text:
Extract iodine from tincture?
What I am trying to do is extract elemental iodine from iodine tincture - that is 5% iodine solution in aqueous ethanol with $\ce{KI}$ as "helper additive", or whatever you'd call it. I don't have povidone-iodine, I don't have iodide tincture, and I don't have $\ce{KI}$. Well, there is $\ce{KI}$ in my tincture, but that is not my point. My point is, there are procedures for extracting $\ce{I2}$ from povidone-iodine, there are procedures for converting iodide salts (in tincture or not) to elemental iodine, and there are those for simply precipitating out $\ce{I_2}$ from tincture, which is my case.
BTW why not just evaporate the ethanol, or is the iodine too volatile - will sublimate with the evaporating ethanol(iodine sublimates at $\ce{113.7 ^\circ C}$)?
I want to use household vinegar - 6% apple vinegar if possible because it is easily available. Also, I want to use 3% pharmacy grade $\ce{H2O2}$ for the same reasons.
I have done this, kind of successfully, but I want to know what is the chemistry behind all that, and how would you calculate the amounts of reagents needed?

Where do the vinegar and the H2O2 come in? What is the point in all this? And what kind of equipment do you have? How did you extract the iodine, if you already managed to do so?

Keep in mind that Iodine is a restricted substance. Trying to isolate iodine from iodine tincture raise a question.

@Karl I have equipment and acetic acid and I have elemental iodine. The point is doing the extraction without equipment, so I can document the whole process. This is what I am asking precisely, where does the acid and H2O2 come in, because I know the procedure works and has been documented, the problem is, they don't explain the reaction mechanism or give quantities.

@Mathew Mahindaratne Wander what this question is... Iodine is not restricted here, you should not assume anyone lives in the US

If you want to save the iodine $\ce{I_2}$ from its solution, you may add enough $\ce{KOH}$ or $\ce{NaOH}$ in the mixture to transform all $\ce{I_2}$ into a colorless mixture of $\ce{KI}$ and $\ce{KIO_3}$. $$\ce{3I_2  +  6OH^- -> IO_3^- + 5I^- + 3H_2O}$$The remaining ethanol can then be evaporated by heating the nearly colorless solution to a temperature higher than $80$°C. When the temperature reaches $100$°C, the ethanol is totally removed from the solution : the hot solution can be cooled down to room temperature, and some acid should be added in order to destroy the excess of $\ce{NaOH}$ ou $\ce{KOH}$, and then to recover the iodine $\ce{I_2}$ according to $$\ce{IO_3^- + 5 I^- + 6 H^+ -> 3 I_2 + 3 H_2O}$$ The totality of the iodine $\ce{I_2}$ dissolved in the original tincture is recovered, without any amount of ethanol, and can be saved by filtration.


text:
Which of the compounds has the highest percentage of enol?
Which of the compounds has the highest percentage of enol?

$\ce{CH3-CHO}$

$\ce{Ph2-CH-CHO}$

$\ce{Ph-CO-Ph}$

Parabenzoquinone

How do I compare these?

How do I compare these?

You could expect higher enol content in the following cases:
1.If there is an active methylene group ( or any other acidic $\ce{C-H}$ bond )
2.Extended conjugation in the enol form
3.Attainment of aromaticity after enolisation
4.Any intramolecular hydrogen bonding possibility in the enol form
Coming to your question:
3,4 - there's no α-hydrogen (no enolisable hydrogen).So, no tautomerisation takes place
1- has 99% keto content
2- is expected to have more enol content due to extended conjugation of the π-electrons of the double bond formed after enolisation with the phenyl rings.
ANS:   2 has the highest enol content.

source: Stack Exchange

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sort: votes


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Why doesn't keto enol tautomerism take place in acid derivatives?
Why does keto enol tautomerism take place only in ketones and aldehydes? Why doesn't it take place in carboxylic acids, esters, amides, anhydrides and acid chlorides?

Well, it does occur with acyl chlorides [https://www.sciencedirect.com/science/article/pii/S0040403901984953] and acyl bromides [https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Carboxylic_Acids/Reactivity_of_Carboxylic_Acids/Hell-Volhard-Zelinskii_Reaction].

Why doesn't it take place in carboxylic acids, esters, amides, anhydrides and acid chlorides?

It does take place in acid derivatives as well.Acid chlorides and esters exhibit enolisation to a decent extent.However, the enol content is lesser than that in carbonyls due to cross-conjugation.
You could expect higher enol content in the following cases:
1.If there is an active methylene group ( or any other acidic $\ce{C-H}$ bond )
2.Extended conjugation in the enol form
3.Attainment of aromaticity after enolisation
4.Any intramolecular hydrogen bonding possibility in the enol form

Would amount of enol form be in the order acid chloride >anhydride>ester>carboxylic acid>amide based on +M and -I effect?

Yes, expected to be.However, I have seen tautomerisation in acid chlorides and esters only!.Generally, the enol content of others that follow esters in the order you have given is very less (< 0.001%) and hence tautomerisation can be neglected

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text:
Methylene "anti" Jahn-Teller effect
Yesterday a Reddit user posted a page from Morrison's Organic Chemistry in which it is said that singlet methylene is less stable than triplet methylene. Another user asked basically the same I'm asking here, as he got no answer and probably won't get any.
I'm used to see radicals (and therefore diradicals) as extremely reactive and lone pairs as stable, when not reacting in some Lewis base fashion.
Also, Jahn-Teller effect explanations usually say that unpaired electrons configurations with each electron on a degenerate orbital are less stable than the paired configuration with some deformed structure, and therefore the structure deforms to "de-degenerate" the orbitals and achieve a lower energy configuration, hence the effect.
What we see here is exactly the opposite, isn't it?
What is it then that makes the triplet state more stable?
Wikipedia confirms the observation and adds a value to the energy difference: 38 kJ.
Wikipedia also says it is normal for carbenes to be more stable in the triplet state.

One big factor is Hund's first rule: you know this from the electron configuration of e.g. boron. I'm not sure that the JT effect is relevant here; I haven't looked too closely, but I don't think there are any degenerate orbitals under consideration.

The specific case of the energy levels of CH2 configurations is discussed in section 8.10 of Orbital Interactions in Chemistry by Albright, Burdett and Whangbo. The small size of the p orbital and the small energy difference between the HOMO and LUMO of the singlet result in a high penalty for confining two electrons to one orbital.

orthocresol: I believe the triplet state has 2 degenerate p orbitals (each with one electron) and the degeneracy is eliminated on the singlet state, which could be the low energy state and therefore result in the Jahn-Teller effect. @Andrew, could you elaborate? I fail to understand. Is the small energy gap intended to suggest a low energy advantage in filling the homo?

Triplet state is not degenerate. Bond angle is about 135 degrees. Needs to be 180 for them to be degenerate.

The Morrison reference say it is 180° for triplet and 103° for singlet. Which fits nicely into a sp and sp² carbon hybridization.

@LocoAsker - every other reference I can find says around 135. Just do a search for "bond angle methylene carbene" and see what you find. Perhaps you'll find something different. Only the 4-electron AH2 type molecules such as BeH2 are 180 degrees, as far as I know.

Then the sp picture is no good.

Short answer is that it requires advanced computational methods to determine which state is lower in energy, but we can rationalize the observed result by investigating the behavior of the canonical delocalized molecular orbitals of the $\ce{CH2}$ carbene.
What follows here is a summary of information found in chapters 7 and 8 of the second edition of Orbital Interactions in Chemistry by Albright, Burdett and Whangbo (Wiley & Sons 2013). I would encourage you to read that text for a fuller understanding.
First, let us consider the Walsh diagram of $\ce{SH2}$ as a representative molecule of the form $\ce{AH2}$, where "A" is any element. In the diagram, the vertical axis represents energy, while the horizontal axis represents the H-S-H bond angle. The shapes of the molecular orbitals at the extremes of bond angle are depicted on each side.
$\ce{SH2}$" />
Although this diagram specifically applies to $\ce{SH2}$, it can be considered as a qualitative representation of any $\ce{AH2}$ molecule.
The most important orbitals for this discussion are labeled "$1\pi_{ux}$" and "$1\pi_{uy}$" on the left side and "$2a_1$" and "$b_1$" in the middle and right side. These molecular orbitals are derived from the $p_x$ and $p_y$ orbitals of the central atom. If the H-A-H bond angle is $180^\circ$, then these two orbitals are equivalent in energy, i.e. degenerate.
When the bond angle is decreased, however, the $p$ orbital that is in the plane of the bend is lowered in energy by its interaction with the H $s$ orbitals. There is also a corresponding increase in energy of the lower $1b_2$ orbital, but its destabilization is less than the stabilization of the $2a_1$ orbital. This loss of orbital degeneracy coupled with a decrease in molecular symmetry is an example of a second-order Jahn-Teller distortion.
As a result, a molecule such as $\ce{BeH2}$, which has only four valence electrons (configuration $(1a_1)^2(1b_2)^2$) becomes less stable as the bond angle deviates from $180$, so it is linear.
As soon as we add one more electron, however, the lower energy of the now partially-filled $2a_1$ orbital is greater than the destabilization of the $1b_2$ orbital, so the molecule adopts a bent shape, as is observed for $\ce{BH2}$ (configuration $(1a_1)^2(1b_2)^2(2a_1)^1$).
Based on this simple analysis, we would predict that the methylene carbene would adopt a bent structure of configuration $(1a_1)^2(1b_2)^2(2a_1)^2$, i.e. a singlet.
What this analysis has ignored, however, is that there is an energetic cost to placing two electrons in the same molecular orbital, as porphyrin described well in another answer. When we add a second electron to an orbital, its energy is raised a bit, a change which is not shown on the Walsh diagram (or typically on any MO diagram). So now we must compare two competing energy changes - the difference between partially-filled $2a_1$ and $b_2$ orbitals (i.e. the triplet state) vs completely filling the $2a_1$ orbital. This comparison is further complicated by the variable of the bond angle. In the singlet state, two electrons are in an orbtial stabilized by an increasing bond angle and two are in an orbital that is destabilized, whereas in the triplet state, only one electron is in an orbital that is stabilized and two are in an orbital that is destabilized. As a result, the optimal bond angle for lowest energy is different in each one. Finally, the starting energies and the slopes of the energy changes as a function of angle are specific to each molecule (e.g. they will be different for $\ce{SiH2}$ and $\ce{CH2}$ even though the number of valence electrons is the same). Altogether, this means that our qualitative analysis is not sufficient to determine whether the triplet or singlet state is lower in energy. We must rely on more complex quantitative methods.
With those quantitative methods, we find that the triplet state is lower in energy for $\ce{CH2}$, as shown in this diagram:

We can see in the diagram that the triplet ($^3B_1$) is always lowest in energy, and the energy minimum is at a bond angle of $133^\circ$. The singlet ($^1A_1$) has a minimum energy at a bond angle of $104^\circ$.
It is important to note that this result is not general. In both of the similar molecules $\ce{SiH2}$ and $\ce{GeH2}$, the singlet state has the lowest energy minimum.
This result is rationalized on the basis that the valence $p$ orbitals of carbon are more contracted than those of Si or Ge. This contraction means that the penalty of putting a second electron into that orbital (as in the singlet state) is greater for carbon than the large elements. Additionally, the stabilization of the $2a_1$ orbital as a function of bond angle is less for carbon than for the larger elments, but this effect is less important.

I'm glad that you have added a Walsh diagram and also the one showing the energy levels vs angle.

The Jahn-Teller effect is not relevant here. The triplet state is observed in many molecules to be of lower energy than the singlet. The Pauli exclusion principle, makes the states symmetric or anti-symmetric to the exchanging electrons.
A triplet state has unpaired spins while in a singlet they are paired. The triplet is lower in energy than the corresponding singlet. One reason for this is that the electrons try to ‘avoid’ one another in the triplet; (see Salem 'Electrons in Chemical Reactions', publ Wiley 1982 for a detailed discussion). This means that the spatial part of the triplet wavefunction is asymmetric to exchange of the electrons;
$$\varphi_T = \psi_1(1)\psi_2(2) − \psi_1(2)\psi_2(1) $$
and the singlet spatial part is symmetric which means that the electrons tend to crowd together raising the energy
$$\varphi_S = \psi_1(1)\psi_2(2) + \psi_1(2)\psi_2(1) $$
The numbers in brackets refer to the electrons’ coordinate. As the total wavefunction is anti-symmetric to exchange of electrons the corresponding spin parts have the opposite symmetry to the spatial parts.
If the spins are on two separate atoms then the singlet and triplet energies can be very close to one another and the order can be changes by external magnetic fields. There is more detail on this effect in this answer How does spin flipping of triplet carbenes occur?

I think the "anti Jahn-Teller effect" is intended to reference the larger H-C-H bond angle in the triplet versus the singlet, which brings the energy of the two p orbitals in the triplet closer to the degeneracy they have in the completely linear molecule. The bending of the molecule away from linear that breaks the degeneracy is considered a Jahn-Teller effect.

Also, while this answer addresses why singlets can have higher energy than the corresponding triplet, it doesn't fully explain why the penalty of two electrons in one orbital is greater than the difference in energy between the two orbitals, ie the "cost" of elevating one electron to a higher energy orbital in order to achieve the triplet. This difference does not always favor the triplet. SiH2 and GeH2 are more stable as singlets.

The methylene is bent in both triplet $\approx 132$ degrees and $\approx 101$ for the singlet $\approx 45 $ kJ/mol higher in energy. Even at $180$ the triplet  is at lower energy than the singlet. (O'Neil et al J. Chem. Phys. v55, 162, 1971)

Do you have a orbital diagram or something like that? How are the electrons organized in the triplet?

if you want to think in terms of  sp$^2$2 orbitals, there is an sp$^2$ type orbital in the plane of the molecule and a p $\pi$ orbital at 90 degrees to this. The triplet is a 'biradical' with one electron in each orbital but the singlet more of a zwitterion with two resonance forms with both electrons in one orbital and none in the other.


text:
Basic Chemistry: Carbon Dioxide Formation Unsure about Reaction
I have the following question description:

Consider a gas composed of  $\ce{CO_2}$,  $\ce{CO}$,  $\ce{O_2}$ and  $\ce{O}$ with 1  $\ce{C}$ nuclei for every 2  $\ce{O}$
nuclei.

I am trying to write the basic chemical reaction but am unsure it's correct. Thus far I have:
$\ce{CO_2}$ +  $\ce{O}$ =  $\ce{CO}$ +  $\ce{O_2}$
Is this the correct format? Pretty sure I have to balance but wanted to make sure I have the correct reaction before I do so. I am just not sure what to do with the extra $\ce{O}$ in the question description...

The question description states: "Reacting Gas Property Mixtures"

Your suggested solution does not have 1 carbon for every 2 oxygen. I suggest, instead, starting with just $\ce{CO2}$.

This would make sense: CO2 = CO + .5*O2. Just the dissociation of CO2. But I still have the O that needs to be considered.., which is throwing me off

This "question" makes no sense. Please give the complete context, and cite the source.

The confusion with the questions is stemming from the fact that 1. $\ce{O}$ gas doesn't exist in nature. 2. You haven't specified the full question, just some of the set up.

If you start with $\ce{CO2}$, you could have one reaction that yields $\ce{CO}$ and $\ce{O2}$, and another where $\ce{O2}$ dissociates:
$$\ce{2CO2(g) -> 2CO(g) + O2(g)}\tag{1}$$
$$\ce{O2(g) -> 2O(g)}\tag{2}$$
Then, you can combined these two in arbitrary ways, or like this with a variable $x$:
$$\ce{2CO2(g) -> 2CO(g) + (1-x) O2(g) + 2x O(g)}\tag{1+2 combined}$$


text:
Change in Entropy and Spontaneity
I was studying thermodynamics and got stuck with some questions. I am pointing them down.

I have studied that $\Delta S_{\text{total}}$
accompanied by a spontaneous process is greater than 0. But is the converse true?
I mean, will any process for which $\Delta S_{\text{total}}>0$ be spontaneous?

Is the $\Delta S_{\text{total}}$ for all reversible processes equal to 0?

Is $\Delta S_{\text{total}}$ for all irreversible process >0?

Please help me with the answers to these with some explanations for the answers.

The answer to all three questions is yes.  This is basically a statement of the 2nd law of thermodynamics.  There are two ways in which the entropy of a system can change: 1.  by entropy exchange with the surroundings and 2. by entropy generation.
Entropy exchange is like exchanging money between two bank accounts.  So any increase in the balance in one account is accompanied by an identical decrease in money in the other account.  So the net change in the amount of money is zero.  In a reversible process, only entropy exchange occurs.  So an increase in the entropy of the system is accompanied by an identical decrease in the entropy of the surroundings (and vice versa).  So the net change in entropy is zero.
Entropy generation is like gaining interest in one or both bank accounts.  So the interest in one or both accounts increase.  In an irreversible spontaneous process, there is entropy generation in addition to entropy exchange.  So the total entropy of the system plus surroundings increases in an irreversible spontaneous process.


text:
How can I analytically calculate the rate of thermal decomposition of calcium carbonate?
I am a first-time poster on Chemistry StackExchange, and not much of a chemist.
I'm trying to determine the rate of thermal decomposition (in mol/s) of calcium carbonate ($\ce{CaCO3}$) at various temperatures (assuming atmospheric pressure).
The best resource I could find was an old handbook about the thermal decomposition of carbonates, which gave the equilibrium constant K for this reaction at various temperatures:

From this data set, it appears that the reaction only proceeds at temperatures above 1100 K. Below this temperature, all values of K are less than 1. A value of K = 1.84 is given at 1200 K, but no further information is provided.
I essentially have three questions:

Is it possible to use K to calculate a reaction rate? Specifically, for this example, would it be possible to calculate what the rate of reaction is, in mol/s, at 1200 K, using the given K value of 1.84?

If #1 is possible, is it also possible to extrapolate the rate of reaction to higher temperatures using the low-temperature equilibrium constants? I'm mainly interested in temperatures in the range from 1200 - 1800 K.

If neither of the above are possible, do I simply need to find (or produce) experimental data on the rate of reaction?

Thank you!

No, the reaction rate is not related to K whatsoever.

First, the reaction rate is not related to the equilibrium constant. Second, you do not mention the unit of K. Third, the decomposition of  CaCO3 happens in a flame at about 800°C, where K is rather low, because there is practically no equilibrium in the flame. CO2 is eliminated as soon as produced.

Since a gas is involved I would be surprised if you couldn't obtain some sort of estimate on the forward/backward rates of reaction from this data, but it would involve many assumptions and require information about the surface properties and reactivity.


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Feasibility of phenyl migration
I am learning the fundamentals of organic chemistry,
here is the reaction which got me stumbling me for hours,
in the below reaction i could see that the amine group would get converted into diazonium salt, but i am clueless how to proceed further..

The compund A(solution) is given below, it seems like a phenyl group has migrated to the adjacent carbon, i have never seen something like this and it is also not clear for me why a carbonyl group is formed in the middle. Overall I dont understand the reaction...it would be helpful if someone points out what did i miss out in reading that made me unfit to solve this problem. Thank you!

It is a carbocation rearrangement, common in organic chemistry. Did you heard Pinacol Pinacolone Rearrangement? This is similar to that.

yeah i got it....but i forgot to mention in the question about the product's sterochemistry, hope u would add on it

This is similar to pinacol-pinacolone rearrangement. Only difference is in pinacol-pinacolone rearrangement, carbocation forms by dehydration while here, it forms through diazonium reaction. Following is the tentative mechanism of the rearrangement:

Since reaction progress through carbocation intermediate, the final product is a racemic mixture. $\ce{CH3-C+-CH3}$ (in pinacol) or $\ce{CH3-C+-H}$ (in given compound) can rotate about middle horizontal bond before alkyl or aryl group migration.

When phenyl is the migrating group there is also the possibility of an electrophilic aromatic substitution mechanism at the ipso position (i.e. forming a cyclopropane intermediate) followed by deprotonation of the O-H bond and breaking the C-O bond.


text:
How to accelerate the evaporation of water at temperature below 5 °C?
How can we make the water to evaporate at low temperature? Does addition of high evaporating solvents help or decrease in the hydrogen bonding by addition of solute will help? Is there any other method to evaporate water at low temperature?

Related: Is there a chemical that when added to water makes the water flash evaporate?

Without working at low pressure, water will evaporate quicker by bubbling it with a currant of dry air, whatever the temperature.

Low pressure is the way to go, adding other solvents doesn't help much
There is only one (easy) way to get cold water to evaporate quickly. That is to lower the pressure. At a pressure of about 1% of atmospheric pressure (achievable with a standard laboratory water aspirated vacuum pump) water boils at about 10 celsius. Even water at 5 degrees celsius will evaporate quickly, especially in a system like a rotary evaporator that maximises the surface area of liquid.
A slightly better vacuum will reduce that to <5 celsius and your water will boil even faster.
Adding other solvents can create azeotropes that boil under normal conditions and atmospheric pressure below the normal boiling point of water, but not by much. And, of course, your water is now contaminated with another solvent.


text:
Nucleophilic substitution of trifluoroiodomethane

Will $\ce{CF3I}$ show nucleophilic substitution? In other words, is the following reaction feasible?
$$\ce{CF3I + NaOH -> CF3OH + NaI}$$

I know that $\ce{I-}$ is a pretty good leaving group. Also due to inductive effects, the carbon already has a $\delta+$ charge. So I think it should react, yet the answer in my textbook is telling me otherwise.
Are there any other effects that I am not considering?

The reaction may not be occurring because solvent may not be correct, as in polar protic or polar aprotic.

If I'm not mistaken, said reaction does not take place because the three fluorine atoms so change the electronics of the system that trifluoromethyl iodide is not nearly as electrophilic as one might think.

@Zhe Could you please elaborate. Is it that the fluorine atoms' electron pairs repel the incoming anion?

No, I think it changes the energy of the σ* orbital.


text:
Conversion of Erythrosine to Fluorescein
I am a home-chemist and I love dyes. I recently got my hands on sodium salt of erythrosine (tetra-iodinated fluorescein). I want to convert erythrosine to fluorescein. I speculate that using dilute NaOH solution might cause formation of sodium iodide and tetra-hydroxylated fluorescein. How can I reduce the alcohol groups without affecting the carboxylic acid groups.
Is there any better way to convert erythrosine to fluorescein?

Would Sodium borohydride work?

@Safrad Faisal Most probably, sodium borohydride will reduce the carboxylic groups.


text:
What happens when an alkyne reacts with ammoniacal Cu2Cl2?
My text book provided me with 2 reactions concerning alkynes (Specifically Terminal Alkynes).
The first one was the dimerisation of an alkyne on passing it through ammoniacal $\ce{Cu2Cl2}$.
All was fine until I encountered the next reaction.
It was a test to distinguish between terminal alkynes and other alkene and alkynes.
It involved the fact that treating a terminal alkyne with  ammoniacal $\ce{Cu2Cl2}$ would result in the alkyne forming a red precipitate as copper replaces hydrogen of one of the terminal hydrogen.
However now I was confused. Which of these reactions actually happens? Or are there specific conditions for each of these reactions that my text missed out?
Please help me in this aspect.

Terminal alkynes (including acetylene) form a $\color{red}{\text{red precipitate}}$ of copper acetylides when it is passed through ammoniacal cuprous chloride solution:
$$\ce{R-C#C-H + CuCl + NH3 -> R-C#C-Cu + NH4Cl}$$
However, in the presence of oxygen, this reaction proceeds further to dimerize the corresponding alkyne. This reaction, the oldest homo-coupling of terminal alkyne reaction,  was first discovered by Carl Andreas Glaser in 1869 (Ref.1, 2), hence the reaction is named Glaser-coupling after him:
$$\ce{2R-C#C-Cu ->[O2][NH3/EtOH] R-C#C-C#C-R }$$
The actual Glaser's reaction was (ref.2):
$$\ce{2Ph-C#C-H ->[CuCl/O2][NH3/EtOH] Ph-C#C-C#C-Ph}$$
References:

Carl Glaser, "Beiträge zur Kenntniss des Acetenylbenzols," Berichte der deutschen chemischen Gesellschaft 1869, 2(1), 422–424 (https://doi.org/10.1002/cber.186900201183).
Chao Liu, "Chapter 1: Introduction," In Lecture Notes in Chemistry 102: Transition Metal Catalyzed Oxidative Cross-Coupling Reactions;  Aiwen Lei, Editor; Springer-Verlag GmbH DE (Part of Springer Nature): Berlin, Germany, 2019, pp. 1-10 (ISBN: 978-3-662-58102-5).

$$\ce{H-C#C-H ->[Cu2Cl2][NH4Cl] H-C#C- Cu+ }$$
This then attacks $\ce{H-C#C-H}$ to give $\ce{H-C#C-CH=CH2}$
Note: This occurs only for terminal alkynes and only in acidic medium(NH4Cl here).
Reference:M S Chouhan Advanced Problems In Organic Chemistry For JEE

May I know the chapter of MS CHOUHAN?


text:
Difference between ΔQ and dQ
What is the difference between $$\Delta Q$$ and $$\mathrm dQ$$ in thermodynamics? I think one is used when the change is very small and the other with bigger changes, but I am not sure.

Difference between Δ, d and δ. The question has little to do with chemistry, it's rather about notations and principles of calculus.

@andselisk, come on my friend. Chemistry cannot exist without mathematics. Thermodynamics is a part and parcel of chemistry curricula.

Also related: Symbols of derivatives and d or Delta?.

This is not about mere arbitrary notation.  It is about fundamental concepts in classical thermodynamics, and the notation flows from these concepts.
So: q and w are not state functions; rather, they are path-dependent. Hence their differential forms are inexact rather than exact. The dyet, $\text{đ}$, is used to indicate that (though typically it is an angled slash rather than a horizontal bar; I don't know how to make that with MathJax).
Since the differential forms of heat and work are inexact, any integration with respect to them must be a path integral. While not all physical chemistry/thermodynamics texts use this specific notation for inexact differentials, I've never seen a physical chemistry textbook that doesn't carefully use some sort of notation to distinguish exact from inexact differentials, because of the conceptual and pedagogical importance of distinguishing between state functions and path functions.
[After seeing the comment about Atkins, I went back and looked at my old Atkins pchem book and, sure enough, he doesn't use distinguishing notation for inexact differentials.  I didn't know this because Atkins' thermo treatment was considered poor, so we only used him for kinetics.  We used Castellan for thermo.]
Hence we have, for a closed system:
$d\text{U} = \text{đ}q + \text{đ}w$
And:
$\Delta \text{U} = q + w$
Or, alternately:
$\Delta \text{U} = \text{Q}  + \text{W}$
Heat and work are path functions that reveal themselves only during a change in state—during a process.  Hence, for a given process, there will be a certain amount of heat flow, q, and work flow, w.  One should not use "$\Delta q$"  or "$\Delta \text{Q}$", as this is conceptually incorrect: It implies that heat is a state function, and thus you could have a change in heat.  No.  There was (or could have been) a change in internal energy, along with some heat flow; there was no "change in heat".
More explicitly:
$$\int_{state 1}^{state2} dU = U_2 - U_2 = \Delta U$$
But:
$$\int_{state 1}^{state2} \text{đ}q \ne q_2 - q_1 = \Delta q$$
because there is no q for state 1 or state 2—q is associated with the path only, not with the state of the system.
Instead:
$$\int_\limits {\text{path from state 1 to state 2}} \text{đ}q = q$$
The only context in which it might make sense to use $\Delta q$  (or $\Delta \text{Q}$) would be if you were comparing two different processes, and calculating the difference in heat flow between those two processes.  This is acceptable, but perilous, since such usage risks people falling into the common misconception of thinking that it is acceptable to use $\Delta q$ to characterize individual processes, which it is not.
Adding something from one of my comments: A key property of exact differentials that both connects them with state functions, and distinguishes them from inexact differentials, is that their cyclic integrals must always be zero. Hence:
$$\oint dU = 0= \Delta U_{cyclic}$$
Conversely, it is not generally the case that:
$$\oint \text{đ}q = 0$$
Interestingly, if we carefully define the path to be reversible, and apply an integrating factor $(\frac{1}{T}) $, we can convert the inexact differential $\text{đ}q_{rev}$ into one of the most recognizable exact differentials in thermodynamics, as follows:
$$\oint \frac{\text{đ}q_{rev}}{T} = \oint dS =0$$

Very relevant and useful answer. I am just wondering about the name of (d-bar) symbol, which is used in many phys. chem texts. I would read it as d-bar. You called it dyet. Can't find this name name by Google.

Okay found it in the book "The Elements of Typographic Style", $dyet$: A basic letter of the alphabet in Serbo-Croatian and Vietnamese.

@M.Farooq For more on the rich and varied history of the dyet, see:  https://en.wikipedia.org/wiki/D_with_stroke
Some textbooks instead use the "eth" -- see:  https://en.wikipedia.org/wiki/Eth

"..._ Atkins' thermo treatment was considered poor, so we only used him for kinetics_..." .  A bit off-topic, but shouldn't it be "we used it"? Since a book is not having gender?

@Micelle, Many languages have genders for non-living things and this problem is brought during mental translations in English writing. Fortunately, or unfortunately English is free from these gender issues ;-)

@Micelle  This is a not uncommon colloquial usage, in which "him" is shorthand for "his approach to the subject".

@M.Farooq Actually, in English, most inanimate objects are properly referred to using the neutral gender, and thus "it" rather than "he" or "she".  There are exceptions (e.g., ships, for which either the neutral or feminine gender is used—or anything a writer wishes to anthropomorphize), but these are, as they say, 'the exceptions that prove the rule.'

There seem to be caveats: (1) $dq_{rev}$ seems to be an exact differential; (2) we tend to say that state functions are not really defined at single points either, only differences or values wrt a reference (such as absolute zero T and zero entropy) even though differences between their values are path independent.

@theorist: Right, like everything there are exceptions. My own language has masculine and feminine and nothing is neutral. Suffice it to say, English is relatively free from these issues and most likely it got rid of it centuries ago.

@BuckThorn  Nope, $dq_{rev}$ (more properly,  $\text{đ}q_{rev}$)  is definitely not an exact differential.  We know it's not an exact differential because, if it were, then $\oint \text{đ}q_{rev} = 0$, always, which is not the case (consider, for instance a reversible Carnot cycle).  You need an integrating factor to convert it to an exact differential, and that integrating factor is, yes, 1/T.  So while  $\text{đ}q_{rev}$ is not an exact differential,  $\frac{\text{đ}q_{rev}}{T} = dS$ is. Hence $\oint \frac{\text{đ}q_{rev}}{T} =  \oint dS =0$.

@theorist good point. The Carnot cycle does it again.

It is rather mathematical than chemical question, as it applies to any continuous and differentiable quantity, not limited to chemistry.
$\Delta Q$ would be macroscopically measurable exchange of thermal energy. But the convention is we use $Q$ is sense of the exchange of heat, not $\Delta Q$.
$\mathrm{d}Q$ is "infinitesimal ( infinitely small) difference aka differential. It is not true aka total differential, as $\int{\mathrm{d}Q}$ depends on path and therefore should not be using "d" - but "đ". I cannot remember the proper MathJax/LaTeX symbol for it, perhaps not available.
So it is $\mathrm{đ}Q$ instead ( copied from the comment ).
So $\mathrm{d}U = \mathrm{đ}Q + \mathrm{đ}W$, where the change of internal energy between 2 states depends only on these states, but exchanged heat and work depends also on the path ( a way how the final state was reached. )
It is used oftenin context of infinitely small changes from equilibrium, that are so small the system can still be considered in equilibrium. An infinite set of such changes makes a reversible process, where the system is at any time in equilibrium.

đQ. Not sure there is a latex symbol for it. Using Unicode might be easier.

How is it inserted on english keyboard ?

Isn't $\Delta Q$ incorrect notation, as $Q$ represents a quantity of energy that is transferred, and since it is not a state function, there is no pre-existing $Q$ that is being changed by the transfer. Thus, we write $\Delta E = q + w$ rather than $\Delta E = \Delta q + \Delta w$.

You are probably right, it is my frequent mistake. But OTOH, it is kind of inconsistent notation, confusing me often, as there can be d, resp. đ, but not $\Delta$, in contrary to other quantities.

In fact, in some sense, $\ce{Q}$ could be considered as the system thermal energy  $\int_0^T{C_\mathrm{v}(T) \cdot dT}$, as the part of $\ce{U}$

I used a compose key... COMPOSE d -

My understanding is that $\Delta Q$ and $dQ$ are both technically incorrect. Only $Q$ and đ$Q$ (rendered in some texts as $\delta Q$) are correct. However, due to the typesetting challenges of đ, many textbooks sloppily use $dQ$ instead (Atkins being a primary example), to the point that $dQ$ seems to be becoming an acceptable notation, even if mathematically incorrect. $\Delta Q$ is still generally frowned upon.


text:
Steric Effects in Friedel-Crafts Reactions
I recently came across a Friedel-Crafts reaction happening between benzoyl chloride and mesitylene with $\ce{AlCl3}$ as a catalyst. Apparently this reaction doesn't occur due to steric effects even though the mesitylene ring is highly activated by the three methyl groups. This got me wondering whether the reaction between benzoyl chloride with two methyl groups at ortho positions and benzene with $\ce{AlCl3}$ catalyst would occur or not. Most importantly, my question is: where all do steric effects become the major factor in the reaction, and how do we identify that? Why is it so?

"Highly activated due to steric effects" This statement is a bit weird. Steric effects at the site of reactivity reduce reactivity because the reactive species have a harder time with productive collisions. The general question is a bit broad because it depends on the nature of the reaction and the scale of the steric bulk.

Ah no I meant something else, I'll edit it at once.

Based on literature, I'd say the unsuccessful Friedel-Crafts reaction between benzoyl chloride and mesitylene is not due to steric reasoning. My argument is based on at least one report of Friedel-Crafts acylation on following polymethylated aromatic compounds (Ref.1):

The mono- and di-acylation have been taken place between 1,3-substitutions in each occasion as indicated in the abstract:

Diacylation of mesitylene, durene and isodurene by the Friedel–Crafts method, using the acylating agent in excess (as both solvent and reagent), gives derivatives which are of potential use as monomers. The bis(p-fluorobenzoyl) derivatives, for example, can function as the electrophilic components in poly(ether ether ketone)(PEEK)-type Polymerisations, and the bisnitrobenzoyl compounds are potential sources of novel diamines. Tribenzoylation [although not tris(p-fluorobenzoylation)] of mesitylene has been observed, but dibenzoylation of m-xylene is a much less efficient process than previously reported.

The reasoning for the resistance to dibenzoylation on m-xylene must be rationalized due to the deactivation of in-plane carbonyl $\pi$-electrons of benzoyl group on aromatic ring after first benzoylation (highly possible 1-benzoyl-2,4-dimethylbenzene). All other instances (2-benzoyl-1,3-dimethyl substitutions), the benzoyl carbonyl group is in out of plane with the corresponding aromatic nucleus (Ref.2), probably due to steric hindrance:
$$
\begin{array}{c|ccc}
\text{Compound} & \nu_\ce{C=O} \ (\ce{CH3C=O}) & \nu_\ce{C=O} \ (\ce{C6H5C=O}) & \text{ECA} \ (c.a.) \\
\hline
\text{Acetophenone}  & \pu{1688 cm-1} & - & -\\
\text{Benzophenone}  & - & \pu{1664 cm-1} & 42^\circ\\
\text{2,4,6-trimethylacetylbenzene}  & \pu{1703 cm-1} & - & 73^\circ\\
\text{2,4,6-trimethylbenzylbenzene}  & -& \pu{1672 cm-1}  & 90^\circ, 40^\circ\\
\text{2,4,6-trimethyl-1,3-diacetylbenzene}  & \pu{1705 cm-1} & - & -\\
\hline
\end{array}
$$
$^a$ ECA: Effective Conformational Angle (whenever two angles are given, first refer to the angle with mesitylene ring).
A comprehensive study of mono- and di-acylation of mesitylene have been done and in all cases, at least mono-acylation was achieved in $\gt 90\%$ whenever anticipated Ref:2). The abstract states:

The Friedel–Crafts acylation of mesitylene has been studied in detail. In the acetylation, propionylation or benzoylation reactions mono- or di-ketones may be formed. Mesitoylation leads only to the monoketone. A two-stage acylation procedure is described for the synthesis of acetylbenzoylmesitylene and acetylpropionylmesitylene. Reversibility has been shown to be a factor in these acylations, especially where an aliphatic acyl group is involved. The relationship between the stereochemistry of mesityl ketones and their u.v., i.r., and $\ce{^1H}$ n.m.r. spectra is discussed.

Overall, these results suggest that Friedel-Crafts acylation reaction would not effected by steric hindrence, at least created by 1,3-dimethyl substitutions. For example, 2,4,6-trimethylacetylbenzene (acetylmesitylene) was achieved in 92% yield; 2,4,6-trimethylbenzoylbenzene (benzoylmesitylene) in 83% yield; 2,4,6-trimethyl-1,3-diacetylylbenzene (diacetylmesitylene) in 95% yield; and 2,4,6-trimethyl-1,3-dibenzoylbenzene (dibenzoylmesitylene) in 89% yield (Ref.2).
It is also noteworthy that the nitration of mesitylene is not effected by steric effects as well. You can find more details on this aspect in here.

References:

Brian N. Hendy,   Kevin H. Patterson,   David M. Smith,   Susan E. Gardner, Neil J. Nicolson, “Polyalkylated aromatic monomers and polymers. Part 1.—Friedel–Crafts diacylation of polymethylbenzenes as a means of monomer synthesis,” J. Mater. Chem. 1995, 5(2), 199-204 (https://doi.org/10.1039/JM9950500199).
P. H. Gore, J. A. Hoskins, “Friedel–Crafts acylations of aromatic hydrocarbons. Part VIII. Mono- and di-acylation of mesitylene,” J. Chem. Soc. C 1970, (4), 517-522 (https://doi.org/10.1039/J39700000517).

So it seems that the reaction does occur at least once. So can it be said that in general Friedel-Crafts reactions aren't affected too much by steric hindrance? Will steric hindrance on the electrophile affect the reaction feasibility ( I know reaction rate will slow down)?

This conclution is based on 1,3-dimethyl substituents. If they are a larger group such as iso-propyl, the story would be different. I know for sury, if we use tert-Bu group, the substitution on ortho-position is nearly impossible. This is a good example the facts for  Zhe's comments (vide supra).


text:
Mechanism of alkaline hydrolysis of cyclic amide

In this mechanism, I got a pretty different mechanism from the solution. From the model answer, there is an intramolecular proton transfer just before the formation of the carboxylate ion. However, in the second intermediate, I think the electrostatic repulsion between nitrogen and oxygen is large, increase separation between nitrogen and hydrogen, hence reducing chance of intramolecular transfer.
My purposed mechanism rely on the H2O given in reaction mixture to undergo proton transfer, forming OH- (matching basic condition) and then pull the proton from R-COOH to form the carboxylate salt (product).
Is my mechanism plausible? What mistakes I have made if the model answer is the only correct mechanism?
Also, please suggest ways that I can improve this type of questions, thank you very much.

Both answers are correct. The intramolecular reaction is fast, but you're in water as a solvent. In reality, you probably have some mix of both mechanisms taking place. If I were grading, I would accept either answer.

@Zhe Thanks a lot. One question - you said the intramolecular reaction is 'fast', can this be explained by "https://chemistry.stackexchange.com/questions/16592/why-is-proton-transfer-so-fast" such that even there is a huge separation (O <-> N electronic repulsion), it occurs very fast?

I don't understand your statement about the huge separation. There may be some charges present, but it's largely going to be attenuated by the solvent. And the intramolecular proton transfer is on the time scale of the conformational change.

I quite don't get your issue with mechanisms. Things like what donates proton are largely irrelevant for mechanistic analysis of reaction - labile protons are transferred all the time.

@Mithoron OK so as a beginner, when there are any protons on any of a molecule, proton transfer can happen provided that it suits the reaction condition? (e.g. cannot generate H3O+ in this basic condition)

You comment isn't particularly comprehensible... If you don't get what labile means, then it's "easily exchangeable" - quite not "any".


text:
I need help checking to see if the answer I got is correct answer
My professor says that the answer is 583.5g. I got 583.0330822g. She says to use her periodic table. Her periodic table says the mass of Ag: 107.87g, N: 14.01g, O: 16.00g, and S: 32.06g. Below is the problem. Did anyone get the same answer as my professor or mine or neither?

Please try to find a reasonable title. While being as concise as possible, it shall indicate, without ambiguity, the subject matter of the question in such a way as to distinguish it from that of other questions, without going into unnecessary detail.

@Evelyn we don't do homework for others.

Noticed that 1167/2 = 583.5

In multiple choice questions, sometimes you have to use the best approximation. If one of the answers was 583.5 g and you were getting 583.0 g, chances are the first choice is correct in the mind of the question writer. It does not have to be the most accurate or factual answer (you have to choose from the given choices using the given data).
Your teacher cannot make a lame excuse that her periodic table must be used. In science there is no authority (his or her periodic table). The International Union of Pure and Applied Chemistry should have the accurate mases. Download the IUPAC period table and and calculate once again using the masses given there. What do you get?
https://iupac.org/what-we-do/periodic-table-of-elements/

Whatever the calculation, finding a result with ten significant figures (583.0330822) is a non-sense when starting from data known with only 4 significant figures like 16.00 for  the oxygen atoms and 14.01 for the nitrogen atom.

@Maurice - I agree that 583.0330822 is nonsense, but the atomic mass of $\ce{AgS2}$ will have 5 significant figures using the data given since both the mass of silver and sulfur are given to the hundredths decimal place.

Your teacher is wrong. Your answer of "none of the above" was correct.

However the value of $\pu{583.0330822 g}$ is nonsense. There are way too many significant figures. The value should be rounded to 5 significant figures, or $\pu{583.03 g}$.

I disagree with user M. Farooq about the periodic table. My perception is that the teacher wanted all the students to use a particular periodic table (hers) to avoid rounding errors between all the students.
First since all the masses were given to the hundredths decimal place, the answer should have 5 significant figures.
$$\begin{align}
\text{molecular mass }\ce{AgNO3} &= 107.87 + 14.01 + 3\times16.00  \\
&=107.87 + 14.01 + 48.00 \\
&= 169.88 \\
\end{align}$$
$$\begin{align}
\text{molecular mass }\ce{Ag2S} &= 2\times107.87 + 32.06  \\
&= 215.74 + 32.06 \\
&= 247.80 \\
\end{align}$$

My calculator shows 247.8 for $\ce{Ag2S}$, but my calculator doesn't understand significant figures. The answer must be 247.80.

$$\begin{align}
\text{mass }\ce{Ag2S} &= \dfrac{799.40}{169.88}\times\dfrac{1}{2}\times247.80 \\
&= 583.03308217565340240169531433953 \\
&\ce{->[Round to 5]} 583.03\\
\end{align}$$
Now the touchy part with such problems is the possibility of rounding errors in intermediate calculations. However the molecular masses are correct to the hundredths place. Also the final calculation was done in one overall calculation so there was no intermediate step. Thus the answer of "none of the above" is correct.
Now if I had calculated the moles of $\ce{Ag2S}$ as a separate step, then I'd used two extra digits in the intermediate calculation to try and avoid rounding errors. Thus:
moles $\ce{Ag2S} = \dfrac{799.40}{169.88}\times\dfrac{1}{2} = 2.352837$
mass $\ce{Ag2S} = 2.352837 \times 247.80 = 583.0330086 \ce{->[round to 5]} 583.03$
I fiddled around a bit with the calculations and I can't determine how the teacher calculated the 583.5 g answer. It seems to obviously be a rounding error, but I don't know how to reconstruct the steps in the teacher's calculation.

think I can point to the calculation error finally. $$\text{mass }\ce{Ag2S} = \dfrac{799.40}{169.88}\times\dfrac{1}{2}\times248(?) =583.5$$


text:
Half life of first order reaction in case of variable volume
I have a doubt about the meaning of half-life of a first-order reaction. Earlier I used to think that it is the time in which concentration of reactant reduces to half of its initial but recently I got a question (I don't have that question right now) undergoing a first-order reaction with variable volume and then my teacher told that half-life is the time when moles of reactant reduces to half and as the volume is constant moles are similar as concentration.
I want to confirm this thought here.

Is volume constant or variable in your reaction?If it is constant, then comparing with number of moles is also correct.

For constant volume reactions, half the reactant moles being consumed = concentration getting halved.

the volume is variable and I have came across two questions where volume is variable and they used moles

@EashaanGodbole you are right my friend but I am confused in case of variable volume.

If the volume is varying with time, you will have to solve the first-order differential equation with $V(t)$ to obtain the rate equation.

For first order reactions (assuming the reaction $\ce{A -> B + C}$):
$$-\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = k[\ce{A}]$$
If the volume is $V(t)$, then $[\ce{A}] = \frac{n}{V(t)}$. You will have to substitute $[\ce{A}]$ in the differential equation and solve it in order to obtain the rate equation in this case.
As for finding the half-life, Wikipedia states that it is the time required for a quantity to reduce to half its initial value. If $V(t) = V_0$, then $[\ce{A}] \propto n$ and the half life for the quantity of substance is equal to the half-life of the concentration of substance. If $V(t)$ is not constant, then you'll have different half-life times for the quantity of the substance and for the concentration of the substance.

sir, I still have some problems as in the case of radioactive decay which is a first-order reaction the half-life is the time required for the decay of half of the number of nuclei.

First order reactions are always about the total count or particles, molar amount or mass.
$$\frac {\mathrm{d}n}{\mathrm{d}t} = -k \cdot n$$
$$\frac {\mathrm{d}m}{\mathrm{d}t} = -k \cdot m$$
The point of using concentration instead is, that it is proportional to them in homogenous, constant volume scenario.
$$c = \frac nV$$
$$c = \frac m{MV}$$
so it can be expressed in terms of $c$  as well:
$$\frac {\mathrm{d}c}{\mathrm{d}t} = -k \cdot c$$
This has great advantage as $c=\mathrm{f}(t)$ can be monitored much easier than $n=\mathrm{g}(t)$ or  $m=\mathrm{h}(t)$
For homogenous, variable volume scenario, concentration alone is not useful, unless multiplied by volume.
$$\frac {\mathrm{d}n}{\mathrm{d}t} = -k \cdot c \cdot V = -k \cdot n$$
$$\frac {\mathrm{d}m}{\mathrm{d}t} = -k \cdot c \cdot V \cdot M = -k \cdot m$$
For non homogenous concentration, concentration must be integrated over the volume
$$\frac {\mathrm{d}n}{\mathrm{d}t} = -k \int_V{c \cdot dV}$$
$$\frac {\mathrm{d}m}{\mathrm{d}t} = -k \cdot M \cdot \int_V{c \cdot dV}$$
or the known molar amount or mass has to be used.
With such approaches, the half-time is scenario independent:
$$t_{1/2} = \frac{ \ln{2}}k$$


text:
Oven safe dessiccant?
Im sure there is a chemist/lab tech that knows a simple solution to my problem, as I cant seem to find a solution.  When trying to strip and re-season cast iron cookware, I occasionally have problems with rust.  I live near the ocean, so there is a lot of humidity in the air.  I strip the cast iron by putting in the oven on the self-cleaning setting for an hour.  This normally works perfectly.  However, sometimes at the end of the hour, I open the door and see the pan has developed surface rust.  Obviously, heat + iron + water = rust.  I am sure there is a way to absorb the moisture in the oven, but I cant find anything.  I looked at desiccants, but they release moisture in the oven when heated - at least the ones I saw.  Is there a desiccant that works at oven temperatures?  Or is there another solution?  Remember, food will be cooked on the cast iron, so the solution has to be food safe in the end.

Could you include more details about how you handle the cookware? Exactly what do you do to it before you put it in the oven?  Also, what kind of oven is it? This site explains that "[GEs] innovative P7 model introduced the pyrolytic cleaning method ... a fancy name for heating the residue until it carbonized and turned to ash. The oven included extra thermodynamic controls to allow it to reach the high temperatures needed. Today, another option is the steam-cleaning oven, which loosens dirt without resorting to extremes of heat."

What is the problem with that rust? You should be able to wipe it off easily.

@BuckThorn I dont do anything.  I am simply putting it into the oven to strip off the seasoning and any burnt on carbon.  The self-cleaning setting is the super high heating, which does turn everything to ash.

@Karl Yes, I can remove the rust after if forms.  However, I was hoping to find a way to prevent that step.

If the rust is hard to get off, it was on there before, hidden by your "seasoning". Otherwise I´m sure it´s not worth the trouble to get the oven interior super dry.

If it is possible to quickly brush off the rust after it has formed this may be much more simple and less tedious than properly preventing it's formation.

@Keltari I don't think there is much you can do to remove the humidity aside from securing a source of dry gas to pump into the oven, which is not typically available in a household kitchen :-) . I would follow the instructions as indicated by AJKoers answer and useful hints from the other answer(s). It is very possible that it is not humidity but any other substances such as fatty acids that are triggering oxidation, but it is hard to say. Another link to a skillet guide: https://www.scienceofcooking.com/science-of-cast-iron-skillet-cooking.html

I like the suggested treatment suggested here which is to basically clean/scrub the pan, dry and coat with vegetable oil, invert the pan, place over aluminum foil on an oven rack and bake. To quote:

Now that you’ve returned your skillet to its base layer, you’ll need to season it for use. Drop a tablespoon of vegetable oil on the pan and spread it around with your fingers or a paper towel. (Take care not to burn yourself on a warm skillet!). Oil the outside and the handle, too. Next, take a folded napkin and wipe off the excess oil. You’ll want the skillet to be lightly greased. Set your oven to 350 degrees and put some aluminum foil on the bottom rack. Then pop the skillet in, upside down, on the upper rack for about 1 hour. Turn the oven off and leave the pan inside until it’s cool. The oil will bake into the pores of the pan, providing a non-stick finish.
Repeat this step a second time and you’ll have an almost glassy surface of seasoning.
Ta-da! Your rusted cast-iron skillet is good as new.

The created carbonized layer is safe, chemically inert and durable.

Reread the question.  I am not asking how to season a skillet.

At the high temperature used to burn off the fatty acids on the surface of the cast iron cookware (pyrolysis) dessicants lose their dessicating ability (at best) and dehydrate (releasing water, which is the opposite of you want to do: capturing water).
Fatty acids oxidize already at low temperatures when highly unsaturated (components of lin seed oil for instance, which can in certain circumstances even catch fire near room
temperature), but most commonly need temperature of 300-500 degree celcius to start oxidizing at a sufficient rate to suit cleaning purposes. At these temperatures: CaCl2, Na2SO4, SiO2, MgCO3, CaSO4, MgSO4, ZnSO4, and K2CO3, which are commonly used dessicants, will all have released the water they can capture, and won't be able to rehydrate until being brought back to lower temperatures.
One exception might be CaO, though it's not able to capture as much water per molecule as others. If pure, minute amounts of CaO on the surface of your cookware are perfectly compatible with the large amount of food you'll put in the pots (it will become calcium hydroxide when it comes in contact with enough water). The downside of a dessicant that works at high temperatures is the problem of recycling (i.e. dehydrating it), for CaO you might have to heat it above 1000 degree celcius before it can be used again as dessicant, which certainly is not the practical recycling you'd want to routinely carry out for household purposes.
The good side is that with increasing temperatures the density of water/steam in your oven too decreases (unless it's a pressure chamber and the pressure strictly accompanies variations of temperature). This means that if you put the cookware in the oven when it's already hot you shouldn't have a lot of rust forming, and thus, as Karl writes, it shouldn't be as big of a problem to remove it as it would be to prevent it's formation.
A solution of dilute oxalic acid at 50-65 degree celcius might help remove the rust (it's often used to remove rust stains on metals as iron oxalate is slightly soluble in water unlike most metal oxalates). It would be worth trying and is in any case compatible with your food as small amounts of oxalates are contained in many edible plants (spinach, most notably).
You can rinse off most of it with hot water. If you were to drink too much of a dilute solution of oxalic acid however, you'd develop kidney stones as it would precipitate in your kidneys mainly as Calcium Oxalate crystals, so if you do this often, definitely have the habit of rinsing the pot with hot water before use.
Citric Acid is another tool that you may try to remove the rust as it too is compatible with food.

Im reading your answer.  First I want to say removing rust is not the problem.  I can scrub it off easily.  My goal is to prevent it from forming.  You are saying calcium oxide will work at oven temperatures and to sprinkle it on the iron?  Im reading up on CaO, its cheap, but it definitely sounds like something I dont want to ingest.

You should not sprinkle it (nor any dessicant) directly on the iron. A dessicant needs to have it's own container allowing it a fair contact surface with the humid air without itself escaping the container. Tiny little traces of the dessicant you will use might though escape the container and end up ont the surface of the pot, which is the worst case scenario you'd face if your setting has been well designed. Very small amounts of Calcium Hydroxide can be very safely used in food (E526 is the code in Europe). Don't handle any chemicals at all if you don't know how to safely use them.

Tiny traces of calcium hydroxide on the surface of your pots not only is compatible with foods, but will also further protect the iron from further rust by passivating its surface.

Don't handle Calcium Oxide, nor the resulting hydroxide without proper chemical protection equipment (skin, eyes, airways, and preventing contamination of the room).

If you want to use it directly on the pot you could do so if isolating the pot from the rest of the oven in another, bigger, oven-safe container (to prevent CaO from attacking non-ferrous metal parts of the oven). You'd then have to rinse the pot with a profuse amount of water before heating it again for drying before applying the fat for the ensuing seasoning step.


text:
Method for regioselective hydroalkylation of terminal alkyne
I'm trying to figure out a method to add a linear alkyl chain R and a hydrogen atom H across the triple bond of a terminal alkyne. The alkyl chain should end up on the internal carbon (C-2):

My attempt :
First, we can hydrolyse the alkyne to produce a methyl ketone. Addition of a Grignard reagent produces the tertiary alcohol. But, now I'm stuck as I can't treat it with $\ce{H2SO4}$, because then ring-expansion will ruin the whole matter.

How to move on?

You want to add R–H across an alkyne? (you need to specify two things for an addition reaction; you can't just add R to one carbon and not do anything to the other carbon, well, unless you're aiming for a radical product. :-) )

@orthocresol: Yes, I want to add R-H across an alkyne, directly or indirectly.

Maybe from the ketone make an enol triflate, then do some kind of Pd cross-coupling. But I'm sure you will get better suggestions from others.

@orthocresol: I'll take a look at it (I heard triflates for the first time actually), and thanks for the edit ;)

What is the nature of R? Alkyl, aryl?

@Waylander: it's a linear alkyl chain

I vaguely remember that dehydration with Tho2 follows an E2 elimintaion and gives the hoffmann product.

To achieve cyclobutylmethyl ketone, you must have to perform oxymercuration reaction. There is a possibility to ring expansion during that process as well. You may need to address that first.

@MathewMahindaratne: thnks for pointig that out, I didn't noticed it.

The simplest route is to treat the alkyne with RCuMgX2 (Normant, 1976) followed by aqueous work up. Schwartz (Zr) and Negishi (Zr, Ti) methodology is also applicable.

Here is a potential solution using old-school chemistry, though with several steps.

Hydroborate the alkyne and work up with $\ce{H2O2}$ to give the aldehyde.

Alkylate the aldehyde enolate with $\ce{RI}$, either directly (LDA or LiHMDS) or via enamine/silyl enol ether

Reduce aldehyde to alcohol ($\ce{NaBH4/EtOH}$)

Form triflate/mesylate and eliminate (DBU/Toluene, reflux)

If you wish to explore @orthocresol's suggestion of enol triflate coupling then coupling the triflate with the cuprate of your alkyl chain looks to be viable according to this Tet. Lett. article from McMurry and Scott1
References

John E. McMurry, William J. Scott, “A new method of olefin synthesis. Coupling of lithium dialkylcuprates with enol triflates,” Tetrahedron Letters 1980, 21(45), 4313-4316, DOI: 10.1016/S0040-4039(00)77845-2

This is the better method, which eliminate rearrangement possibilities.

Also, the last step can be achieved by converting alcohol to halide followed by $\mathrm{E2}$ elimination (using $\ce{t-BuK}$).

Alkylation of primary, saturated halides with enamines does not work well. Stork and Dowd made an improvement by enolizing t-butylimines with Grignard reagents. https://pubs.acs.org/doi/pdf/10.1021/ja00897a040


text:
Forming ester from amide using protonated alcohol as catalyst
In my level of understanding, if I want to make an amide from an ester, I should add:

$\ce{H3O+}$ for hydrolysis to make carboxylic acid
$\ce{SOCl2}$ to form acid chloride
$\ce{ROH}$, pyridine to form the ester as product

I think this is a good way since it is difficult to form a more reactive product compared to the reactant, but in one of the question, I really want to try to make the 3-step process 1-step:

My attempt is to use a protonated $\ce{ROH}$ as acid-catalst, this time, to form a methyl ester. I used methanol, and somehow I can get to the end.
[for the 4th intermediate, I searched online and found that $\mathrm{p}K_\mathrm{a}$ of $\ce{(CH3)2NH = 10.73}$, $\mathrm{p}K_\mathrm{a}$ of $\ce{CH3OH = 15.56}$, this means that $\ce{(CH3)2NH+}$ is a better leaving group (lower $\mathrm{p}K_\mathrm{a}$ --> lower basicity --> more stable in the reaction mixture)]
Can someone please kindly tell me what is wrong with my mechanism? I know it should be wrong, since the 3-step process is the most suitable way and yield the most.

There are several points that I'd throw out:

As a technicality, you probably can't buy protonated alcohols; you'd need to use the alcohol plus an acid like HCl in a non-aqueous solvent.

Now, once the amine drops off, it will mop up your catalytic acid, so you had probably better use at least stoichiometric acid, and possibly quite a bit more.

Amides are more stable than esters; you know that already. So, the reaction thermodynamics are not in your favour to begin with. Thus, in order to get a decent yield, you probably also need to use a big excess of alcohol (Le Chatelier's principle: more alcohol → more ester).

You'd probably also need to heat it quite a bit to get the reaction going. (The same is true for hydrolysis with aqueous acid.)

All in all, probably you could get it to work, but it might not necessarily be fun, and if your reactants had any other sensitive functional groups on it they would pretty much be destroyed by heating in strong acid.
In general, this is a tough reaction, and I suspect that if you could find a general way to do this cleanly then it would be quite a substantial achievement. Recently, Hie et al. have developed a Ni-catalysed procedure (Nature 2015, 524 (7563), 79–83), but it only works on aromatic  amides, so isn't applicable to your example substrate. (Essentially, both the group attached to the carbonyl carbon, as well as one of the groups on nitrogen, have to be aromatic. Despite that obvious limitation in scope, it's still in Nature, which maybe suggests something about the difficulty of the transformation.)

A search of "alcoholysis of amide to esters" produces a lot of hits with a range of catalysts. This one with catalysis by CeO2 looks a good one  http://www.shokubai.org/16kjsc/pdf/Youth_oral/YO_D11.pdf

@Waylander still 200 °C for 15 h! but scope looks better for OP's substrate...

@orthocresol there's a good reason why this is not a common reaction!


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What happens on heating an ester?
While solving a few questions a came across a problem which asked me what would happen when sec-butyl ethanoate was heated.
I had never read about this reaction before and took to researching through a few of my textbooks, but could not find an explanation as to what would happen.
I also thought along the lines free radical formations and tried to make an attempt.
However according to the answer the reaction was a syn elimination and it would give a Hoffmann product. But I couldn't get any more information out of it.
But now I was stumped. So please help me out of you know anything about this.

Edit:
After googling for sometime I came across a paper which said that this particular compound would decompose to give cis butene.
However I could not understand the reason for this. Here is where I saw it:
https://pubs.acs.org/doi/pdf/10.1021/i360047a014

Depending upon the zeolite employed with sec-butyl acetate in your link, the cis/trans ratio varied between less and greater than one. cis-Butene is not the exclusive butene formed.

@user55119 Okay! Noted.

This is an example of a Thermal syn elimination. The Wikipedia article is a good place to start: here

image from here
From ref 2

Unsymmetrical Acetate and xanthate esters yield a mixture of all the possible alkenes, but there is usually a predominance of the more highly substituted alkene.


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Identification of the organic compound

An organic compound A $(\ce{C10H14O})$ exhibits following characteristics.
A) It reacts with metallic sodium to give a colorless odorless gas.
B) It is oxidized by $\ce{KMnO4}$ to benzoic acid.
C) It can be resolved.
D) It does not give precipitate with iodine in presence of $\ce{NaOH}.$
E) It changes the color of acidified potassium dichromate from orange to bluish
green, and gives a chiral compound.
How many of the following compounds satisfy all the five criteria from A to E?

The given answer is the compounds 4, 5, 10 satisfy the criteria but I think the compounds 5, 6, 12 satisfy the criteria.
From statement A, we come to know it's an alcohol (which is common to every compound).
From statement B, we come to know that it has a benzylic hydrogen (which rules out the compound 11).
From statement C, we come to know that it has a chiral carbon. (which rules out options 4, 7, 9).
From statement D, we come to know that $\ce{OH}$ group is not attached to a methyl group (which rules out options 3, 10).
For statement E, I guess potassium dichromate will oxidise 2° alcohol to ketone and 1° alcohol to carboxylic acid and the compound still remains chiral (which rules out options 1, 2, 8).
So where's the mistake?

I think that correct answer should be 5 and 6 only, because:

12 will produce terephthalic acid, instead of benzoic acid as it possess two benzylic-H (also given by 7).

4 and 10 doesn't satisfy statement C and D, as you've mentioned.

I agree; the answer given is clearly incorrect. 4 and 10 fail the criteria

Thanks for the answer. I didn't notice the extra methyl group in 12 :).

The given answer is wrong in many ways:

4 did not satisfy both criteria C and E, because it is not a chiral compound to begin with.
10 did not satisfy the criteria D, because it does give a precipitate with iodine in presence of $\ce{NaOH}$.

That gives only 5 in given answer to satisfy all criteria. Let's eliminate given compounds in systematic order:

That leaves only 5 and 6 to satisfy all criteria.

How do you know that #3 and #10 precipitate?

@Stanislav Bashkyrtsev: 3 and 10 are oxidized by $\ce{I2}$ to give methyl ketones first, which give positive idoform test.

@ Mathew Mahindaratne, why don't the carbanion form at the benzylic carbon, and react further giving a negative iodoform test

@Govind: Can you elaborate your question a bit? I didn't understand your angle.

I think 10th should also be in the answer and should satisfy condition D. As the carbanion intermediate formed while iodoform test would be more stable at benzylic carbon

@Govind: Compound $\bf{10}$ will give a precipitate with $\ce{I2/NaOH}$ not because your reason. First, $\ce{I2/NaOH}$ oxidizes $2^\circ$-alcohol to ketone, which would be a methylketone, satisfying the condition for iodoform test.

@ Mathew Mahindaratne It is not necessary that all methyl ketones give iodoform test. Take for instance Methyl acetoacetate


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Hydride shift and Ring Expansion
Below is a problem I got stuck at in a test .

The first compund in this chain was easy to figure out , it was something like this :
Then I had this intermediate compound for B.(I didn't add the bromide anion because this is where my confusion starts)
Now ,here it shouldn't be correct to do a hydride shift outside the ring as it would result in a much less stable carbocation . But if I proceed using that idea and solving further for part D and E , I am getting option (b) as my answer and the correct answer is option (d).
Although if I do a hydride shift there then that opens a way for a ring expansion and hence a correct answer (according to the answer key) .

Consider if D is the primary alcohol from HCHO addition to the tert Grignard. The HI conditions could give A or a primary cation that would undergo hydride shift to the methylcyclohexane cation i.e the precursor to D

Ring expansion takes place in the step $\ce{D->E}$.
You were going right till the intermediate. Your intermediate will be attacked by $\ce{Br-}$ anions (as temp. is normal, therefore $\ce{S_N1}$ mechanism). Product will be B.
Then, it will form grignard reagent as C, which will react with aldehyde to produce alcohol D.
After that, it will react with HI to form a carbocation intermediate, which will rearrange to give product E.
The whole mechanism is as follows,

Looks correct, but E is the iodide not the bromide

@Waylander: Thanks, I've edited that.

Your answer A is correct. For B, I don't understand why you want to make stable carbocation to relatively high energy carbocation. Nonetheless, B is a compound, not a intermediate. Sure, dehydroxylation of A by $\ce{HBr}$ would give $3^\circ$-carbocation as indicated. But that intermediate would further proceed to react with remaining $\ce{Br-}$ to give B (1-bromomethylcyclopentane) as the product.
B reacts with $\ce{Mg}$ in ether to give the Grignard reagent C (methylcyclopentyl magnesium bromide). This react with $\ce{HCHO}$ (formaldehyde) to give the $1^\circ$-alcohol D (1-methylcyclopentylmethanol; the structure in answer $(A)$ with $\ce{OH}$ in place of $\ce{I}$).
Protonation of D by $\ce{HI}$ would give a $1^\circ$-alkylhydroxonium ion, which further subjected to dehydrate with concomitant ring expansion to give $3^\circ$-methylcyclohexyl carbocation. This carbocation would further react with remaining $\ce{I-}$ to give E (1-iodomethylcyclohexane; the structure in option $(D)$) as the final product.

I think you mean D dehydrates to give the cation, not decarboxylates.

@Waylander: Yes, in deed. Thank you for pointing out that.


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Changing pressure on equilibrium
Recently I heard my teacher say that when increasing the total pressure of the system, the reaction which is "more dependent on collisions" is favoured.
This seems to make sense, although I've looked at many resources online and on SE, and people have either given a mathematical explanation or simply talked about Le Chatelier's principle (increasing pressure favours reaction to produce less moles)
My question is: is it correct to say that increasing the pressure favours the reaction which involves more molecules colliding?
Take for example the forward reaction here which is favoured upon pressure increase. (4 molecules need to collide for forward and 2 molecules need to collide for reverse reaction)
$$\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}$$
I'd appreciate an intuitive explanation about collisions and interactions between molecules rather than something mathematical. Thanks in advance!

Be aware molecularity of a particular reaction step ( number of colliding molecules ) is not the same as the sum of the stoichiometric coefficients of reactants. Reactions, where 4 molecules have to collide, are very rare and their rate must be minimal unless under high pressure. But what you teacher says is rather right, even if said by somewhat obfuscated way. The side where is more molecules, has steeper dependence of the reaction rate with the increasing pressure. Also, if youwrite the equilivrium equation, you can see which way the reaction shifts with pressure.

The concentration of a species in an ideal gas is directly proportional to the pressure, which means increasing the pressure (while keeping mole fractions constant) will increase concentrations.

reaction shift in the direction,where no. of gaseous molecules are more

Your teacher might have a weird way of expressing himself, or not know what he´s talking about, in any case that statement is imo not worth giving it another thought. Collisions are important for kinetics, not for thermodynamics, at least not on a high school level.

Yes, I agree with your teacher's comment that upon increasing the total pressure of a system, there are select reactions "more dependent on collisions", and such reactions have been a matter of study.
For example, per this 2008 work reported in 'The Journal of Physical Chemistry',  The Temperature and Pressure Dependence of the Reactions H + O2 (+M) → HO2 (+M) and H + OH (+M) → H2O (+M), where '+M' here denotes the catalytic presence of a surface, as can be provided by another gas molecule.
To quote from the abstract:

The reactions $\ce{H + O2 (+M) → HO2 (+M)}$ and $\ce{H + OH (+M) → H2O (+M)}$ have been studied using high-level quantum chemistry methods. On the basis of potential energy hypersurfaces obtained at the CASPT2/aug-cc-pVTZ level of theory, high-pressure limiting rate coefficients have been calculated using variable reaction coordinate transition state theory. Over the temperature range 300−3000 K, the following expressions were obtained...Available experimental data on the pressure dependence of the reactions were analyzed using a two-dimensional master equation.

Note, reactions are being studied via surface collisions involving the addition of  Ar and N2 gases. The selection of the gas relates to chemical inertness and the total collisional cross-section of the particular gas particles, all of which are components of more complex theories surrounding collision frequency and collision theory (see, for example, discussion here).

I know you wanted something non-mathematical, but I think the best way to understand this intutively requires some simple math.
Suppose you have the following elementary reaction:
$$\ce{A(g) + B(g) <=> C(g)}$$
By "elementary reaction" I mean that this shows the actual reaction mechanism, such that the rate equation can be obtained directly from it.
Let the rate constants for the forward and reverse reactions be $k_f$ and $k_r$, respectively.
Since A and B are produced from C, the rate at which A and B are produced is proportional to [C].  And since A and B are consumed by a biomolecular reaction involving both A and B, the rate at which each is consumed is proportional to $\ce{[A]\times[B]}$.  We can thus express the rate at which [A] and [B] change as follows:
$$\ce{\frac{d[A]}{dt}=\frac{d[B]}{dt}}=k_r[C]-k_f[A][B]$$
And, using the same reasoning for C, we have:
$$\ce{\frac{d[C]}{dt}=k_f[A][B]-k_r[C]}$$
At equilibrium, the rates of the forward and reverse reactions are equal, i.e.:
$$\ce{k_f[A][B]=k_r[C]}$$
Let's suppose you are at equilibrium, and suddenly double the pressure (before the reaction has a chance to respond).  In that case, the rate of the forward reaction ($\ce{k_f[A][B]}$) will initially increase four-fold (because both [A] and [B] double, and $2 \times 2 = 4$), while the rate of the reverse reaction ($\ce{k_r[C]}$) will only double.  Consequently the reaction will shift to the right (towards the product side). Thus you can see what your teacher (probably) meant when he or she said that the reaction shifts away from the side more dependent on collisions—because the rate at which the species on that side are consumed will be greater.
Eventually, the reaction will requilibrate.  The equilibrium constant will be unchanged, but the relative concentrations of species will have changed to favor the product.
Note this would not apply to liquid-phase reactions, since an increase in pressure will have only a negligible effect on concentration.  In this case, you need to consider the relative volumes of the reactants and products.


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Why is CH4 Nonpolar but CH2Cl2 is polar?
I keep reading that the reason why $\ce{CH2Cl2}$ is polar because due to its tetrahedral shape, the dipoles can not cancel each other out but doesn't $\ce{CH4}$ also have tetrahedral shape too? I assumed, since the reason for $\ce{CH2Cl2}$ being polar is apparently due to the 109.5° angles from the tetrahedral shape, that this means that the surrounding atoms in all tetrahedral-shaped compounds are not placed directly opposite of each other and that is why the dipoles can't cancel each other out. Am I understanding this wrong? Or if this is the case, then shouldn't the dipoles of the $\ce{H}$ atoms in $\ce{CH4}$ not be able to cancel each other out either because they aren't directly opposite of each other? Is it different because all the surrounding atoms in this case would have the same dipole?

Look at the picture here: https://chemistry.stackexchange.com/a/118202/79678 and you will see that the bond dipoles cancel out for methane or carbon tetrachloride. But not for dichloromethane!

There are two reasons why dichloromethane is polar and the tetrahedral shape is only one
The reasons why any molecule is polar–which is usual chemistry talk means that the molecule has a dipole moment–is that the individual dipole moments of its constituent bonds don't balance out. That means that we need to know both the polarity of the constituent bonds and the 3D shape of the molecule.
All the bonds in methane are the same (and are fairly non-polar to start with). So, even if methane were flat with bonds forming a square with carbon at the centre, it would be non-polar. It isn't flat: it is tetrahedral but, since all the bonds are the same, even a tiny dipole on each would still cancel out. carbon tetrachloride has 4 polar C-Cl bonds but is still non-polar as those dipoles cancel out exactly.
Dichloromethane, were it a flat molecule, could be either polar or non-polar (the dipoles of C-Cl bonds on opposite sides would exactly cancel out, but on the same side would not, obviously). But it isn't that shape, it is tetrahedral. But there is no way to organise two C-Cl bonds in a tetrahedral structure so their dipoles balance out. Build a 3D model if this isn't obvious.
So you need to know two facts to judge molecular polarity: the bonds making up the molecule and the shape of the molecule. Methane isn't non-polar because of its tetrahedral shape: it is non-polar because all the bonds are the same. Dichloromethane is polar because it has different polarity bonds and its shape cannot arrange those bond dipoles to cancel out.

Note that polarity can be considered for the whole molecule, functional group or particular bonds.E.g. $\ce{CO2}$ has zero permanent dipole moment, as bond dipoles cancel each other. But the bonds themselves are polar enough due their dipole moment, being able to involve the molecule in polar intermolecular interactions.
Bonds of  both $\ce{CH4}$ and $\ce{CH2Cl2}$ have small dipole moments ( $\ce{C-Cl}$ bigger than $\ce{C-H}$ ). But $\ce{CH2Cl2}$ have nonzero dipole moment as a molecule, having its bond dipoles oriented asymetrically. $\ce{CH4}$ has dipole zero moment due molecule symmetry. Even though, $\ce{CH2Cl2}$ is rather somewhat polar than polar ( like e.g. methanol or water)

IT is not correct to describe carbon dioxide as "polar" in almost any sense in chemistry. Its use as an example here is both wrong and deeply misleading.

OK, it may be better to describe it as non polar with polar bonds :-)


text:
Why aren't nitriles very good electrophiles?
I was wondering why nitriles do not react as electrophiles like carbonyl unless they have to, given that the C-N triple bond is still pretty polarized. The LUMO will be higher in energy than a carbonyl due to the higher electronegativity of oxygen, but I don't see why nitriles are far worse in electrophilicity.

Who says they don't? They are not as reactive as carbonyls, but there are plenty of reactions with nucleophilic attack at the nitrile carbon - example: http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch20/ch20-3-5-3.html


text:
How do you separate microcrystalline cellulose from amino acids in a powder mixture?
I am looking to separate out the microcrystalline cellulose that has been mixed in with amino acids in a powder mixture.
I have tried suspending the powder mixture in water and then filtering out the microcrystalline cellulose afterwards. The amino acids in the mixture dissolve cleanly without a trace in water after stirring. Although microcrystalline cellulose is supposed to be insoluble in water, which mostly does form a layer in the solution once the fine particles settle after a while, somehow I find that a slight amount of it seems to gets dissolved as well. Either that or it requires an even finer filter, although I have already tried using a filter of 0.1 micron pore size, which should have been enough to filter out any microcrystalline cellulose particles that are still suspended in the solution, but it seems that some traces still remain after filtering.
I would like to cleanly separate out the microcrystalline cellulose from the amino acids in the mixture. Any suggestions as to how I would be able to do this, such as using water or other solvents, or through other means to separate out the microcrystalline cellulose from the amino acids in the powder mixture?

Can you centrifuge it?

https://www.nature.com/articles/s41598-020-68642-7

I would try to use a 5-10 cm layer of fine SiO2 particles above a slow filter in a Buchner funnel. I'd be surprised if it didn't do the trick.
Consider that the cellulose particles you see after filtration could come from the paper filter as well (I sometimes see such tiny tiny particles floating around after filtration), in which case it may help to submerge the filter in a beaker of hot water before use, and in the worst case, carry out the SiO2 filtration AFTER the filter paper (in a column with a fritted disk for instance, above larger sized SiO2 not to waste the fritted glass with tiny particles).


text:
Do pipettes always outclass glass syringes?
I was previously working in a lab preparing lots of samples for gas chromatography that exclusively used glass pipettes and was able to get really high precision between replicates. Now I'm working on my own project doing a lot of liquid dilutions that need to be very accurate with an older pipette that I am looking to potentially replace for better results. I'm specifically looking at 1 mL volumes.
Tolerances for glass syringes seem to be about +/-1% (not as much info), where pipettes are +/-0.6% accuracy and +/-0.2% precision. There were definitely some other reasons to use syringes vs pipettes at my old lab, but are syringes really always outclassed by pipettes? Can an experienced syringe operator beat a pipette?
I realize tip choice, needle, calibration, and ambient temperature controls probably have larger effects than these differences (though seemingly favor syringes), so I am wondering more about the relative optimal performance of each.

If evaporation is not a significant influence (depending on solvent's boiling temperature / concentration / exposed liquid area, temperature around, etc.) you might consider to determine the $\Delta{}m$ of tare of the syringe and filled syringe to substitute volume by mass and gravity.

Always? It is difficult to use a pipette to inject into a port for a GC, or GC/MS.

Glass syringes and glass pipettes are high precision and high accuracy tools of an analytical chemist. Like any tool at home, such as a cloth cutting scissor and a plant pruning clipper do the same job of cutting but their applications are different.
As an analytical chemist, I have not seen anyone using microsyringes for standard preparations in volumetric flasks. One uses a class A pipette for volumetric transfers and to make different standards.
Now you wish inject known microliter or nanoliter volumes in a GC injection port or an HPLC rheodyne, can we use glass pipettes? You would choose a glass syringe to pierce the GC septum.
The tip is that you would use the highest quality glass pipette (class A) and highest quality syringes in each case. class A 1 mL pipettes are commercially available.
If you really wish to compare, you would have to do a calibration yourself on a analytical balance under tightly controlled conditions.


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What is the main contributor of the color of honey?
I was told that the color of honey contains so much chemistry and that was interesting to me and after doing a couple of research I found that the color of honey depends on the presence of plant pigments including carotene, xanthophylls, anthocyanins, flavonoids, polyphenols, as well as amino acids and mineral salts. So as we can see there are many factors to the color of honey but my question is which out of all of the listed above is the main contributor and is the most affected by the increase in temperature (for example for acidity the main contributor is gluconic acid).

Barring your question being seen by someone with expertise in the matter of honey compositions, this appears to be too broad to get an answer.

Didn't you answer the question yourself based on your own findings: color of honey depends on the presence of plant pigments including carotene, xanthophylls, anthocyanins, flavonoids, polyphenols, as well as amino acids and mineral salts. This paper should suffice for your high school project. https://www.tandfonline.com/doi/abs/10.1080/0005772X.1969.11097248

As I said in your previous posts, these questions are PhD level projects. Do you know how components are there in chocolate? Literally thousands. Natural products are very complicated.

I totally agree with you @M.Farooq but there is no specific one that is mostly affected by the temperature?

The paper is here which discusses color deterioration with temperature. Again it is dependent on honey source. It is related to HMF production. Read it https://www.tandfonline.com/doi/abs/10.1080/00218839.1967.11100165

Years ago, the area I lived in had a wonderful independent small honey business, selling more than a dozen specific types of honey (sourced from different flowers). They had a wonderful range of colors specific to the type of flower. So, clearly the plant origins had a major impact. But, since they probably innumerable molecules that resulted in a given color would all age differently under temperature, attributing any color change to a few specific molecules seems impossible.

I appreciate all your valuable comments and @M.Farooq, I don't think I can access the article do I have to log in?

Also if anyone has any articles or journals that involve honey's acidity and color that would be very helpful and thanks again to everyone that tried to help.

You can investigate the dependence of the color of the different micronutrients you list on pH. Consider the sugars in honey as representing a relatively inert background. Consider also that the pH of honey does not vary all that much about its median value.

I found that the color of honey depends on the presence of plant
pigments including carotene, xanthophylls, anthocyanins, flavonoids,
polyphenols, as well as amino acids and mineral salts

Yes, you are right about that. The color differences is due to the difference in levels of minerals, anti-oxidants, and other trace elements that are present. In particular, levels of polyphenols, flavonoids, and carotenoid have been observed to have high correlation with color. Polyphenols are found in high concentration in darker honeys, whereas flavonoids are found in higher concentrations in lighter honeys.
References

Antioxidant and antimicrobial capacity of several monofloral Cuban honeys and their correlation with color, polyphenol content and other chemical compounds by Jose M.Alvarez-Suarez Sara Tulipani, Daimy Díaz, Yadiley Estevez, Stefania Romandini, Francesca Giampieri, Elisabetta Damiani, Paola Astolfi, Stefano Bompadre, Maurizio Battino, Volume 48, Issues 8–9, August–September 2010, Pages 2490-2499, DOI: https://doi.org/10.1016/j.fct.2010.06.021
Total Phenol, Antioxidant Activity, Flavonoids, Anthocyanins and Color of Honey as Affected by Floral Origin Found in the Arid and Semiarid Mediterranean Areas by
Taha M. Rababah  Mohamad Al‐Omoush  Susan Brewer  Mohammad Alhamad  Wade Yang  Mohammad Alrababah  Abd Al‐Majeed Al‐Ghzawi  Muhammad Al‐ u'datt  Khalil Ereifej  Fawzi Alsheyab  Ranya Esoh  Ali Almajwal, Volume38, Issue3, June 2014, Pages 1119-1128, DOI: 10.1111/jfpp.12071

Thank you very much, Nilay Ghosh, for the answer but do you mind by any chance sharing the references with me because I cannot access them.

@MoKamal I also do not have access to these papers. I only cited them for reference. If you like the answer, make sure to accept it(click the checkbox). It will be appreciated.


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What is the purpose of subtracting the blank when constructing an analytical calibration curve?
When building a calibration curve for quantification of some analyte using any spectroscopic technique, what is the purpose of subtracting the intensity of the blank from the intensity of each point of the curve?
I can build two calibration curves using the same measurements, one plotting the total intensity recorded at each point (F) and the other plotting the total intensity recorded minus the intensity when the concentration of the analyte is zero (F-F0), both against analyte concentration.
When making a linear regression for each curve, the slope will be the same and the difference between the intercepts will be the intensity of the blank (F0). If I try to quantify an analyte in a given sample with a different matrix than the one used in the standard (different solvent, for instance) maybe the "blank of the sample" so to speak would be different than the blank of the standards (F0). In that case, the result would be equally wrong, using either of the curves, because I would either be comparing the response of the sample-F0 to analytic curve based on F-F0 or the total response of the sample to analytic curve based on F.
It would make sense to me if I could, when analyzing an unknown sample, measure an exact blank of the sample, and use this measurement to compare with the analytical curve built using F-F0. Then I would be comparing two intensities associated only with the analyte, but that is't possible.
Am I missing something? Does subtracting the blank helps in any way when the objective is to quantify the analyte in samples?
I've checked two different undergraduate level analytical chemistry books and found no answer. They just say it is done this way.

(+1) for a perceptive question! You do not have to subtract the blank for each point on the curve. The analytical blank, and how it should be estimated and dealt with, is an unexpectedly subtle issue that has eluded the determined harmonization efforts of ISO and IUPAC for years. I recommend looking at these four publications: 1. L.A. Currie, “Detection: International update, and some emerging di-lemmas involving calibration, the blank, and multiple detection decisions”, Chemom. Intell. Lab. Syst. 37 (1997) 151-181. Continued next comment.

2. L.A. Currie, for IUPAC, “Nomenclature in evaluation of analytical methods including detection and quantification capabilities”, Pure Appl. Chem. 67 (1995) 1699-1723.  IUPAC © 1995.     3. K. Danzer, L.A. Currie, for IUPAC, “Guidelines for Calibration in Analytical Chemistry”, Pure Appl. Chem. 70 (1998) 993-1014. IUPAC ©1998.      4. ISO 11843-2, “Capability of Detection - Part 2: Methodology in the linear calibration case” ISO, Genève, 2000.

Thank you very much for your comment, @EdV. I will be checking these publications as soon as possible.

Good to see interesting analytical chemistry questions after a long time.
I can give a qualitative example. Suppose you wish to collect the spectrum of a star. What time would be the best to carry out the measurement? You would agree that night time would be the best, because constant unwanted sunlight background is missing at the time of measurements. Subtraction of the blank serves a similar purpose, you wish to remove the background signal from your analytical signal so that you can assess the magnitude of your analytical signal. The blank also helps to serve to establish the minimum concentration that can be detected reliably. This is why blank subtraction is carried out
One of the commentators on your question has written a whole book on this subject.
You did not mention the analytical instrument you are using, it would have helped to discuss more. Anyway, calibration curve is another story. In this case, you certainly can make two calibration curves, and you are very correct, their slopes are the same. The sample concentration will turn out to be the same from both.
With blank correction, your equation should have a form y= mx with negligible intercept.
Without blank correction, your equation has the form y=mx+c, where c is the intercept, the average blank reading.
I may ask another thought provoking question which is certainly not discussed anywhere.
For example, what if I square all the signal intensity and then plot a calibration curve as a function of concentration. The calibration curve will still remain linear.
a) What will happen to the magnitude of slope?
b) Will the calibration curve sensitivity increase? "Apparently, yes!"
c) If the sensitivity increases by this simple mathematical manipulation, why shouldn't we do it?
Coming to the second part of the question:

If I try to quantify an analyte in a given sample with a different
matrix than the one used in the standard (different solvent, for
instance) maybe the "blank of the sample" so to speak would be
different than the blank of the standards (F0).

In this case, you should not/ perhaps cannot use a simple calibration curve. One of the key assumptions of a calibration curve is that the matrix of the sample is the same as the standard. If you violate this rule, then we cannot apply this technique.
Consequently, you will have to use a technique called Standard Addition Method. It produces very accurate results, when the sample matrix is significantly different from the standards.

"if I square all the signal intensity and then plot a calibration curve as a function of concentration. The calibration curve will still remain linear" - hm.. did you mean a function of concentration$^2$? "night time would be the best, because constant unwanted sunlight background is missing" - but what OP is asking is about "measure the signal w/o star, then measuring star signal during day, then subtract one from another" - which removes the systematic error, but it also increases sd(noise). So while it allows replacing $mx+c$ with $mx$ it seem to also decrease the precision.

Could you please elaborate on how blank subtraction helps establishing the minimum detectable concentration?