question remains to some degree unanswered, at least the part "how do the old and new conventions relate?"

I've attached a figure to show the dihedral angles. Hope this helps.

Part of a peptide chain. The four atoms defining the $\psi$ (N to N) and $\phi$ (CO to CO) torsion angles are shown on the right. As these chains are drawn as if completely extended, the angles are $180^\text{o}$. When the main chain atoms are eclipsed, the angles are zero. The 6 atoms in the shaded areas are in the same plane.
Ramachandran plot together with the structure for bacteriorhodopsin.

Ramachandran plot of the $\phi$ and $\psi$ torsion angles in the protein bacteriorhodopsin (pdb 1FBB), which contains extensive $\alpha$ - helices. Most angles cluster around the values $\phi = -56^\text{o}, \psi = -53^\text{o}$ typical of an $\alpha$ - helix. A regular or ideal helix would have angles $-60^\text{o}$ and $-50^\text{o}$. The area of $\beta$-sheet structure, of which there is very little in this protein, is in the top left corner bounded by approximately $-55$ and $+60$ degrees.
The structure of the protein shows extensive helical structure. The retinal chromophore, which is positioned in the centre of the column of helices, is shown also.

Thanks for your answer but can you please add how do I define phi=zero? on which conformation the phi would be called zero? similarly for psi, and other positive and negative values?

My questiom is even more basic. Its about how we are counting phi and psi; from which frame of reference.

I think this is what you asked, in the top figure 'As these chains are drawn as if completely extended, the angles are 180 degrees. When the main chain atoms are eclipsed, the angles are zero. The 6 atoms in the shaded areas are in the same plane.'

as to positive negative you will just have to try with a structure such as the one above. If you use python I have code that will draw diagrams and can addd maths  about dihedral angles if needed.

I dont know python but I am trying different online simulators. although recognising C, N , O is difficult in most of these simulators.

The protein data bank has a good simulator.

This is not really an answer. This is some conclusions or assumptions.
1. The ϕ (old) and the ϕ (new) dihedral angle:
It is quite clear that the graph look is unchanged along horizontal axis.

Where ϕ (old) corresponds to ϕ (new) as follows:
ϕ (old)   .................   ϕ  (new)
0 deg    ......... ..........   -180 deg
180 deg  ....... ..........    0 deg
360 deg  .... ..........   +180 deg
Conclusion:
ϕ (Old) - 180 degrees = ϕ  (new)
or, ϕ (old) = ϕ (new) + 180 degrees.
2. The ϕ' (old) and the ψ (new) dihedral angle
Note that diagram is scroll shifted.

Where ϕ' (old) and ψ (new) are related as followes:
ϕ' (old) ..... .....  ψ (new)
0 deg ............   0 deg
180 deg ........... 180 deg, -180 deg.
360 deg ........... 0 deg
It means we do not have a frame shift between ϕ' (old) and ψ (new) , although there is the difference in direction of counting the angle, from where the negative sign comes.

[OP] On today's convention; how we define the zero value, positive value and negative values of phi and psi dihedral angle?

In general, you have four atoms (1, 2, 3, 4) in a chain to define a torsion angle. You rotate around the bond connecting atom 2 with atom 3 to change the torsion angle (bond angles and lengths stay constant). The conformation where 1 and 4 are closest is called zero degrees (cisoid). The conformation where 1 and 4 are furthest apart (transoid, extended conformation) is called 180 degrees. In both cases, all four atoms are in a single plane.
To measure the torsion angles of other conformations, you consider the plane through atoms 1,2,3 compared to the plane through atoms 2,3,4. The angle between these planes is the torsion angle. For a given angle, there are two conformations related by mirror symmetry. The right-handed one is given the positive torsion angle while the left-handed one is given the negative one (see figure in OP's question).
For a mathematical treatment with figures, see https://www.math.fsu.edu/~quine/MB_10/6_torsion.pdf
For the protein main chain, there are multiple ways to define a torsion, e.g. N-CA-C-N or N-CA-C-O. Switching the definition will lead to a shift of torsion angles by 180 degrees. The current convention, however, is to use N-CA-C-N and C-N-CA-C, and this is reflected in the current version of the Ramachandran plot.
The authoratative reference for the definition of torsion angles in general is the IUPAC gold book, citing DOI:10.1351/pac199668122193.

The authoratative reference for which atoms are used to define phi and psi seems to be Richardson, J.S. (1981). "Anatomy and Taxonomy of Protein Structures". Advances in Protein Chemistry. 34: 167–339.

The Cisoid and Transoid is a good way to understand.


text:
Why does the addition of heavy water cause OH and NH peaks in NMR spectra to disappear?
When $\ce{D2O}$ is added to a sample, the OH and NH peaks in a $\ce{^1H}$ NMR spectrum disappear. This is the basis of (for example) the $\ce{D2O}$ shake test, and is also the reason why peptide/protein NMR is commonly run using $90\%~\ce{H2O}/10\%~\ce{D2O}$  as a solvent (or else the NH peaks would be unobservable).
What causes them to disappear from the spectrum?

The $\ce{D^{+}}$ exchanges with the proton and the exchange process leads to line broadening; usually the proton signal becomes so broad that it is no longer visible.

Even if there is no broadening you attenuate the signal according to the concentration of protons and deuterons in the solvent, which exchange quantitatively on the OH.

@BuckThorn   I read the question such that the OP was asking about an OH proton signal from an organic alcohol sample (e.g. hydroxy or hydroxyl, not hydroxide).  In such a case the concentration of the hydroxyl proton does not change $relative$ to the protons in the organic backbone of the alcohol. So while the chemical shift of the hydroxyl proton will change with addition of $\ce{D2O}$, the hydroxyl proton signal should still be visible (if the backbone protons are still visible) if exchange could be slowed compared to the timescale of the nmr experiment.

@ron I'm not sure I agree: $\ce{ROH + D2O <=> ROD + HOD}$ the hydroxyl proton concentration is decreased, with a corresponding increase in the concentration of HOD. My interpretation of the question is the same as yours, although the first edit muddled the question a bit, I must admit. I rolled back to the original.

@ron In fact I was doubly unclear in not writing explicitly "deuterium oxide" or deuterated solvent with exchangeable protons/deuterons ("protic"). I also meant that OD is not going to show up in a proton spectrum (although might have an effect through couplings). Exchangeable protons (including alcohol OH) will see their signals attenuated although the mechanisms and extent may differ depending on exchange rates and concentration of exchangeable protons in the solvent.

@orthocresol Yes, when I said "the concentration of the hydroxyl proton does not change relative to...", by hydroxyl proton I meant [$\ce{RO-H +DO-H}$].

R-OH + D20 interact to equilibrium. So you get R-OD + H2O. The more D2O you add the less R-OH you get until this is not longer detectable .

To be perfectly honest here, this doesn't add much to the topic and it's been discussed in the comments above.


text:
Does the composition of a solution keep changing when being boiled?
The mole fractions of the components of vapours aren't the same as those of the liquid-liquid solution (unless the solution is azeotropic), this implies a change in the composition of the liquid solution when vapours arise. Consequently, the partial vapour pressures of the components would also change and there would be a messed up cycle. There has to be something that I'm understanding wrongly and I'd like to be told what it is.

Yes, the composition of the liquid does change. However, if you carry out it in a closed vessel the compositions would reach an equilibrium

But a liquid cannot boil in a sealed container..?

Well you can heat it in a closed container. If you mean "boil" in the sense of boiling off vapour, then yes it would have to be an open container, and definitely the composition of the liquid will change over time.

Distillation, right?

Here is a textbook chapter: https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Physical_Equilibria/Fractional_Distillation_of_Non-ideal_Mixtures_(Azeotropes)

It is not a messed up cycle. Let's consider what happens with an example.
If you heat up a mixture containing $10$% ethanol + $90$% water, the mixture boils at $92$°C, producing a vapor containing $50$% ethanol + $50$% water. With this operation the liquid loses more ethanol than water. So its concentration in  ethanol decreases, and it is necessary to heat it up to a higher temperature to maintain the process of evaporation. When about one tenth of the liquid has distilled, the residual liquid contains still $5$% ethanol, and it is necessary to heat it up to $95$°C to obtain and continue boiling. At $95$°C, the vapor emitted contains $35$% ethanol. At $98°$C, the liquid contains still $2$% ethanol and the vapor $20$% ethanol. When the last drops are vaporized, the temperature is $100$°C and the last drops are made of pure water.

I was actually studying fractional distillation when the problem arose. When distilling ethanol, the solution of a given concentration is boiled and it's vapours extracted (and condensed). What confused me was this- how do we get a particular composition of vapours while boiling a solution (which is essential to obtaining a higher concentration of ethanol)? Or to do so, are the vapours extracted from a closed vessel rather than from a constantly boiling solution? Your answer gives an outlook of moving 'up the curve' in a positive azeotrope phase diagram, which I now fully understand.


text:
Is is potentially harmful to have a vitreous-enamel item in a small poorly-ventilated room?
I am planning to purchase two enamel photo-frames for a room in the attic which is short in height and not well-ventilated. I would like to know if enamel like that is subject to deterioration that may leach fumes or such in the setting.
I have read a lot of pieces online about its composition often involving harmful chemicals like lead and cadmium, and I wish to know if there is any chance of potential adverse exposure in a finished enamel product.
EDIT: This potential percolation of such toxic chemicals on contact-surfaces is an established fact, and has seen food-security agencies revise standards several times. Of course, food-contact surfaces are an instance where the deterioration is rapidified for obvious reasons.

True vitreous enamel has any toxic chemicals embedded in glass, locked in place as is used for nuclear storage. Furthermore, there are few volatile compounds of heavy metals, but certainly any volatile compound would have been removed at the temperatures used in vitrification, 1000°C and higher.
However, don't eat the vitreous-enameled photo frame... since enamel is glass, and the sharp slivers are likely to harm your alimentary and digestive systems!

So you’re saying there is definitely never any prospect of such seepage as I suggest, irrespective of treatment and composition of the enamel?


text:
How to melt and roll silver chloride into sheets
I want to make silver chloride cathodes for a magnesium salt water battery.
Are there any tips on how to proceed?
I hope to melt the silver chloride in a covered ceramic crucible in an electric furnace, then cast a small ingot.
This ingot will be pressed and rolled into a thin sheet.
Silver chloride is nonconductive and is made conductive by superficially reducing the surface to silver by immersion in a photo-graphic developing solution.
By forming a silver outer surface that is porous, no base grid need be used with silver chloride.
Any help or direction to information is appreciated so I don't have to "reinvent the wheel".

Silver chloride is not a metal, you can't roll it.

The wikipedia article on silver chloride says it can be hot pressed into windows and lens shapes.

@EdV Jup, just like KBr (at RT even), but then you end up with a solid piece, nothing porous. I'm not clear how this battery is supposed to work.

It looks trickier than using a support grid. It should be the other way around, perhaps. I.e. silver grid / foil with some AgCl deposited onto. That can be easily done electrochemically, and if one stops the deposition while the process is ongoing, the contact to the underlying metal electrode is guaranteed. Look for quasi Ag/AgCl electrode.

@Karl Yeah, so like making a KBr pellet in a press. I also have no idea how the proposed cell would work.

There are different ideas on addressing the impermeability of the rolled AgCl sheet.  Some say you should add the AgCl to a weak solution of photographic developer for some unspecified period of time until a small portion of each AgCl particle has been turned into metallic Silver.  Then this mass is dried and pressed to form a conductive porous block. Others say that the AgCl should be formed into a sheet of the desired thickness and then the outside is reduced just enough to be conductive but still be porous enough to allow Cl- ions of the electrolyte to penetrate.  An "inside out" electrode.

@Ed V, do you have experience making KBr in a press?  If so what sort of pressure would I need to exert to form a pellet?

@Alchimista You say in your comment that it is done easily, but I am a bit clueless. How would I electrochemically deposit AgCl onto a mesh?

@user85003 silver mesh plus NaCl

The last time I made KBr pellets was long ago, using a simple lab pellet press. There are also even simpler presses that use a couple of wrenches and elbow grease. I don’t know the actual pressure because the process was done until you got a good quality pellet for the IR spectrometer. As for making Ag/AgCl electrodes, e.g., reference electrodes, that is just a matter of electrolytically oxidizing Ag in a KCl or NaCl solution. The comment by @user85003 is interesting: perhaps chat them them?

@EdV Typical IR press tools are rated to 2-10 kbar. 2kbar at 45mm dia. are of course already 30t, 400kg at 5mm dia. At some lower pressure limit one will notice that it actually takes time for the KBr to cold flow and fuse into a solid piece. ;)


text:
On the relative ease of E2 dehydration
Concerning alcohol dehydration, I get that it's E1 in secondary and tertiary alcohols, that it's E2 in primary alcohols, and also why E1 reaction is easier for tertiary alcohols as compared to secondary ones.
Why is the ease of dehydration for primary alcohols less than both of these? It would have been obvious it was reacting through E1, but it's doing E2, which is favourable for primary alcohols.
Why then, is it harder to dehydrate primary alcohols than either of the other two?

Because most dehydration conditions are geared towards E1. That is, the acidic conditions create an oxonium ion that is more amenable to generating a carbocation, but a primary alcohol isn't very likely to generate the cation.

@Zhe: oh, ok, so E2 requires strong bases, but the medium's acidic and that favours E1. Is that how it works?

That's more or less how I'm seeing it.

The rate of reaction depends on how stable the transition state is. In both E1 and E2 good leaving group is present along with presence of a weak and strong base ,respectively. This leads to an alkene type of transition state in E2 mechanism and a cationic type transition state in E1 mechanism . Both these transition states gets stabilized by the electron donating ability of the attached alkyl groups (i.e By Hyperconjugation and Inductive Effect).This leads to the increase in rate of reaction and thus the secondary and tertiary alcohols are easier to dehydrate than the primary alcohols.
Hope this helps!!

Acid catalysed dehydration of primary alcohols is primarily by an E1 mechanism hence slower than dehydration of secondary and tertiary alcohols. The competing reaction is SN2 formation of ethers by attack of an alcohol molecule on the hydrated molecule. Bidentate reactions such as the Oppenauer oxidation, oxidation by quinones, or even chromate oxidation result in alpha attack and give ketones or aldehydes. The various mechanisms could possibly involve beta abstraction to form enolates with subsequent isomerization. I don't know of a direct E2 elimination to give alkenes involving alcohols in acidic solutions.


text:
Are (1R,2S)‐1,2‐dibromocyclopentane and (1R,3S)‐1,3‐dibromocyclopentane positional isomers?

Are these positional isomers or not? According to me they are, since Br substituent is located at different positions in both the pictures, but the answer key in one of my mock test websites says they're not. Can someone please guide me?

Does the the website you are using specify why these are not isomers?

No, unfortunately not

They are definitely position isomers.. .No doubt in that .The answer in the website must be wrong.


text:
What reagents are used to selectively reduce 2,2,3‐trimethyloxirane to secondary and tertiary alcohols?
Problem

Answer

 (c) $x = \ce{LiAlH4},~y = \ce{LiAlH4}/\ce{AlCl3}$

Questions
I know $\ce {LiAlH4}$ is a reducing agent because it gives four $\ce{H-}$ and will attack on our reactant on carbon with less substituents, i.e. in this case on $\ce{-CH-}$ and on attack the $\ce{-O-}$ will go to other carbon and on protonation we get 3° alcohol.
But what about 2° alcohol? How can we get that?
What does $\ce{AlCl3}$ do such that we get 2° alcohol?
I thought maybe the oxygen is giving its lone pair to $\ce{Al}$ being deficient in $\ce{e-}.$ Then what happens if it does so? Will now $\ce{H-}$ attack 3° carbon? Does that mean it's following SN1 mechanism? Why?

LiAlH4 + AlCl3 gives Alane AlH3 which is the reducing agent.  https://pubs.acs.org/doi/10.1021/ja01038a042

Ok, it gives Alane  but according to link I think it's only saying that it will give a trans compound on attack i.e attacking reagent which is $\ce {H-} $ and -$\ce{OH}$ will be opposite to each other .

I've struggled to find a clear mechanism for alane reduction of epoxides, but here is my (slightly) educated guess. Alane has more Lewis acid character than LiAlH4 so when it binds to the oxygen a cationic centre develops at the tertiary carbon which is then reduced by hydride from the alane. In contrast with LiAlH4 the hydride attack is a more nucleophilic process so it attacks the least hindred centre

As Waylander has pointed out in the comments, the mechanism would proceed as follows in case of $\ce{LiAlH4/AlCl3}$ as a reagent to produce a secondary alcohol (SN1), whereas in case of just $\ce{LiAlH4}$, the reaction proceeds via SN2 to give the tertiary alcohol by attacking the less hindered carbon.

Alane has more Lewis acid character than $\ce{LiAlH4}$ so when it binds to the oxygen a cationic centre develops at the tertiary carbon which is then reduced by hydride from the alane. In contrast with $\ce{LiAlH4}$ the hydride attack is a more nucleophilic process so it attacks the least hindred centre

Reaction with $\ce{LiAlH4}$ is base catalyzed ring opening while that with $\ce{LiAlH4/AlCl3}$ is acid catalyzed ring opening.
In base catalyzed ring opening the nucleophile attacks less substituted (and therefore less sterically hindered) carbon atom of an unsymmetrical epoxide, while in acid catalyzed ring opening the nucleophile attacks at the more substituted carbon atom of an unsymmetrical epoxide.


text:
How do I write alchohol formulas?
I have a task to write a formula for 2,4 dimethyl-hexane-1-ol.
If I am incorrect, please suggest me how to write it, but I believe it should look like this:
CH3-CH2-CH2-CH2-CH-CH2-CH2-CH2
/
OH
So it has 8 C, 16 H and 1 O, but I am not sure if that is where the OH should go (4th CH in the drawing as in 2,4).
Also does "1-ol" mean there is only one OH?
In short, how do I count where does a OH goes?

No, that's not how pretty much all of it works. What do you think "hexane" means, to begin with?

1-ol means the alcohol substituent is on the first carbon in the carbon chain. Similarly, 2-ol means -OH is on the second carbon.

First up, no '1-ol' doesn't mean there is one OH, the 1 signifies the position of the OH group in your parent chain (here ,hexane) which in this case is the C1 carbon. How many OH groups are present is represented by adding di/tri... as a prefix to '-Ol'.
You need to learn how to identify and number a parent chain in organic compounds first in order to make the structures of orgo compounds using their IUPAC names .
Anyway, OH group will just attach to a terminal alkene of hexane and whichever terminal you choose, you need to number it 1. This will cover the hexan-1-ol part.
The carbon in your parent chain, adjacent to the carbon you just numbered , would be numbered 2 and so on and so forth until you have covered all the carbons of your parent chain.
Now just attach a methyl group on the carbon numbered 2 and another methyl to the carbon numbered 4 and that will cover your 2.4 dimethyl.

How to ask and answer nomenclature questions?

@andselisk thank you, i am new to this site, are there any more such posts about how to answer specific type of questions?

No prob, the general advice is to include a citation for the current IUPAC recommendations since they are the authoritative source for chemical nomenclature. There not so many Meta posts regarding Q&Aing on a particular topic. For example, there is FAQ update: How to ask (or close) questions that involve potentially controlled or dangerous substances?, and you can always search Meta if you want to inform yourself.


text:
Nanocellulose suspension in water being dialysed in ethanol
I am working on nanocellulose. To prepare the sample for silanization, the first step is to dialyse the preparation in ethanol overnight. I wonder why this is a necessary step. Sorry, I am a physicist working on something which involves a lot of chemistry. I am alone in my lab so I don't know who else to ask. Thank you very much!

Where did you get the preparation instructions?

Hello. My prof gave me the protocol.  I didn't ask but my guess is that it is a solubility issue w/ the silane. Thanks for the reply


text:
S(p)-C(p) vs. O(p)-C(p) overlap
In Grossman, The Art of Writing Reasonable Organic Reaction Mechanisms, he provides the following explanation. The question asks to explain why the difference in pKa values between PhSH and EtSH is much smaller than PhOH and EtOH. In the solutions manual he writes:

Why is p orbital overlap dispositive for the strength of resonance stabilization exactly? I don't get it.
Furthermore, is this something I can simply take for granted about
the size of atoms or does this derive from a more general
principle about orbital overlap it would be useful to take forward?
In other words, its there a way I may have deduced this from
something else?
Could you show me any other and perhaps trickier examples where p orbital overlap modulates resonance structures in this way, in terms
of size or otherwise? I'm just always looking for more practice
material that's out of my wheelhouse so I can get exposure. Thanks
guys.

Umm, I think you should be less verbose and more specific. A bit formatting to highlight important parts, would be useful, at least.

In general, bigger p-block atoms tend to make weaker bonds, especially pi bonds. That's a rule of thumb. Well, maybe it's not taught properly, but that's beside the point, like any "how to know it in the first place"

There is no such book as "The Art of Mechanisms" written by Grossman.

@andselisk I believe he's referring to Grossman's "The Art of Writing Reasonable Organic Reaction Mechanisms", albeit in a contracted form.

@C_Lycoris I figured, but what's the point? A citation must be accurate and complete set of information, not a riddle.

@andselisk no opposition to that logic, at least not from me.

I agree with you guys too. Editing the answer now.


text:
Reaction of alkynes with sodium in liquid ammonia
Problem

The hydrocarbon which can react with sodium in liquid ammonia is?
(A) $\ce{CH3CH2C#CH}$
(B) $\ce{CH3CH=CHCH3}$
(C) $\ce{CH3CH2C#CCH2CH3}$
(D) $\ce{CH3CH2CH2C#CCH2CH2CH3}$

Answer

 (A) $\ce{CH3CH2C#CH}$

Question
I am confused why the answer cant be (C) or (D) since these two could undergo reaction with sodium in ammonia to give me trans-alkenes as products.

Well, I guess they missed this possibility.

Similar to this (even poorer) question Similar to https://chemistry.stackexchange.com/questions/94716/which-of-the-following-react-with-sodium-in-liquid-ammonia. The one marked in red in the picture that you have included is the only which doesn't occur.

@Alchimista Ah so i think the question itself has ambiguity. Also yes, i know the second reaction wont undergo birch reduction, i mis clicked option lol.

@Alchimista Nicely found. Well, I guess we can safely close this as a dupe and solve the $\ce{#CH3}$ mystery once and for all.

Poorer in my first comment referred to the quiz not really to the SE questions. @Andselisk yes its a kind of duplicate.

You are correct.
Na/NH3 is a standard method for reducing internal alkynes to trans (E) alkenes.
More reading here and here


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Why is the intensity of the IR spectra of IBr higher than that of ICl?
Why is the intensity of the IR spectra of $\ce{IBr}$ higher than that of $\ce{ICl}$?
I know that the IR spectra intensities are proportional to the derivative of the dipole moment with internuclear distance. However, how would one be able to deduce (without looking at the IR spectra) that the IR spectra for $\ce{IBr}$ is higher than $\ce{ICl}$ in the gas phase?
I would have thought it would be the other way around since $\ce{Cl}$ has the higher electronegativity.


text:
Which atoms in indole possess the greatest amount of electron density?
Presumably the amine acts as an electron donor, but that would suggest extra electron density of the adjacent carbons, making them more shield, which isn't seen in the results in the image below.

Specifically I'm wondering what causes the carbon with a chemical shift of 111 ppm to be less deshielded than the carbon with 123 ppm, as they look like they should be similar electronically, and I can't work out a resonance form that gives different results. Any help with this, and a more general rule set for chemical shift in aromatic compounds, would be much appreciated.

+1 Good question. Although note that in my experience, carbon shifts are much more difficult to analyse and interpret with our usual intuitive notions of electronegativity, resonance etc. You may want to look at the proton NMR and see if you can interpret that with the resonance structures, inductive effects etc.

Why so many close votes? This does not look like a homework question in any way.


text:
Names of the f suborbitals. (Reference Request)
Today, I have been browsing the internet to know the names of the "f" orbitals. But I have varying answers on the web. Different sources seem to be stating different names. Like here,

and here

Besides the first 2 pictures, this question on Quora also gives different names.
By names, I mean names of suborbitals like $d_{xy}$, $d_{z^2}$ etc.
Right now, I am really confused. An answer with a solid reference would definitely make my day.

What does "name" mean here? And please don't educate yourself on the internet with random Google images.

By saying "name", I am talking about like d(xy) and d(z^2) etc. It wasn't included in my textbook, that's why I googled it.

The reason you see different labels is that the set of real orbitals is not unique. It depends on what combination of the spherical harmonics you use to form each orbital. See here: https://en.m.wikipedia.org/wiki/Cubic_harmonic


text:
Is there a computer program capable of showing protein conformational changes based on pH?
As it is known, changes in pH change the attractions between the groups in the side chains of the protein.
Acidification can, for example, cause protonation of the $\ce{COO-}$ end to $\ce{COOH},$ reducing interaction forces and provoking the unfold of the protein.
I would like to know if there is any software (or any resource) that, given the quaternary structure of a protein, allows you to visualize or obtain an analysis of how this structure will vary according to different pH values.

Knowing whether residues are solvent exposed and what other residues (groups) are nearby suffices to allow prediction of state of protonation. Beyond that the problem you outline becomes a bit hard to understand. How would you "give the quaternary structure of a protein"? What do you mean by that?

The closest I can think of is protein-protein docking software. I would start with a web search for that.

@BuckThorn, that´s exactly it. I want to perform a MD with the protein in the different states of protonation in order to quantify the affinity with different ligands. By the quaternary structure I mean the protein in it´s physiological active form, which is the structure used in the docking. How can I check whether the residues are solvent exposed?

Run the sim and look for solvent-protein contacts. Read the simulation software documentation to figure out how, or ask over at material modelling SE.


text:
About the acidity of phenytoin

Can you explain why phenitoin is weak acid even though nitrogen is base?
Does carbonyl group always make structure acid?
If this structure doesn't have carbonyl group, would it be base not acid?

The NH between the carbonyls is in a similar position to succinimide. The negative charge on losing the proton can be delocalised across both carbonyls.

@Waylander So shouldn't it's acidity be more as the negative charge is in resonance with two carbonyls? It should be a good acid in my opinion, although that is a relative term.

Depends on your definition of a good acid. Succinimide has pKa 9.3, Phenytoin is 8.3. That makes it a stronger acid than phenol (pKa 10). It forms a stable sodium salt.

I see. Then why doesn't the lone pair on nitrogen move?
Plus I could understand this structure can resonance so which can it be acid or base?

When pheytoin lose proton as acid and the resonance structure is stable, isn't it strong acid? Because the strong acid is made when the resonance structure is stable(can resonance)
Sorry, if I'm wrong though..

Two carbonyl group pulls the electron strongly so I think it's a strong acid though..


text:
About the acidity of Sultiame

Why is sulthiam base even hough sulfur can accept the electron by resonance from nitrogen
like carbonyl group do?

Aniline is still basic. It's only that it's basicity is reduced due to resonance effects. Same applies here.

Amides are not usually considered as base though.

So, you're asking about acidity, or basicity? Both are pretty weak in sulphonamides.


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Hybrid lobes of sp3d2 and sp3d3
It's known that $\ce{sp^3d}$ hybridization has 5 lobes; 3 equatorial with $\ce{sp^2}$ lobes and 2 axial with $\ce{dp}$ lobes. Do we have such splitting in $\ce{sp^3d^2}$ and $\ce{sp^3d^3}$ hybridization states too?
What I think is $\ce{sp^3d^2}$ splits into $4 \ce{dsp^2}$ lobes and $2 \ce{dp}$ lobes.
Is this correct?

We seem to continually get questions about such out-dated (not to mention wrong) theories.
https://chemistry.stackexchange.com/questions/54737/how-are-the-hybrid-orbitals-of-sulfur-hexafluoride-shaped,
https://chemistry.stackexchange.com/questions/66698/theoretically-which-d-orbital-participates-in-sp3d-and-sp3d2?r=SearchResults,
https://chemistry.stackexchange.com/questions/75009/number-of-f-br-f-angle-of-90-degree-in-brf5?r=SearchResults,
https://chemistry.stackexchange.com/questions/129358/does-the-hybridization-of-sp3d2-type-exist?rq=1

@C_Lycoris Please don't use $$..$$ for links, you can just paste the links, and it will do the job. Using $$ just makes your comment appear large as each link is displayed on a new paragraph. Also, hybridization is still taught in high schools and the first year of university; I would be cautious before calling it out-dated and wrong.

@ShoubhikRMaiti It makes it by far worse that this garbage, deprecated, outdated, wrong stuff is still taught at universities. It is wrong, there is no doubt about it. Absolutely wrong. Is as wrong as the flag earth theory. If here is not the place to mention this, we could pack up shop immediately, there would be absolutely no Use for this site.

@Martin-マーチン I am sorry, but I have to disagree. If you look at any organic textbook, most of the explanations still invoke hybridisation theory. A lot of other outdated stuff is also taught in high schools and universities as a pedagogic tool (like wet tests of ions etc.). I agree that it's not completely right, but as long as its taught in some level, I feel that people should be able to ask questions about it here.

@ShoubhikRMaiti You can disagree with me, I'm fine with that.  I've never said that I want to disallow these questions, but I object vehemently to the notion that we have to be courteous about our replies. If the theory is wrong, we have to be able, no it is our duty, to point that out. Also, the is a huge difference between outdated methods and completely wrong theories. And by the way, hybridisation is not a theory, it is a descriptive mathematical tool. What is described in organic chemistry texts is often enough okay; however, anything in this question certainly is not.

To answer this, consider the purpose of the mathematical exercise we call "hybridization". The goal is to carve up the total electron density of the molecule into discrete segments (bonding orbitals) that are localized mostly to the area between two nuclei and which have some type of symmetry (usually either sigma or pi) with respect to the axis connecting the two nuclei.
A final important criterion is that these hybrid orbitals should be consistent with our intuitive understanding of equivalent bonds. For example, the four C-H bonds in methane are indistinguishable. Thus, the four hybrid bonding orbitals should also be indistinguishable.
With that in mind, consider that the six metal-ligand bonds in an octahedral complex are also indistinguishable (assuming all six ligands are identical). It does not make sense, then, to use four dsp2 hybrid orbitals and two dp, since that is not match six indistinguishable bonds. Instead, we extend the methane example and make six identical sp3d2 hybrids.
With odd numbers of ligands like 5 or 7, the metal ligand bonds are not all equivalent, so it makes sense to have a separate hybridization construct for each type of metal-ligand bond. For example, in a trigonal bipyramid, the equatorial bonds form one group and the axial bonds another.

While this is mathematically probably correct, I do reject this answer as it enforces the delusion that there is actually any physical meaning in this.

@Martin-マーチン What is "physical meaning"? I was careful to specify that there's no difference in the physical result (ie net electron probability density) of a hybrid orbital description vs canonical delocalized orbitals (which I assume have "physical meaning" for you?). Given the indistinguishability of electrons, I'm not sure that any model that places electrons in discrete orbitals has physical meaning.

I'm not sure I want to dive down this rabbit hole, but I also cannot let this accusation stand. The keyword in my original comment should be any. Can canonical orbitals be interpreted in a physical meaningful way? Maybe, but that wasn't my point, and I gladly refer to Observability of Orbitals and Orbital Energies and Evidence of orbitals? for further discussion. My point is (was) that it is impossible to utilise spd/dsp hybrid orbitals in any physically meaningful way.


text:
On the use of hexane in the organic reduction by DIBAL-H
In some variations of the organic reduction by DIBAL-H, I've seen hexane used, as in this equation from this website;

I've looked it up, but I haven't seen any explanation for why hexane's used there. It's also seen in the action of DIBAL-H on alkyl nitriles to give aldehydes with ammonia.
Can anyone help me with why it's used in both cases?

It violently reacts with water and is soluble in hydrocarbons like hexane. Hexane is just a solvent here

@SafdarFaisal: that works for the first case, thanks. For the alkyl nitrile reaction, apparently there's this idea of an acid being formed instead if you don't use hexane first or something. Can you elaborate how that works? It wasn't very clear where I got it from.

Hydrolysis of alkyl nitriles lead to formation of acids, hexane prevents that side reaction from happening since there is no water to hydrolyse

You should give a reference for your second case.

Also, reduction by DIBAL-H is not a type of Rosenmund reduction.

@SafdarFaisal: That answers it, thanks. MathewMahindaratne, it's from the notes sent to us after class, which are handwritten and kinda indecipherable. And I do know they're not an example of Rosenmund reductions, just mentioned it as the other place where I saw hexane being used.

@Harry Holmes: I changed the title and some parts of text body according to correct reaction type. You are welcome to change it back if you didn't like it.

@MathewMahindaratne: it's better this way, thanks.

Unlike other reducing metal hydrides (e.g., $\ce{NaBH4}$ and $\ce{LiAlH4}$), diisobutylaluminum hydride (DIBAL-H) is a liquid at room temperature and dissolve in many hydrocarbons such as toluene and hexanes, which also have very low freezing points. For example, hydrocarbons toluene and hexanes both have freezing points around $\pu{-95 ^\circ C}$. Thus, they are good solvents for reaction involving reduction by DIBAL-H at low temperatures such as the example given in the question ($\pu{-70 ^\circ C}$). As shown in the equation, the low reaction temperatures help stop the reduction at aldehyde level. At room temperature with excess DIBAL-H, the reduction may continue further to give corresponding alcohols or even corresponding alkanes.
DIBAL-H reduces alkyl or aryl nitriles to their corresponding imines at low temperatures and resultant imines converted to corresponding aldehydes upon acid work-up. The same principle mentioned above apply here as well. Here, at high temperatures, imines reduced further to give corresponding amines.
Note: To see the mechanism of the reduction, see here.


text:
Why does (2-methoxyethyl)benzene undergo nitration at the ortho position (major product)?
This is an option in a question in which we needed to mark all the correct reactions:

According to the answer key, this is a correct option.
As an alkyl group is directly bounded to the benzene ring, I thought that the reaction will have para product as the major product.
I have a similar doubt with the nitration of toluene. I am unclear as to why ortho product is the major product.
It's be highly appreciated if anyone could please explain the reasons behind the major products of both the reactions.

Inductive effect is distance dependent and thus the electron density at ortho position will be greater than that at para position.

Ok, but I considered hyperconjugation (because it is more dominating than inductive effect), and then steric hinderance factor - so due to hyperconjugation, the group will be ortho-para directing, and then to avoid steric hinderance, the major product would have been para product. I read that in case of ethyl and other bulkier alkyl groups on the benzene ring, para product is the major product.

The positions are similarly activated - you get more ortho product because there are 2 ortho positions and only 1 para position.

Alright, but then why are the major products of nitration of aniline, anisole, and other bulkier groups on benzene para products


text:
Which oxygen is favored for protonation when hydrolysis of an acetal occurs?
My question is about which oxygen is favored for protonation in acidic conditions. So, R1 and R2 are different in the following diagram (for example, one is propyl but the other is ethyl). I looked up my organic chemistry materials, but only symmetric cases (R1 = R2) were suggested.

I believe that there will be a competition between two pathways, and it is related to the stability of intermediate in the reaction mechanism. But how can I estimate which pathway is favored or the stability of the intermediates?

I don't see a practical application of this concept, all acetals that I could find were those where R1 = R2 since it's part of a reaction that happens in one media.

@SafdarFaisal I see why the symmetric case is common, though I can also find articles about a synthesis of unsymmetric acetals.

A somewhat complex example is a glycosidic bond, for example a 1,4 linkage between two glucose molecules. The "1" position involved in the bond is then an acetal with two very different substituents.

Possible dupe? Its a specific case but still..https://chemistry.stackexchange.com/questions/18588/hydrolysis-of-cyclic-acetal?rq=1


text:
What is the reaction mechanism between boric acid and 2-ethyl-1-hexanol
Here is a photo from an article, Hydrometallurgy 2016, 160, 129–136

This indicates the reaction mechanism as esterification. However, the written reaction equation ($\ce{H3BO3 + nROH -> H3BO3.nROH}$) suggests the mechanism as solvation.
Shouldn't the reaction be $\ce{H3BO3 + ROH -> H2RBO3 + H2O}$ or something like that based on the scheme?

The research paper sighted in the question (Ref.1) is about the extraction of boron (as boric acid: $\ce{H3BO3}$) from salt lake brine using 2-ethylhexanol/kerosene as organic phase. It is true that the given diagram in this paper (as shown in the question) is showing an esterification, but the authors refer it to two references, one of them is Ref.2. However, the association used to calculate the distribution of boric acid between aqueous and organic phases in Ref.1 is not the esterification. Accordingly to the paper, the association reaction is as follows:
$$\ce{H3BO3 + nROH <=> H3BO3.nROH}$$
Since there are no water molecule elimination during the reaction, this is not an esterification. It is a solvolysis reaction as OP correctly recognized.
Extraction of boron from existing sources is a challenging problem (Ref.3). Boron can be extracted as neutral ester $(\ce{B(OR)3})$, as boric acid complexes $(\ce{(H3BO3)_x.(ROH)_y})$ or as tetrahydroxy borate $(\ce{R+B(OH)4-})$. However, the extraction of boric acid by monohydric alcohols is caused by purely physical actions and the amount of extracted boric acid is roughly proportional to the concentration of an extractant (Ref.3). Accordingly, it is safe to say the extraction process is based on complexation of boric acid with intended alcohol.

References:

Ran Zhang, Yingming Xie, Jianfeng Song, Lixin Xing, Dingfeng Kong, Xue-Mei Li, Tao He, "Extraction of boron from salt lake brine using 2-ethylhexanol," Hydrometallurgy 2016, 160,  129–136 (DOI: https://doi.org/10.1016/j.hydromet.2016.01.001)(PDF).
Jianhua LÜ, Jidong LIU, Yujie SUN, Chunli LI, "Kinetics of Forward Extraction of Boric Acid from Salt Lake Brine by 2-Ethyl-1,3-hexanediol in Toluene Using Single Drop Technique," Chinese Journal of Chemical Engineering 2014, 22(5), 496-502 (DOI: https://doi.org/10.1016/S1004-9541(14)60059-8).
Adam Balinski, Volker Recksiek, Norman Kelly, "Solvent Extraction of Boric Acid: Comparison of Five Different Monohydric Alcohols and Equilibrium Modeling with Numerical Methods," Processes 2021, 9(2),  398 (15 pages) (DOI: https://doi.org/10.3390/pr9020398).

Many thanks for your comprehensive answer. Here are my following problems:

I know 2-ethyl-1-hexanol extract vanadium as vanadic acid properly and silicium as salicylic acid relatively. Please Check: https://www.sciencedirect.com/science/article/abs/pii/S1383586617317823
It seems 2-ethyl-1-hexanol tends to extract acidic complexes, which can happen via esterification because the reaction is between an alcohol and an acid.
Does my argument seem valid to you?
What equation fits the scheme I shared? Is my suggestion an appropriate one?
Thank you again.

Unfortunately, I don't have access to said reference at this time. Yet, I suggested that your suggestion is appropriate according to the given image. However, it is clear from multiple reports from various authors that monohydric alcohols give complexation under the reaction conditions (at least at room temperature) while diols and higher hydric alcohols tends to esterify the reactant. That need extra striping step(s) to recover intended materials (here boric acid).


text:
When an aqueous extract of a substance is said to be administered at a certain quantity, is the quoted mass analogous to one as original substance?
I'm trying to read studies about the possible toxicity of the leaves of some plant often said to possess positive medicinal properties.
In the studies, performed on rats, it is usually said that an aqueous and/or ethanol extract of the plant was administered at $x \ \pu{mg/kg}$ of mice mass, and from that data possible lethal or detrimental dose is extrapolated.
What I don't understand and is not explained in the texts (therefore I assume there is standard practice I'm not aware of) is how concentrated is $\pu{1 g}$ of the aqueous extract compared to $\pu{1 g}$ of the original fresh leaves. For example, one study simply states, referring to the administered aqueous extract: "The $\mathrm{LD_{50}}$ was estimated to be $\pu{1585 mg/kg}$." Should one infer that the amount of chemical substances administered in this scenario is equal to that found in $\pu{1,585 mg}$ of fresh leaves? Or is it possibly less/more concentrated?

The denominator in LD is the unlucky animal mass, nothing about the original concentration in the leaves or the extract.

I agree with @Alchimista about animal mass. However, numerator of $\mathrm{LD_{50}}$ is 50% of the lethal dose of plant extract or fresh plant material. We cannot say exactly what it is until we look at the procedure. Hence, It'd be beneficial if you can put the reference.

@MathewMahindaratne yes but for precision: From the data LD we only know the concentration as amount of X per mass of the animals causing the death of half (this is the meaning of 50% ) of them (in absolute or within a frame time, this I don't know). For acquatic organism it can also be a normal concentration (of X in the water where the organisations live), though.

The concentration in the leaves is sort of irrelevant if you are administering an extract, no? I would not infer anything about the amount of fresh leaves that roughly corresponds to the amount of active substance in extract. One important reason is that the amount may vary between leaves and with season, plant, growing conditions...... Even if explicitly stated that an extract was obtained from a certain amount of plant material, there is no guarantee that the yield will be the same next time. It is similar to the yield of a chemical reaction, which may vary batch to batch.

*above organisation reads as organisms. Keyboard :(

@MathewMahindaratne Here is, for example, the redacted study I made a reference to in the question: https://pubmed.ncbi.nlm.nih.gov/22138517/

By reading that paper is clear to me that medical (pieces) of research are everything but not science :) Of someone want to laugh s/he should search for "integral rediscovered" .

@Alchimista So you agree there is information missing from this paper? Perhaps the free version is missing the important information that is available in the complete paper?

@TLSO one has to read the full. But they saying nothing. In chemistry one could design free experiments like that for almost everything. It is like reporting absorption spectra of transparent things or the synthesis of "colourless dyes". I am not saying that data aren't useful. Just that it is not Science. Indeed is Medicine. When I think of many senseless experiments and the likes, and I think that some are doing the same with animals, well, I really wish there is a strong control over the experiments. But I am afraid a dose of HNMR is more evaluated than a mice.

You probably get the $\mathrm{LD_{50}} = \pu{1585 mg/kg}$ value by reading the abstract of the paper. That's why you got confused by that value. However, it is not $\pu{1585 mg}$ of fresh leaves. It is actually $\pu{1585 mg}$ of dry leave extract. I mean dry residue obtained by removal of water from an aqueous extract of dried leaves. The authors have made a stock solution of that dry residue in  $\pu{200 mg/mL}$ concentration by dissolving in enough distilled water ($\mathrm{pH} = 6.7$). Thus, $\pu{7.9 mL}$ of that solution is lethal to a rodent weighing $\pu{1.00 kg}$. On the other hand, since they have used rodents averaging $\pu{100 g}$ of body weight, each of them need only $\pu{0.79 mL}$ of that solution to be lethal.
Hope this explain what you are looking for.

Reference:
Olufunsho Awodele, Ibrahim Adekunle Oreagba, Saidi Odoma, Jaime A. Teixeira da Silva, Vincent Oluseye Osunkalu, "Toxicological evaluation of the aqueous leaf extract of Moringa oleifera Lam.(Moringaceae)," Journal of Ethnopharmacology 2012, 139(2), 330–336 (DOI: https://doi.org/10.1016/j.jep.2011.10.008).

You accessed the full article in order to have that information, I assume, because I can't see any mention of this method in the redacted version. So to be certain, they basically completely (or close enough to) removed the water component from the fresh leaves (possibly then turning them into powder), and then when administering them to the rats they dissolved the dry leaves in distilled water at a 20mg per 1ml concentration? Did they also mention by what factor were the fresh leaves concentrated through dehydration? Thanks.

They air-dried leaves and then added certain amount of water. After $\pu{24 h}$ or so, they remove solid leaves particles (say filtration). The water from liquid filtrate was removed in an oven. What remain was a solid residue. That residue they dissolve in water to get $\pu{200 mg/mL}$ concentration. That is the solution they administered to animals.

So they air-dried the leaves in order to have new water more easily filter the chemical components through the dried leaves, and after that re-dried the solution in order to be left only with the dry components they managed to extract from the leaves (and then dissolved them once again in water in order to have a standardized solution to give the rats)? So from that I understand two things: first, it's not explained what is the exact chemical makeup of the dry extract and at what ratios, so even that processed substance might be different from one leaf batch to another.

Second, the 1,585 mg/kg result involves 1,585 milligrams of a certain concentrate of chemicals from fresh leaves, and not 1,585 milligrams of fresh leaves, and thus even though we don't know by how much were the leaves concentrated during the dehydration->aqueous extract->dehydration process (quite important), we at least know the approximate LD50 in fresh leaves would probably be greater than 1,585 mg/kg, perhaps several times as much. Yes?

You understand it correctly. Actually this is very crude determination. And also, I think its some what sloppy work and the value might be way off. For example, according to their research, $\pu{158.5 g}$ of leave extract is lethal for a man of $\pu{100 kg}$. I know Moringa is eaten in tropical countries as a vegetable (both leaves and fruit). The leave extract is $\approx 20%$ of leave weight. Hence, if a certain person eats a one pound of leaves for lunch he may endup in a hospital! :-)

This is what I was trying to understand, because they mentioned this plant (and more specifically its leaves) in the news a couple of months ago with raving claims about its benefits, but then I also found researches suggesting it's relatively toxic. But who knows, it might potentially be this dangerous all the while most people who consume it traditionally aren't aware of it, just usually not eating this much at once so they're not at the deadly dosage (but is it still harmful eating for example 100 grams each day? Who knows). Read about konzo disease for a different traditional food example.

By the way, how did you have access to the complete information in the research paper? Is it through some subscription? Because I often want to read up about some subject but the sources I find aren't freely available other than as redacted tidbits. Thanks.

It is possible to have a long-term toxicity as warned by the authors. But I didn't find any toxic effect by local users of "Moroonga" (what locals called it) as vegetable. I think it is safe except for one sloppy work. Well, The university I'm working in has subscribed the journal. Only I have to do is go through the university website to get access even if I'm working at home these days. Good luck on your future searches!

At the core of your question there is misunderstanding of what the papers you refer to are discussing. The LD per se in not related to the extract concentration nor to the naturally occurring principle(s) in the plants.
You seem interested in a relation between the original concentration and the extracts concentration. This will depend on several factors from the molecular properties to the the extraction technique used.
Furthermore, note that the extracts can be further concentrated or diluted according to uses and convenience.
In short, the information you are looking for would be typically found in the experimental part.
LD50 is the concentration (whatever expressed depending on the case)* leading to the death of half of the test animals**. These are the data listed in the papers as per your resume. There is not relation to the concentration of the extract except for trivial reason, ie a more concentrated extract will require a smaller amount to reach the same LD.
*it can be the amount of X per kg of the body and thus a practical concentration rather than a chemical-physical one; the amount of X per volume of water (as in the case of fishes); the amount of X per amount of gas, etc.
**unfortunately used even in routine test, when chemical analysis could be adopted instead.

I understand what LD50 itself refers to, but it must be correlated with a specific form of the substance tested in order for it to be meaningful, no? As you said, if extract B is more concentrated than extract A, the lethal dose of extract B should be smaller than that of extract A. I simply can't find any sentence within the several studies I've read (although not all are entirely available for free) and these basically discuss a substance (plant matter) that is commonly taken as a traditional medicine in its leaf-form. Providing an average extract-to-fresh leaves ratio seems important.

That is the point. It is LD of X, not LD of a specific extract. If, say, the daily alcohol intake should not exceed Y, it does not matter (for this discussion) if you drink whisky or beer. Approaching Y alcohol you must stop. Of course one could define a limit to beer wine etc. And the same for the extracts. If you know what LD is, than it is meaningful for sure. Let me add that speed of administration etc can influence toxicity, but this ceases to be matter for our format/scope.

@TLSO of course many things are or can be important. It would sound like "the LD50 for X is Y, this corresponds roughly to three cups of an infusion of Z grams of fresh leaves...". All those aspects cannot be deduced from LD data. In a way I don't see what is your question. One can prepare differently concentrated cyanide solutions. The LD of CN- won't change. And still can be compared to that of another poison. From this comparison, how the solution were prepared is irrelevant.

The way the leaf solutions were made in the study, or perhaps what are the concentration levels compared to the source matter (fresh leaves in this situation) does seem relevant, if it could mean "1,585 mg containing 50% fresh leaves 50% water", or perhaps "1,585 mg containing 70% dry leaf powder 30% water" etc. I didn't understand what were they referring to in the redacted paper, but Mathew Mahindaratne apparently has access to the full paper and he explained what was the mentioned process for creating the doses and what were they composed of.

And thus I now understand that the milligrams given refer to a completely dry matter extracted from the leaves through air-drying, extracting in water and then dehydrating the aqueous extract. Perhaps that was the standard practice one might be familiar with, but I didn't see anyone mention this up until Mathew cited it from the full paper. I still would have been interested in knowing how concentrated was the dry mass versus the original mass of fresh leaves, even though such a processed concentrated extract could likely still differ from one batch of leaves to another.

@TLSO then there was no basically question except for what the LD values refer too. Of course the way how the extract was made is crucial. As concluded in these thread, than the data are meaningful in relation to tea preparation but not very so in a molecular way.

Perhaps if the complete data also compares extract from several different batches, it could infer how chemically consistent are the leaves from different yields. Also, I don't know if they analyzed the chemical content or not. Perhaps the extract didn't successfully remove from the leaves all of the substances that are expected to be found.

@TLSO in this respect I fully agree. But it is still valid for getting an idea in the context given, ie people making various "teas".

Yes, but unless it was verified in the study, I wonder if the presumption that the concentrated extract was about 5 times as concentrated as the original mass of leaves is accurate when referring to its chemical profile. The water was removed, but probably not all of the solids went into the extract, and perhaps some of the missing content also causes a biological reaction.

To close the discussion. If your meaning was "How do they know about the concentration of the extract administered" then it should have been the question. I do not have full accessed but reading comments it seems that they did not really know the specific actual content of the active principle(s). In other words it can be that the data gives a kind of qualitative. It still makes sense in the context probably. In the same way we can estimate the content of caffeine in a long coffee cup. This is to say that all my discussion was about LD comparison and not on the robustness of this specific work

Yeah, but I meant more in the vein of — if you filtered water through coffee but then removed the rough solids from the coffee beans, perhaps that resulting extract misses many of the "active ingredients" in the plant matter. An analysis of the chemical content in a fresh leaf mass, then compared to an analysis of the leftover solids following the aqueous extraction, is needed to take place in order to know what percentage of the important chemicals was actually extracted.

Yes without knowing the exact extract composition the paper (which I don't access) would be bullshit. As I said, nobody would do that vague experiment with high quality chemicals. Likely individual mices are cheaper, so they do whatever it pass in your mind. One can do the same saying "lot of people like mustard" let's have a look look. (surely done, perhaps useful, but not a scientific work).@TLSO

They have to analyse extracts for the plant drug concentration. They have idea about typical drug content in plant and use accordingly the ratio between solvent volume and plant leaves mass, together with various concentration techniques like vaporization or solvent/solvent extraction.
The drug concentration in final solution is then multiplied by the used extract volume to get the contained mass of the drug. Finally is calculated the drug/mice mass ratio.
Alternative is calculation leaves/mice mass ratio, if the direct drug content analysis is not available. If you do not know what the drug concentration is, you can relate only to the plant mass. Like plant mass / solvent volume, or plant mass / mice mass.

Perhaps I'm misunderstanding, but you're saying that unless an analysis of the chemicals within the extract + a comparison to the chemicals' concentration within the plant was explicitly said to take place in the study, one cannot say what is the correlation between the chemical concentration of the aqueous extract and that of the original plant matter? edit: I can see you've updated your answer, so I need to read it anew.

I guess you got it. If you do not know what the drug concentration is, you can relate only to the plant mass. Like plant mass / solvent volume, or plant mass / mice mass.

OK, so to my understanding they are supposed to standardize the chemical concentration to some manner, but you suggested what's quoted in the results is a drug (as in mass of actual studied chemical rather than the entirety of the aqueous solution)/mice ratio, while at least when I'm reading the studies it seems to refer to the water containing the leaf extract, which as I said isn't clarified in regards to its relative concentration.

It would have been fine if I could infer it parallels plant mass/mice mass (even if regarding plant matter chemical concentration could vary widely), but as I said I'm really not certain from the wording this is the case. Unless it has to be the case due to standard practice.

Well, particular interpretation obviously depends on the particular wording. It may be they calculate mg of extract per g of mice, with well defining the extract elsewhere.

@TLSO it is you that try to infer data which are not attainable from LD. How the various extract and their concentration has to be known from the preparation method. Of course it is possible that there is a common practice to work routinely. Comparing LD can be done anyway, the only point is to know how much of the principle was given. But LD50 or 100 can be compared without previous/other knowledge.