text:
Which of the following has a stronger hydrogen bonding: F-H- -O or F-H- -F?
I thought of two approaches to solve this question . Since the electronegativity of Oxygen is lower than Fluorine it would be a better donor than Fluorine to the electron deficient Hydrogen . Another approach was since the ∆- charge of fluorine would be more than oxygen it would attract the ∆+ charge on hydrogen leading to more stability . I could not decide which approach was right . I saw this question in a book in which the correct ans was given as F-H- -F

F---H---F is stronger. This is an example of hydrogen bonding. More the partial charges generated on each atom better the interaction. The hydrogen bond is an effect. The cause is the high seperation of partial charge, which is maximum in H---F bond. Similarly comparing corresponding say O-H and and F-H the latter will be more polar leading to stronger interaction.

@AdityaRoychowdhury Hydrogen bonding is not just dipolar interaction. I think the answer might be a bit of a matter of opinion if you have to answer it with textbook knowledge.

https://www.mdpi.com/2073-4352/6/1/3/pdf This free paper might help.. It might take some time to find what you want, I haven't gone through it properly

I do not understand this question. Are we talking about $\ce{[FHF]-}$ and $\ce{[FHO]-}$? I wouldn't expect the latter to be stable at all, and I would not want to guess (if it was stable) which one the hydrogen bond was, where the radical density is located, etc., and the overall bonding situation. If this is an exercise from a book, please quote it verbatim (in the body) and cite the source. In general: $\ce{[FHF]-}$ is often used as a benchmark case for the strongest hydrogen bond. See Do symmetric hydrogen bonds in neutral molecules exist?

Naturally related: Is KHF2 an ionic compound or a covalent compound?

Experimentally determined , the $\ce{F^-\cdots HF}$ bond has energy 163 kJ/mol and $\ce{F^-\cdots HOH}$, 96 kJ/mol , according to Jeffery 'Introduction to Hydrogen Bonding', (OUP 1997).

A hydrogen bond is defined as the attractive interaction between an X–H unit (where the electronegativity of X is greater than of H) and a donor atom, or a Lewis base (B).
Hydrogen bond strength is a combination of factors:

An electrostatic contribution
The formation of a covalent interaction (charge transfer from B to H–X)
Dispersion forces

The electrostatic component (the attraction of a bonded hydrogen atom to a region of relatively high electron density of the base (B)) is considered the dominant contribution to most hydrogen bonds.
Regarding the X–H unit, the greater the electronegativity of X, the stronger is the hydrogen bonding donor (higher hydrogen bonding acidity).
However, regarding the donor atoms, electronegativity influences two factors that go in opposite directions: greater electronegativity = stronger dipole (stronger electrostatic interaction), but smaller charge transfer (less electron pair donation and less covalent bonding).
Gilli developed a pKa equalization approach for predicting the strength of hydrogen bonds in aqueous solution. It tries to rationalize, for instance, the fact that a water-water hydrogen bond is six times weaker than a water-hydronium hydrogen bond, despite both being O–H···O hydrogen bonds. According to this approach, a $\Delta$pKa value should be calculated to estimate the strength of the hydrogen bond
$\Delta$pKa = (XH···B) = pKa(XH) - pKa(BH+)
The closer $\Delta$pKa is to 0, the stronger the predicted hydrogen bond in aqueous solution. Therefore, considering:
$$\begin{array}{c|c|c|}
\text{Species} & {pK_a} \\ \hline
\text{HF} & \text{3.15} \\ \hline
{H_3O^+} & \text{-1.7} \\ \hline
{H_2O} & \text{14} \\ \hline
\end{array}$$
$\Delta$pKa(H-F···F-) = 3.2 - (3.2) = 0 (strong)
$\Delta$pKa(H-F···HOH) = 3.2 - (-1.7) = 4.9 (medium strong)
$\Delta$pKa(H-F···HO-) = 3.2 - (14) = -10.8 (medium strong)
Therefore, regarding your question, it seems that F- would form a stronger hydrogen bonding interaction with HF due to a strong electrostatic contribution, even though hydroxide is a better donor than fluoride.
Source:
Inorganic Chemistry 5th ed - Miessler, Fischer, Tarr
Please note that in neutral organic molecules, basicity outweighs the dipolar contribution; therefore, amines are the strongest hydrogen bond acceptors (or electron pair donors), followed by alcohols and fluoro-compounds
More on this: http://fbdd-lit.blogspot.com/2017/11/hydrogen-bonding-and-electroengativity.html

Such analysis may have some value, but neither bifluoride is an usual h-bonded entity nor hydrofluoric acid really weak.

https://chemistry.stackexchange.com/questions/34818/is-hf-the-least-acidic-hydrogen-halide/34829#34829


text:
Proton NMR of tert-amyl alcohol
I just started learning about proton NMR.
According to Molbase, the HNMR data for tert-amyl alcohol (2-methylbutan-1-ol) shows four kinds of protons at 0.9 ppm, 1.24 ppm, 1.44, and 3.65 ppm.

However, it seems contradictory to what I learnt, since the protons on secondary carbon (1.44 ppm) are downfield to the protons on tertiary carbon (1.24 ppm), which confuses me, since I learnt that protons on tertiary carbons experience more de-shielding.
Is there something I have missed out in my thought process?
EDIT: Thanks to @orthocresol and @Buttonwood for pointing out my error. My revised question would then be:
Between the two types of protons on secondary carbons (1.24 and 1.44 ppm), why is it that the one at 1.44 ppm is more deshielded?
From what I understand, it is because carbon is more electronegative than hydrogen, therefore causing the deshielding effect, but at the same time from organic chemistry I learnt that alkyl groups are electron donating, which should cause shielding. Is there some misconception in my thinking?

(1) Since you're just starting, it's a great time for you to unlearn the terms "downfield" and "upfield". These are ancient terms which come from the era of continuous-wave NMR and for some reason have not been erased from the literature yet. Stick to "deshielded" and "shielded"; they're much more easily understood. (2) In tert-amyl alcohol, I don't see any protons attached to tertiary carbons; only a secondary carbon, and two different types of methyl carbons. Could you clarify?

In addition to @orthocresol's comment, your reference accounts for four different types of protons.  Both in the visual representation as well as in the molecular structure-like one, there is the one shown for 3.65 ppm (which, depending on the experimental circumstances, either a) equally could be recorded at a different position, or b) could be invisible).

As a beginner to NMR, you should be aware of how important integration values in $\mathrm{^1H-NMR}$. If you knew and had them in your NMR, you could have easily identify these peaks.

@orthocresol ,Buttonwood Hi, I have updated the question, would it be possible for you to re-look at it

@MathewMahindaratne Hi, yes I understand the utility of such integration values, but here I was wondering about the explanation for the chemical shifts of these peaks, hope this clears things up :)

You can just remove the older part of your question.

I recommend you add to your question a list with all peaks, their integral, and the group it belongs to. The description in your text is not correct, the molecule has only one secondary carbon atom.

There are only two kinds of protons: those are on primary carbon(s) and those on secondary carbon.

Let's be careful when we talk about "kinds," since that's not, to my knowledge, a standard term. There are 12 hydrogen atoms, and these are in chemically equivalent groups of 3, 2, 6, and 1. In principle these groups have different shifts, but in practice, the peaks may overlap or be so broad as to be difficult to see in a spectrum.

tert.-Amyl alcohol is 2-methylbutan-2-ol not 2-methylbutan-1-ol.

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White snow-like powder below sodium hydroxide solution
I have created sodium hydroxide solution with half cell brine electrolysis but, finally I got white snow-like insoluble powder sank below sodium hydroxide.
What should I do keep or remove it?

Sodium hydroxide is easily contaminated with sodium carbonate. The free base continuously reacts with atmospheric carbon dioxide to form the salt.When a relatively concentrated solution of NaOH is prepared, sodium carbonate slowly precipitates out as a white gelatinous mass.
a) How long did you expose it to air?
b) What is the expected concentration of the base?
c) What are your electrodes made of?

a) over 10 hour with 24 volts  b) to be 35% c) my electrode are made from graphite(carbon)

Okay, that is likely to be sodium carbonate.

what should i do make new one or filter sodium carbonate

Sorry for delay, it may take a while to sink                     here is the picture https://photos.app.goo.gl/SRccbZpoPZg1gp5f6

If your water is hard, it could be carbonate of calcium, otherwise it could be sodium carbonate. What is the pH?

the water has almost 7.4 ph

Another possibility:  the brine may have impurities involving salts of other metals.  Calcium salts could produce a precipitate of calcium carbonate or calcium hydroxide, the latter rendered poorly soluble in a concentrated alkali solution.  Magnesium and most transition metals are likely to produce hydroxide precipitates, with magnesium hydroxide in particular also being white like the observed precipitate.

but I am using Non-Iodized pure sodium chloride and distilled water , mix it and i got brine solution

(+1) Per my EDITED answer,  I am in agreement with Oscar.

For the record, I have witnessed the described fluffy whitish salt in a galvanic cell with iron and carbon electrodes in a mix of MgSO4 and H2O2 (acidified with phosphoric acid). My best guess is that it is a basic salt of magnesium.
In the current context of electrolysis of  brine, per a source (click on the article, Hypochlorite Generation from a Water Softener Spent Brine), to quote from Page 6-7:

It is widely known that in the alkali-chlorine generation process the brine quality is the main factor in achieving an efficient process. The challenge of this process was to treat spent brine with high concentrations mainly of calcium, magnesium, sulfate, and alkalinity. The brine generated at stage 3 presented an average amount of 2.7% of sodium chloride (Figure 1) and was characterized in terms of calcium, magnesium, and total hardness.

Also, the solubility of CO2 is much reduced in salt water, so I offer an alternative proposition of a basic salt of magnesium, and less likely, given the textual nature of the precipitate, a calcium salt.
[EDIT] Just saw an alternate opinion by Oscar Lanzi, who also suggests a Mg salt, albeit I am not as definite on its identity.
Also, saw the response that the starting NaCl was table salt, which can contain an anti-caking agent. Per a source:

The most widely used anticaking agents include the stearates of calcium and magnesium, silica and various silicates, talc, as well as flour and starch.

So, a basic magnesium salt or possibly calcium still remains a possibility.

so what should i do

Try repeating using crystals of Rock Salt and see if you get a cleaner product, and I agree, it is wise to limit air exposure. However, for most applications of NaOH, likely not material, so just decant to remove insoluble residue.

thanks for your support i am making a new one , but if i get it again it will contaminate the  sodium hydroxide solution or just mixtrure

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text:
Internal energy for Phase Transitions
Can we justify that "For sublimation of a solid at 1 atm  $\Delta U>0$ at low temperature and $\Delta U<0$ at high temperature?"
I read several answers and came to know that $U=f(T)$ is valid only for ideal gases at $\pu{1 atm}$ pressure where $T$ is temperature.
Internal energy is the sum of kinetic energy of molecules and chemical potential energy of molecules.
For phase transition even if the temperature remains constant, its internal energy changes.
I tried  to justify  by considering that while sublimation as solid changes to gases its degrees of freedom increases and we may say internal energy is a function of degrees of freedom so  as temperature increases so $U_2 > U_1$ so $\Delta U >0$ but this contradicts what we have to prove.
Also similarly what can  we say about internal energy for other phase transitions ?

Answer given is: ABCD
source:Physical Chemistry (Vol. II) (Class XI) Chemistry Module For JEE (Main+Advanced) Paperback – 1 January 2018
by Resonance Eduventures Limited

Can we justify that "For sublimation of a solid at 1 atm  $\Delta U>0$ at low temperature and $\Delta U<0$ at high temperature?"

No.
$\Delta U>0$, always, for sublimation, because of the energy needed to separate the atoms or molecules in changing from the solid to the gas phase.
As for the enthalpy,
$$H = U +PV \Rightarrow \Delta H = \Delta U+\Delta (PV)=\Delta U+ P\Delta V+ V \Delta P$$
Thus at constant pressure:
$$\Delta H = \Delta U+ P\Delta V \approx  \Delta U+ nRT$$
[Here I've used the ideal gas law to approximate the volume of the gas, and ignored the volume of the solid, which (at 1 atm, room temperature) is about 3 orders of magnitude smaller.]
I.e., $\Delta H$ will be even more positive than $\Delta U$, because of the pV-work required to make space for the gas.
The reason why sublimation becomes more favorable as the temperature increases is because sublimation has a positive $\Delta S$, and $\Delta G = \Delta H - T \Delta S$.  But, irrespective of temperature, at any temperature and pressure at which the solid can exist,  $\Delta U_{solid->gas} >0 $.
$\Delta U >0$ for liquid->gas and solid->liquid phase transitions as well (with the possible exception of the solid->liquid phase transitions for helium-3 and helium-4 at extremely low temperatures which, at least according to https://en.wikipedia.org/wiki/Enthalpy_of_fusion, have $\Delta H < 0$; but solid helium doesn't exist at the 1 atm pressure specified by the OP).
*Yes, you might be able to come up some extreme hypothetical mechanical system where the solid is under such great pressure that the intermolecular forces are so far into the repulsive part of their potentials that $\Delta U_{solid->gas} <0 $.  But that's clearly not what the OP had in mind, since s/he specified 1 atm.  And, in addition, such an arrangement would require that only the solid, but not the gas, be at that extreme pressure.  So, strictly speaking, that $\Delta U$ would not be for the sublimation alone, it would be for the sublimation plus the pressure change.

btw I got this question in my textbook which was a multiple choice question which said "for sublimation of a solid at 1 atm,which of the following may be correct" and options    $\Delta U >0$ at low temperature,$\Delta U < 0$ at high temperature were given correct,so did they necessarily meant extreme hypothetical conditions?

@user69608  Can you give a complete reference to the textbook, and a screenshot of the problem and answer?

edited the question

Can we justify that "For sublimation of a solid at 1 atm  $\Delta U>0$ at low temperature and $\Delta U<0$ at high temperature?"

A justification can be suggested by writing
$$\Delta_{\mathrm{sub}} U = \Delta _{\mathrm{sub}} H - RTn$$
(but in retrospect it has a flaw, see below). If you assume the enthalpy of sublimation is approximately constant (weakly dependent on T) then it is clear that to sublimate a constant amount of substance n requires an input of energy at low T (in the form of input heat $q_p = \Delta _{\mathrm{sub}} H$ to break bonds in the solid lattice) but results in a net reduction in internal energy of the system (due to work done on expansion) at sufficiently high T.
As hinted at in another answer, however, this justification has one flaw*. The flaw is that (in the constant pressure scenario) heat input provides energy not only to break bonds, but also to drive the expansion of the gas (perform work). In the isochoric (constant volume) scenario there is no work and $\Delta U = q_V$.  In the isobaric (constant pressure) scenario the sum of the additional energy input as heat (>0) to perform expansion work and energy loss due to the expansion work (<0) cancel, so these changes do not alter $\Delta U$.
It should be noted that $\Delta _{\mathrm{sub}} H>0$ (sublimation is endothermic).
It is interesting to note that in the case of vaporization the enthalpy drops to zero as T nears the critical point. Similarly the expansion work required drops to zero near the critical point.
*I thought there was an additional flaw, but haven't figured it out entirely.

What if for a solid  at high temperature $\Delta  H$ is sufficiently greater than  $RTn$ then $\Delta U$ would come positive which contradicts.?

Can you answer the  other questions in the post?

You can argue along the same lines for the liquid-vapor transition. In that case the $\Delta H$ also decreases with increasing T because of the relative heat capacities of the two phases. For the liquid-solid transition it is a little less obvious what to expect. The pV expansion term for melting is much smaller than for vaporization or sublimation. It can be negative (eg water) or positive (most pure substances). The pV term is presumably dominated by the enthalpy term which increases with increasing T (because the liquid has the greater heat capacity).

And why degrees of freedom thing  fails?

As your question shows, at low T condensed phases are lower energy states then vapor phases, but they are also lower in entropy (I suppose that's what you mean by degrees of freedom, the number of microstates available to the system, which is usually described using the entropy).

@BuckThorn  This is incorrect.  The extra energy needed to do pV-expansion work at higher temperature does not reduce $\Delta U$, which is the amount of energy needed to separate the molecules plus their increase in kinetic energy due to the increased degrees of freedom as they go from solid to gas. Rather, the expansion work is added to $\Delta U$, which is itself always positive for sublimation.

@theorist during an expansion the system does work, so the change in energy is negative

@BuckThorn There are two issues here.  The first is $\Delta H$ vs $\Delta U$: This is a sublimation, so it's at constant T.  Imagine an endothermic process (reaction, phase change, doesn't matter) at constant T & V.  q_v = $\Delta U$. Now suppose we switch to constant p, and further suppose the reactant is a solid and the product is a gas.  More heat (q_p) will need to flow into the reactant mixture (equal to $p \Delta V$) because now we additionally need to do the work to expand the system. Look at the math:  $\Delta H = \Delta U + p \Delta V$.  The pV term is positive, not negative.

But the first issue ($\Delta H$ vs $\Delta U$) is incidental to the question, which is what is happening with $\Delta U$.  Think about it physically -- you are pulling molecules apart.  $\Delta U$ can't be negative, irrespective of whether you have to add or subtract pV to get $\Delta H$ (though you do have to add pV:  $\Delta H = \Delta U + p \Delta V$).   The pV term only affects  $\Delta H$.  It has no effect on $Delta U$.

@theorist Well, I've come around to agreeing that the "justification" in my answer has a flaw, and you point in the right direction, but I think you miss that $\Delta U = q + w$ so that w very much affects $\Delta U$. Looking at $\Delta U = q_v$ is an interesting alternative but difficult to imagine: how do you transition from liquid to gas without providing molecules with more room?

BTW I think that's what happens at the critical point (there is no change in density so no work is done).

We're doing this at constant temperature, so any pV work required by doing this at constant p is cancelled out by increased heat flow into the system, leaving $\Delta U$ unaffected by the pV work.  That increased heat flow due to pV work at constant p accounts for nearly all the difference between $\Delta H$ and $\Delta U$*. [*There will be a slight effect on$\Delta U$ from the p-dependence of the internal energy of the solid, but that is small.]
And you can carry out sublimation at constant V by putting  a solid into an evacuated container.

But the most important thing is to think physically about $\Delta U$ --think of these intermolecular associations that hold the molecules in a solid together as potential wells, and you are bringing the molecules out of those potential wells as you separate them into the gas phase. So the internal energy can't go down when you do this (assuming this is purely a phase transition, and you aren't changing the identity of the molecules themselves).

@theorist the question asks for a justification for or against the proposition, and the choice remains whether to use a purely thermodynamic argument or to appeal to questions about microscopic behavior. While you can unquestionably argue based on stat mech (number of microstates etc) I like that thermodynamics can provide a justification. I admit that my initial one was misleading but it's interesting why. The reason is that $\Delta U<0$ would violate the second law.

Your initial justification wasn't "misleading", it was simply incorrect.  I believe characterizing it as "misleading" obfuscates things, making it confusing for other readers to follow this.  And I'm afraid your new justification is also incorrect.  This has nothing to do with the 2nd law. The fundamental flaw in your argument is that you "assume the enthalpy of sublimation is approximately constant (weakly dependent on T)", and then proceed to infer changes in $\Delta U$ based on a changing (with temp) value for $nRT$ subtracted from an approximately constant (with T) $\Delta H$....

....But this is completely backwards!  The $nRT$ term is a pV-work term, and thus affects $\Delta H$, not $\Delta U$!  Hence it makes no sense to talk about changing T as having an effect via an $nRT$ term, and then to say $\Delta H$ is approximately constant with T.  Again, your entire argument is based on the $nRT$ creating a temperature-dependence in $\Delta U$, but that temperature-dependence is in $\Delta H$, not $\Delta U$.  I.e., you are creating a fallacious T-dependence in $\Delta U$.

@theorist You bring up some interesting points worth taking into consideration. Thanks for your helpful input!

Perhaps we should ask @ChetMiller for arbitration. I know my original justification was flawed (but your arguments against it do not seem quite right either) and I am still working to construct a cycle to show that the second law is violated but it's dawning on me that this will be a challenge. Your own justification is partly based on a hand-wavy molecular description ("degrees of freedom"?) which is intuitively satisfying and yet not rigorous. Chet displays a clear mind about these problems and has broad experience based on which to judge. Maybe he can be of help?

You are right that my degrees of freedom argument was not a good one, so I've deleted it.  I will endeavour to come up with a better one, based on the temperature-dependence of $\Delta U$ for sublimation, perhaps this weekend.  I sometimes disagree with Chet Miller on thermodynamics, so I would not view him as an acceptable arbitrator.  I might suggest Karsten Theis, but let's hold off until I work on this more.


text:
Ion Representations
Some books put dots around ions, enclose them in braces and then put an integer over them to signify the negative charge, in such representations, the integer is only a way to represent the charge and is not supposed to be added with the other valance electrons when calculating the total no. of electrons, right?
Eg: O with 8 dots around it, in braces with -2 over it, means that ion has only 10 electrons and not 12, right?

Read up about Lewis Structures. That may be what you are looking for. the dots signify the electrons.

The neutral oxygen atom has 8 electrons, 2 on the inner shell K, and 6 on the outer shell L (or valence shell). So this neutral atom is represented by the letter O with six dots around it. It is possible to add two more electrons. This makes the oxide ion $\ce{O^{2-}}$. This ion can be represented with all its outer electrons, with 8 dots. This symbol $\ce{O}$ with its eight dots is enclosed in braces, before adding the total charge (2-) over it.


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Potential energy difference with different force fields
I am modeling a simple 2D Graphitic Carbon Nitride (g-C3N4) nanosheet with 54 atoms. I used several force fields with both LAMMPS and GROMACS and the potential energies of the system in npt at 1K and 1 atmosphere are:

LAMMPS using ffield.reax.CHOSFCl = -6,024 Kcal/mol
LAMMPS using tersoff = -7,148 Kcal/mol
LAMMPS using ATB server - gromos54a7 forcefield = -270 Kcal/mol
GROMACS: using ATB server - gromos54a7 forcefield = -228 Kcal/mol

We know that the zero for potential energy is not the same for all the force fields but is it usual to be an order of magnitude difference between REAX and GROMOS force fields?
P.S. All the force fields were recommended and used by the literature for the g-C3N4.

I would be less interested in the absolute energies and more interested in a relative energy. Do you have any relative energies benchmarked against previous computational studies?

You might get a better answer asking this in https://mattermodeling.stackexchange.com/


text:
Electrolysis with baking soda solution, a penny as the cathode and graphite for the anode
I set up an electrolysis experiment to show my kids and wanted to explain to them what is happening.
My setup has a penny as the anode and graphite for the cathode. The solution is water with baking soda mixed in.
I see bubbles at each of the electrodes which I assume are hydrogen and oxygen bubbling out. I also see some cloudy substance coming off the penny, which I assume is some sort of copper compound, but I'm not sure.
Can someone help me figure out what's going on in the setup at the two electrodes?
Edit : corrected the electrode descriptions. I earlier stated penny as the cathode and graphite as the anode.

For a demo like this, one should rather use graphite electrodes. There is a reason why reactive metals are not employed in electrochemistry as electrodes.

What voltage did you apply?

I had a 9V battery attached.

Yes, I agree you are getting hydrogen an oxygen as you expected.
However, the copper anode is undergoing anodic corrosion and forming CuO (which is characteristically black) and possibly also some Cu2O (yellow or red, depending on the size of the particles, where the color combination could be described as orange).
Now, in the presence of HCO3- and O2, a naturally occurring mixed salt, copper patina (which forms on the copper roofs of houses) could also be forming. Wikipedia on patina, to quote:

patina is a coating of various chemical compounds such as oxides, carbonates,... formed on the surface during exposure to atmospheric elements (oxygen, rain, acid rain, carbon dioxide,...

Patina has been described as a blueish-greenish naturally forming on a copper surface (here, your copper penny electrode).
A closely related version of your electrolysis experiment has been previously discussed elsewhere.

The OP said that a graphite anode was used; the copper penny was the cathode. "Cloudy" substance coming off the penny was not described as colored - but perhaps so faint as to be undetectable. And yet, oxidation should not be occurring at the cathode.

The patina did show up after a while. I also noticed that the penny was pretty eaten into at the end. So I suspect that some of the copper made it into the solution - perhaps as the white cloudy stuff in the solution. Could this be a copper + carbonate compound? I also just realised that I had the electrodes backwards (copper anode and graphite cathode)


text:
Why are Tyrosine and Tryptophan considered hydrophobic?
Since Tyrosine and Tryptophan are amino acids, their polarity is determined on their side chains or R groups. If their R groups are polar, the amino acid is polar. Both Tyrosine and Tryptophan are listed as non-polar molecules. However, Tryptophan has nitrogen in its side chain and Tyrosine has oxygen, both of which have a high electronegativity compared to the carbon that makes up the bulk of both molecules, which is something commonly seen in polar amino acids like Serine, which has oxygen in its side chain. This should mean that Tyrosine and Tryptophan are polar, so why are they non-polar, even with an electronegative atom in their side chains?

Unlike that in organic chemistry, polar and non-polar compounds determination is done differently in biochemistry, probably because most biochemical reactions are done in aqueous medium. Therefore, the classification of them as polar and non-polar is somewhat different from source to source. In general, amino acids, which can make hydrogen bonding in aqueous solutions are called polar, and ones repel from water are non-polar. According to Ref.1:

All polar amino acids have either an $\ce{OH}$ or $\ce{NH2}$ group (when in aqueous environment), and can therefore make hydrogen bonds with other suitable groups.

Following image shows 20 amino acids categorized to polar (10) and non-polar groups (10):

All 10 non-polar amino acids are hydrophobic and 10 polar amino acids are hydrophilic. Polar amino acid residues have a tendency to be on the outside of a protein, due to the hydrophilic properties of the side chain (Ref.1). Keep in mind that tryptophan has indole function, but its lone pair of nitrogen is involved in the aromatic system. Thus, it make only weal H-bonding, which are not good enough to categorize as "polar."
References:

Bruce Alberts, Alexander Johnson, Julian Lewis, Martin Raff, Keith Roberts, Peter Walter, In Molecular Biology of the Cell; 5th edition, Garland Science: New York, NY, 2008, p. 125-129 (ISBN-13: 978-0815341055).

Tryptophan is less hydrophobic than phenylalanine and as a whole tryptophan is considered non polar although it has an additional N-H group at its side chain.. Look at the side chain of tryptophan which is more hydrophobic and large so if a tiny part (N-H) in its side chain is polar than still it can not influence the hydrophobicity of the entire side chain which is so large.. and secondly, although it has N-H group at its side chain so we accept that it will offer the hydrogen ion their lone pair electron to get charged but that is not happing actually ..because the lone pair electrons of tryptophan are busy in the resonance of the two fused ring so these electrons do not even try to look at the hydrogen ions in water so its neutral


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Why doesn't propynal give cannizzaro reaction?
Cannizzaro reaction is a disproportionation reaction which is given by aldehydes and ketones having no α-H, forming alcohol (reduction) and carboxylic acid (oxidation) as its products. The compound, Propynal has no α-H and it should be giving Cannizzaro reaction but, my textbook says that the reaction is not given by Propynal. Can someone please explain to me why Propynal doesn't give Cannizzaro reaction?

My suspicion is that while it has no alpha hydrogens, it does contain an acidic alkyne C-H proton that can be removed by base to produce an anionic conjugate base. This may thus hinder the Cannizzaro reaction.

@Tan Yong Boon:  You are correct. That was my first instinct as well. Acidities of enolizable hydrogens and terminal alkknyl hydrogens are compatible.

I think this is assisted by the electron-withdrawing carbonyl group making the alkyne-H more acidic

The anion is not resonance stabllized. Is propynal recovered or does base destroy it? E.g., formation of malonaldehyde anion. The lack of formation of propargyl alcohol or propargylic acid may just mean other events occur.

It is known that terminal alkynes reacts$1$ with $\ce{NaNH2}$ and produces salt, much like an acid-base reaction, hence it is a fast reaction.
Furthermore, the aldehyde group is an EWG, which makes it easy to lose a proton, hence drives the reaction!
Whereas, cannizzaro reaction happens through multiple reversible steps as seen in the following image,

Therefore, acid-base reaction in this case has much higher probability than Cannizzaro reaction.

References
$1$: http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch09/ch9-2.html

Do you have a reference for the reaction of alkynes with NaOH?

@Waylander, I've added a reference for that

@Rahul Verma, NaOH doesn't react with terminal alkynes (your answer's first line), it is too weak to react with strong bases. Even the Reference you mentioned, has mentioned alkynes won't react with strong bases. Reference: https://chemistry.stackexchange.com/questions/8448/why-does-acetylene-having-an-acidic-proton-not-react-with-strong-bases-like-na

I confused NaNH2 with NaOH, @Waylander. This happens with me, often.

Yes @HariSriram, I didn't even noticed that.

@RahulVerma The proton of the terminal carbon reacts with NaOH and your premise is, the carbonyl groups enhances the acidity of the terminal carbon. Am I right?

Yes @HariSriram, you're right.


text:
Why I get acidic solution at brine anode?
I have decomposed $\ce{NaCl}$ solution (brine) with electrolysis and yield $\ce{NaOH}$ at cathode, at anode I got acidic while expected was $\ce{NaClO}$ (alkaline) if it is $\ce{HOCl + HCl}$. Where does sodium ion goes, and if it doesn't, then why?

The sodium ions go where the electric field drives them. How could they not to?

so the sodium ion goes to sodium hydroxide solution and make hypochlorus acid at the anode

Well, basically yes, that's about it.

what if i put the anode and cathode in the same beaker and the same brine solution

Then the separation will be smaller, and if you stir the solution, it won't develop at all.

i am feared why acting like bleach(sodium hypochlorite) it stains black cloths


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Chemical reaction of radicals with different multiplicity
I am studying the chemical reaction of radicals by computational chemistry software.
I just want to know what is the rules for multiplicity.
I have two radicals for which, the ground state of one is a doublet and the other is a triplet.
A chemical reaction between them is possible or not, if a chemical reaction is possible then in what multiplicity the reaction will proceed.
what multiplicity should I use in my input file?

There are no general rules. it depends on the nature of the radical.

With respect to the question "if a chemical reaction is possible or not", here is some background material on oxygen species.
In particular, Singlet oxygen apparently reacts with oxygen (also referred to as dioxygen) creating a very strong (but transient) oxidizing atomic oxygen.
Examples of possible reactions of interest, including atomic oxygen, can be found, per Equations (1) to (5), from a work discussing graphene patterning here: http://olab.physics.sjtu.edu.cn/papers/2017/29.Huan%20Yue_PCCP_2017.pdf , where apparently, O(3P) is created from severe collision quenching of O(1D) atom with air or oxygen and acts as the major oxidant in the work by Huan Yue and colleagues ‘Exploring the working mechanism of graphene patterning by magnetic-assisted UV ozonation’.
Note, O(3P) is also known as highly reactive ground-state 3P oxygen and a form of atomic oxygen.
Also be mindful, in the real world if any water vapor is present, Singlet oxygen reacts with H2O forming H2O2, where O(1D) has a half-life of around 45 minutes.


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Why active mass of solids and pure liquids taken as unity
There is a similar question already asked here and answers provided say that activity of solid is taken as unity because their density dosn't changes... but I still have some doubts in it.
Suppose we have a solid for example $\ce{NH4Cl}$ which decomposes into gases $\ce{NH3}$ and $\ce{HCl}$ then $K_{\mathrm{eq}} = [\ce{NH3}][\ce{HCl}]$.
Now if we take in a container which initially has no $\ce{HCl}$ and $\ce{NH3}$ and put some amount of $\ce{NH4Cl}_\mathrm{(s)}$ in it then after some time it will achieve equilibrium. Let the amount of solid left be $x$ mol. In equilibrium the  rate of production of $\ce{NH4Cl}$ is equal to the rate of its decomposition.
Question:
Now if we add another $x$ mol of $\ce{NH4Cl}_\mathrm{(s)}$ in the container then the rate of production of $\ce{NH3}$ and $\ce{HCl}$ should be double and hence would exceed the rate of production of $\ce{NH4Cl}_\mathrm{(s)}$ which will alter the equilibrium but the equilibrium constant equation predicts this would not be the case. Where is the mistake?

Related: https://chemistry.stackexchange.com/questions/14058/why-are-solids-and-liquids-not-included-in-the-equilibrium-constant-what-about

It will be good if someone can actually give explanation that why his argument (based on relative rates) is wrong, but not just telling why the correct concept is correct, unless contradiction between the two is apparent. I always have the same question too, why won't more gas produced as the equality between forward and backward rate got broken.

Maybe the following reasoning may help Param 1729. The thermal decomposition of $\ce{NH_4Cl}$ may be pedagogically seen as occurring in two steps. First, solid $\ce{NH_4Cl}$ is simply vaporized producing the hypothetical gaseous species $\ce{NH_4Cl(g)}$
$$\ce{NH_4Cl(s) <=> NH_4Cl(g)}\label{rxn:nh4cl}\tag{1}$$
And this transition is described by an equilibrium constant $\ce{K_1}$ which is similar to Henry's law
$$\ce{K_1 = \frac{p(NH_4Cl)(g)}{[NH_4Cl(s)]}}$$
Immediately after this vaporisation phenomena, a gas phase decomposition happens with the following equation :
$$\ce{NH_4Cl(g) <=> NH_3(g) + HCl(g)}\label{run:nh3+hcl}\tag{2}$$
with an equilibrium constant $\ce{K_2}$ defined by :
$$\ce{K_2 = \frac{p(NH_3)·p(HCl)}{p{(NH_4Cl)(g)}}}$$
So what is really observed is the sum $(1) + (2)= (3)$ $$\ce{NH_4Cl(s) <=>  NH_3(g) + HCl(g)} \label{run:dnh4cl}\tag{3}$$ with a constant $\ce{K_3}$ equal to
$$\ce{K_3 =  \frac{p(NH_3)·p(HCl)}{[NH_4Cl](s)} = K_1·K_2}$$
This last expression shows that the pressure of the gases $\ce{NH_3}$ and $\ce{HCl}$ does not depend on the amount of $\ce{NH_4Cl}$ in the solid state. And, as $\ce{[NH_4Cl](s)}$ is a constant, it may be included in $\ce{K_3}$ giving a new constant $\ce{K_3'}$
$$\ce{K_3' = p(NH_3)·p(HCl) = K_1·K_2·[NH_4Cl](s)}$$ This is "equivalent" to stating that the concentration of $\ce{NH_4Cl(s)}$ is equal to $1$.

Maurice, sorry but I don't get how the last expression shows that the pressure of gases does not depend on the amount of ammonium chloride in solid state.

@ The99sLearner. The concentration $\ce{[NH_4Cl]}$ in the solid state does not depend on the amount of salt in the container. It is the same in $1$ gram or in $10$ grams pure $\ce{NH_4Cl}$

It is exactly the same for water. The concentration of $\ce{H_2O}$ in pure water is $55.55$ mol/L in any sample of water, big or small.

@ Maurice. Why is the concentration of solid taken as unity? Why do we take its concentration proportional to its density? Why don't we define its concentration to be its no. of moles/volume of container?

@ Param. Of course you may define the concentration of the solids as you did. But this concentration never changes, whatever the experiment. It is equal to $28.5$ M. So why bother about a useless constant ? You better introduce it into $\ce{K_3}$ to make $\ce{K_3' }$ = $28.5$ $\ce{K_3}$, which is a useful constant. This K'3 can be immediately used for studying mixtures of HCl and NH3 gases.

@ Maurice. If we define concentration of solid as no. of moles/volume of container then it will change as the reaction proceeds it will not be constant. I want to know that why rate of forward reaction (from solid to gases) is not dependent on amount of solid. It seems like if we increase the amount of solid then the rate of forward reaction should increase and hence the amount of solid should alter the equilibrium but it doesn't.

@ Param. The concentration of the solid is not the number of mole/volume of the container. No ! It is the number of moles /volume of the solid. It is exactly like Henry's law : the pressure in the gas phase is proportional to the concentration in the liquid phase (not in the whole container)

@ Maurice. Is dissociation of solid into gases a surface phenomenon? Does the reverse i.e. conversion of gases into solid also occurs on the surface? If it is a surface phenomenon i.e. the rate of both the forward and backward reaction dependent on the surface area of solid then I can see why concentration of solids is defined the way it is.               If the rate of reactions is not dependent on the surface area but on the amount of solid then my question still persist.

@ Param. The rate of the reaction is dependent on the surface of the solid. But the equilibrium constant is not. The final state may be obtained sooner. Why don't we continue this private discussion in chat ?

Let us continue this discussion in chat.


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Home foundry — melting aluminum, mitigating undesirables
I am going to be melting aluminum cylinder head and some soda cans in the home foundry. I am aware the sulphur and the hydrogen present in the melt can make aluminum prone to cracks and brittle.
Are there recommended ingredients that can be put inside the molten aluminum besides flux that precipitates or remove undesirables?

welcome to SE, remember that soda cans would have carbon due to coatings and carbon tends to make aluminium brittle. I am not sure how can you get rid of carbon because passing oxygen to form $CO_2$ might also oxidize your aluminium. I would suggest adding pure aluminium ingot to atleast dilute your melt of the undesirables.

well I didn't know that. This is crucial information. I am more interested in not having to buy pure aluminum ingot because that will un-motivate me for the purpose I am recycling the aluminum.

I never encountered S or C as problems in Al. The problem with hydrogen is that it can cause porosity as it is dissolved in the molten Al and comes out as a gas during freezing. Engine blocks and heads are usually Si alloys ( 5 to 12 %) like the wheels. They may have some Cu to make it hardenable. I expect contamination problems would be more likely Fe and Cu ( getting too high ). The cans usually have very little alloy.

acknowledged. Is there any remedy to mitigate the hydrogen? Also Si,Cu are good in my opinion for giving aluminum some strength so they are not considered undesirable right? Also what is the source of hydrogen in Aluminum, is it the coatings on aluminum cans?

I was just pointing out that H does not embrittle Al, it causes porosity.  If it is present ,I would suspect moisture in the furnace atmosphere ; oxidizing the Al and releasing H to dissolve in the melt.

alright. I think this answers the question about the fact S or C isn't undesirables.


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Iodobenzene reaction with copper(I) ion?

Does it reduce it to benzene like $\ce{LiAlH4}$ does? Any source for this reaction?

https://www.organic-chemistry.org/namedreactions/ullmann-reaction.shtm

I saw that but it says copper solid not positive ion unless I’m not seeing it can you tell me the section ?

Under the "Mechanism of the Ullmann Reaction" section, The active species is a copper(I)-compound which undergoes oxidative addition with the second equivalent of halide, followed by reductive elimination and the formation of the aryl-aryl carbon bond. Also, there must be some Cu present

You need a nucleophile to replace iodide here. For example: $$\ce{Ar-Br + CuCN ->[solvent][\Delta, \gt \pu{125 ^\circ C}] Ar-CN}$$ I have done this with 60% yield.

For a room temperature Ullmann reaction, see: https://pubs.acs.org/doi/abs/10.1021/ja00522a058


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Is tertiary-butyl same as neo-butyl?
By definition neoalkanes have two-times branching of methylgroup at second last carbon.
But in no book which I could find, it is not alternatively named as neo-butyl.
Is the definition right, or am I missing something?

What is the source of your definition? As you define neo, it seems to me you would need a minimum of 5 carbons. "Neo-butane", aka isobutane, has only one methyl branch from C2 of propane.

@user55119 but that is what I learnt in lecture.Can you please provide a better definition.

There is no neo-butane. Only two isomers of butane are known : n-butane, and isobutane (or méthylpropane). The 2,2-dimethylpropane you are mentioning is not a butane. it has 5 C atoms. I must nevertheless mention that your question is not related to its title. In the title you speak of butyl. In the text you speak of butane.

Thanks. Thank you. Question's description now corrected.

Neo is a trivial prefix, which in the context of hydrocarbons, means two carbons bound to  the same carbon (geminal) of the longest chain. The only hydrocarbon that commonly uses neo is neopentane. 2,2-Dimethylbutane could be called neohexane but I have never heard it. After neopentane, the nomenclature gets too complicated. By then, IUPAC nomenclature prevails.

The prefix neo- is only used with pentane. It has a historical origin. In the $19$th century, the first $\ce{C_5H_{12}}$ that has been discovered was our n-pentane. At that time it was simply called pentane. After some time, a second $\ce{C_5H_{12}}$ was discovered. Today we know that it is methylbutane. As it was an isomer of pentane, the chemists decided to name it iso-pentane, without knowing exactly its structure. Later on a third $\ce{C_5H_{12}}$ was obtained in the lab, which is presently known as dimethylpropane. Here they had no idea about the way of naming it correctly. That is why somebody proposed to create the prefix neo-. This proposition was accepted with enthusiasm. Anyway the nature of the isomers was a puzzle at this time. So these isomers were intensely studied and their structure became slowly explained. This is why it was not necessary to use these prefixes (iso-, neo-, and maybe others - who knows?) for describing the numerous hexane and heptane isomers.


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Does there exist a black coloured metal?
Most of the metals are just painted/coated black. Does there exist a black coloured metal, either natural or artificial, such that even when we scratch it, it will still remain black?

Have you looked for "Vantablack"?

Well, platinum black is a thing.

Neither vantablack nor platinum black are metals in the sense relevant to the question.

Do you consider graphite a metal? It does conduct...

Highly and degenerately doped semiconductors and semi-metals can be at least dark in visible light in many cases yet have good electrical conductivity. However it would be a stretch to call them either "metals" or "black" without qualifying those labels.
They appear dark in visible light for (superficially at least) the same reason that obsidian or black marble appears dark; they mostly transmit light entering the surface rather than reflect it, then slowly absorb it internally over microns or millimeters. The same is true for many semiconductors, but since their index of refraction in visible wavelengths is in the 2.5 to 5 ballpark their specular Fresnel reflection can be 20 to 50%! Compare to glass with an index of 1.4 or 1.5 and Fresnel reflection of ~4%.
See all the excellent answers to Why are most metals gray/silver? in Physics SE.
There's a short discussion of plasma frequency and semiconductors in this answer which begins:

Metals have a high density of free carriers that can be though of as a plasma. The plasma frequency of metals is usually in the ultraviolet (except for Born & Wolf; Alkali metals transparent to UV? Cesium transparent to blue?), meaning any electromagnetic wave below the plasma frequency would be rapidly absorbed and re-emitted (fancy words for "reflected" by the metal). Just like Earth's ionosphere reflecting HF frequencies and lower but transmitting most VHF and higher, a metallized film that reflects visible light and infrared light (heat) would also reflect all RF radiation; anything below the plasma frequency.
$$\omega_{pe}=\sqrt{\frac{n_e e^2}{m^* \epsilon_0}}$$

A polished surface could look different depending on what is being reflected, but a roughened or ground surface will look grayish.
This piece looks fairly smooth and so it has a shiny appearance due to Fresnel reflection, but it only reflects about 40% of the light, whereas a polished metal can be 95% or higher in some cases.


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How can I bring down the pH of corrosive substance leaving its physical properties intact?
I am developing geopolymers with kaolinite clay and calcium hydroxide activated with sodium carbonate solution. The alkalinity of the mix is too high to handle.. what can I do to bring it down?

Try neutralizing part or all of the Ca(OH)2 (with CO2), and employ more of the mildly alkaline Na2CO3.

Geopolymers are synthesized at very high pH. You are not supposed to use glassware. What is it corroding??

It is Na2CO3 that turns into caustic soda when mixed with water which makes it corrosive.. Also i believe the silicates in the kaolinite also turns caustic at when mixed to this.. i want to make it a material which can be handled by workers.. hence i need to bring the pH to less than 12 i guess..

Add sodium bicarbonate. The pH will soon go down under 9.

When creating sodium hydroxide from sodium carbonate, I would not wonder why it is so much alkaline.


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Function of sodium amalgam and CO2 in formation of aldohexose
There is this following conversion in my book:

$$\gamma-\text{Lactone}\ce{->[Na/Hg +CO2]}\text{Aldohexose}$$

I'm not sure how this reaction takes place. I searched for mechanisms and couldn't find one.What's the function of CO2? Any help is appreciated.

It's a reducing agent. Does this help? https://chemistry.stackexchange.com/questions/119881/fischers-proof-of-the-structure-of-glucose/121294#121294

I couldn't find a detailed mechanism for this specific reaction. Also what is the function of CO2?

CO2 is likely a buffer. The reaction probably gives overreduction of the aldehyde to an alcohol with simple lactones. A 2-electron reduction of the carbonyl group followed by protonation from solvent leads to a hydroxy aldehyde, some of which survives further reduction by forming the intramolecular hemiacetal. I’m sure the rxn is insufficient but Fischer only need some product to prove his point.

Oh that makes sense. CO2 is probably a buffer because the reaction is carried out at round 3-5 ph. Thanks!

I meant inefficient, not insufficient. ;)

I guessed it. No probs!


text:
What kind of filter can remove zeolite dust from ethanol?
I only really have high school level knowledge of chemistry, and also I don't really own any lab equipment.
I'm using 3A zeolite molecular sieves to dry ethanol (and sometimes isopropanol) and they seem to be working great.
However, I'm always left with a bunch of dust.
I can let the dust settle over night and then pour off the top in the morning, but it's still got an orange tint to it. I can keep repeating that process but it starts to take a really long time.
I tried running it through a coffee filter but the particles are too small for the coffee filter to catch.
What should I look for in a filter paper / what other material could I use to filter the remaining dust out of the ethanol?
I don't want to distill it, I don't have the setup for that.

Better filter paper. Whatman #1? Of course you will be picking up moisture as you filter.

Thanks; I just looked up Whatman #1, it says it has a filter size (terminology?) of 11 microns. I just did a cursory search trying to find the size of zeolite dust but I couldn't make sense of it, I found size distribution graphs with numbers ranging from 2 to 20 microns in different complicated contexts. Would the Whatman #1 paper definitely work? (I'll order some anyways and see how it goes)

If you have a lot of dust it may mean that your zeolite is degrading because of excess water in your ethanol. Was the dust present before drying? What is the source of the ethanol? In research labs, sieves are often flame-dried in a flask under vacuum. They do pick up water sitting around in a can that is continually opened and closed.

Could it also be because of cheap zeolite? The ethanol was clear before drying. The ethanol is waste ethanol from an SLA printer with polymers removed but still contains other impurities (leftover unknown photoinitiator [possibly TPPO], unknown binder). Originally it was 95% ethanol I bought from a local lab supply place. I've been drying the sieves in a kitchen oven, I don't have the ability to do it in a vacuum. The sieves could definitely pick up water from opening and closing the jar (I keep them in a mason jar).

I think it is too much to ask for the sieves to remove 5% water. One way to do this in a lab is to add some sodium to 95% ethanol and diethyl phthalate. The mixture is heated to to allow hydroxide from water saponify diethyl phthalate. Then 100% ethanol is distilled out. Not to be attempted in your kitchen. Good Luck!

Most lab-used molecular sieves come with very little dust and are usually supplied as small spherical lumps. If you handle them well you don't get a lot of dust. Perhaps you have cheap sives that are very dusty.

Yeah... I went for the cheapest source I could find, and I'm starting to feel like I got what I paid for... I'm going to try to find better quality sieves, and play with finer filters too.

You can use a sintered funnel with a small bed of celite.

An explanation of filter paper grades is here https://www.simada.co.il/images/filters/Whatman_filter_paper_guide_en.pdf

@Beerhunter Thanks, yeah I saw that. I'm waiting on grade 1, 6, and 597. I also threw together a DIY vacuum filter with a paint can, a funnel, some caulk, and a shop vac a couple days ago but just purchased an actual sintered funnel w/ vac connector, some flasks, and a cheap vacuum filter pump.
And actually, I had great luck yesterday just running the solvent through some bunched up cotton rags (hardware store "bag of rags") in a funnel.
Planning on posting the results once I do some proper tests.

I found it useful. Thank you for coming back and annotating your original answer. Super cool. I'm kinda in the same boat; I have some lab experience and a few chemistry classes, Virology, ect. But not much exp with zeolite or mol sieves.
I too, bought the cheapest jug of 3A mol sieve on Amazon. my god, man, the horror I felt when I dumped a 40$ bottle of Everclear into the filter and it came out dusty orange color. - like a hobo threw a cigar butt into it. - I ended up using <1um paper under a diatom cake and I came out with some pristine 99% anhydrous ethanol to use as a solvent for an extrac

Since this never got a proper answer I thought I'd come back with a follow-up.
Essentially the root cause was low quality sieves (I had just picked the cheapest I could find on the internet). There are probably other things that work but all of the following captured the dust when used alone:

Whatman #6 filters do a decent job. I have #597 too but I don't recall if they worked here. (Whatman filter grade chart).
A bed of sodium bentonite over a #597, which  I was using to filter other stuff, also incidentally captured the sieve dust.
I never actually tried celite; but it was recommended in the comments so that could be a solution, too.

And the following also helped with the issue:

Simply letting the solutions sit for a day then pouring off the top left a significant portion of the dust settled at the bottom. I'd do this anyways before running it through a filter, to lighten the filter load.
More importantly, they were just cheap sieves. The second time I used them after baking them there was significantly less dust, and by the third time the issue was basically solved.

The next time I get cheap sieves I think I might wash them in the solvent first just to get the initial dust load out, recharge them, and then use them for their intended purpose. Although I suspect that if I balance the time, cost, and energy of doing this with the extra cost of better sieves, it's probably a good idea just to pay for the better sieves.
Pretty obvious in retrospect but maybe a fellow non-chemist will find it useful some day.

This is a great answer (and your original question was a great question).  Thanks for taking the time to post it.


text:
Reaction Of Grignard's Reagent with Acyl Chloride
My organic chemistry book says that Grignard reagents react with acyl chloride just as hydride reagents.
Further in an in-text question

now using above theory, my answer is

But Answer in Solution Manual Is

It seems to be a mistake on their front.. What you have gotten is correct.

Safdar is right. It is not possible to add a vinyl group with a Grignard containing three C atoms

This is clearly an error in the book. Your answer is correct.

You may write the author or the editor of the book at the publishers about your find to consider a correction for future editions of the book.  Some keep a compilation of errata available to the public (e.g., http://www.uky.edu/~rbgros1/textbook.html).  A check if there is a newer edition than the one in your hands may be useful, though.

Please add a citation for the textbook you are using. Also consider applying [tag:erratum].


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A Spectrophotometric technique to evaluate Vitamin C concentration in Citrus Fruit
Is there a visible spectrophotometric technique that can gage the concentration of vitamin C in solution? I plan on boiling different citrus fruit in water, and taking aliquotes of the boiled solution to determine total leached vitamin C. I’ve found a couple of UV spectrophotometric techniques, but wasn’t certain if those techniques could carry forward using a visible spectrophotometer. Those sources also didn’t specify what the vitamin C concentration range was for a Beer Lambert’s Law assay.
Would spectrophotometry be possible? Or would it be more feasible to perform a redox titration of vitamin C (ascorbic acid) using iodine?

Yes, it is possible to determine vitamin C by spectrophotometry. It is very tedious and requires sample preparation! I will not recommend it.

It is not quite correct of the statement that "ascorbic acid cannot be determined by Beer Lambert's law, because it does not absorb visible light." Although the second part of the statement is true, the highlighted part is incorrect since Beer Lambert's law applies for ultra-violet range of the spectrum. According to Ref.1, ascorbic acid (Vitamin C) absorbs light in UV range of the spectrum:

The result revealed that $\lambda_\mathrm{max}$ (peak wavelength of spectra) of L-ascorbic acid at $\mathrm{pH} \ 2.0$ was $\pu{289.0 nm}$ whereas at neutral $\mathrm{pH} \ 7.0$, $\lambda_\mathrm{max}$ was $\pu{295.4 nm}$. In alkaline $\mathrm{pH} \ 8.6$, $\lambda_\mathrm{max}$ was $\pu{295.4 nm}$ and at $\mathrm{pH} \ 7.4$ the $\lambda_\mathrm{max}$ of L-ascorbic acid remained the same as $\pu{295.4 nm}$.

Thus, if OP has access to a UV/vis spectrometer, assuming no interference from other components in the fruit extract in the UV range, OP can try to determine ascorbic acid percentage in fruits of interest using a direct method with calibration curve of known ascorbic acid concentrations.
In the event of having interferences, there is a better method of getting absorbance in visible region of the spectrum (Ref.2). This is a well-established method, which is called the 2,4-dinitrophenyl hydrazine method (Ref.3).
The method involves oxidation of ascorbic acid to dehydroascorbic acid by bromine water followed by preparation of the corresponding osazone using 2,4-dinitrophenylhydeazine (2,4-DNPH). The osazone upon treatment with 85% $\ce{H2SO4}$ forms red colored solution, which absorbs light at $\pu{521 nm}$. The procedure was successfully used to determine vitamin C content in various fruits and vegetables (Ref.3). The calibration curve of the red dye prepared from standard ascorbic acid samples is shown in the following diagram (LHS; the plots are from Ref.3):

This method need you to prepare a working solution (5% metaphosphoric acid-10% acetic acid): Fifteen grams of solid metaphosphoric acid were dissolved in mixture of $\pu{40 mL}$ of glacial acetic acid and $\pu{450 mL}$ of distilled water in a $\pu{500 mL}$ volumetric flask.
The standard vitamin C (ascorbic acid) solutions for calibration curve: First, prepare the $\pu{500 ppm}$ ($\pu{500 mg/L}$) standard stock solution by dissolving $\pu{0.05 g }$ of standard crystalline ascorbic acid in $\pu{100 mL}$ of working solution. From that stock solution, you can prepare $5, 10, 15, 20$, and $\pu{25 ppm}$ standard test samples by serial dilution (using working solution to dilute) to prepare the calibration curve.
In addition, you need 10% thiourea solution, 2,4-DNPH solution, and 85% $\ce{H2SO4}$ solution in hand.
Procedure: To the equal volumes of filtered fruit solution (prepared by using working solution to extract and dilute) or standard ascorbic acid solutions, add few drops of bromine water until the solution became colored (bromine water color; this is to confirm the completion of the oxidation of ascorbic acid to dehydroascorbic acid). Then few drops of thiourea was added to it to remove the excess bromine and the clear solution was obtained as result. Then 2,4-DNPH solution was added equal volume to all these oxidized ascorbic acid samples (both standards and fruit extract samples) and they are kept in $\pu{37 ^\circ C}$ water bath for $\pu{3 h}$ (Ref.4). When the coupling reaction of 2,4-DNPH dye with dehydroascorbic acid has been completed, add equal volumes of 85% $\ce{H2SO4}$ solution ($5\times$ volume of 2,4-DNPH) followed by spectrophotometric determination.
It should be noted that any other compound in the extract also reacts with 2,4-DNPH and 85% $\ce{H2SO4}$ solution, if they contain carbonyl groups. For example, common component in fruit is glucose (and other sugar), dye from which are giving absorbance at around $\pu{450 nm}$ (see the complete scan of dye prepared by D-glucose on the RHS of the above diagram). As indicated in the diagram, these interference is minimal at $\pu{521 nm}$.
Note: This method is good to determine total ascorbic acid content (ascorbic acid + dehydroascorbic acid). The full text of the relevant paper is attached for the convenience.
References:

Shaheen A. Maniyar, Jameel G. Jargar, Swastika N. Das, Salim A. Dhundasi, Kusal K. Das, "Alteration of chemical behavior of L-ascorbic acid in combination with nickel sulfate at different $\mathrm{pH}$ solutions in vitro," Asian Pacific Journal of Tropical Biomedicine 2012, 2(3), 220–222 (https://doi.org/10.1016/S2221-1691(12)60045-8).
R. Riemschneider, M. Z. Abedin, R. P. Mocellin, “Qualitäts und stabilisierung prüfung hitze konservierter Nahrungsmittel unter verwendung von Vit C als kriterium-Mitt 1 (Quality and stability evaluation of food preserved by heat utilizing vitamin C as a criterion),” Alimenta 1976, 15, 171-174.
M. Rahman Khan, M. M. Rahman, M. S. Islam, S. A. Begum, “A Simple UV-spectrophotometric Method for the Determination of Vitamin C Content in Various Fruits and Vegetables at Sylhet Area in Bangladesh,” Journal of Biological Sciences 2006, 6(2), 388-392 (DOI: 10.3923/jbs.2006.388.392)(Full Text).
Anal Parimal Desai, Shuchi Desai, "UV Spectroscopic Method for Determination of Vitamin C (Ascorbic Acid) Content in Different Fruits in South Gujarat Region," Int. J. Environ. Sci. Nat. Res. 2019, 21(2),  IJESNR.MS.ID.556056 (4 pages)(DOI: 10.19080/IJESNR.2019.21.556056)(PDF).

Ascorbic acid cannot be determined by Beer Lambert's law, because it does not absorb visible light. The best way is to perform a redox titration with iodine. Usually a citrus fruit is $45-50$ mg ascorbic acid in $100$ g juice. But you should know that there is also $3840 - 6080$  mg citric acid in $100$ g juice.
Ref.: Konrsd Diem, Scientific Tables, CIBA-Geigy SA, Basel, Switzerland, $1972$

At least a similar post about the topic was posted here on ChemSE, too: https://chemistry.stackexchange.com/questions/84436/how-to-find-the-vitamin-c-content-of-orange-juice/135936#135936.  Equally have a look at publications like the one in J. Chem. Educ. (2016): https://pubs.acs.org/doi/10.1021/acs.jchemed.6b00344 or the older in 1940: https://pubs.acs.org/doi/abs/10.1021/ed017p180.

Maurice, yes direct determination is not possible but indirect methods including the official fluorimetric methods exist!

@ Buttonwood. Thank you for the recent reference about the use of iron and 1,10.phenanthroline complex. It was really interesting

The redox titration may not work with real fruit juices because of the juice matrix!

@M. Farooq. You are right. A real fruit juice may contain other reducing substances, apart from ascorbic acid. Carotène for example. But there is only $ 0.012$ mg carotène in $100$ g juice. I don't know which proportion of proteins are reducing substances. But there is only between $0.5 - 1.1$ mg protein in $100$ g juice.

@Maurica, another problem is the color. End point location is difficult.

@ M. Farooq. Titration with iodine and starch is rather precise and not really affected by the color, if the juice is pale yellowish.


text:
Help with mechanism similar to Robinson Annulation

I need to figure out a mechanism for this reaction. All I know is that it is supposed to have a similar mechanism to Robinson Annulation.
This is the best I could come up with, but I've been told it is incorrect.

When trying to correct this, the only thing I could think of is that maybe in the cis configuration the ethoxy group provided too much steric hindrance for the second bond to be formed. In that case, I could have

I am far from sure about this. Can you guys help point me in the right direction?
Edit:
Apparently the correct mechanism is

As far as I can tell, I was marked off for not adding and then immediately abstracting a proton at the same position, and I was marked down for this twice. I thought the point of a mechanism was to show the most direct way to reach the product-why would I bother to add a proton to an intermediate, just to take it away the next step, leaving me with the same intermediate, and the same next step? Maybe someone here can explain why my omission is significant.

Great job!!! There may be an Z to E isomerization as you described. I'd give you an A, if I could! PS: Info on a solvent would have been useful.

The reason I asked about solvent was that the product is a vinylogous, non-enolic -diketone. Ethoxide can add to the cyclopentanone carbonyl and the fragment then ring to give an ester. Secondly, Chem. Abst. doesn't know about this reaction.

I don't know the solvent. I assumed it was ethanol because the catalytic base was just showed as EtO-, instead of OH- or something else


text:
Preparing Hypochlorous acid from salt water
I am a school student and I want help with a project
My aim:
Preparing Hypochlorous Acid (Suitable for sanitation).
I searched across the internet and found an article at https://www.hypochlorousacid.com/hocl-chemistry  in which Salt Water electrolysis was used to make HClO.
But I have two problems here:

This process entirely relies on the ion-exchange membrane.
Titanium anode and nickel cathode are used in this process.

These make the process a very costly affair.
My question :
Is there any alternative way in which I can prepare HClO? (Probably at home).
I could not find the suitable tags for this question. So if you can add suitable tags, please edit this question

Given the risks involved, you should not do this type of experiment at home without teacher's supervision. Gone are the days when teenagers owned a home lab and eventually won Nobel Prizes. If it is a home project, I can give you other interesting project ideas related to spectroscopy.

I didn't mean that. I mean that I want to do this with commonly available and low-cost items (like salt and water as described in the question). And for the electrolysis part, I can do that in adult supervision. It's nice to see that you can also give me new ideas, but I would like to continue with the present project. You can definitely give some solution to the problem given in the question.

Have you seen UNESCO book of 700 Science Experiments for Everyone. This was my favorite book. Go to the electrochemistry section to see the set-up suitable for school experiments. It is available here (register to borrow the book) https://archive.org/details/700scienceexperi0000unes/page/n7/mode/2up


text:
What Chemical Could This Represent
Update
I found someone else who has been trying to solve the same problem and they said that my drawing was not correct. We had a discussion and I conceded that his monochrome picture is more square on and what he was saying had merit. I have updated the drawing to reflect the changes. All black is the monochrome image the red is just notational information.I have this drawing that looks like a chemical but I have no idea what it is.
It is a bit of a detective story because I have very little information to go on.
Could this be a chemical? If so what chemical/s?
If not any other Idea's?
Each circle is the same size.
120Deg index.
All equal distance from each other.
From the original picture it looks like the distance between each circle is a circles distance. So if a circle was 20 units in diameter then from center to center would be 40 units or 20 units from edge of one to the edge of the next. The image I have is at an angle so making this difficult but I believe this is correct. I have re-drawn the image as accurately as I could.
The drawing is all I have to go on. I am 56 years young and have not looked at chemistry like this since school so I am not much good to myself so need help.

It is just like the ancients who saw Gemini, a little dipper and a big dipper in the stars in the sky. I don't think it represents anything. It is a nice design.

Protonated guanidine, maybe?

It can be too many things, if each circle represent a more complex units than an element

I did find boron amide, $\ce{B(NH2)3}$. See pdf here: https://pubs.rsc.org/en/Content/ArticlePDF/2011/NJ/C0NJ00517G?page=Search

While further researching this I have come across additional information that proved my drawing incorrect. This caused the post to be closed. I started the post again with just the updated information. https://chemistry.stackexchange.com/questions/138766/what-chemical-could-this-be

@Craig Instead of asking a new question, update this question with new diagrams. I am sure editing might re-open your question. BTW, did you went through my linked pdf in my above comment? Does that compound suit your needs?

It is impossible to be definitive with such a simple picture.
But, if we assume the colour code is black for carbon and that hydrogens have been omitted, the simplest answer is (to use a non-systematic name) tri-isopropyl methane. See this (very basic) description for systematic names.

Do you think the "molecule" is not flat?

@JamesGaidis Chemists draw flat, simplified pictures of 3D molecules all the time (especially if they can leave out the hydrogens). On my assumed structure the molecules is clearly not flat.

I also would assume that it is not flat but a representation of a chemical. The moderator closed this post after I updated it with new information that became available to me and my original drawing was proven to be incorrect.

The new post can be found here https://chemistry.stackexchange.com/questions/138766/what-chemical-could-this-be


text:
Would a voltaic cell still work if you replaced both the cathode and anode with a carbon/graphite electrode?
for example, in this cell:
Zn(s)|Zn2+(aq)||Ag+(aq)|Ag(s)
if you replace the silver cathode with a carbon electrode, i know it would still transfer electrons, but if you also replaced the Zinc electrode with a carbon one, would it still work?
I saw some examples where inert electrodes were used, and one would be carbon, the other Platinum. This is (what im asking about) the reason why you need separate ones?

What net chemical reaction you suppose would be the source of electrical energy ?

the zinc? but isn't this cell dependent on the fact that electrons need to react, but based on that logic, neither of the electrodes can be changed?

Is zinc a chemical reaction ? There must be ongoing 2 spatially separated redox half-reactions, one providing electrons, once consuming electrons. Generally, electrodes may, or may not react.

If you replace the silver electrode by a carbon one, the cell would still work, as $\ce{Ag+}$ can still be reduced and deposit itself on carbon( that becomes a silver electrode). OTOH, if  the zinc electrode is replaced by carbon one, there is no zinc to be oxidized to $\ce{Zn^2+}$ and the cell would not work, as $\ce{Zn^2+}$ cannot be oxidized either.
As any galvanic cell needs chemical processes providing electrons at their anodes and absorbing electrons at their cathodes. These processes may or may not chemically involve the material of the electrodes. Both inert electrodes can be used for electrolytes, that can be reduced at the cathode and oxidized at anode.
There is the commercial vanadium cell used for  energy storage, using 2 redox half-reactions below. The interesting fact is vanadium has there four different oxidation numbers, $\ce{V^{II}/V^{III}}$ in one of compartments and $\ce{V^{IV}/V^{V}}$ in the other:
Charging
\begin{align}
\ce{V^3+ + e-  &->V^2+}\\
\ce{VO^2+ + H2O &->  VO2^+ + e- + 2 H+ }\\
\end{align}
Discharging
\begin{align}
\ce{V^2+ &-> V^3+ + e-}\\
\ce{VO2^+ + e- + 2 H+  &-> VO^2+ + H2O  }
\end{align}


text:
Comparison of energy release from different types of fuels
I've recently been reading about the different amount of energy release for different types of fuels/hydrocarbons.
Anyways, I was wondering why butanol would release more energy (kj/mol) than methanol? I'm guessing the same theory applies for methane vs butane?

Consider the primary alcohol serie. Methanol as the first serie member is like half-water, while higher members converge to hydrocarbons. I missed originally you ask about the molar value, thinking about the specific value MJ/kg. The reason related to this is very trivial, butanol has bigger molecules. 3 big wooden logs in fireplace give more energy than 3 small wooden logs.

This is essentially a question concerning the trend of molecules in the same homologus series that have differing chain lengths. When we increase chain length, the number of C-C bonds increase in proportion to C-H bonds. Looking at the bond energies, the bond energy of a C-C bond is $347 kJ/mol$, while for C-H it is $414 kJ/mol$. This means to break a C-C bond compared to a C-H bond, less energy is required. Also when increasing the number of C-C bonds, while producing the same products ($CO_{2(g)} + H_2O_{(l)}$), more carbon dioxide is produced in proportion to the amount of water. The enthalpy of formation of carbon dioxide, results in it releasing more energy than that of water. Taking this into account with the lower energy requirement to break the reactants, the chain length is generally directly proportional to the energy produced. Thus, the COMPLETE combustion of Butanol would have a higher energy out put than Methanol and likewise Butane > Methane.
It is important to note that, as the compounds get more energy dense, i.e release more energy upon combustion, that it becomes harder for the compound to reach complete combustion; what I mean is that more oxygen is required for the reaction to release the full energy output. If not enough oxygen is supplied to the system soot ($C_{(s)}$) & Carbon Monoxide ($CO_{(g)}$) will be produced and the energy produced will be less than the energy produced in the complete combustion. (The degree to which this works, depends on how many moles of oxygen is supplied).

(+1) Your comment on incomplete combustion being generally inefficient in oxygen with longer-chain molecules is certainly correct.  However, am I correct to suspect that combustion in an atmosphere of chlorine gas where the products would be HCl and various carbon chlorides/carbon-hydrogen chlorides may actually tend to be more efficient?  However, obviously the latter products are extremely problematic in an earth atmosphere, but perhaps a 3rd stage rocket in space should not be using O2 based fuel mixes.


text:
What is the effect of putting salt on water during electrolysis
I want to decompose $\ce{H}$ ion and $\ce{O}$ ion by putting carbon electrode in the water, but it look so slow. What if I put sodium chloride? Would putting $\ce{NaCl}$ have effect on output gas($\ce{H}$ and $\ce{O}$)?

If a question is asked on Chemistry SE site, then, in contrary to sites like Quora, it is expected from the author to elaborate the topic of the question by doing at least basic own topic review, writing what he/she has found and understood, and what is the stumble stone. The quick questions without explicitly expressed particular effort are not very welcome, and may be closed.

Yeah there is an effect putting salt during electrolysis. it give you chlorine ion at the anode which is dangerous at higher amount. in the place of sodium chloride you can use sodium hydroxide you can get it from stores it is also safe or sodium bicarbonate from baking soda.
Good luck !!

The concentration of the added sodium chloride is not specified. If small quantity of NaCl is added it will not generate chlorine at the anode i.e. no overvoltage will occur. Only in highly concentrated solutions like brine can on electrolysis generate chlorine gas at the anode, as oxygen release is kinetically hinderd.


text:
What are some materias/substances that can be applied to a ball such that the ball will stck to a wall,if thrown
I am trying to make a prototype for the "valkyrie camera" seen in the video game Rainbow 6 Siege
This is how it works
https://gfycat.com/impressivegrippingauklet
the black ball seen is the camera,as you can see the camera is thrown and it sticks to a surface.
are there any materials that are capable of such action?

The trajectory of the ball seems non-physical, I wonder how they simulated that. In any case, the ball itself seems to be a hollow plastic sphere containing the camera, and the system within is activated by pressing somewhere on its surface. This is a non-stick ball when in your hand and yet it sticks upon contact with other surfaces, remaining attached under the weight of camera plus sphere. It might be coated with a thin layer of transparent adhesive (otherwise the camera would not clearly resolve whatever is outside the ball).

Seems impossible. I would suggest making the ball morph into a "sticky" form after casting, either using a timer or a sensor. An impact sensor which causes some attachment mechanism to extend seems a reasonable solution. I would not use an adhesive.

Memories of https://en.m.wikipedia.org/wiki/Wacky_WallWalker come to mind. Our cats would go nuts watching them slowly ‘walk’ down walls until they could pounce...

What if we designate an area as "hold-able" while the rest of the surface is covered by the "sticky material X" .X will have a cover which can be removed before deployment.I am mostly curious whether an X exists that is sticky enough to as has been noted before to sustain                                                                                Weight of ball + equipment.                                                                                impulse generated on impact.

A magnet, perhaps?

I am afraid a magnet cannot be used as the ball must stick to nonmagnetic surfaces.


text:
Arrange in order increasing reactivity towards nucleophilic substitution

Arrange the following compounds in order of increasing reactivity towards SN2 reaction

According to me, the order should be P>Q>S>R as more hindrance should result in lesser reactivity but answer given in the book is P>R>Q>S. Why is the reactivity of R greater than that of Q and S even though the leaving group is on a $2^\circ$ carbon.

Possibly the stability of the benzylic cation is more important than hindrance is determing the rate between P and R

Where is this from?

@Safdar Himanshu Pandey

@Waylander Easy to miss that this question is specific to $S_{\mathrm{N}}2$ reactivity.

The reactivity of the SN2 reaction is determined by the stability of its transition state. More stable the transition state (hereby referred to as TS), lower will the activation energy be for the reaction, and therefore, the reactivity increases.
Let's first have a look at the transition state of an SN2 reaction.$^{[1]}$

What you see inside the square brackets is the transition state, Nu represents the incoming nucleophile, while X represents the outgoing leaving group (both bearing a partial negative charge).
In this transition state, the carbon atom in the middle has a p orbital that shares one pair of electrons between the old and the new bonds. This suggests that the transition state for an SN2 reaction has a more or less planar carbon atom at the centre with the nucleophile and the leaving group arranged at $180$° to each other.
The point to be noted here is that the p-type orbital of carbon in the TS stays directly perpendicular to the plane formed by its three substituents, which is the same as the plane of the ring (in this case, all the three hydrogens shown in red). This means that sterically bulky groups which are present in the plane of the ring, and directly protrude towards the site of reaction (which is still perpendicular to this plane) and therefore will have a much bigger effect in inhibiting the entry of the incoming nucleophile as compared to those outside it.
This will be a general idea for the stability of TS (and hence the reactivity towards the SN2 reaction) in other cases as well.
Now, although all the options have adjacent benzyl groups, it might be worthwhile to see how benzyl and allyl groups stabilize the TS. These compounds react rapidly by the SN2 mechanism because the π system of the adjacent double bond can stabilize the transition state by conjugation.
The p orbital at the reaction center (shown below in brown, and corresponding to the brown orbital in the previous diagram) has to make two partial bonds with only two electrons — it is electron deficient, and so any additional electron density it can gather from an adjacent π system will stabilize the transition state and increase the rate of the reaction. This is shown here for allyl bromide:$^{[2]}$

The benzyl group acts in much the same way using the π system of the benzene ring for conjugation with the p orbital in the transition state. This stabilization will be present in all the four options,and I am making no attempt to compare them here
So, the geometry of the transition state and steric consideration will put S at the end due to the highly directed bulkiness of the di-ortho substituted benzene group, while P will jump ahead due to the absence of such clashes. S will still have benzylic interactions, but the incoming nucleophile will be hindered by the methyl group which points towards the perpendicular p orbital
Out of R and Q, R might achieve favourability over Q because again, a steric clash in the plane of substituents by a benzene ring having a methyl group directed towards the site of reactivity (the p-orbital perpendicular to plane of the ring) will generate much more hindrance than a single methyl group smaller than o-methyl benzene, which is directed away from the site.
Finally, another small reason that may tip the scales in the favour of R could be an interaction between the σ $\ce{C-H}$ bonds and the p orbital of carbon, in a kind of hyper-conjugative fashion similar to what happens in a carbocation (although this interaction would most probably be weak) as follows:

This weak σ $\ce{C-H -> p+}$ might lead to a lowering of the energy of the p-orbital, and thereby slightly increase the feasibility of attack by lowering the activation energy.
The visualization for this interaction may be seen in a similar fashion as hyperconjugation for a carbocation:$^{[3]}$

Such an interaction is not quite feasible in the TS of Q, where any interaction with the ortho-substituted methyl group will require complete delocalization over the benzene ring, which is not very probable as the TS is a short-lived species, and is not a full-fledged carbocation (especially since the leaving group is chloride, which is a good leaving group and doesn't lead to excessive charge development by a very early or late departure)
Hence, you get the order P>R>Q>S.
References:
$[1]$ Organic Chemistry Second edition, Jonathan Clayden, Nick Greeves and Stuart Warren,
ISBN $9780199270293$ Nucleophilic substitution at the carbonyl group, [p.$340$]
$[2]$ Organic Chemistry Second edition, Jonathan Clayden, Nick Greeves and Stuart Warren, Nucleophilic substitution at the carbonyl group, [p.$341$]
$[3]$ Molecular orbitals and organic chemical reactions, Fleming

I think there are some problems with this answer. Note that orbital interactions with the benzylic system are strongest in Q and S because steric constraints force the C-Cl bond out of plane (particularly in S).

@Zhe I have made no comments about the benzylic interactions. I guess my answer has been grossly misinterpreted. When I said that the the sterically bigger methyl group protrudes towards the site of reactivity,I meant that it does so by staying im the trigonal plane of substituents(same as plane of the ring) but the p orbital,the actual site of reactivity is always perpendicular to this plane. I have clearly stated that in my 5th paragraph. I have also nowhere made any comments about how strong are the benzylic interactions anywhere

You spent some time discussing the allylic interaction, which I interpreted to be an analogue to the electrophiles being benzylic here. But that orbital overlap with the anti-bonding orbital is a consistent feature across the substrates, so I was confused that this feature was highlighted.

In any case, you added a note, so I think it is better.

I drew the 3d structures of R and Q in MolView but still can't see why would the hinderance in Q be greater than that in R

This is a really good question.
The reason you got it wrong is because you only identified one feature that leads to faster rate of bimolecular substitution, and that is secondary electrophiles react more slowly than primary electrophiles. That's good because you are trying to identify patterns of reactivity.
The reason that the secondary electrophile reacts slower is because the site of substitution is more crowded. But you have to think about the structure for the two electrophiles on the right (building a model will really help). The C-Cl bond is going to be perpendicular to the ring to be most reactive in a benzylic electrophile. But that means that the ortho-substitution is starting to crowd the site of reactivity. The more substitution you have, the worse the steric crowding gets, so S definitely reacts the slowest. In this case, the ortho-substition is a big enough deal that it's a more important factor than being a secondary electrophile.

Well, I get your point. But if someone comes up to me on the street and gives me a similar question the compounds of which have hindrance in some other form, how am I supposed to decide that some lesser important factor will overpower this other major factor in that specific question?

The steric inhibition of resonance is attributed here. Thats not very difficult to notice.

@AdityaRoychowdhury Where are you going to invoke resonance for an $S_{\mathrm{N}2$ transition state?

@RobinSingh That is the ultimate question in organic chemistry. Frequently, the answer, like here, is build models. But in general, only experience and examples and reality will give you the answer. If you had asked me with no information where R or Q reacted faster, I would say there's a significant chance I'd get it wrong.

@AdityaRoychowdhury Note also that orbital alignment is strongest (enforced by steric constraints) in the examples that react slowest.

I think the question wants you to consider the SN1 mechanism as well in your rankings. As Q and S each have a more electron rich ring than R, they generate a more stable carbocation on loss of the chloride ion than R. Therefore, the rate of SN1 has increased considerably and so a smaller proportion of the molecules go through the SN2 route. The carbocation forms on the carbon bonded to the ring therefore it is easily stabilised by the mesomeric effect.
The fact that the question doesn't provide a specific nucleophile makes me believe that they're probing electronic effects rather than steric hinderances.

OK, but the question says to order these electrophiles in terms of their reactivity towards bimolecular substitution.