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Are alkaline conditions required for Maillard reaction?
The Maillard reaction, a form of non-enzymatic browning, refers to a cascade of chemical reactions that frequently occurs in the cooking of foods (e.g. baking of pastries, roasting of meats). It primarily starts as a reaction between amines and carbonyl compounds. It is stated on the Wikipedia page that the reaction is accelerated by alkaline conditions:

This process is accelerated in an alkaline environment (e.g., lye applied to darken pretzels), as the amino groups ($\ce {RNH3^+ → RNH2}$) are deprotonated, hence have an increased nucleophilicity.

Similar tips on using alkali to hasten the Maillard reaction have also been propagated on various cooking websites, even on cooking.stackexchange. In this post, I would like to investigate the chemistry behind this piece of advice.
The use of alkali is chemically sound since the first step of the initial stage of the reaction is the formation of a Schiff base from the amine and the carbonyl compound. However, is it a bit far-fetched to say that since the first step is accelerated, the entire reaction cascade is similarly sped up significantly?
More importantly, as I have mentioned previously in this post: Conditions for imine formation, the reaction is primarily catalysed by acid and not by base. Yes, nucleophilicity of the amine increases. However, the alkali medium does not facilitate the leaving of certain groups. Thus, the overall effect of the alkali may not be exactly favourable.
The fact of the matter is that alkalis, such as baking soda, have been used in cooking for a long period of time. Thus, it certainly does improve the taste of food by some mechanism. However, I question whether this mechanism is indeed the hastening of the Maillard reaction.


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Why is hydrogen not very electronegative?
I looked at my reactivity series and I have just realized that hydrogen and carbon are both included in the series, even though they are non-metals. So why is hydrogen more electropositive than negative? I assume that it is because of its single proton in its nucleus. But on the other hand, hydrogen only has one energy level. Doesn't that mean that the valence electron is closer to the proton in the nucleus, and isn't this more significant than the number of protons by looking at Coulomb's law?
So my question is, what factors made hydrogen less electronegative and more electropositive? In addition to this, it would be great as well if anyone can explain the existence of carbon in the reactivity series?

Hydrogen is less electronegative than some metals and more electronegative than many others. Carbon is not there, unless you are talking about some obscure alternative series.

@IvanNeretin yeah i noticed, it would be great if Dr. can help me out with my main question, what makes Hydrogen more electropositive. Thank you Dr.

It's small and has 1 proton. Why should it be attractive for electron pair, if hydrogen is fine with its one electron?

Why would you expect it to be?

Well I'd say this is not that easy to answer.
First of all, how much does the activity series really relate to the electronegativity? For elements like Fluorine it fits pretty well. But according to the activity series, Gold is quite noble as well, and while its electronegativity might be high among the metals, in comparison to the non-metals it is still quite low.
The second thing is, Hydrogen is a very special element. As a Hydride chemist I can only tell you there are so many strange phenomena when working with hydrogen and often you can't simply explain it in a few sentences. I was recently listening to a talk by Hiroshi Kageyama on metal hydrides and he puts the hydrogen in the periodic table not above Lithium but above Carbon since they match in terms of electronegativity much more. And you can confirm that once you are working with C-H bonds, they tend to cleave in a radical reaction rather than a polar one (unless you use superacid systems that are currently designed to functionalize methane).
I guess this electronegativity comes from a few aspects. If we take the most simple definition, that it is the sum of the electron affinity and the ionization energy (divided by two), we can see that hydrogen has the affinity to take up another electron (only 1 electron in the s-orbital); hydrides are a thing that can be prepared and usually they are quite stable if kept dry as well. Nonetheless we know that it has quite a low electron affinity. Perhaps not as low as Fluorine or Oxygen but comparable to Arsenic for example. So there is already a negative contribution in the equation that will affect the total electrongativity.
About the ionization potential: Well Hydrogen is quite a case. H+ is a thing yes. As a protonated species in water or other solvent systems, but at least from what I have been taught a real H^+ is hard to get. Without the help of the solvation or hydration energy of the solvent or water that stabilizes a H^+, a real gas phase species of H+ or a free proton is mostly approximated with calculated values.
And if you take a look in comparisons for the first ionization energy for all elements you will see that hydrogen is quite high.

This also makes sense. For other elements we usually take the valence electrons. To excite the inner 1s electrons for heavier elements we usually have to impact electrons of high kinetic energy (e.g. an x-ray tube). For hydrogen there is only one proton but still the single electron has to face the full positive charge which creates a much stronger connection than for example Lithium or Sodium.
So in summary there is a mediocre to low electron affinity combined with a more or less high ionization potential. The sum of this gives a mediocre value much like we find it for hydrogen. It's not as low as in metals but for non-metals it's quite low and therefore much like carbon.
About your second question. What do you mean with existence in the acitivity series? Carbon has redox-chemistry as well. If you burn Carbon you will get the oxide which is an oxidized Carbon.

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Gas solution - State of matter
When a gas is dissolved in water does the state of gas changes ? We write it as aqueous but what is the state that exists ? does it has the properties of gases when it is dilluted ?

Good question! For a solid dissolved in liquid, obviously the solid behaves as a liquid rather than as a solid, since its particles are separated and moving freely.
A gas, though, is defined as free to fill its container, unlike a liquid with a definite volume. Since a dissolved gas (as opposed to one chemically bound to a liquid) is free to disperse, e.g. $\ce{CO2}$ bubbling out of seltzer, one might still consider it gaseous, rather than liquid, just loosely bound to the liquid.
That said, in a closed container in equilibrium, or over a short time, dissolved gases do behave as liquids. One can pour $\ce{NH4OH}$ ($\ce{NH3}$ aq.) into $\ce{H2SO4}$ ( $\ce{SO3}$ aq.), and the majority of the dissolved gases stay in solution -- though some $\ce{NH3}$ gets out, leaving one tearing and coughing, if not done under a hood.
It appears to be a matter of definition, then if dissolved gases are considered as gas or liquid.
BTW, gases dissolved in solids, such as methane clathrate, are more tightly locked in place and may behave as solids until warmed.

so when we have a reaction in water solution that produces co2 we put the sign (g) because the container is open ?

Yes... most of the CO2 bubble out quickly as a gas. On the other hand, if NH3 is evolved, it stays in solution longer, though it, too, comes out of solution. React CO2 with NH3, and it stays dissolved, as ammonium carbonate, so the answer is not always obvious.

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NMR: How the relaxation times T1 and T2 depend on the correlation time / amount of molecular tumbling.
Why do nuclei that have a smaller / faster correlation time have a higher / slower T1 / T2?
From my understanding:

Fast Brownian motion creates a wide range of B.local (local magnetic field created by dipole of spin X near spin Y) fluctuation frequencies
Slow motion gives rise to fewer frequencies but more intense fluctuations
Fluctuations at the right frequency cause transitions between energy levels / relaxation / re-establishment of equilibrium
The larger a molecule the slower it tumbles (correlation time is slower)
The slower the tumbling the faster the relaxation

To me, logically, it'd make sense that if a molecule tumbles faster then the relaxation should be faster, not the other way round. Can anyone explain this?
Thanks

Careful: It's not so much about the intensity of the fluctuations, but about how their influence integrates over time.

Fast Brownian motion does not usually create larger fluctuations, they just have a wider frequency spectrum.

First recall that the T1 and T2 times reflect different things in the NMR experiment. T1 is the decay of population back to equilibrium and T2 is the decay of the coherence between the spins so we expect them to behave differently in different solvents.
Because of the interaction with other nuclear spins in the molecule, and with electrons in the solvent molecules as they tumble around in solution, the individual magnetic moments making up the magnetisation experience slightly different local magnetic fields. These fields change in time (because the molecules are tumbling in solution) and induce random transitions between the NMR spin states and these random transitions must cause the magnetisation to decay back to equilibrium. The main causes are
(a) Rotational Diffusion. Molecules interact via dipole-dipole coupling which changes as the molecules rotate and/or diffuse. Rotational diffusion is a major cause of decay.
(b) Spin-Spin interaction. This is a direct effect caused by any paramagnetic species, eg. O$_2$ and can be dominant in aerated solutions. It occurs because the magnetogyric ratio ($\gamma$) for paramagnetic molecules and ions is approx 1000 times greater than for nuclei.
(c) Chemical Shift Anisotropy. The shielding observed in chemical shifts is due to the locally induced magnetic fields. As a molecule tumbles in solution, its position with respect to the external field $B_0$ changes and this acts to reduce the magnetisation because, from the molecule's viewpoint, the external magnetic field now appears to be randomly varying.
Solvent motion is characterised by a correlation time $\tau_c$ describing the angle that a solvent molecule has at any time $t$ compared to the angle it had at time zero, clearly this will depend on the nature and viscosity of the solvent. The correlation time can be considered as the tumbling (rotational diffusion) time of a molecule in solution, and is proportional to the solute size and solution viscosity.
Population decay T1 occurs fastest when motion of solvent is of same frequency as transition frequency, so is most effective close to transition (approx RF) frequency, say, 400MHz and so T1 times are short. (You can think of this, classically, as two oscillators of the same frequency becoming in resonance with one another and so are able to interact maximally).
In solvents with long correlation times motion is too slow to cause transitions and the solvent can be likened to an effective static average field and so interact only slightly and T1 times are long.
In solvents with short correlation times these can be so short that they present only an average field to the nuclear spin i.e. they move very fast compared to the solute's motion, and so interact only slightly and T1 is long. The effect this has is shown in the figure.
T2 dephasing time is unaffected by fast solvent for the same reasons as for T1 just described, but because T2 depends on dephasing between spins, which has a range of low frequencies, therefore T2 is sensitive to slow solvent motion (i.e. is in resonance with slow motion) and decreases as $\tau_c$ increases.

Sketch of log T1 and log T2  times vs log correlation time in an NMR experiment.

Relaxation is often very counterintuitive and isn't a particularly easy subject to understand, so it's not always useful to argue based on what seems to be logical. I'm not sure where you're studying this from, but it seems to be slightly lacking in the detail needed to properly understand what's going on. Here is a brief overview of what's going on, but I strongly recommend consulting a textbook which will explain it in more detail. Chapter 9 of James Keeler's Understanding NMR Spectroscopy (2nd ed.) is a very good reference. You can find a pdf of the corresponding chapter in the previous edition here.
The link between rate of tumbling (correlation time, $\tau_c$) and the "amount of fluctuations at the right frequency" is given by the spectral density
$$j(\omega) = \frac{2\tau_c}{1 + \omega^2\tau_c^2}$$
This function tells you how much the random fields are fluctuating at each frequency $\omega$. For relaxation, the important frequency to look at is the Larmor frequency $\omega_0$, i.e. the central quantity is
$$j(\omega_0) = \frac{2\tau_c}{1 + \omega_0^2\tau_c^2}$$
By differentiating this with respect to $\tau_c$ you can find that this is at a maximum when $\tau_c = 1/\omega_0$. The rate of spin-lattice (longitudinal) relaxation is proportional to $j(\omega_0)$, so it is fastest when $\tau_c = 1/\omega_0$. That means that $T_1$ will be smallest when $\tau_c = 1/\omega_0$. When $\tau_c$ decreases below this, or increases above this, spin-lattice relaxation will be slowed down and $T_1$ will increase.
On the other hand, spin-spin (transverse) relaxation depends on the sum $j(\omega_0) + j(0)$. $j(0)$ is simply equal to $2\tau_c$, and you can show (again with calculus) as $\tau_c$ increases, the total sum $j(\omega_0) + j(0)$ increases continually. So $T_2$ decreases when $\tau_c$ increases.


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Is the neutral di-µ-oxo-bridged manganese dimer with eight ammonia ligands a reasonable structure?
In its guide to model antiferromagnetic couplig, Gaussian uses $\ce{Mn2O2(NH3)8}$ as a model compound.[1] It argues that manganese is in II oxidation high spin state; in other words, in both manganese are all d orbitals are singly occupied. Furthermore it creates a structure, where the metal centres are octahedral coordinated. The level of theory used is U-B3LYP/3-21G, which is in my opinion not enough to discuss these complexes in the first case.
I could not find any evidence of a structure proposed by Gaussian, the closest I have gotten is given as a cation $\ce{[Mn2O2(NH3)8]^z+}, (z\in2,3,4)$.[2]
I have reason to doubt the structure given in the tutorial,[3] as in the $\frac{5}{2}$ high spin case there are only four orbitals in manganese left which could accept bonding partners. Given that two of them are used by the bridging oxo ligands, only two ammonia should bond to the manganese centres. This would lead to the formula $\ce{Mn2O2(NH3)4}$. The situation changes once you start removing electrons or pair them up. Then the question changes to whether ammonia is a strong enough ligand to induce lower spin states.
Summarising: Is $\ce{Mn2O2(NH3)8}$ a reasonable structure?
For convenience, here is the structure as proposed by Gaussian (they do create a different one within the tutorial though):

References & Notes:

Gaussian Website: http://gaussian.com/afc/
Mcgrady, J. E.; Stranger, R. Redox-Induced Changes in the Geometry and Electronic Structure of Di-μ-oxo-Bridged Manganese Dimers. J. Am. Chem. Soc. 1997, 119 (36), 8512–8522 DOI: 10.1021/ja964360r.
In my optimisations on a different level of theory I was unable to reproduce the structure.

Looks reasonable. Mn (II) complexes have no problem with being high spin and hexacoordinate.

@Mithoron Actually I'd supposed it would be quasi tetrahedral.

Nah, should be octahedral just like manganese(II) chloride and other complexes. Counting orbitals seems to be only confusing for this stuff.

@Mithoron I guess confusing enough to ask a question here. Need to look up the chloride though, seems interesting enough.

What do you mean ‘four orbitals left’?

@Martin-マーチン I think you might want to check out http://dx.doi.org/10.1002/(SICI)1521-3765(19990201)5:2<577::AID-CHEM577>3.0.CO;2-2, namely compound 4. I searched ICSD and CSD for neutral complexes with Mn(N4)(O)2Mn(N4) skeleton, and this turned out to be the only match.

@Jan High-spin manganese(II) has $\ce{[Ar] 3d^5 4s^0 4p^0}$, which means there are 4 orbital unoccupied, which could accept ligand bonds; or in other words: left = empty.

The structure is extremely reasonable to the point where I would ask why any other di-oxido-bridged manganese(II) ammine complex would not show this structure if it were proposed. To describe why, I am going to be lazy and just copy my molecular orbital scheme of a general octahedral 3d-metal complex that I have used many times before.

Scheme 1: Octahedral $\ce{[ML6]}$ complex with no π interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.
As you can see, the four empty orbitals ($\mathrm{a_{1g}}$ and $\mathrm{t_{1u}}$) are indeed present and part of the bond-forming orbitals, but the fact that the $\mathrm{e_g}$ orbitals are also half-filled does not prevent them from participating, too. In fact, these orbitals turn into $\mathrm{e_g^*}$ orbitals in the resulting complex, formally antibonding. This, however, doesn’t matter, because there are still many more electrons in bonding orbitals ($\mathrm{t_{1u}, a_{1g}}$ and $\mathrm{e_g}$) that this small amount of lost energy is more than offset. Typically, for a high-spin $\mathrm{d^5}$ complex, the bond order would be somewhere around 5/6. Still well within the overall bonding range.
If one looks at the kinetic stability of various complexes, one will find that $\mathrm{d^3}$ and low-spin $\mathrm{d^6}$ often give very inert octahedral complexes as no antibonding orbitals (of the simplified scheme) are populated—but even a $\mathrm{d^{10}}$ or $\mathrm{d^0}$ metal will form these complexes, their structures just won’t be as long-lasting. (Zinc has been shown to equilibrate between tetra, penta and hexacoordinated complexes in aquaeous solution as I was told in my coordination chemistry lecture.)
So yes, this octahedral complex is perfectly fine.


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Does berylium gain or lose electrons?
Does beryllium want to gain or lose electrons? I know it has 4 protons, 2 valence electrons, and it will gain or lose 2 electrons, but what is it?

Lose.
Look at the column it's in on the Periodic Table of the Elements.  It's above Magnesium.
Look at where in the row it's in.  There are six elements to the right of it in that row.  It would have to gain six electrons to fill up the valence shell, or lose two to vacate the shell.
There's more than just those two things to consider, but that answers the question most of the time.
And, even though many teachers fall into this trap, be careful with saying "want" in terms of gaining / losing electrons.  It is more stable for Beryllium to lose two electrons than it is for it to gain six.

FYI answering such questions is discouraged here.

@Mithoron Is there a guide / reference / community standards page?  Or can you please provide more clarification as to why this question shouldn't be answered?

answered my own question: https://chemistry.stackexchange.com/help/on-topic

https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange https://chemistry.meta.stackexchange.com/questions/3181/lets-decide-on-what-we-should-close-as-homework https://chemistry.meta.stackexchange.com/questions/2991/when-is-it-safe-to-answer-homework-questions


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Is the water in an aqueous solution undergoing a reaction considered part of the system or surroundings?
System: In chemistry, a system is a chemical reaction. A system operates within its surroundings. Energy can move between the two.
Surroundings: The environment around a particular chemical reaction. The chemical reaction is the system.
Above are my textbook's definitions for system and surroundings.
For example, mixing two aqueous solutions to produce a precipitate. Although the water doesn't undergo any permanent chemical changes, it is still enabling the reaction (much like a catalyst). So is water then considered part of the system?

Water is definitely part of the system for aqueous reactions.

The boundary between system and surroundings is defined by you the scientist. Still, some choices make more sense than others, depending on the question you want to answer. When you perform a calorimetric experiment, for instance, you monitor the heat generated or absorbed during a reaction (such as formation of a precipitate). The heat originates or is consumed in the local combination of reagents to form the bonds between precipitate atoms and during segregation of solvent molecules. That heat then dissipates into the rest of the vessel and is registered, for instance as a temperature change. In the final analysis, what you call the system here would probably include only what has changed and you are specifically interested in, namely the molecules involved in the actual reaction. This would include solvent molecules associated with the solubilized reagents (the involvement of solvent is made explicit when reagents are labelled as aqueous, for instance). The behavior of the bulk solvent, which serves an "auxiliary" role, for instance by transferring heat across the vessel, is important during calibrations, but is not the question of interest, and would usefully be regarded as the surroundings.
Water can be regarded as a catalyst in many aqueous reactions, but when a reaction occurs in a solvent that associates strongly "catalyst" is not generally a proper description for the solvent, as the energy of both reagents and products may be stabilized/destabilized by the solvent, and to different degrees (and not just the energy of an intermediate in the reaction).


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Acidity comparison between cyclopropene and cyclopropane
There many sources on the internet saying that cyclopropane is more acidic than cyclopropene because the conjugate base in latter would have antiaromaticity which would destabilize it.
But in this explanation we are assuming that the central $\ce{C-H}$ bond in cyclopropene ionizes. I argue that the $\ce{C(sp^2)-H(s)}$ bond should ionize because of two factors:

The $\ce{sp^2}$ carbon would better accommodate the negative charge than the central $\ce{sp^3}$ carbon.
Consequent conjugate base would not contain anti aromaticity.

My logic can also be rationalized by considering acidity order between straight chain alkanes and alkenes.
This would mean inversion of the order which is incorrect according to experimental data. What is the flaw in my reasoning?

https://chemistry.stackexchange.com/questions/51467/is-the-cyclopropenyl-anion-antiaromatic

This reference does in fact indicate that cyclopropene ($pK_a=29$) has a much higher a acidity than cyclopropane ($pK_a=44$).  Deprotonation at the vinylic hydrogen would account for this.  That cyclopropene is deprotonated at a vinylic carbon is briefly summarized here with a primary reference given.

Many sources say that pKa of cyclopropene is 61.

Say you have a molecule of citric acid.  If you could magically prevent the carboxyl protons from dissociating you'd find that there is a certain $pK_a$ for the remaining hydroxyl (alcohol) hydrogen, which in practice chemists could get from extrapolating data for citrate esters.  But in the real acid the carboxyl protons beat the alcohol protons to the punch and you see only the former dissociations.  Similarly, $pK_a=61$ for cyclopropene picks the methylene hydrogens but the molecule really deprotonates at the vinylic hydrogens with $pK_a=29$.


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Calculate the transition dipole moments between the 1s inner shell and the high-lying states of a molecule
I am trying to calculate the transition dipole moments between the 1s inner shell and the  high-lying states of N2, namely, $<\psi_{1s}|{\bf r}|\psi_{high-lying}>$.
For this, I installed GAMESS (US) and calculated molecular orbitals of N2.
However, I only got HOMO-6 to LUMO+2.
I want LUMO+3, 4, 5... and if possible, 1s.
How can I do this in GAMESS? Or do I need other softwares?

Might have found the option you are looking for. GAMESS controls MO printing with the NPREO keyword. If you can specify ranges of orbitals either by their number (e.g. NPREO(1)=1,10 prints the first 10 orbitals) or relative to the HOMO (e.g. NPREO(1)=-3,-5 prints from HOMO-3 to LUMO+4). You can also give 3rd and 4th numbers to list the range of orbital energies to print.
These print options are described on p.23 of the input descriptions document on the Gordon research group page.

Thank you for your comments.  I checked that option but the default value gives all the calculated MOs. Actually, I kind of solved the problem myself. By changing the basis from STO to N311, and added NDFUNC, NPFUNC, DIFFS, DIFFSP options in the $BASIS group, I obtained up to LUMO+62. (although I'm very skeptical of the vailidity of those MOs.)

@N.Saito What makes you sceptical of the MOs? It may be worth asking another question to try to figure out whether something is wrong with the MOs. Just be sure to include the level of theory and basis set you are using in that new question.

The energies of the higher LUMO orbitals are too high. (e.g. LUMO+62 has 42.9 a.u. (1167 eV).) Also, these orbitals looks very localized to each nitrogen atom. I will describe the details in another question.


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Regioselectivity in iodolactonisation of γ,δ-unsaturated carboxylic acid

In the last step of the iodolactonisation of 3,3-dimethylpent-4-enoic acid,[1,2] why does the carboxylate attack the iodonium to form a 5-membered ring, when a more stable 6-membered ring is possible? Both of the sites seem to be equally favorable sterically. I can't come up with a reason favoring the formation of the 5-membered ring.

References

Takano, S.; Sato, N.; Akiyama, M.; Ogasawara, K. A Synthesis of trans- and cis-Caronaldehydes. Heterocycles 1985, 23 (11), 2859–2872. DOI: 10.3987/R-1985-11-2859.
Ding, Y.; Jiang, X. A Novel Highly Stereoselective Synthesis of the A-Ring of Taxol Via Two Aldol Reactions. J. Chem. Soc., Chem. Commun. 1995, 1693 DOI: 10.1039/C39950001693.

There are several arguments. Firstly, five-membered ring formation is generally kinetically more favourable than 6-membered ring formation. As ring size increases, the entropy of activation also increases – this is explained in greater detail in Clayden 2ed, pp 805–807. It is true that six-membered rings are less strained, but that is generally more of a thermodynamic consideration, and from a kinetic point of view the entropy factor happens to outweigh the enthalpy factor.
Secondly, for cationic ring-opening reactions (e.g. reactions of halonium ions, or epoxide opening under acidic conditions), it is usually the more substituted end that preferentially reacts. See e.g. Regioselectivity of bromination of alkenes. This is also explained in Clayden 2ed, pp 436–438.
However, probably the most relevant factor here is stereoelectronic in nature. Stereoelectronic effects refer to effects which depend on the spatial orientation ("stereo") of orbitals ("electronics"). In the case of an SN2 reaction, the key requirement is that the lone pair of the nucleophile (Nu) must be able to reach the σ* orbital of the C–X bond being broken. Since the σ* orbital points out of carbon diametrically opposite the leaving group X, this is equivalent to saying that we need a collinear Nu···C···X geometry in the transition state.
For ring-forming reactions, there are a set of guidelines – Baldwin's rules – which empirically describe whether these reactions are stereoelectronically favoured (i.e. whether the nucleophile and electrophile are capable of orienting themselves in a geometry that allows for reaction). In the case of six-membered ring formation via substitution at a tetrahedral carbon, the reactions are labelled 6-exo-tet or 6-endo-tet depending on whether the leaving group X is outside the 6-membered ring (exo), or inside the ring (endo). [Technically, 6-endo-tet reactions are not ring forming.]

It turns out that although 6-exo-tet reactions are perfectly OK, 6-endo-tet reactions simply do not happen, as the nucleophile cannot stretch itself to reach the C–X σ*.
The opening of a three-membered ring is somewhat intermediate between these two cases. Because the C–X bond is constrained in a small ring, the reaction leans towards being a disfavoured 6-endo-tet reaction, even though formally it is 6-exo (as the leaving group is outside the ring being formed):

The lack of overlap isn't obvious from the 2D diagrams here, but bear in mind that these compounds are not flat – especially in the case of the epoxide, where if the epoxide is in the plane of the paper, the alkyl substituent must be pointing up towards us (or down away from us).
As a result of this, ring opening of the iodonium ion to give a five-membered ring is much more favourable.

A similar example is seen in acid-catalysed epoxide opening.[1,2] The formation of the 6-membered ring (the tetrahydropyran) is not observed at all:

References

Janda, K.; Shevlin, C.; Lerner, R. Antibody catalysis of a disfavored chemical transformation. Science 1993, 259 (5094), 490–493 DOI: 10.1126/science.8424171.
Na, J.; Houk, K. N.; Shevlin, C. G.; Janda, K. D.; Lerner, R. A. The energetic advantage of 5-exo versus 6-endo epoxide openings: a preference overwhelmed by antibody catalysis. J. Am. Chem. Soc. 1993, 115 (18), 8453–8454. DOI: 10.1021/ja00071a067.

Although both 5- and 6- membered rings are thermodynamically stable, 5-membered rings are kinetically faster to form. See the referenced JACS paper and figure for an example. You may also be interested in reading Baldwin's rules for ring closure.
Baldwin, J. E. J. Chem. Soc., Chem. Commun. 1976, 734–736.

Casadei, M. A.; Galli, C.; Mandolini, L. J. Am. Chem. Soc. 1984, 106, 1051–1056.


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What is the product of the following series of reactions?

.

The answer given to this question is .
This is how I attempted solving the problem.
Considering that my answer and the author's answer is differing only by a methyl group, is this a publishing error or am I wrong somewhere?

Himanshu pandey, Advanced Problems in Organic Chemistry, 9th edition page 219 question number 4

Yes, it is an error and your answer is correct. I have the same book and I have also got the same answer as you

It is evident from counting the carbons and the types of reactions performed that they are missing a carbon. Nice work, anyway.

Not sure there is much further to be said, maybe @ringo could convert that into an answer? Anyway there's an anon VTC flag on this, and it doesn't constitute homework, and indicating that is the purpose of this comment. Ugh, awkward.


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What will be the possible conditions to react bromo alkane with diethyl phenylmalonate?

How to react 1-bromopentane with diethyl phenylmalonate?
I used NaH (1.5eq) but the yield is less than 30%. I used KOH but yield is low. I used tBuOK and it did not work, I used K2CO3, but not working.
I tried it, K2CO3 2eq with 18-crown-6 and KI 0.1eq, in acetone but starting material is not consumed, almost only 30% has been finished after more than 72 hours and TLC shows a different spot, it does not look like desired product. Starting material is recovering when I use K2CO3, but not in case of NaH 1.5eq with DMF or THF but yield is less than 30%. Toluene with NaH under reflux condition is also behaving like K2CO3 and my next step is hydrolysis and decarboxylation.
I need guideline about what I am missing here.

Ethoxide in ethanol would work

i used NaH(1.5eq) but yeild is less then 30.

i used KOH but yeild is low. i used tBuOK not work, i used K2CO3, but not working

Are you recovering starting material? K2CO3 in MeCN with some added KI is worth trying

i tried it, K2CO3 2eq with 18-crown-6 and KI 0.1eq, in acetone but starting material is not going to consume, almost only 30% hasbeen finished after more then 72hours and TLC shows a different point, its not looks like desired product.

I suggest you edit your question to put all this information in it and then it  can be reopened.

yes, starting material is recovering when i use K2CO3,  but not in case of NaH 1.5eq with DMF or THF but yeild is less than 30.

What is your next step in the synthesis? Hydrolysis and decarboxylation of one of the ester groups?

How good is your NaH? If the quality is OK, you may be getting elimination.

You have tried a number of sensible approaches and nothing is working well. I think the problem here is that the anion you are creating is very stable and hindered and no amount of tweaking will significantly improve it. You either accept the yield as it is and scale the reaction accordingly, or change your strategy.
Your next step is hydrolysis and decarboxylation to give a substituted phenylacetic acid so I suggest you start with ethyl phenylacetate. You could alkylate it with NaH (1eq) in DMF with KI then 1-bromopentane. That should alkylate well, but there will be a percentage of di-alkylation so I'm going to suggest you condense ethyl phenylacetate with pentanal and then hydrogenate (or otherwise reduce the double bond). I think this will give you a cleaner reaction mixture and higher yield.


text:
Do xenon di- and tetrafluorides react with glass?
A question came in my test asking which of the compounds are not stored in glass. The answers were $\ce{HF},$ $\ce{XeF6},$ $\ce{XeF4},$ $\ce{XeF2}$.
I know about $\ce{HF}$ and $\ce{XeF6},$ but not about the other two. As far as I remember, xenon does not have any oxide or oxyfluoride in the $+4$ and $+2$ oxidation state.
Is the given answer correct, and if so, what are the reactions involved?

Interestingly, when a professor brought a small amount of $\ce{XeF2}$ into the lecture hall to pass around as a demonstration, I recall it being in a glass vial. However, if it was he could have just filled it into a glass one from a plastic one to prevent the plastic smell from obstructing the $\ce{XeF2}$ smell.

@Jan - it isn't like adding some HF to a glass bottle will instantly dissolve the glass. Similarly, the glass will probably last just fine with a bit of gas in it.

@JonCuster XeF2 isn't gas in r.t.. Also it shouldn't attack glass unless there's been for ex. significant hydrolysis or high temp.

Firstly I'll note the title and the body of the question ask slightly different things; just because substance X doesn't react with glass doesn't mean substance X should be stored in glass.
As far as I can tell $\ce{XeF4}$ and $\ce{XeF2}$ don't react with glass, at least not quickly. Greenwood and Earnshaw says of $\ce{XeF6}$, when comparing it with the other two fluorides, "the compound cannot be handled in glass or quartz because of a stepwise reaction" (see Why is xenon hexafluoride not stored in glass?). Housecroft and Sharpe says "At 298K $\ce{XeF6}$ reacts with silica (preventing the handling of $\ce{XeF6}$ in silica apparatus) ... while $\ce{XeF2}$ and $\ce{XeF4}$ do so only when heated". So it appears the lower halides do not react with glass at s.t.p.
Whether they should be stored in glass depends, I am guessing, upon the importance of hydrolysis by water in the atmosphere; this forms HF which will attack the glass. I can't answer this - I only mix compounds in computers.
Note an oxide of Xenon is known in the +4 state (see https://en.wikipedia.org/wiki/Xenon_dioxide) as is at least one oxyfluoride (https://en.wikipedia.org/wiki/Xenon_oxydifluoride)


text:
How do you calculate the delta H for multiple reactions?
I am having difficulty in a question:

Given that
1. $\ce{N_2(g) \rightarrow 2N(g), \Delta H = 941 kJ/mol}$
2. $\ce{N_2(g) \rightarrow N_2^+ + e^-, \Delta H = 1501 kJ/mol}$
3. $\ce{N(g) \rightarrow N^+ + e^-, \Delta H = 1402 kJ/mol}$,
Find the energy interval that produces atomic nitrogen in the flask but not any ions

I think that the energy interval should be $ 941 \leq E \leq 941 + 1402$, but the textbook says that the energy interval should be $ 941 \leq E \leq 1402$.
I think I am correct because $\ce{1402 kJ/mol}$ is needed to be supplied after the initial $\ce{941 kJ/mol}$ that is used to break the $\ce{N_2}$ bond.
Can someone please see the flaw in my reasoning?

A reaction can proceed with reactants have energy greater than or equal to the activation energy. The enthalpy change of reaction does not need to be taken into consideration. In this case, considering the reaction energy profile of an endothermic reaction, it is likely that the activation energy required is larger than the delta H values.

Find the energy interval that produces atomic nitrogen in the flask but not any ions

This is the key. The question wants you to determine how much energy you need to supply just to only have atomic nitrogen, no ions.
If you supply $941~kJ/mol$, then you have supplied enough energy to move reaction 1 forward. In your flask, you would have $N(g)$ inside. Now, if you further supply energy up to $1402~kJ/mol$, then you have enough energy to make reaction 3 proceed, and that will convert your atomic nitrogen to nitrogen ions, so that's a no-no.
So you have to maintain $E$ that is in between $941~kJ/mol$ to $1402~kJ/mol$ to prevent reaction 3 from proceeding.
If you supply $1402~kJ/mol$ after breaking the bond, then you will definitely produce $N^+$ ions as you have met reaction 3's $E$ requirement.


text:
Dependence of Osmotic Effects on Particle Size
van't Hof's equation for osmotic pressure is derived with the assumption, among others, that the particle size of the solute is comparable to the particles size of the solvent. It holds well so long as they are at least within an order of magnitude or two of each other -- but clearly it breaks down eventually, because you don't get much in the way of osmosis with a suspension of sand, or (to be silly) a handful of ball bearings.
I'm interested in the middle range, here -- in particular, I'm assuming there is a middle range, where particles contribute somewhat to osmolarity but at a lower rate than unity. Is there a decent derivation -- or a formula, I'd accept a formula but I'd prefer a derivation -- for the dependence of osmotic pressure (or potential, or otherwise closely related property) on particle size?

Cross-posted by OP at https://physics.stackexchange.com/questions/449203/dependence-of-osmotic-effects-on-particle-size


text:
Reversible and irreversible cyclic processes
The answer in Arihant's book is given as (b), but wouldn't (c) be correct as well? This is a question of IIT JAM Chemistry.

A system undergoes two cyclic processes 1 and 2. Process 1 is reversible and Process 2 is irreversible. The correct statement relating to the two processes is
  (a) $\delta S=0$ for Process 1, while $\Delta S\neq0$ for Process 2
  (b) $Q_\text{cyclic}=0$ for Process 1 and $Q_\text{cyclic}\neq0$ for Process 2
  (c) More heat can be converted to work in Process 1 than in Process 2
  (d) More work can be converted to heat in Process 1 than in Process 2

Yes, you are correct

Answer b is incorrect for the reversible Carnot cycle.  The book has given the wrong answer. Answer c is correct if only one cycle of each process is carried out.  You can obviously convert more heat to work if you run process 2 through the same cycle several times, which also counts as a cyclic process.

Well Sir, it's given that from Clausius inequality, for reversible process, q(cycle)=0 and for irreversible process , q(cycle) less than 0 . I agree with you on the latter.

That is NOT what the Clausius inequality says.  Are you saying that you think that Qcyclic is zero for the reversible Carnot cycle.

So, what do you think?  Do you still stand by what you said about the Clausius inequality indicating that, for a reversible process q(cycle) = 0?

Yes sir, actually this answer given in the book got me a bit confused. I understood that q(cyclic) is not 0 for a reversible cyclic process. But, would q(cyclic) for an irreversible process be less than q(cyclic) for reversible process?

Depending on the reversible path selected, it could be less or it could be more. But, the change in entropy for any reversible cyclic path is zero.   Please understand that I am asserting that, contrary to what your book says, answer b is incorrect.

Also, of course, the change in entropy for any irreversible cyclic path is also zero.

Yes sir I understand that answer b is incorrect. But sir, (with reference to your last comment that ∆S for any irreversible cyclic change is also zero), doesn't the Clausius inequality say that it's less than zero?

No.  Entropy is a function of state.  So, in a cyclic process, where the final state of the system is the same as the initial state of the system, the change in entropy is zero for both reversible and irreversible processes.  The Clausius inequality says that the change in entropy for the combination of system and surroundings is greater or equal to zero.  So, for an irreversible cyclic process, all the entropy generated in the process is transferred to the surroundings.

A is correct because
B can't b correct as heat is a path dependent process.
Nothing can be commented over C and D because we are not sure about the type of thermodynamic process mention in the question.
Here A can be inferred for sure as entropy for cyclic reversible process is constant.


text:
Topic in Salt hydrolysis that I couldn't understand
Salts can be made up of all combination S of ( SA ,WA) and (SB , WB)
I understand how to deal with salts of strong acids with both strong and weak bases and also of strong bases with weak and strong acids but I want to understand how can we find the pH for the salts of weak acids and bases ? Also If you could explain the relationship of hydrolysis constant and the dissociation constants it would be appreciated.
Thank you in advance

You solve for the same equilibrium in the same way, except there are more variables and more constraints.

What constraints please elaborate

https://chemistry.stackexchange.com/questions/75047/ph-of-ammonium-acetate-solution/99988?sfb=2#99988

The two equilibria are coupled because the hydronium ion concentration is the same in both, since it's the same species.


text:
Calculate molecular orbitals of high-lying or Rydberg states in GAMESS (US)
I am trying to calculate the high-lying or Rydberg orbitals (i.e. unoccupied orbitals with their energies just below the ionization energy) of a nitrogen molecule.
I use GAMESS(US) with the following input file.
!   File created by the GAMESS Input Deck Generator Plugin for Avogadro
 $BASIS GBASIS=N311 NDFUNC=3 NFFUNC=0 NPFUNC=3 DIFFSP=.TRUE.
 NGAUSS=6 DIFFS=.TRUE. $END
 $CONTRL SCFTYP=RHF RUNTYP=ENERGY DFTTYP=SLATER MAXIT=99 $END
 $SYSTEM TIMLIM=1440 MWORDS=500 $END

 $DATA
Title
C1
N     7.0     0.00000     0.00000     0.00000
N     7.0     0.00000     0.00000     1.12001
 $END

The output file contains 70 orbitals, but the high-lying orbitals seem to have too big energy.
For example, LUMO+62 has 42.8618 Hartree (or 1166 eV), which is far above the ionization energy of a nitrogen molecule.
How can I correctly get the high-lying or Rydberg orbitals?

one recommendation I would make is to use a larger basis set. I don't have a formal theory behind this, but I imagine to get a decent description of the $n^{th}$ MO you need significantly more than n MOs. In your case, I would think you might need at least 100 or so. Also, I don't know GAMESS well, but where is the spin set? I know N2 can give some difficulties due to the choice of spin state.

I found a large basis set called PCseg-4. This basis gives more than 200 MOs. For the spin, the MULT option in the $CONTROL group specifies the spin multiplicity of the electronic state (i.e. singlet or triplet). I use the default setting, which means I only calculate the singlet states.

There was a Molcas presentation talking about special Rydberg basis sets; I don't know much about that but maybe you need such a BS.


text:
Can benzene ring be the part of a acyl group
Can benzene ring attached to c=o i.e carbonyl group be considered as a acyl group?

Yes, acyl groups can be derived from all kinds of oxoacids.
According to the Glossary of class names of organic compounds and reactivity intermediates based on structure (IUPAC Recommendations 1995), acyl groups are defined as follows:

Groups formed by removing one or more hydroxy groups from oxoacids that have the general structure $\ce{R_$k$E(=O)_l(OH)_m}\ (l\neq0)$, and replacement analogues of such acyl groups. (…) In organic chemistry an unspecified acyl group is commonly a carboxylic acyl group.

Especially concerning acyl groups derived from carboxylic acids in organic chemistry, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) defines:

Carboacyl groups are $\ce{R-CO-{}}$, $\ce{-OC-R-CO-{}}$, or $\ce{-OC-R-[R'-CO-]_$x$-R''-CO-{}}$ groups and their functional replacement analogues, where $\ce{R}$, $\ce{R'}$, and $\ce{R''}$ are chains, rings, or ring systems, derived from carboxylic acids by the removal of the hydroxy group from each carboxylic acid group that is expressed by the suffix, and $x = 1 ,2,3$, etc.
  (…)

In particular, benzoyl is explicitly mentioned as an example for an acyl group.


text:
Can sp2 hybridised carbon show chirality?
I know that any carbon to show chirality has to be bonded to 4 different atoms (groups of atoms). So a double bond is
considered as that the carbon is achiral. I learnt from my lecturer that dissubstituted allene is an exception. If it is true that sp2 hybridised carbon cannot show chirality, are there any other exceptions?

Hexahelicene (and similar molecules) consist entirely of sp2 carbons but are chiral because of steric effects that constrain their overall shape.

Then strictly we do not have sp2 hybridization.  You have to allow some latitude in the interpretation,

Carbons (or any other atoms) don't show chirality; molecules do.

Ivan Neretin then what about chirality centres

Adesh: Then you have to speak about a structural part of a molecule or even its whole. Not about a C atom. And strictly speaking @Ivan Neretin is correct. A single atom can be seen at best as a chiral center.

Chirality is not necessarily produced by chiral centers.

You seem to be confusing asymmetric carbon (centres) with overall chirality (of molecules).

You are confusing two different concepts—and I don’t blame you because many chemists don’t cleanly separate the two concepts.
One concept is that of an asymmetric (carbon) atom. To establish whether a carbon atom is asymmetric, compare it to its mirror image. If you can overlap perfectly only by rotation and translation, the atom is symmetric; if the two mirror images act as left and right hands, it is asymmetric. Since $\mathrm{sp^2}$ carbon atoms are, by definition, in a planar environment, they will always be symmetric because a plane of symmetry will always traverse them.
The second concept is overall chirality of a molecule. For this, you basically do the same thing as you did for an individual carbon except you mirror the entire molecule and attempt to overlap it with the original molecule. This, of course, depends on the overall 3D structure. Again, if the molecule as a whole contains a plane of symmetry, it is by definition achiral.
The two definitions of asymmetry are nowhere near equivalent. It is possible for a compound to be composed entirely of asymmetric carbon atoms and yet be achiral, e.g. meso-1,2-dichloro-1,2-difluoroethane. Likewise, it is common for a molecule that consists entirely of symmetric carbon atoms to be overall asymmetric like BINOL, allenes or helicenes. Thus, one should always carefully separate the two concepts to not accidentally cause a mix-up.

As correctly stated in a comment by @Ivan Neretin, chirality is not bound to stereogenic centers (e.g. the textbook examples of carbon atoms with four different substitutents, or chiral sulfoxides) only.
Returning to your question, an other option to build chirality is of course axial chirality, where the allenes -- formally built from $sp^2$-hybridized carbon atoms -- offer enantiomeric forms.

(source)


text:
Is there a way to check reactions?
I'm very new to organic chemistry.
I would like to practise simple reactions (HX + Alkenes, X2 with Alkenes and so on) with examples i create myself. Is there a way to check if my products are correct?
I have access to Chemdraw and ChemSketch, however I don't know how and if these tools can be used to do that, i.e. draw a reaction , or two educts and let them show the product.
Is there any software available which can do that?

I don't think so. There are a lot of introductory OC books with exercises.

Not really exactly what you're looking for, but if you have access you can look up a reaction up on SciFinder with its drawing functionality. If the products are too simple you might just find that people will buy it as starting material from a supplier, though.

In August 2018 IBM launched a free cloud service called IBM RXN. This neural network is quite young, but it is constantly being trained, and the prediction should get better over time. I haven't used it all that extensively, but it seems it works quite adequately for simple reactions, including the OP's case. For example:

They also published a great paper [1] which describes on which base their AI operates.
References

Schwaller, P.; Gaudin, T.; Lányi, D.; Bekas, C.; Laino, T. “Found in Translation”: Predicting Outcomes of Complex Organic Chemistry Reactions Using Neural Sequence-to-Sequence Models. Chemical Science 2018, 9 (28), 6091–6098. https://doi.org/10.1039/C8SC02339E.
(Open Access)


text:
Is it possible to convert methane and carbon dioxide to acetic acid?
Please forgive a bumbling amateur.  I have been looking up bond strengths and it seems to me that it should be possible to combine methane with carbon dioxide to make acetic acid:
$$\ce{CH4 + CO2 ->[\color{red}{?}] CH3COOH}$$
One has to break a C=O bond (745 kJ/mol) and a C–H bond (413 kJ/mol), but one gets back a C–O bond (358 kJ/mol), an O–H bond (467 kJ/mol) and a C–C bond (347 kJ/mol) so one comes out ahead.  I know that bond energies vary slightly but this doesn't sound too far out of whack.  Wouldn't it be rather nice if one could catalyse this reaction and soak up lots of these readily available gases? Could we induce some bacterium to do it for us?

@ThomasForster The problem with reactions like this is that, while they might look good on paper, they are practically hard to do not least because they require complex processes (which are expensive) and large energy inputs (though biological systems can do similar things, but those are hard to scale up).

In particular, methane is very hard to selectively activate for such reactions (burning is easy, selectively reacting it is not).

One might as well just use methyl lithium which actually easily reacts that way. But still … problems!

Can't you activate methane to methanium in a superacid?  What would methane plus CO2 do in a superacid?

Seems you still not get something. There may be a dozen of methods to make acetic acid from methane and CO2 - all of them workable, but al have nothing to do with lowering CO2 emission, rather the opposite.

@ThomasForster The reason things are not done on an industrial scale is usually economics. The expense of the materials, equipment and the energy inputs have to make sense given the price you can get for the products. Industry uses processes like Saabre (syngas to acetic acid) and Cativa (methanol to acetic acid) which use readily available inputs. Using carbon dioxide would require much costlier plant and a more complex process.

@matt_black, could I encourage you to turn your comments into an answer?

Already methanotrophic bacteria can oxidize methane to formate and the Wood-Ljungdahl pathway can take this formate and a carbon dioxide molecule and coenzyme A and produce acetyl coenzyme A so the catalysts are already there.

So this process could be scaled up...?


text:
Is volume of strong base used with strong acid same as volume used with weak acid?
At the equivalence point, why do we take volume of strong base used for titration of strong acid the same as the volume used for titration of weak acid? Shouldn't there be a difference? For example, in case of titration of CH3COOH against NaOH, even though as the small amount of H+ ions(initially) react with OH- ions and take the dissociation of CH3COOH in the forward direction, shouldn't the common ion effect of CH3COO- ion from CH3COOHNa suppress the dissociation of CH3COOH equally?

I meant CH3COONa.

Related: https://chemistry.stackexchange.com/questions/60407/why-will-a-strong-acid-neutralize-as-much-base-as-a-weak-acid/60409#60409

A weak acid exists in an equilibrium and once you start adding alkali $\ce H^+$ is removed disrupting the equilibrium meaning more acid molecules will dissociate - the equilibrium position shifts right.
$$\ce {H_xA <=> xH^+ + A^{x-}}$$
So by the time you've added all the alkali you will have neutralised every acid molecule - even if it wasn't dissociated in the beginning. So as long as you add the same number of moles of acid and alkali no matter if they are strong or weak, at the equivalence point the number of moles of acid and alkali neutralised will be equal.

What causes this to happen? Shouldn't the common ion effect of CH3COO- ion from CH3COONa suppress the dissociation of CH3COOH equally?

When the $\ce {NaOH}$ reacts with the acid it is removing $\ce {H_3O^+}$ from the mixture which due to le chatelier principle drives the rate of the forward reaction replacing the $\ce {H_3O^+}$ which has been used up, meaning the moles of $\ce {H_xA}$decrease. this process continues until there is no $\ce {H_xA}$ left. Also if no more $\ce {H_xA}$dissocates the moles of $\ce {H^+}$ in the mixture would be zero stopping the reverse reaction meaning the $\ce A^{x-}$ acts as if it isn't there. The $\ce {H^+}$ is what is important here not $\ce A^{x-}$


text:
Do redox reactions always contain pure elements?
According to a video by the Organic Chemistry Tutor, a quite well-known chemistry channel on YouTube, you can easily identify a redox reaction by seeing if there are atoms in their elemental states on one side of a reaction and form compounds (or compounds decomposing to elemental atoms) on the other. Is this accurate?

Not quite. True, all such reactions are redox. But not all (and not even most) redox reactions are like this.

Do not try to learn chemistry via moronic youtube videos. You can be sure they are going to tell you some supersimplyfied nonsense, which breaks the  second you try to use it on sth else.

Counterexample : K+Na -> NaK https://en.wikipedia.org/wiki/Sodium-potassium_alloy

The problem is that until you have learned a good deal of chemistry, you can't tell moronic videos from the good ones. Ditto for teachers and textbooks.

Some textbooks are however known to be good. I wouldn't say the same about youtube channels. ;-)

@Karl Then again, maybe some youtube channels are also known to be good, just not to you or me? I'd rather give them a benefit of the doubt. (I don't watch any, and don't plan to anytime soon.)

@IvanNeretin Neither do I. ;-)

I'm voting to leave this open. It's based on a common guideline for identifying redox and the misinterpretations that surround it. Perfectly valid question IMO, and not homework.

Here is a redox reaction that does not contain any zero-valent atom:
$$\ce{2FeCl3 + SnCl2 + 2 HCl -> 2 FeCl2 + H2[SnCl6]}\tag{1}$$


text:
How can I get self-diffusion coefficient from MSD (mean sqaure displacement)
I run some molecular dynamic simulations for polymer system (liquid), and I have data of MSD vs t(time). MSD-t relation looks linear proportional. Now, which one is the proper way to calculate the self diffusion coefficient?
1) D = MSD / 6t (for 3D system) and get the asymptotic value at large t.
2) Slope of normal plot of MSD-t
3) Slope of log-log plot of log(MSD)-log(t)
4) Slope of semi-log plot of MSD-log(t)
I searched for literature's but there are too many different methods, not sure what to choose here.

multiple-choice? homework?

In three dimensions, assumed to be isotropic, the equation is $\langle x^2\rangle=6Dt$ so the slope of the plot of $\langle x^2\rangle$ vs $t$ will give the diffusion coefficient. However, this may vary as time proceeds depending on just how well the average is calculated and of course on the system you are studying.

I think when the slope of number 3 (log-log) is near or equal to 1, the result that you get by number 2 is more accurate, even you can use only a part of the number 2 plot that has a slope in a log-log manner near to 1.

I would usually recommend the second option (fitting to the slope of the MSD at long times) over the first (dividing MSD by the time). If you compute $\langle \Delta r^2(t)\rangle/6t$, there is a residual effect of the velocity correlations at short time.
Suppose the velocity correlation function decays exponentially in time
$$
\langle \mathbf{v}(0)\cdot \mathbf{v}(t)\rangle = \langle |\mathbf{v}|^2\rangle \exp(-t/\tau) .
$$
This is related to the diffusion coefficient by
$$D= \frac{1}{3} \int_0^\infty \langle \mathbf{v}(0)\cdot \mathbf{v}(t)\rangle \, dt
=\frac{1}{3}\langle |\mathbf{v}|^2\rangle\tau .
$$
Since
$$
\Delta \mathbf{r}(t)=\mathbf{r}(t)-\mathbf{r}(0)=\int_0^t \mathbf{v}(t') \, dt',
$$
it follows that
$$
\frac{d}{dt} \langle | \Delta \mathbf{r}(t)|^2 \rangle =
 6D\, [1-\exp(-t/\tau)] ,
$$
and this converges towards the correct answer, $6D$, exponentially fast in the time $t$.
On the other hand
$$
\frac{1}{t} \langle | \Delta \mathbf{r}(t)|^2 \rangle =
 6D\, \left[ 1 - \frac{\tau}{t} + \frac{\tau}{t}\exp(-t/\tau)\right]  .
$$
This converges more slowly, proportional to $t^{-1}$.
Looking at a log-log plot may be useful, as it may highlight changes between different regimes of behaviour (for example, at long times) that you would miss otherwise. This might be especially true of a polymer system like yours. But your option (3) "slope of log-log plot"  does not give you the diffusion coefficient! It just gives you an idea of whether the behaviour is diffusive (slope $=1$) or not (slope $\neq1$). And option (4) "slope of semi-log plot" does not give you $D$ either.

Thank you. Do I need to divide the slope of MSD-time from linear plot by 6, for 3D system consideration? Or just slope of MSD-time from linear plot is Ds?

Divide the slope by $6$, assuming the MSD is the sum of the squares of all three $x,y,z$ mean squared displacements. The result should be quite close to the result you get by dividing the MSD by $6t$, but as my answer says, it has better convergence properties.


text:
How is occupation number of lowest unoccupied natural orbital defined?
I have run some calculations to analyse the occupation of natural orbitals. Since it is needed to run in triplet state, I get two orbitals with occupation number 1. Now, which orbital occupation number is to be taken as the occupation  number of lowest unoccupied natural orbital as it is needed to define the diradical character in the system ?

Why would the degeneracy in occupied orbitals raise any ambiguities for the unnocupied orbitals?

I am facing issue in defining it. The reported values and that from my calculations do not match. May be my procedure is wrong but I am not able to identify where is the fault. @Feodoran

I don't know what values you are talking about, but if you have two singly occupied orbitals (SOMO), then the LUMO should be the one next lowest in energy. For natural orbitals it probably has some occupation number slightly larger than 0. But I am not aware of any strict definition here.

thank you @Feodoran that is also my line of thought but then my diradical character values are not matching with the reported values (e.g. for polyacenes)


text:
How do I neutralize 1.2 % muriatic acid (134.4 liters) residue with 99 % caustic soda flakes?
I cleaned my commercial water purification system with muriatic acid but I overdid it. Now there is excess muriatic acid residue. I used 6 liters of  29 % muriatic acid and further reduced with 33.6 gallons (134.4 liters) water as solvent at 1.2 % concentration of muriatic  hcl.I believe the mole for the solution is 3.29 moles, is that right?
I have  99 % caustic soda flakes that I can use to neutralize muriatic acid residue. How do I go about it? How do I mix caustic soda at the right amount to neutralize the muriatic acid residue?
I'm not a chemist or a student so  I would really, really appreciate your help guys. I'm on the verge of losing my small business if I don't fix this.

Better using backing soda and you just add it until gas evolution ceases. Don't forget standard safety behavior.

I can't. I have to mix this in the tank and run through the whole machine provided I allow the caustic soda to settle down a bit after a few minutes. Besides that baking soda leaves a residue. I tried it once, didn't like it.

Hi rich, I'm inclined to believe that the community closed this for the wrong reason (it does look like a homework question, funnily enough, but I'm sure that this is an actual problem you are facing).  Your edit has put it into the reopen queue so that people can vote to reopen it.  I'm not a chemist of any import, but I would say that the amount of flakes that you put in would depend on the amount of the residue, not necessarily the amount of solution that you originally put in.

Thank you Jonsca for your reply. I measured the water. It measured 5.10 ph. Usually chlorinated water measures around 6.30 ph. And I could really taste the acidity of the water though it has lessened somewhat after a few days. The acidity even sticks to my gums when I taste it. Isn't there a way to calculate  how much caustic soda I would need to adjust the ph judging from the measurement I took ?

Of course there is a way. Problem is I cannot trust the dara you provide

For a rough calculation assume density is about 1. The amount of hcl is the 1% of about 134 kg. So 1340 g of HCl. This divided by the molecular way gives about 37 moles. Then multiplying by the soda mw gives you about 1480 g of caustic soda. Use goggles.

Alchimista, theres no need to be rude to me.                                                     The ph meter is the only instrument I have to measure the water.

Rich I wasn't rude!  I wanted to say that I take the conc with care as otherwise you could have answered yourself. However at about 1 percent whatever you refer to mass/mass or msss/volume there is w much change and I decided to give you the receipt. You will end up at more or less the original pH of your water.  Br careful handling caustic soda.

Ok, I 'm a bit stressed right now. I'm a small businessman, not a chemist so you got to a bit patient with me.

The 134.4 liters is the water I mixed with 6 liters of  29% Hcl. Bringing down the concentration to 1.2 % Hcl.  Isn't 1480 grams ( or more than a kilo) a bit too much ?

I refer to title!. I will look for what you have done step by step but later. ...

Let us make it simple.  The current molarity of your solution is 10 to the power -pH so it is about 10 to minus 5.2. You have this amount of moles HCl for each liter. 0.000052 times Volume of water times 40 gives you the amount of caustic soda in grams.

:))) actually 10 to minus 5.2 equals to 0.0000063. The rest is unchanged :)))

Eh.. you lost me. How many grams ? I studied history back in college, not math because math confuses me.Alchimista I really appreciate your help.

Is that 3.3 grams  of caustic soda needed ?

I think it would be a bad idea to calculate a certain amount of caustic soda and then add this total calculated amount. Your assumptions for the calculation could easily be wrong. If replacing the water in the system is not an option, I would rather add small amounts of diluted NaOH and measure the pH after each step. Be sure that it has mixed with the water in the system properly to avoid false measurements.

0.0000063 × 40 × actual V in liters = grams of NaOH.

@Aventurin gave a good advice too. That brings all back to my first comments ;)

Alchimista, I just want to clarify. The answer to your calculation is .00338688. Does that mean I move the decimal point twice? That's .33 grams of caustic soda needed ?


text:
Find order of reaction with respect to A and B and find the rate law expression for 2A+B -> C

Please help me in this chemistry question

Try to look at two entries where either concentration of A or B is constant, and observe how the the rate and the concentrarion of the other reactant varies

Can you please give more information, this is little to begin with

For example, in experiment 1 and 2, conc of A is constant. Also, you can see that as concentration of B changes, rate of reaction remains unaffected by that change. So this means that the rate would be independent of concn of B. Now similarly,try for A

1- Write the rate law expression :
 $r=k[A]^x[B]^y$
2-Take experiment one and two :
 $$5.07\times{10^{-5}}=k[0.2]^x[0.3]^y\tag{1}$$
$$5.07\times{10^{-5}}=k[0.2]^x[0.1]^y\tag{2}$$
3- Divide equation one by equation two:
$$\frac{5.07\times{10^{-5}}}{5.07\times{10^{-5}}}=\frac{k[0.2]^x[0.3]^y}{k[0.2]^x[0.1]^y}\tag{3a}$$
$$1=3^{y}\tag{3b}$$
$$3^{0}=3^{y}\tag{3c}$$ ,so:y=0(the order of reaction with respect to B), now : the rate law expression :
 $r=k[A]^x[B]^{0}=k[A]^x$
4-Take experiment two and three:
 $$5.07\times{10^{-5}}=k[0.2]^x\tag{4}$$
$$1.43\times{10^{-4}}=k[0.4]^x\tag{5}$$
5-Divide equation five by equation four:
$$\frac{1.43\times{10^{-4}}}{5.07\times{10^{-5}}}=\frac{k[0.4]^x}{k[0.2]^x}\tag{6a}$$
$$2.8=2^{x}\tag{6b}$$
$$log2.8= x log 2\tag{6c}$$,so:$x\approx{1.5}$(the order of reaction with respect to A), now : the rate law expression :
 $r=k[A]^{1.5}[B]^{0}=k[A]^{1.5}$


text:
Strong or weak force between protons and neutrons?
I had learned that what keeps the protons and neutrons bound is the strong force and the weak force has to do with radioactive decay. but today I read an article from Science Daily that contradicts this. Here is the quote:
“Protons and neutrons are made of smaller particles called quarks that are bound together by the strong interaction, which is one of the four known forces of nature: strong force, electromagnetism, weak force and gravity. The weak force exists in the tiny distance within and between protons and neutrons; the strong interaction confines quarks in neutrons and protons.”
Can this be clarified by the community? Thanks!

It is saying the weak force can exist between protons, neutrons and quarks but it isn't saying it is the force that holds them together.

This is outside the realm of chemistry and would be more appropriate on Physics SE

@Tyberius - I'll skip my usual comment about Glenn Seaborg, the field of nuclear chemistry, and winning chemistry Nobel prizes... (Or not as it may be!).

While I don't like closing nuclear questions, because it is about particle, structure, not related to reactions, I am inclined to say this is off topic in regards to chemistry and better suited for Physics.SE.

@JonCuster I don't mean to deny nuclear the existence of nuclear chemistry, but I think when you get into discussing quarks, you are no longer dealing with chemistry. Really, I would say a lot of the more recent transuranium elements synthesized are borderline chemistry at this point because their short lifetime has basically precluded analysis of their chemical properties, at least for the time being.

@Tyberius - I kind of kid, but the reality is that undergraduate chemistry courses typically do touch on protons and neutron and isotopes and whatnot. While I'm active on both Physics and Chemistry, it does seem to me that having answers to simple nuclear physics questions that arise in chemistry classes here should not be out of the question. For more in-depth answers there is always Physics. But just because it is on-topic there (perhaps closed as homework!) does not imply it is off-topic here.

@JonCuster I can agree with that, I'm all for having an answer from multiple fields where it's possible and keeping Chemistry's scope fairly inclusive.

I think it is a poor phrasing. The strong force is responsible for keeping quarks together inside a proton or a neutron, but also between quarks belonging to different proton/neutron (but I am not sure for proton-proton interaction because of the repulsive electromagnetic force).
By contrast, the weak force is responsible for turning one type of quarks into another type or quarks, thus transforming a proton into a neutron in the case of beta+ radioactive decay, with emission of a positron (the overall charge is always conserved).

Strong force holds protons together. It is actually called the residual strong force. It is unlike the strong force inside protons and neutrons where if you pull a quark you can never pull it out of a proton or neutron. Nuclear strong force has a range. The weak force is what makes protons and neutrons turn from one to another.


text:
How to detect equivalence points?
How do we detect equivalence point from a titration of an acid with a basic titrant from thу titration curve ($\mathrm{pH}$ vs. Volume of titrant added)?
Can we predict an equivalence point just by seeing the shape?
Some people say that

when $$\frac{\mathrm{d(pH)}}{\mathrm{d}V} = 0,$$ equivalence point has been reached.

Is it correct? I doubt if it is correct. It is also written that there is a distinct "break" in the curve at equivalence point. What is this break?

At equivalence point dpH/dV shows a peak, not 0.

This should provide all you need:  https://en.m.wikipedia.org/wiki/Titration_curve

You do not detect equivalence points. What you detect is the endpoint. This is a very critical distinction.
You have some feature you are explicitly measuring, be it pH or color or opacity.
Once this feature changes you have reached the endpoint. The endpoint is not the equivalence point because you will never have exactly the right amount. Frequently, you overshoot a bit.
Also, the reason that the endpoint is detectable is because the transition is very sharp and sudden. That is because near the equivalence point, the pH (and whatever pH derived feature you have) changes very rapidly. The derivative is definitely not zero. In fact, the derivative you've provided is never zero but monotonically increasing or decreasing depending on whether you're tritrating with acid or base. It has to. If you add more base, the pH is only going to increase in value.


text:
Calculating enthalpy change for a real gas

Calculate $∆H$ for $1$ mole of real gas undergoing change of state from $\pu{10^5 Pa}$, $\pu{300 K}$ to $\pu{2e5 Pa}$, $\pu{600 K}$. The values of real gas constants are $a = 0$ and $b = \pu{50 mL/mol}$.

Given: $C_p = 20$ (in SI unit),
$$ \left(\frac{∂H}{∂P}\right)_T = T\left(\frac{∂V}{∂P}\right)_T + V$$

My approach has been like this:
First, I wrote the real gas equation. Then I partially differentiated this whole equation w.r.t. $P$ keeping temperature constant so as to find $∂V/∂P$ and substituted the value of $∂V/∂P$ in the equation given in the question. My actual aim was to use the formula
$$\mathrm{d}H = nC_p\mathrm{d}t + \left(\frac{∂H}{∂p}\right)\mathrm{d}p.$$
Next, I substituted $(∂H/∂p)$ from the given equation in the the given formula, but now I am stuck here.
The given equation in question is incorrect (even dimensionally!) as stated in comment.
Real equation: $ \left(\frac{∂H}{∂P}\right)_T = V -T\left(\frac{∂V}{∂T}\right)_P $

Shouldn't there be a minus sign in from of the first term in your equation for partial of H with respect to P?

The equation should read $$\left(\frac{\partial H}{\partial P}\right)_T=V-T\left(\frac{\partial V}{\partial T}\right)_P$$

It is ∂V/∂P and not ∂V/∂T

No it's not.  That would give it the wrong units compared to the other terms.

OK that part u guessed from dimensional analysis but how are you so sure about the negative sign

I didn't guess anything.  I've worked with that equation many times before, and presented its derivation in several threads on various forums.  The equation is derived in every thermodynamics book that I am familiar with.

Okay now I see where all the trouble began from. It was the question that was wrong. I worked with the corrected equation and easily reached to the solution.It was the incorrect equation that was making all the mess. Working with the incorrect equation I got terms of temperature and while I knew I had to put the limits on pressure, I was totally clueless about which value of temperature to use. I was stuck here for 2 days. Thank u so very much. I think I should delete the question now. (Or should I correct the equation and leave it?)

I wouldn't do a thing if I were you.  Just leave it so other members can benefit if they encounter the same kind of situation.

Bdw how would have we dealt if the terms of temperature didn't disappear?I mean which value of temperature should be used, 300K or 600K

Then we would have had to use the 3 step procedure I discussed in my Answer, involving constant temperature and pressure path segments.

You started out correctly by getting the relationship for $\mathrm{d}H$ for your van der Waals gas algebraically. The usual approach to evaluating a change like this for a real gas is to break the change into three steps (since you will typically know $C_p$ only in the limit of ideal gas behavior at low pressures):
Step 1: Calculate $\Delta H$ for the change from $\pu{10^5 Pa}$ and $\pu{300 K}$ to $\pu{0 Pa}$ at $\pu{300 K}$.
Step 2: Calculate $\Delta H$ for the change from $\pu{0 Pa}$ and $\pu{300 K}$ to $\pu{0 Pa}$ and $\pu{600 K}$ (i.e., using the ideal gas heat capacity).
Step 3: Calculate $\Delta H$ for the change from $\pu{0 Pa}$ and $\pu{600 K}$ to $\pu{2e5 Pa}$ and $\pu{600 K}$.
Based on Hess' law, add the three changes together.
For this particular problem, the 3 step procedure is not necessary because of the following:
$$\left(\frac{\partial H}{\partial P}\right)_T=b$$
So,
$$\mathrm{d}H = C_p\mathrm{d}T + b\mathrm{d}P$$
Since $b$ is independent of temperature, $C_p$ is independent of pressure. So for this system, the equation integrates immediately to
$$\Delta H = C_p\Delta T + b\Delta P$$


text:
Raman spectra in Gaussian
I want to simulate Raman spectra of benzene in Gaussian 09.
Calculated spectra for a single ring Benzene ($\ce{C6H6}$) in Gaussian differs about $\pu{40 cm^{-1}}$ for each peak, compared with experimental Raman's databases.
How can I solve this problem?
Gaussian file created with GaussView6:
MP2 config

DFT config:

Parameters: (for optimization)

When discussion computational chemistry (and particularly when asking for advice), it's important to say which method(s) and basis sets you are using. Gaussian 09 is a software package with many very different quantum chemistry methods.

Did you scale it? Did you compare it to gas phase data?

@Persian_Gulf PJR is correct. The methods and basis set used are crucial for understanding what wrong with your results. If you could include your input script, that would be helpful in answering the questions.

@PJR I tried DFT and MP2. But it still is differs from experimental results. (MP2 is a little better).

@DSVA In Gaussview, I entered a single Benzene molecule. Is it possible to consider liquid phase in calculation?

For the DFT please also specify the functional (e.g. b3lyp ) since that also makes a huge difference. As @Tyberius said, the input script would really be the most helpful. The basis set you are using will also make a difference and the default is not necessarily suitable. In general, I would recommend increasing the basis set to something of at least triple zeta quality (potentially quadruple since benzene is so small). You also have the option of using an ab-initio method like CCSD or CCSD(T) but those will take considerably longer.

@PJR I added the config files made by GaussView

Liquid phase is possible at least as implicit solvent models via the scrf keyword.

+1. Sorry to see your question was closed. I'm trying to create a Stack Exchange just for computational chemistry and materials modeling, and questions like this would be very welcome. Do you think you can help support us by clicking commit?
https://area51.stackexchange.com/proposals/122958/materials-modeling?referrer=MGFkOGRlMTg1ODg2NTUxZDQ1MDVjZmQyOTg1ZDcxODQzMjc0OGIyMjBhZDM4ZDQ5MzM0MWE1YzZhNDBjMzhlYfHJBfp5EvPDPybryqjeDgXYucHU-wUv4jp4j6imjIfF0 Only 2.7% of committers fulfill commitment anyway, but it would be very very helpful if you could support us.

Thank you for attaching your input file.
The first place to start for improving these calculations is your choice of basis set. 6-31G is a rather small basis set and is likely a source of significant error here. If you're fond of using the Pople type basis then you would be better off using something like 6-31++G* so that you include polarization and diffuse functions. With your current basis I suspect that you are missing important polarization of the atoms upon distortion.
If you really wanted a very accurate answer I would recommend the following scheme. Optimize your geometry at the DFT level and then use that geometry as a starting place for a optimization and frequency calculation at a higher level of theory. For example use the cc-pvtz basis set (a correlation consistent basis set that's more efficient generally than the Pople type) and CCSD(T). This will provide a very accurate gas phase raman spectrum.
Edit:
The only additional Gaussian specific parameter you should need is the Geom=Checkpoint command for the CCSD(T) optimization and frequency run. This will use your existing checkpoint file as a geometry starting point.
The other keywords (basis and method) are exactly what they should be cc-pvdz and ccsd(t). For more information on basis keywords I recommend the users manual.

Thanks a lot for your answer. You mean I set the optimization (DFT/ccpvtz) and for frequency CCSD. may I ask you to  specify all parameters?

I added the parameter for optimization below my question.


text:
Organic Chemistry - Reduction
Can anybody help me with the product? Seems like Birch reduction, but Birch reduction is used for conjugated systems and alkynes.

Looks like isomerising the tri-substituted double bond

Can't find any such reaction in Reaxys...

Stupid question: Is the top of that structure cut-off to obscure a oxygen in the form of a carbonyl?

There exists a similar rxn Bouveault–Blanc reduction used for reduction of esters to primary alcohols and this reagent can reduce $\alpha,\beta-$unsaturated ketones

@Zhe it is not a carbonyl... I found the question as it.


text:
Is Al2O3 (aluminum oxide) diamagnetic or paramagnetic?
The title about covers it, I am an amateur and I have found that checking 'molecule orbitals' can reveal the answer here but I cannot locate such an orbital. The reason for this question is I am trying to sort aluminum oxide beads from aluminum oxide beads coated in palladium with a magnet via the Lenz Effect.
I have researched magnetic properties. It appears that I can determine them by seeing how many unpaired valence electrons there are. I have no idea how to figure out what the valence orbital of a molecule is, even though such information is widely available for individual elements. For molecules, this information is referred to as molecule orbitals. I expect that it is diamagnetic because both Oxygen and Aluminum are not noble, but this sentence should reveal my level of chemistry knowledge.

What exactly are you trying to do?

We have a policy against just asking for answers to maintain quality. Please show some effort to understand the principles underlying your question such as providing what you think the answer is with the logic behind it.

Al2O3 has 9 electrons so is paramagnetic, but Pd has 10 so is diamagnetic. Looks like will just need to drop these all past a magnet and see if the difference in force is enough to sort them.

Your reasoning is so wrong that you mayn't put it straight in half a dozen follow-up questions. For starters Al2O3 isn't molecular solid!

Actually the Lenz effect doesn't rely on paramagnetism to sort metallic objects from nonmetallic. See for example this video, where the moving magnets actually impel the nonferrous metals, permitting separation from nonmetallic components. If you take a strong magnet, like the NdFeB magnets you can get at the hardware store or even the Dollar Store, and move it rapidly over the bottom surface of an aluminum frying pan without it touching the surface you can actually feel the force between the two. This dynamic force (dynamic in that you can't feel the force between pan and magnet if they aren't moving relative to each other unless the pan is actually iron) is sufficient to impart more momentum to the metal waste in the video above and so achieve separation.
Looking at this table we can see that $\text{Al}_2\text{O}_3$ is in fact diamagnetic with $\chi_{\text{m}}=-37\times10^{-6}\text{cm}^{3}/\text{mol}$ and $\text{Pd}$ is paramagnetic with $\chi_{\text{m}}=+540\times10^{-6}\text{cm}^{3}/\text{mol}$. I have a strong pyramid magnet that can pick up some of the granules from crushed Cheerios and has a visible pull on some of the lettering on U.S. currency. I suspect it could separate out Pd-coated from noncoated beads statically.


text:
Lipophilicity computational prediction
Is there a web based (or otherwise) computational chemistry tool which is able to calculate lipophilicity of a given compound?

http://ilab.cds.rsc.org/ is good but I think requires institutional access

www.SwissADME.ch will calculate several molecular properties, including LogP and LogS.


text:
Question during radius of gyration calculation from XYZ trajectoris of atoms
I have molecular dynamic simulation result of 100 polymer molecules (liquid), and I'm trying to calculate the radius of gyration using atomistic mass, molecular mass, and xyz coordinate of atoms. I'm trying to follow what LAMMPS is doing to calculate the radius of gyration $R_\text{g}$:
$$R_\text{g}^2 = \frac{1}{M} \sum_i m_i (r_i - r_\text{cm})^2$$
Here is my step:
1) Calculate distance from each atom to COM of molecule, for X, Y, and Z direction.
2) Square the distances (for each X, Y, and Z)
3) Multiply distances (for each X, Y, and Z) by atomic mass for each atom
4) Repeat 1~3 for all atoms of corresponding molecule and sum up the values.
5) Divide the value from 4) by molecular mass
6) Now we have the square of the radius of gyration, but it has 3 components (X,Y,Z). Sum the three components together (they are already squared) and then square root. This would be a Rg of single molecule.
7) Perform 1~6 for all other molecules. (and average in the future)
My calculation result is different from what I get from LAMMPS calculation result. Which part of my script is wrong? I'm not sure what is wrong with my calculation process. Do I need to exclude mass, only consider Ri-Rcom?

Your method seems to be ok but you could have a typo somewhere in your code .The radius of gyration $R$ is defined for each axis just as is the moment of inertia $I$ and this is defined for each axis $\alpha$ as $I_\alpha=MR_\alpha^2$ where $M$ is the total mass. You might be better calculating the moment of inertia first.  If you want a single $R$ then you have to decide what you need, average, average squared etc.

Thinking again, your equations should produce $R_{\alpha,\beta} $ where $\alpha,\beta$ are  combinations of x, y and z, i.e (x,x) (x,y),  (x,z) etc making 9 in all. These form a square matrix that should be diagonalised to find eigenvalues which are the $R$'s you seek. As mentioned above it is normal to find the moments of inertia first in the matrix method just outlined.

@porphyrin Thank you, Right now, my script calculate dx (Xi-Xcom), dy (Yi-Ycom), and dz (Zi-Zcom), so it seems I only consider Rg-xx, Rg-yy, and Rg-zz. Should I need to consider Rg-xy, Rg-xz, and Rg-yz as well, to calculate overall Rg?

yes, see my answer below

You don't say exactly what the difference is between your calculation and the LAMMPS calculation, but I'm going to make a guess that you have neglected to account for the periodic boundary conditions properly. The LAMMPS documentation for the compute gyration command says

The coordinates of an atom contribute to Rg in “unwrapped” form, by using the image flags associated with each atom. See the dump custom command for a discussion of “unwrapped” coordinates. See the Atoms section of the read_data command for a discussion of image flags and how they are set for each atom. You can reset the image flags (e.g. to 0) before invoking this compute by using the set image command.

If you are trying to duplicate this calculation "by hand", you need to "unwrap" the atomic coordinates. This means that, for every molecule, all the atoms in the molecule need to belong to the same "copy" of the molecule, i.e. not to different periodic images of the molecule. You need to ensure this before you even calculate the position of the centre of mass of each molecule.
I'll illustrate one way to do this for a simple linear chain molecule of $N$ atoms. Something similar should work for a more complicated molecule, but you'll need to think about it yourself. I'm going to assume that the atoms are labelled in order $1, 2, 3, \ldots N$, and that $1$ is bonded to $2$, $2$ to $3$, and in general $j$ bonded to $j+1$. Let the atomic positions be $\mathbf{r}_j$. You may want to copy these to a separate working array, to avoid changing the originals.
We'll take the position of atom $1$ as a reference point. In the following loop, start by setting $j=2$.

Compute $\mathbf{d}_{j}=\mathbf{r}_{j}-\mathbf{r}_{j-1}$.
Apply the minimum image correction to $\mathbf{d}_{j}$.
Redefine $\mathbf{r}_{j}=\mathbf{r}_{j-1}+\mathbf{d}_{j}$.
Increment $j\rightarrow j+1$ and return to step 1, unless $j=N$ in which case stop.

Now you should be safe to compute the centre of mass of that molecule, and work out the radius of gyration, in the way you describe in your question. You can repeat the calculation for every molecule independently.
A further point about the LAMMPS compute gyration command is that it not only returns the scalar value $R_g$ (which should correspond to what you describe calculating), it also returns the separate $xx$, $yy$, and $zz$ terms, and also the off-diagonal $xy$, $yz$ and $zx$ terms. This is because the gyration tensor is, like the inertia tensor, a second-rank symmetric tensor. This may be of interest if your polymer configurations are significantly non-spherical in shape. Using these values, you can determine the principal axes of the gyration tensor for each molecule, and the corresponding three principal radii of gyration. It's basically a question of diagonalizing the $3\times3$ symmetric matrix whose components are returned by the LAMMPS command, along with $R_g$. I'm assuming, though, that this is not what you are worried about.
[EDIT: I've just noticed that @porphyrin made this same point in a comment while I was typing in my answer].

EDIT following OP comment.
It seems that "unwrapped" coordinates are already being used for this calculation, in post-processing. So it is possible that my answer does not  explain the difference between the LAMMPS calculation and the one implemented by the OP.
However, different definitions of "unwrapped" may be used for different purposes. In calculating atomic mean-squared displacements as a function of time, to get the diffusion coefficient, it is sufficient to make sure that periodic boundary corrections are never applied to the atomic positions as they evolve in time. However, for that purpose, there is no need to consider which molecule an atom belongs to. In order to compute the molecular centre-of-mass, and from that the radius of gyration $R_g$, the atoms must be grouped together by molecule. This means that, at some stage, a calculation similar to the one described above must be carried out. Maybe this has been done, in producing the "unwrapped" trajectories, maybe not.
So, it is still worth either carrying out the correction described above, or at least checking that bonds between atoms have reasonable lengths in the "unwrapped" trajectories. For example, for a linear chain of $N$ atoms, compute $d_j^2=|\mathbf{r}_j-\mathbf{r}_{j-1}|^2$ for $j=2,3,\ldots N$, and just store the maximum $d_j^2$ in the molecule. Extend this calculation to cover every polymer molecule, and print out the maximum value of $d_j^2$ discovered for the whole set of molecules. How does this compare with the square of the bond length between adjacent atoms, defined by the molecular force field? If there has been no error with the periodic boundaries, the two numbers should at least be similar.

thank you so much. My script does not consider off-diagonal parts, only consider xx, yy,  and zz. Do I need to include them as well to calculate overall scalar Rg? Regarding LAMMPS, I print out unwrapped coordinates, and all calculations read unwrapped xyz. Problem of LAMMPS is the limit of number of groups, so I can't print out Rg or other data for all molecules, so I need to post process

No, you don't need to consider the off-diagonal elements in order to compute the scalar $R_g$. Your existing procedure looks OK. If you are already using unwrapped coordinates, I am not sure that my answer deserves to be accepted! It may be that someone else can think of the right answer.

I have had a further thought about this, and have edited my answer to suggest a further check that you can carry out if you wish.

It is easier to start from the beginning and explain how to calculate the moments of inertia. The equations below are a bit different from the one you quote.
If all the atoms are rigidly connected together, the $k^{th}$ atom and its velocity vector $\hat v_k$, are related to the angular velocity of the molecule $\hat \omega$ about the centre of mass as
$$\hat v_k =\hat\omega \times \hat r_k$$
where $\hat r_k$ is the position vector from the centre of mass and $\hat \omega$ is a vector but does not carry an index. This is because in a rigid body, all atoms move with the same angular velocity. The angular momentum for the $k^{th}$ atom is defined as the vector cross product
$$ \hat J_k =\hat r_k \times \hat p_k$$
where $\hat p = m\hat v$ is the momentum vector and the total angular momentum is the sum over all $n$ atoms and is
$$\hat J = \sum_{k=1}^n m_k(\hat r_k \times \hat v_k )=\sum_{k=1}^n m_k(\hat r_k \times (\hat\omega \times \hat v_k )$$
Expanding the triple product into dot products gives
$$\hat J = \sum_{k=1}^n m_k\left( (\hat r_k \cdot \hat r_k)\hat\omega -(\hat r_k \cdot \hat\omega)\hat r_k \right) = \sum_{k=1}^n m_k\left(r_k^2\hat\omega-(\hat r_k \cdot \hat\omega)\hat r_k    \right)$$
The vector $\hat J$ has components $x, \,y$, and $z$ so it represents three equations. This can be written as a matrix equation but note that $r_k^2$ is a number; it is the perpendicular distance of atom $k$ from an axis, $x,\, y$ or $z$, but $\hat r_k$ is the vector $\hat r_k = (x_k\, y_k \,z_k)$ describing the position of atom $k$.
$$\hat J_{(x,y,z),k}=m_kr_k^2
\begin{bmatrix}\omega_x \\ \omega_y\\ \omega_z \end{bmatrix}-m_k \left( [x_k \,y_k \, z_k]
\begin{bmatrix}\omega_x \\ \omega_y\\ \omega_z \end{bmatrix} \right)
\begin{bmatrix}x_k\\ y_k\\ z_k \end{bmatrix} $$
The $x$ component for the $k^{th}$ atom is found by expanding the dot product as $x_k\omega_x + y_k\omega_y + z_k\omega_z$ and then multiplying by $x_km_k$ and rearranging a little
$$J_{x,k} = m_k\left((r_k^2 - x_k^2)\omega_x - x_ky_k\omega_y - x_kz_k\omega_z \right) \tag{1}$$
There are similar equations for the $y$ and $z$ direction components.
By comparing coefficients of $\omega_x$, and those of $\omega_y$ and $\omega_z$ the  equations the diagonal terms in this matrix are
$$I_{x,x} = \sum_k m_k(r_k^2-x_k^2)$$
and similar equations for $y$ and $z$.
Because $r^2=x^2+y^2+z^2$ we can rewrite $I_{x,x} =\sum_k m_k (y_k^2 + z_k^2)$  and similarly for the two other diagonal terms. These terms are called moments of inertia coefficients and cannot be negative as they are the sum of squared terms.
The cross terms $I_{x,y}$, for example, are called products of inertia
$$I_{x,y}=-\sum_k m_kx_ky_k, \qquad I_{x,z}=-\sum_k m_kx_kz_k, \qquad I_{y,z}=-\sum_k m_ky_kz_k   $$
Equation 1 can be rewritten for each atom $k$ using the inertial coefficients
$$\begin{align}
J_x &= I_{xx}\omega_x + I_{xy}\omega_y + I_{xz}\omega_z \\
J_y &= I_{yx}\omega_x + I_{yy}\omega_y + I_{yz}\omega_z\\
J_z &= I_{zx}\omega_x + I_{zy}\omega_y + I_{zz}\omega_z\end{align}$$
and, of the nine coefficients, only six are different because of symmetry; $I_{xy} = I_{yx}$  and so forth. In matrix form these equations are
$$\hat J=\hat I\hat \omega =\begin{bmatrix}
I_{xx} & I_{xy} & I_{xz} \\
I_{xy} & I_{yy} & I_{yz} \\
I_{xz} & I_{yz} & I_{zz} \end{bmatrix}\begin{bmatrix} \omega_x \\ \omega_y \\ \omega_z\end{bmatrix} $$
The matrix $\hat I$, is also sometimes either called the moment of inertia dyadic or the inertia tensor, but, more importantly, it is symmetrical and Hermitian so has real eigenvalues and orthogonal eigenvectors.
The next step in the calculation is to perform a principal axis transform, which we can view as a rotation of the inertia matrix to remove all the off-diagonal terms that become zero on forming a diagonal matrix, i.e find the eigenvalues of the matrix which produce moments of inertia about the principal axes.
Some python code is given below: it could be refined but should work
# coordinates in xyz[i,j], i is atom, j=0,1,2 = x,y,z ;
# masses in mass(atm[i])
# tmass is total mass of molecule
# coordinates in angstrom, mass in amu

import numpy as np
from numpy import linalg as LA

amu  = 1.6605e-27
angst= 1e-10

com=    [0.0 for i in range(3)]
ss =    np.zeros((n,3),dtype=float )   # n = number atoms,  3 -> x,y,z
for i in range(n):
    ss[i,0]=  xyz[i,0]*mass[atm[i]]
    ss[i,1]=  xyz[i,1]*mass[atm[i]]
    ss[i,2]=  xyz[i,2]*mass[atm[i]]

com = np.sum(ss,0)/tmass
print('{:s} {:10.5g} {:10.5g} {:10.5g}'.format( 'cente of mass', com[0],com[1],com[2]) )

ss = xyz - com
print('CoM based coordinates \n',ss )
r = [0.0 for i in range(n)]
for i in range(n):
    r[i]=np.sqrt((xyz[i][0]- com[0])**2+(xyz[i][1]-com[1])**2+(xyz[i][2]-com[2])**2)

Ixx= np.sum( mass[atm[i]] *(r[i]**2  - (ss[i][0])**2 ) for i in range(n)  )
Iyy= np.sum( mass[atm[i]] *(r[i]**2  - (ss[i][1])**2 ) for i in range(n)  )
Izz= np.sum( mass[atm[i]] *(r[i]**2  - (ss[i][2])**2 ) for i in range(n)  )

Ixy= np.sum( -mass[atm[i]] * ss[i][0]*ss[i][1]  for i in range(n)  )
Ixz= np.sum( -mass[atm[i]] * ss[i][0]*ss[i][2]  for i in range(n)  )
Iyz= np.sum( -mass[atm[i]] * ss[i][1]*ss[i][2]  for i in range(n)  )

M= [[Ixx,Ixy,Ixz],[Ixy,Iyy,Iyz],[Ixz,Iyz,Izz]]
eigvals = LA.eigh(M)                             # eigenvalues

print('\n eigenvalues\n', eigvals[0])
print('\n moments of inertia kg.m^2', eigvals[0]*amu*angst**2)
print('radius of gyration /m', np.sqrt(eigvals[0]*amu*angst**2/(tmass*amu) ) )


text:
Confused about energy change diagrams
I have been reading about endothermic and exothermic reactions, as well as change in enthalpy.
From what I have read, an exothermic reaction is where more energy is given out as bonds are made when products form than energy taken in as the bonds of reactants are broken.
Wouldn't this mean that the products have more enthalpy? However, a diagram online shows this (https://www.bbc.com/bitesize/guides/zg84y4j/revision/2):

Why does this diagram show that more energy is taken in as the bonds of reactants are broken? Isn't it incorrect?
I am only just getting started with this topic so I'm just learning concepts. I am not yet familiar with the formulas.

Is the energy level of reactants and products that counts. The diagram is clear and energy release is schematically shown by the green arrow. The products are more stable. They have more energy in the sense that you would have to input it to the system in order to reverse the reaction. Look at it with an analogy of a mass at different eight. But I am not sure what you are asking for, the question isn't that clear

Wouldn't this mean that the products have more enthalpy?

The products have less enthalpy, maybe the following diagram remediates misconception :

$ΔH=Σproducts-Σreactants = {-965.5 -(-74.87)}=-890.63$

Is the energy of reactants and products in the ground state of all states (vibration, rotational, electronic)?