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Ortho vs para alkylation of chlorobenzene
I have a question to the solution regarding the product formed in a reaction with chlorobenzene with the use of (i) $\ce{CH3CH2Cl}$ and (ii)$ \ce{AlCl3}$. Since the -Cl substituent is ortho- para-directing shouldn't the answer then be a mixture of 1-chloro-4-ethylbenzene and 1-chloro-2-ethylbenzene? - I suppose that the para-orientation is the major product. But how is that determined?

https://chemistry.stackexchange.com/questions/47088/reaction-between-chlorobenzene-and-methyl-chloride

Thanks. The answers in the given link are very helpful. Though i still dont understand, why the mayor formed product is the para-orientation instead of ortho. Is the para vs ortho only determined experimentially? Because I dont know which orientation to choose in other excersises that include reactions with benzenes that have an ortho- and para-directing substituent.

Chlorine is $\sigma$ acceptor, thus para > orto. But in general orto/para orientation it poorly controlled and separation step is usually needed. Notable exceptions are reactions with sterically hindered orientants and and cases when both aromatic compound and active particle coordinate to the same Lewis acid. (acilation of phenols with TiCl4, for example)

In general para isomera are thermodynamically more stable than ortho isomers (except in cases when ther is hydrogen bonding in ortho isomer). There are several reasons:-
1. Steric hindrance makes para isomer more stable.
2.Benzene substituents rotate and change thier plane due to steric hindrance. It stops delecolaization of lone pair of electrons (if present in substituents) causing elevation in energy.
Both proper orientation (steric hindrance) and appropriate energy (temperature) is required for formation of product.
Para isomer satisfies these conditions readily and hence are major product.


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Do molecules change shape when they become vibrationally excited?
In quantum mechanics we learn the the classical picture of a vibrational excitation is wrong. The atoms will 'vibrate' only while the molecule is in a coherent superposition of two vibrational states, and will find itself in a stationary state once this coherence is broken.
Let's say I have a carbon monoxide in vacuum, and I excite it from v = 0 to v = 1. In order to describe the electron distribution of the molecule, I would expect the electron distribution to be given by the sum of the electron distributions for the CO at each bond distance, weighted by the nuclear wave function.
So... An excited CO in vacuum would look similar to a stationary molecule of C-o-o, where the little o's indicate that the electron density is smaller than that of a regular oxygen. Is this interpretation correct?

Vibrational excitation changes the bond geometry in a molecule. Definitely, always. I'm not  sure where you want to go with your question however.

What I am thinking here is that it changes more than the geometry. The only parameter to change the geometry of CO is to change the bond length. I am thinking that a vibrational excitation actually makes the CO look like a three-atom linear molecule. I am trying to understand if this interpretation is correct.

And why exactly must atom positions always be stationary while not in a transition? Born-Oppenheimer approximation? Not saying you're wrong, but I've never looked at it that way.

This is how I interpret the nuclear wave function. For example, an electron in a 2p orbital is considered to be 'stationary' because the expectation values associated with its position (and dipole moment) does not depend on time. The state v = 1 is a stationary eigenstate, so the expectation value of the dipole moment does not change in time.

Well of course " an electron in a 2p orbital is considered to be 'stationary'" because the energy of the electron is stable, not because the electron isn't "moving." You seem to be conflating wave models of quantum mechanics and particle models of classical physics and wanting consistency between the two which just doesn't exist.

@Max That doesn't make much sense. The expectation value is constant even for a classical oscillator if there is no excitation (the energy of the system is constant). It doesn't mean it is not moving.

Thank you all, I think I did not manage to convey my question properly. I have edited my question and tried to make it a bit clearer - hopefully I can get some more feedback!

Rather than essentially changing your question and in course invalidating existing answers, please ask a new question linking back to the other establishing context.

Sorry, I will edit it back and do as you suggest. Thank you!

Yes they change 'shape'. A diatomic only  in the sense that the bond extends and contracts, but for  a triatomic or polyatomic molecule there are bond stretches and bends. The number of different ways these vibrations change the 'static' shape of a molecule, such as may be measured by x-ray diffraction, depends on the symmetry and this is described by the point group. Only certain combinations of bond stretches and bends are allowed and these are called 'normal modes'.
Some other points.
A molecule will vibrate even when it has zero point energy, i.e. when $v=0$; by the Heisenberg principle this must happen. If it were stationary then it would have infinite energy because its position is determined exactly. The electron density, if it were to be measured would thus be an average of that at the zero point energy. Calculations of this made at zero potential energy would thus be slightly different. Slightly because the bond extension is only a few percent of the bond length at this energy, and also most nuclear wavefunction density is located at the minimum energy.
We assume that the Born-Oppenheimer condition applies, briefly that electrons being far lighter than nuclei instantaneously adjust to the slow nuclear motion as a molecule vibrates. This allows us to draw the potential energy profile, for example a Harmonic or Morse potential. When we draw the familiar wavefunctions for vibrational motion these are the nuclear wavefunctions at each energy and are the stationary solutions to the Schroedinger equation and we interpret these as meaning that the molecule has a probability of being at each position given by the square of the wavefunction.
In a long time spectroscopic experiment (e.g. using a spectrophotometer) only the spectrum is measured, i.e. energy gaps between levels.
In experiments using a femtosecond laser pulse$^*$ it is possible to excite several vibrational wavefunctions almost instantaneously, i.e. in a time shorter than a vibrational period, in this case a 'wavepacket'$^{**}$ is produced and this 'particle' moves left and right inside the potential. This motion has been observed by using a second femtosecond laser pulse as a probe. As the energies of the vibrational levels are not all the same the wavepacket, which starts off in phase, will become out of phase then 'in phase' again and so goes on repeatedly recurring until it is disturbed, say by a collision with another molecule. (In these time resolved experiments Fourier transforming the data gives the spectrum.)
$^*$ As a femtosecond pulse is short it has a wide spectral width and can therefore excite several energy levels.
$^{**}$ A wavepacket is the direct summation of the vibrational wavefunctions multiplied by a time dependent term related to the vibrational frequencies. Thus the wavepacket evolves in time. The wavepacket recurs at regular time intervals due to the time dependent terms coming into phase just as they were at time zero. This is a property of a collection of oscillators; children on swings would show the same effect.

Thank you! But this is not precisely what I had in mind. I have edited my question to make what I mean a bit clearer :-)

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Does a constrained wave function method exist similar to constrained density functional theory?
I am trying to find out if a constrained wave function method exists for running quantum chemistry calculations. I know that constrained DFT exists and can be used to constrain the charge density to certain parts of molecule but I was wondering if a similar method exists for Hartree Fock, if not why not? Are there limitations that prevent similar constraints from being applied to Hartree Fock?

"if not why not" that is always easy: because nobody bothered to make it

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Why is spin-contamination undesirable?
Why is spin-contamination, introduced by e.g. Unrestricted Hartree Fock, undesirable, and to which problems further on can it lead?
The only problem I currently see myself is that spin-contamined orbitals cannot be used in spectroscopy, because the selection rules don't hold.

Not an expert on this, but my understanding is that if you have spin contamination, you know your wavefunction must be wrong in some way as the true wavefunction is an eigenfunction of the total spin operator.

But then 'restricted' methods such as RHF usually give you only an approximate wavefunction too, so why is the fact that it is approximate worse with UHF?

In a nutshell, if you have spin contamination, your energy-optimised solution is a mixture of two or more distinct spin states, which will probably have quite different properties for anything related to spin, so any derived quantities that you compute from such a wavefunction are suspect and may have little or nothing to do with the actual wavefunction and derived properties for the ground state of that system.

Thanks. Could you please give me examples of any properties related to spin and how they are useful in chemistry?

Just to clarify, I’m trying to get a sense of the scope of chemistry problems for which I can use UHF.


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Bond dissociation energy order for C−C, N−N, O−O and F−F single bonds
Data from Chemistry LibreTexts — Strength of Covalent Bonds, Table 8.8.1: Average Bond Energies (kJ/mol) for Commonly Encountered Bonds at 273 K:
$$
\begin{array}{cr}
\hline
\text{Bond} & D/\pu{kJ mol-1} \\
\hline
\ce{C-C} & 346 \\
\ce{N-N} & \approx 167 \\
\ce{O-O} & \approx 142 \\
\ce{F-F} & 155 \\
\hline
\end{array}
$$
I couldn't understand the trend of bond dissociation energy of various single bonds. Why
$$D(\ce{C-C}) > D(\ce{N-N}) > D(\ce{F-F}) > D(\ce{O-O})?$$
I had mugged up the trend the now, and according to me BDE should continuously increase while going from left to right in a period due to decrease in size and increase in ENC. Why this irregular trend?

A simple explanation would be using the idea of lone pair-lone pair repulsions. As we go from C-C to F-F, there is an increasing number of lone pairs on the bonding atoms. Thus, this increasing destabilising interaction results in the bond energies following that particular trend.

In chemistry there is nothing like always. You have to see any element characteristics than decide its property. The case you are asking about:

Noteworthy exceptions are single bonds between the period 2 atoms of groups 15, 16, and 17 (i.e., $\ce{N, O, F}$), which are unusually weak compared with single bonds between their larger congeners. It is likely that the $\ce{N–N, O–O,}$ and $\ce{F–F}$ single bonds are weaker than might be expected due to strong repulsive interactions between lone pairs of electrons on adjacent atoms.
$\ce{N-N}gt \ce{F-F}$, in $\ce{F}$ there are greater number of lone pairs than in $\ce{N}$ so it experiences strong repulsion between its electrons that make it vulnerable to be broken.
$\ce{F-F} \gt \ce{O-O}$ it's so because $\ce{F}$ atom is very small that enables strong electrostatic attraction between nucleus and electrons
For further information on 3) you can follow this link.


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How to know about radical character of a diradical molecule?
I understand that the radical character is quantified by the occupation number of the lowest unoccupied natural orbital obtained from UHF calculations or from CASSCF calculations.
My issue is that I already have an organic diradical whose ground state is high-spin state. In this case how can I quantify the radical character ?
Is there a way to know about the mixing percentage from higher spin states which in turn could give us information about the radical character of the molecule ?

I don't understand what you mean by quantifying the radical character. In a triplet state, arguably a diradical, that 'character' obviously does not only depend on the LUMO, because more than that orbital will be partially filled. Are you looking to find the 'position' of the unpaired electrons? If so, you are probably interested in the spin density.

I have a plannar molecule with two radical centers, one at each extreme. That electrons are reflected in SOMO's. Now I have increased the distance in these radical centers by increasing the spacer length. The problem now arises in the S_2 value for the complete system which is more that 2 (i.e. S(S+1)). There seems to be mixing from higher spin-states. But I am not able to quantify it.

What level of theory is that on? That sound like spin contamination and appears that the wavefunction you have is not a sensible one. In CAS, you have a pure spin state, so for a triplet S² is 2. If you have a biradicaloid singlet, then the spin is 0, as is S². You can then quantify the character of the wave function by looking at the coefficients of the biradicaloid determinants.

I have performed broken-symmetry  calculations, and in that both HS and BS states do not converge to required S_2 values that is 2.0 and 1.0 respectively. Then I performed optimization in higher spin state i.e. with S=2, and compared relative final energy values of different spin states (i.e. S=1, S=0, S=2). With the increase in coupler length I observe the gap between S=1 and S=2 final energies is decreasing. From this I am concluding about the mixing from higher spin states and S_2 values from broken-symmetry calculations.

You have not answered what level of theory you are doing that on. Anyway, a broken-symmetry ansatz is the poor-mans way to decribe a multireference system. Except from some qualitative interpretation, there is no way you could quantify anything. If your S² is different from the expected value (How much anyway?) then that is an error in your methodology. It means that your level of theory does not describe your system. Apart from that you cannot learn anything from it, especially not the mixing of higher spin states. Please [edit] in additional information.

I have used functionals B3LYP, M06 and PBE0 with TZVP basis set. My $S^2$ values are ranging from 2.14 to 3.33 (for the HS state) from somewhere in middle to highest diradical in series.

Please [edit] your question to include any and all information we have gathered in the comments section.

Take a look at this is the article, which provides quantitative characterization of radical character:
M. Head-Gordon. Characterizing unpaired electrons from the one-particle density matrix. Chem. Phys. Lett., Vol. 372, Issues 3–4, 2003, p. 508-511
Based on natural orbitals, the analysis in this paper is applicable to any correlated method - it needs only one-particle state density matrix.

thank you so much for the reply. I will read the paper and get back if I have any doubts regarding it.


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Franck-Condon factors in photoelectron spectroscopy
Franck-Condon factors have been used to predict / assign the vibrational structure of photo-electron spectra.
$FCF=|\langle\psi''_{vib}|\psi'_{vib}\rangle|^2 = |\langle\psi''_{vib,1}\psi''_{vib,2},\ldots,\psi''_{vib,n}|\psi'_{vib,1}\psi'_{vib,2},\ldots,\psi'_{vib,n-1}, \psi'_{vib, emitted}\rangle|^2=|\langle\psi''_{vib,1}|\psi'_{vib,1}\rangle|^2\ldots|\langle\psi''_{vib,n}|\psi'_{vib,emitted}\rangle|^2$,
Question: what is $\psi'_{vib,emitted}$? Is it just a free particle, with its wavefunction spanning all space, hence it doesn't have any effect on the integral?

Yes it seems that you are correct that the final wavefunction  is a plane wave. This presentation gives some detail that you might find interesting ; http://photon-science.desy.de/sites/site_photonscience/content/e62/e189219/e189559/e190086/e190091/e190093/infoboxContent203186/xray-14_eng.pdf

$\psi^\prime_{vib, emitted}$ is an energy normalized continuum wavefunction. The Franck–Condon factors involving bound ($\psi_v$) and continuum states ($\psi_E$) are actually Franck–Condon densities that are functions of the energy.
The Franck–Condon factors and densities are related by a closure relation
$$ \sum_{v^{\prime\prime}}|\langle \psi_{v^{\prime\prime}}|\psi_{v^{\prime}}\rangle|^2 + \int |\langle\psi_{E^{\prime\prime}}|\psi_{v^{\prime}}\rangle|^2 dE^{\prime\prime} = 1$$
References:
[1] J. Tellinghuisen, The Franck-Condon Principle in Bound-Free Transitions, in Photodissociation and Photoionization, edited by K.P. Lawley (John Wiley & Sons, Ltd, 1985).
[2] R.J. Le Roy, Comput. Phys. Commun. 52, 383 (1989).


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Why is the melting point of iodine so high compared to ammonia? What about Zn and C(diamond)?
Increasing order of melting point by searching the temperature on the web:

$$\ce{NH3} <\ce{I2} <\ce{Zn}<\ce{C_{diamond}}.$$

The idea I have read on "Chemistry: The Central Science" by Theodore L. Brown, H. Eugene LeMay Jr., Bruce E. Bursten on Chapter 11 says something related:
We need more energy to break intramolecular bonds than overcame an intermolecular bond. So, as much stronger is the bond, most energy is required to overcome it.
I want to know why $I_2$ has a higher melting point than $NH_3$. I suspect that $I_2$ has the strongest bond (so a bigger melting point). $I_2$ has intermolecular bonds such as dispersion force (It has big polarizability because of its larger size so his intermolecular bond is strong) and $NH_3$ it has Hydrogen bonding force (the strongest intermolecular force) and also dispersion force. So, I think, but I am not very sure that iodine's force is bigger than NH3 force because of its size (and polarizability).
Is it correct? Must I have into account covalent force in $I_2$?
And why Zinc has an upper melting point in relation with C? Is it because Zn has a covalent bond and Zinc metallic bond or it does not matter?

Comparing the melting points of things with radically different structures isn't going to reveal useful patterns. Nor is comparing ammonia to iodine on the basis of bond types. An iodine molecule is 15 times heavier than an ammonia molecule; why expect them to be the same based on bonding alone?

We prefer to not use MathJax in the title field, see here for details. Also @manooooh

I am trying to correct the question to be clearer.

It is clear actually and should be rather closed as duplicate. It's just that big molecules bind stronger with vdW then small with hydrogen bonds. That's it.

@Mithoron Thanks! What about Z and C(Diamond) ?


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What percentage of a 20 volume solution is actually hydrogen peroxide?
Not too sure how to go about answering this question. I know that volume Strength is the volumes of O2 released on the decomposition of one volume of H2O2 at STP. However, I'm not certain where to go from here, nor if I'm headed in the right direction.

If your hydrogen peroxide solution is 20 vol% hydrogen peroxide, then for every 1L of solution, 200 mL of that is hydrogen peroxide. Now use their respective densities to convert everything to masses, then moles.

@a-cyclohexane-molecule confusingly, "20 volume" doesn't necessarily refer to 20% v/v: https://en.wikipedia.org/wiki/Hydrogen_peroxide#Availability

When a 3% solution of H2O2 decomposes completely, it produces 10 times as much volume of gas as the initial volume of liquid. This is the common hydrogen peroxide available in stores. I buy it often.
So, it would seem intuitive that "20 volume Hydrogen Peroxide" would be a 6% solution.
To check, let's take 1 liter of H2O2 and let it decompose: 20 liters of O2 is 20/22.4 = 0.893 moles of O2. To get 0.893 moles of O2, you must have started with 2 x 0.893 = 1.786 moles of H2O2, which weighs 60.7 grams. It looks like it's close to 6% (ignoring the density of the 6% solution).


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Order of lewis acidity for BBr3, BCl3, BMe3, BH3, BPh3?
In case of $\ce{BCl_3}$ vs $\ce{BH3}$, which one is the controlling factor?
1) Is it p-pi p-pi interaction in $\ce{BCl_3}$ that makes it weaker than  $\ce{BH3}$?
2) Or is it the -I effect of $\ce{Cl}$ dominating over others, despite of p-pi p-pi interaction that makes it stronger than $\ce{BH3}$ as well as $\ce{BMe3}$?

I know you didn’t mention it specifically, but here’s something related that’s pretty interesting: https://pubs.acs.org/doi/abs/10.1021/ic990713m

BBr3 is more acidic than BCl3 due to weaker pi back bonding in BBr3 as compared to BCl3.BCl3 is stronger acid than BH3 due to electron withdrawing effect of Cl make B becomes more electron deficient despite of pi back bonding in BCl3. Due to steric reason BMe3 is stronger acid than BPh3.


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The nature of Chemical Bonds

It was nearly a year ago when I appeared for my first interview through the KVPY (a research oriented scholarship exam) when one of the questions the panel asked me was :
What is a chemical bond?
I had always pondered about the same question for ages and I had never come up to a satisfactory answer. Every book I have read keeps on talking about ionic bonds, which aren't 'bonds' at all but rather electroststic forces of attraction, and how covalent bonds are 'sharing' of electrons between two nuclei. This gets even more confusing when some books state that the 'nature' of bonding between certain atoms is around 50% ionic and 50% covalent.
Amiss all the confusion I had convinced my self that there is truely speaking no such thing as a 'bond'. All there existed was electro static forces of attraction which were either strong or either weak. (because there are only 'three fundamental forces' that I know of. )
Electrons when they are bound to multiple atoms together (like how planets around binary stars work) seemed to explain satisfactorily to me what are bonds.
However the panelist (mind you but I have heard that these panelists are actual scientists) who asked the question was not very happy with my answer and kept on trying to force me to talk about ionic and covalent bonds, by asking me how we classify them and which is stronger.
Is my interpretation correct? Why did the panelist not like my explanation?
An answer with highschool (and slightly higher) mathematics is okay, if not an explanation would be nice, otherwise it would be perfectly fine if you could point out mistakes in my argument and tell me if I was on the right track for interpretation.

In reference to the above context, would it be possible for someone to brief over a few different approaches developed to understand chemical bonding, the forces behind them, and the pros and cons of each theory.

Needless to say, I didn't qualify for the scholarship, and still get mocked by friends for answering 'there is no such thing as a "bond"'.

related https://chemistry.stackexchange.com/questions/230/what-is-the-difference-between-physical-and-chemical-bonds https://chemistry.stackexchange.com/questions/32533/why-is-an-ionic-bond-a-chemical-and-not-a-physical-bond

https://chemistry.stackexchange.com/questions/86918/the-impossibility-of-100-ionic-bond

And this is probably duplicate of https://chemistry.stackexchange.com/questions/710/fundamental-forces-behind-covalent-bonding

In short, you can argue that there is 'no such thing as a bond' if you define a bond as some kind of entity itself - however there certainly is such a thing if you define a bond as two atoms bound by a force which results in a stable outcome. You may be interested in reading into QTAIM, developed by Bader and coworkers over many years, which rationalises this line of thought. Also read "The nature of chemical bonds" (or along those lines) by Pauling.

Also - the closure of this question doesn't limit the number of interesting answers which can be presented - I'd suggest trying to form a less broad question if you want to open up the opportunity for a good answer.

Correction to my first comment: the atoms bound do not necessarily have to be bound by a 'force' in the classical sense, with respect to quantum mechanical considerations.

I've edited the question as to make it less broad, I hope it opens up to the community for answers.

"Why did the panelist not like my explanation?" My guess is that it's because you overthought things a little and turned a chemical question into a philosophical one. It'd be a little like asking someone how you would ride a horse and being told that horses don't exist--there are only arrangements of tissues that make a creature we call a "horse," but there's nothing inherently special about the horse itself as an entity. I mean, sure you might be correct, but I clearly want to test you on your horsemanship knowledge, not your philosophical ability.

@chipbuster Hmm well, I was sort of forced myself into because of lack of knowledge, and I don't think that analogy is what I understood of the question. If a kid asks me what is a what is a horse, my answer would be what I had said. If he would have asked me, talk about the chemical bond of say, H2, then it would have been the situation as you have presented.

@Debaditya It's definitely not a perfect analogy, and I can't entirely fault you for responding the way you did (I think everyone has this kind realization at some point). I'm just pointing out that the questioner was expecting a particular type of response and got something different, which is probably what threw them off.


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Would an (NH3)2+ molecule be trigonal planar like BH3 rather than trigonal pyramidal?
I've been learning about using MO theory to explain why $\ce{BH3}$ and $\ce{NH3}$ have different geometries and by following the line of reasoning used to rationalise the differences in geometry I came to wonder whether ionisation or the addition of electrons to a molecule can change its geometry.
Eg: $\ce{NH3}$ is trigonal pyramidal but an $\ce{NH3^2+}$ ion would have the same electron configuration as $\ce{BH3}$, which is trigonal planar; does this mean $\ce{NH3^2+}$ would be trigonal planar?
Edit: My bad when I first posted this I wrote $\ce{NH3^2-}$ but meant $\ce{NH3^2+}$, this was just an example, my main objective was to find out whether changing the geometry through ionisation was possible.

The short answer is: yes. The slightly longer answer is: a molecule always reacts to a change within itself or the environment.

It would likely launch off proton and become H2N+

Not only is changing shape due to ionization possible, it has been observed.  Not with ammonia as suggested in the question, but with cyclooctatetraene, which is nonplanar in its neutral state but switches to a planar, aromatic ring structure when reduced to its di-anion.


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distillation of aqueous CAN solution for the purpose of a reversible and strong endothermic reaction in cool packs
I know calcium ammonium nitrate (CAN) is used for cold packs. By bending a small metal plate the salt gets dissolved in water (separated pack) and this process is endothermic.
I would like to heat this calcium ammonium nitrate solution to get back anhydrous CAN. Is there a way I can do it without safety issues?

Source 1 advices to heat it up to 210 °C.
Source 2 mentions the possibility to decompose at 60 °C into nitrogen oxides

Do you think I can heat the mixture up to 100 °C?

https://chemistry.stackexchange.com/questions/78776/in-a-reaction-between-calcium-ammonium-nitrate-or-urea-with-water-does-nitrogen

Can I heat this icepack made of calcium ammonium nitrate up to get a reversible process? Are there alternatives with higher reaction enthalpy?

This is an extremely difficult problem. The answer is worth a lot of money to people who would like to use this for something other than personal comfort cold pack.

What we think is irrelevant.  Try heating a very small amount to $100\pu{^\circ C}$ and see what happens.


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1H splitting pattern of benzyl CH2 protons
During my PhD, I prepared the two compounds below (there was actually a slightly longer chain but it's not inherently relevant).
When looking at the 1H NMR spectra of these compounds, something slightly weird occurred.
In one of them, the CH2 group of the benzyl appeared as a singlet (s, 2H). In the other, the CH2 group of the benzyl was observed as a multiplet (an AB quartet like structure, to be specific).
Is there any rationale for why the two cases would be different, given the distal nature of the closest stereocentre (i.e. why are the two CH2 protons diastereotopic to begin with given how distal they are)?

Is the left compound the singlet and the right compound the multiplet? I drew out some chair conformations, and if I'm on the right track, then hopefully, I've correctly identified which is which.

Yes.   The 1,2-anti 1,4-anti is a singlet and the 1,2-syn 1,4-syn is a multiplet

They should both be multiplets due to coupling with the unequivalent aryl protons and between each other. Were both of them run on the same instrument? Was the singlet a bit too broad? The CH2 protons are diastereotopic since there is no plane of symmetry or Cn rotation that can relate them due to the asymmetry of the right part of the molecule.

@AMM I think he's saying that the chiral environment that results in those protons being diastereotopic is very far away.

@Zhe They are not that far away  in my opinion (i guess this is determined by ones experience). Also it is better to interpret NMR spectra using symmetry rather which protons look like they are in the same environment. Take for example this: ClP(NPhBn)2, the two CH2 protons of the Bn look like they have the same environment. Nevertheless, they resonate as two dd due to coupling between them and coupling to the P because based on symmetry they are not equivalent. In the above, i suspect the singlet is actually an apparent singlet due to coupling to many protons with similar J.

Well, it has to be an apparent singlet. It's diastereotopic like you said... But the fact that's under consideration is that it is an apparent singlet rather than not.

This is a bit speculative. I'm still thinking about it, and maybe this shouldn't even been an answer. I drew out the left structure and the enantiomer of the right structure because I want the configuration at the carbon with the OMOM group to be the same.

I think you're calling these the 1,2-anti-1,4-anti and 1,2-syn-1,4-syn, respectively. The main focus here was on the conformation about the 1,2-carbon-carbon axis. I put the two hydrogens anti to each other. In the anti-anti case, the alkyl chain extends off away from the benzyl moiety while in the syn-syn case, the alkyl chain is pointed more towrads the benzyl moiety.
I'm not in love with this explanation. It's a bit speculative, but I used the tools I have and at least, it's consistent with your observation about splitting.

Nice rationalisation! I set some calculations going to see what the low energy conformations in chloroform look like, will update to let you know if they match your predictions

I dont understand this one guys, in both structures you will have rotation around the CH-CH bond where the grey protons are drawn (and all other bonds as well). Therefore in both cases the alkyl chain will spend the same amount of time close to the aryl for both structures drawn.

@AMM That's only true if all of the conformations are equally likely. The issue is that there is enough substitution on the 1- and 2- carbons that you might have a strong conformational bias. Again, this is speculative at best, so the real answer may be something else. It's also likely solvent dependent, so if the spectra were taken in a different solvent, I suspect you might not get the same results.


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Understanding Equivalent Wyckoff positions
I'm somewhat confused by the concept of Wyckoff positions in crystal structures. From Quantum Chemistry of Solids by Robert Evarestov, the definition of a Wyckoff position is "all the crystallographic orbits which have the same (not only isomorphic, but the same) site-symmetry group". I was okay with this, but he then went on to define equivalent Wyckoff positions as positions such that "equivalent sites have the same point-group symmetry and the same orientations of symmetry elements with respect to the lattice".
What I fail to understand is how the equivalent Wyckoff positions don't collapse to a single Wyckoff position, as they seem to meet the criteria of having the same site-symmetry. Is there a clear explanation of what distinguishes two Wyckoff positions enough to be equivalent, but not just a single position?
To clarify my question a little more, here is the specific example given in the text for the space group 136, also called P42/mnm or $D_{4h}^{14}$.

Pairs of Wyckoff positions a-b, f-g, and i-j have isomorphic site-symmetry groups
  ($D_{2h},C_{2v},\text{ and } C_s$ respectively). As is seen from the table, the pairs of points a-b and f-g not only have isomorphic, but also equivalent Wyckoff positions. The equivalent sites have the same point group symmetry and the same orientations of symmetry elements with respect to the lattice.

Somewhat related old question:https://chemistry.stackexchange.com/questions/65430/equivalent-wyckoff-position-in-different-groups

I believe I have identified what my misunderstanding was. I was incorrectly conflating point group symmetry with site symmetry, which is a subgroup of the overall point group symmetry that leaves the given position unchanged. So, it is possible for two points to have equivalent site symmetries while not having the same site symmetry, e.g the points might both have mirror plane symmetry, but not the same mirror plane. The orbits from these generating points would not be conjugate and so they would be put into separate Wyckoff positions. However, depending on where you choose the origin to be, the different Wyckoff positions can interchange the generating points of their orbits, making them equivalent positions.
An explicit example of this is Rock salt $\ce{NaCl}$ (space group N225), where either $\ce{Na}$ or $\ce{Cl}$ can placed at Wyckoff position $a$ while the other is placed at $b$. Formally, equivalent Wyckoff positions are those which can be transformed into each other by applying the generators of the Euclidean normalizer. I recently found Symmetry Relations between Crystal Structures by Ulrich Müller and it has been very helpful in clarifying my understanding of space groups.

It seems to me that care has to be taken with the words 'equivalent'  and 'same' in the definitions above. Equivalent positions do not mean the same position. In the case, say, of a 4 fold axis any point say in the first quadrant can be replicated to the other three quadrants and these are equivalent points but not the same point, clearly. In some cases such as with reflections in a plane, two sets of identical (or equivalent) points may be produced and so and the number of equivalent points is reduced. The number of equivalent sets is called the multiplicity. These have been called special positions when they lie on a symmetry element
In the Wyckoff notation equivalent positions are labelled consecutively $a, b, c \dots $ starting with $a$ for the highest symmetry. A position $2f$, for example, indicates two equivalent atoms at position given by symmetry $f$. For each space group a table of general and special positions can be made.  This includes point symmetries, positions multiplicities and Wyckoff symbols. The table below shows examples.

Table copied from  page 75 of 'Introduction to Crystallography' by  D. Sands, publ. Dover

I've edited my question to include the specific example give of equivalent Wyckoff positions, which seems to differ from the definition you are using.


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Is there a compound that contains water as a subcomponent?
Is there a molecule that contains water chemically bonded to another molecule/element? Could someone provide an example or two?

That is a hydrate

Define "chemically bonded". Are you talking about ionic, polar, dipole-dipole, or dispersion forces, or something else?

I am talking about ionic/covalent.

Off the top of my head: Coordination complexes that have water ligated to a (transition) metal atom.

See corresponding Wikipedia article
A familiar example would be (hydrated) copper sulfate $\ce{CuSO4.5H2O}$, which has 4 water molecules ligated to the copper atom. (The fifth water molecule mentioned in the chemical formula is associated copper sulfate via hydrogen bonds)

 The picture is sources from the Wikipedia article for copper sulfate, and the diagram is sourced from this Quora post

If you're new to the idea of complex ions, this chemguide article offers a good intro to the topic.

Many examples such as this. You can have a look at montmorillonite as well (hydrated sodium calcium aluminium magnesium silicate hydroxide), even if the exact nature of the bond with water is a bit unclear (there is a balance between the bond with hydroxide and the associated water molecules).


text:
Nomenclature of the following aromatic compound
What is the name of the compound $\ce{(C_6H_5)-CH=N-CH_3}$  ?
I doubt,  it is  n-methyl benzalamine . Is it correct ? Besides , what could be other names of it?

Benzalamine seems to be an ocassionally and formerly used general name for compounds of the formula
$$\ce{C6H5-CH=N(R')R}$$
(where C6H5- is phenyl; R′ may be H). This name probably came from the fact that they can be prepared by a reaction of benzaldehyde with an amine:
$$\ce{C6H5-CH=O + R-NH-R' -> C6H5-CH=N(R')R}$$
$\ce{C6H5-CH=NH-CH3}$ could be, in this manner, named methylbenzalamine or N-methylbenzalamine. You can also find name α-methylbenzalamine, which would be a different compound, $\ce{C6H5-CH(CH3)=NH2}$
However, according to the IUPAC nomenclature, the compound class is and imine, or aldimine.
The correct name of $\ce{C6H5-CH=NH-CH3}$ would be

N-methyl-1-phenylmethanimine (preferred IUPAC name, 2013)

Also note that there is cis/trans (Z/E) isomerism possible.

That is called: N-methyl-1-phenylmethanimine or N-Benzylidenemethylamine or N-methyl-benzaldimine. CAS number is: 622-29-7.


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Acid neutralization
I am working on a biogas plant,
digester volume 3000 liter,
right now pH condition is acidic at 4.00
my requirement is to maintain pH to 6.50 to 7.20
what should I do?
some suggestions are there like, add NaOH or CaCo3
but I am confused about the how much amount should I add to increase pH

There has to be information on web about biogas digesters. What kind of gunk are you even digesting? // Off the top of my head I'd think adding anything willy nilly to the digester would be tedious at best. Seems like you need to take a sample, treat that, and then use sample size to scale up additions. // Part of the problem here is $\ce{Na+}$. I'd suspect that microbes aren't going to like high $\ce{Na+}$ concentrations from NaOH or NaHCO3. So you need a biofriendly base.

Add a base (whichever is suitable for your application) and then use pH paper (eg. https://www.coleparmer.in/i/hach-ph-strips-ph-paper-test-strips-0-to-14-ph-range-100-strips-pk/0554642) to test the pH in a sample of a specific volume of the mixture. Keep adding base (keeping track of the total amount added) and measuring the pH using pH paper until you are in the desired pH range. Just be careful not to add too much base and end up at higher pH than you need. Also make sure to give enough time to the solution to stir after each base addition before testing the pH. Once done, scale the amount of base used to the total volume in the digester. Ideally i would use an inline pH meter to continuously monitor the pH if possible (could contact mettler toledo who has really good quality products but wont be cheap, eg https://www.mt.com/int/en/home/products/Process-Analytics/pH-probe.html).
It seems that here: https://www.biogasworld.com/biogas-plant-monitoring/ a procedure is described that addresses your question where the ratio of volatile organic acids to alkaline buffer capacity is determined and therefore the risk of acidification of a biogas plant.

Down voted - Doubt NaOH is good for a biodigester. Need to take a sample and scale up. It doesn't seem that pH paper strips would work. Not sure if there are any specialty strips that would let you measure pH to 6.50 to 7.20 accurately.

@MaxW The question was to maintain the pH in the specified (broad) range and how to monitor the pH, not which base to add (actually NaOH was the OP suggestion and i used it as an example of a base, hence said 'say NaOH'). I assume it is obvious that he needs to sample and scale up (it is 3000 L after all).  Here are the strips: https://www.coleparmer.in/i/hach-ph-strips-ph-paper-test-strips-0-to-14-ph-range-100-strips-pk/0554642, they measure within 1 unit. Alternatively a simple pH meter can also be used. Also, 6.50 to 7.20 is not accurate, it is a range. Thank you for the comment i ll edit.


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Are there any reducing agents for reduction of nitrobenzene to aniline other than Sn/HCl?
I was taught that Sn/HCl should be used for reduction of nitrobenzene to aniline. Can I ask if there are any alternative reducing agents (for example, LiAlH4?) to do the same task?

Hydrogenation of aromatic nitro groups (over, for example, $\ce{Pd/C}$) usually results in reduction to the corresponding aniline and is probably easier to carry out and work up as compared to $\ce{Sn/HCl}$.
Wikipedia has a list of several conditions for reduction of $\ce{ArNO2}$ to $\ce{ArNH2}$.  It also states that $\ce{LiAlH4}$ reduces it to the azo compound $\ce{ArN=NAr}$, which is corroborated by March's Advanced Organic Chemistry (7th ed.).
In any case, from a practical perspective, you would probably want to avoid using $\ce{LiAlH4}$ if there's a simpler alternative. It's dangerous and the workup is not fun.

Hydrogenation would always be my preferred conditions because of the ease of running and working up. My second choice was tin (II) chloride in EtOAc at reflux.

Can I ask if you have any idea why do my school's notes state that H2(g) with Pt(s) is not able to reduce nitrobenzene? Has there been a shift in the consensus brought by, for example, the publication of certain papers)?

@TimingLiu they're probably simply wrong, as hydrogenation over Pd or Pt is very common. https://doi.org/10.1016/B978-0-12-395532-6.50014-3 "Aromatic nitro groups are easily hydrogenated to the corresponding
aromatic amine. Palladium or platinum catalysts are by far the most used
in these reductions; rhodium and ruthenium catalysts have as yet found
only limited and specialized use. The reductions occur readily under mild
conditions and usually without complication." This was written in 1967.

I agree with @orthocresol, the notes are just wrong. I've used Pt on C many times to reduce a range of aromatic nitro groups. AFAIK this is done on an industrial scale.


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Haloform Reaction of Furfural
My instructor pointed out that furfural undergoes haloform reaction - I don't understand how. I couldn't figure out the mechanism on my own, and couldn't find much online either. (regarding the products, or the mechanism) Could someone please help me out?
As far as I know, only methyl ketones and alcohols  oxidisable to the same undergo haloform reaction - so something unexpected must be happening here.

It'd be great if you could provide a detailed mechanism and explanation, and if possible provide sources.

Are you sure it was "furfural undergoes haloform reaction" or was it maybe "furfural undergoes reaction with haloforms"? The latter happens in methanolic KOH to give methoxy-(2-furyl)-acetic acid.

I think it's the latter, the reaction of furfural with haloforms in alkaline medium. Could you highlight the mechanism?

https://pubs.acs.org/doi/abs/10.1021/ja01500a061?journalCode=jacsat I don't have time to draw it out, but hopefully you can just figure it out from this.

@Zhe I see no reason for it to be such kind of misunderstanding, uch more probable is a case similar to https://chemistry.stackexchange.com/questions/96106/can-1-3-5-trihydroxybenzene-perform-haloform-reaction

@Mithoron I don't understand what you are saying. To my knowledge there's no misunderstanding on anyone's part. Maybe mishearing...

@Zhe Your paper is about synthesis and your earlier comments suggests misunderstanding on OP's part. I say furfural does react in environment used for haloform reaction, just in different way.

Related: Does Br2/H2O oxidize all aldehydes?

Aldehydes can be oxidized by bromine under acidic conditions per the mechanism is given below.  The haloform reaction, however, occurs under basic conditions and requires a methyl ketone.  Your professor probably forgot that the haloform is for methyl ketones in basic conditions and called the reaction below a haloform even though it isn't. Given the similarity in the oxidation of the carbonyl group to a carboxylate with a halogen, (and the thousands of organic reactions to remember,) it seems like a simple easy mistake.


text:
Why is the residual proton resonance in deuterated benzene a singlet?
In deuterated benzene ($99\%$) the residual protio peak is a singlet at $\pu{7.16 ppm}$. This however doesn't make sense since the protons will be randomly distributed among various benzene molecules. Therefore there won't be a mixture of $99$ $\ce{C6D6}$ molecules and $1$ $\ce{C6H6}$ but $94$ $\ce{C6D6}$ and $6$ $\ce{C6D5H}$. Considering that deuterium has a spin of $1$ the residual solvent peak in D6-benzene should be a septet. Any thoughts?

To a very good approximation, $J_\ce{HD}$ in a deuterated molecule is related to $J_\ce{HH}$ in the corresponding protonated molecule by the gyromagnetic ratios:
$$\frac{J_\ce{HD}}{J_\ce{HH}} = \frac{\gamma_\ce{D}}{\gamma_\ce{H}} = \frac{\pu{41.066 * 10^6 rad s-1 T-1}}{\pu{267.522 * 10^6 rad s-1 T-1}} = 0.15$$
In benzene where $^3J_\ce{HH} = \pu{7.6 Hz}$ and $^4J_\ce{HH} = \pu{1.4 Hz}$ (Reich NMR tables), we would thus have $^3J_\ce{HD} \approx \pu{1.2 Hz}$ and $^4J_\ce{HD} \approx \pu{0.2 Hz}$. The four-bond meta coupling would be all but invisible, and the three-bond ortho coupling if resolved would give rise to a $1:2:3:2:1$ quintet.
On a lower-field spectrometer this relatively small coupling is usually not properly resolved. A suitable window multiplication, though, will reveal all. I don't have a d6-benzene spectrum on hand, but I do have a $\pu{500 MHz}$ d8-toluene spectrum here. The two spectra on the left are the aromatic residual peaks, but at the bottom-left a Gaussian window function has been used to improve resolution (at the cost of signal:noise ratio). This allows us to resolve the three-bond ortho couplings of $\pu{1.1 Hz}$.

At the bottom-right is the signal corresponding to residual $\ce{C6D5CD2H}$. Since the two-bond H–D coupling has a larger magnitude ($^2J_\ce{HD} = \pu{2.2 Hz}$), it is much easier to resolve, and can be seen even with the normal exponential window function (which sacrifices resolution for signal:noise ratio improvement).
Similarly, in d6-DMSO (shown below at $\pu{500 MHz}$), the two-bond coupling is more easily resolved.

Really nice answer. +1


text:
Does this particular molecule (pyridine-I-pyridine) have a common name?
Here's a pic of the molecule:

I've been investigating this system to test a computational chem method, but I've realized I don't really know a name for this system. $I-pyr_2$? $py_2I$?
In my mind I call it "pip" for pyridine-I-pyridine, but I don't want to publish and look silly finding out this has a common non-systematic name later (ex: quinine, caffeine, etc).
I'm not worried about getting an IUPAC name exactly right, becuase nobody actually does that. But if there's a more common way of calling it, that's what I want.

You understand it would need to be cation?

Yes it's a cation, but the stability isn't relevant here. I'm using it to benchmark computational methods, not use in a lab experiment

Ok, more stable then I suspected: https://en.wikipedia.org/wiki/Bis(pyridine)iodonium(I)_tetrafluoroborate

Wow, I'm amazed I didn't find that wiki article previously!
However that seems to be a salt of this particular molecule, not the thing itself.

There are plenty of databases that can be searched for that compound. ChemSpider and PubChem are only the top two choices. And then there is google ... Also you failed to specify a charge (or state whether it is neutral). You shouldn't worry about looking stupid when publishing, you should worry about looking stupid in the peer review process. If you're unable to name a molecule, it won't get published anyway.

Please also see our guidelines on how to ask and answer nomenclature questions.


text:
precipitation reaction - cellulose in schweizer reagent reacts with acid
In the regeneration of dissolved cellulose (in a cuprammonium solution) in an acid bath, say sulphuric acid, what is the chemical reaction?
There is a lot about the chemistry of the production of the copper ammonia complex but nothing about the precipitation step.
Is the precipitation possible with any acid?
Apologies about the tag - can't find any existing ones that are appropriate

It's possible to regenerate cellulose from Schweizer's reagent by dilution in plain water, though neutralization of the high pH+ cuprammonium solution with an acid is more efficient. $\ce{HCl}$ works as well as $\ce{H2SO4}$. Economics probably drives the choice.
The chemistry is based on breaking cellulose, a polysaccharide (polymer)  β(1→4) linked D-glucose, into smaller, soluble, segments, and reconstituting the longer polysaccharide through neutralization.


text:
Is the stoichiometric coefficient equivalent to the mole fraction?
I have a small problem to understand (or distinguish between) two concepts of free mixing enthalpy.
In several textbooks, I found the following formula for the molar free mixing enthalpy (binary system):
$\Delta_{mix}G_m = RT (x_A \ln{a_A} + x_B \ln{a_B}) \tag{1}$
When it comes to free reaction enthalpies, the given formula is:
$\Delta_{r}G_m = \Delta_{r}G{^\circ}_m + RT \sum \nu_i \ln{a_i} \tag{2}$
Where $RT \sum \nu_i \ln{a_i}$ represents the influence of mixing of the products and reagents. As far as I understand this, the second term of Eqn. (2) , e.i. $RT \sum \nu_i \ln{a_i}$, should be equal to $\Delta_{mix}G_m$ (Eqn. (1)).
But this would only be true, if $x_i = \nu_i$. My problem is to verify this.
If one considers Eqn. (3):
$d\xi = \frac{dn_i}{\nu_i} \tag{3}$
With $x_i = \frac{n_i}{\sum n_i}$ and, therefore $dn_i = dx_i\sum n_i$, follows that
$\nu_i d\xi = n dx_i \tag{4}$
with $n = \sum n_i$. To solve Eqn. (4) an integration is necessary, but I am not sure, how to set the integration boundaries. My guess is:
Since $\xi$ goes from 0 to 1, what means at 0 no substances were converted and at 1 all (= n) substances have transformed, the boundaries would be as follows:
$\nu_i \int_{0}^{n} d\xi = n \int_{0}^{x_i}dx_i \tag{5}$
In fact, this would lead to $\nu_i = x_i$. Unfortunately, I think this might be a fallacy because I might have chosen the boundaries to fulfill my expectation ^^
What is the correct way to interpret this?

$\nu_i$ and $x_i$ are not equal (in general). This can simply be verified by understanding that in general the $\nu_i$ can be negative (for educts) and greater than one which are both properties $x_i$ does not have since it's values are by definition confined to the range from zero to one.

And what causes the "difference" in the two expressions for the free mixing enthalpy?

The difference comes from the very definition of these two enthalpies. While $\Delta_{mix}G_m = G_{after} - G_{before}$, the reaction enthalpy is defined as $\Delta_rG_m = \left(\frac{\partial G}{\partial \xi}\right)_{p,T}$. As you can see one is a difference whilst the other is a differentiation. Therefore it makes sense that the corresponding expressions differ. If you happen to understand German, I could give you a link to a script dealing with this in more detail.

Thanks, I understand german, so this script might be helpful.

I forgot: This might be covered by the script, but I am aware of the difference in definition. But both concepts describe the mixture of several components. Why should the mixing part of the reaction Gibbs energy differ from a "normal" mixture?

Based on your commend, I came over another possibility which might solve the issue: Since $\Delta_rG_m = \left(\frac{\partial G}{\partial \xi}\right){p,T}$ the following should be valid: $\int{before}^{after} dG = RT\int_{before}^{after}\ln{a_i}dn_i$, with $d\xi=\frac{dn_i}{\nu_i}$. The problem is now, that I don't know $a_i(n_i)$ and therefore I cannot integrate the right side. If one could consider $\ln{a_i}$ to not depend on $n_i$, it would lead to the equality of the both mixing terms, but this is not true, as far as I know.

Here's the link to the script: https://www.dropbox.com/sh/zs8q3ptg8xm78ju/AACQRH7Og5hmA1GWyqdz7_gAa/Chemie/PC_I.pdf?dl=0 The relevant section are 2.4 and 2.5.

Well I still think that one being a differentiation and the other a subtraction is the origin of the difference. In the first equation you have the mixing effect as a difference and in the other you have the 'differentiated mixing effect'. I believe that I do not have a better explanation for this. And the whole thing with the $a_i$ comes into play when viewing real systems instead of ideal ones. But the difference in the equations apply also for ideal systems. Therefore the difference can't be rooted in the $a_i$...


text:
Relative molecular mass of an unknown gas in a mixture

A gaseous mixture contains oxygen and another unknown gas in the molar ratio of 4:1 diffuses through a porous plug into 245 seconds, under similar conditions same volume of oxygen takes 220 seconds to diffuse. What is the molecular mass of the unknown gas?

I used the Graham's law of diffusion to find the molecular mass of mixture.
i.e.
$$\frac{r_\text{mix}}{r_{\ce{O2}}}=\frac{220}{245}=\sqrt{\frac{M_{\ce{O2}}}{M_\text{mix}}}$$
$$M_\text{mix}=39.6$$
But what's next? How do I calculate the $M_\text{gas}$?

You know the molar ratio of oxygen to the unknown gas is 4:1, so you can write an equation for the molar mass of the unknown gas in terms of $M_\text{mix}$.

How do I do that?

Think about what $M_\text{mix}$ is. If I have 0.8 moles of $A$ and 0.2 moles of $B$, what is the (molar) mass of the mixture in terms of the molar masses $m_A$ and $m_B$ of $A$ and $B$ respectively?

After getting relative molecular mass ie $39.6$,
$$\mathrm{M_{net} = \frac{(n_1\times M_1 + n_2\times M_2)}{(n_1 + n_2)}} \tag{i}$$
Given that the molar ratio of gases is $4:1$, the molar ratio of oxygen is $\mathrm{4x}$ and that of the unknown gas is $\mathrm x$. Plug $\mathrm{n_1 = 4x,\ n_2 = x,\ M_1= 32,\ M_2 = M,\ M_{net} = 39.6}$ in equation (i) and get $\mathrm M$.

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text:
COR para directing? - Organic chemistry
Why does COCH3 become para-directing? I thought that COR was meta-directing. Is it because of the AlCl3, or is there another explanation?

COR is meta directing. NH2 however is o,p directing and the reaction you have drawn is the acylation of Ph-NH2

Waylander says it all.  A carbonyl-bearing group is deactivating and meta directing when it is on the substrate ring.  But here the carbonyl-bearing group comes from the electrophile and instead, an amino group is attached to the ring.  The direction comes from the amino group, which is activating and ortho/para directing, and as with most such groups steric effects favor the para product.


text:
Comparing Acidic Strength of H3PO4 and H3AsO4
The $\mathrm{p}K_\mathrm{a}$ values for $\ce{H3AsO4}$ and $\ce{H3PO4}$ are $2.30$ and $2.12$ which indicate that $\ce{H3PO4}$ is a stronger acid compared to $\ce{H3AsO4}$. I looked up their electronegativity values to find that they are almost the same for both arsenic and phosphorus. I feel $\ce{H3AsO4}$ should be the stronger acid because arsenic atom is larger in size and the charge is spread out in much more volume which should make the conjugate base more stable. What am I missing out?

Charge is mostly on oxygens, BTW such a difference as this is negligible .

When comparing acidic strength for oxoacids with the same number of oxygens, typically the acid with the central atom of higher electronegativity - in this case P, will be more acidic. This is because the electrons are slightly withdrawn from the OH bond as the electronegativity of the central atom increases. As bonding electrons are pulled closer to the OH bond, the molecule becomes more polar, and so the molecule becomes a stronger acid.

Wikipedia lists the electronegativity of phosphorous as 2.253 and arsenic as 2.211 on the Allen scale.  Wikipedia's values for the pKa1, pKa2 and Ka3 of phosphoric acid are 2.148, 7.198 and 12.319. For arsenic acid the values are 2.19, 6.94, and 11.5. Consider the pka2 and pKa3 values I find the electronegativity argument dubious.

The stability of an acid is judged by the strength of it's conjugate base. If the conjugate base is stable, it implies that it is a weak base and hence the corresponding acid is stronger.
Now consider $\ce{H3PO4}$. The structure has one $\ce{P=O}$ bond and $\ce{3 P-OH}$ bonds.

The conjugate base would be one Hydrogen short from an $\ce{OH}$ group. Similarly with $\ce{H3AsO4}$. However, the conjugation (resonance) will be better in the case of $\ce{P=O}$ bond, since it is an orbital overlap of oxygen's $\mathrm{2p}$ and phosphorus' $\mathrm{3p}$ as against arsenic's $\mathrm{4p}$. $\mathrm{2p}$ and $\mathrm{3p}$ are of comparable size and account for a better overlap and better resonance, distribution of electronic charge and hence a more stable conjugate base.
Consequently, $\ce{H3PO4}$ is better than $\ce{H3AsO4}$ as an acid.

H3PO4 has more EN atoms than H3AsO3 first of all. So it wins that first (checkmark more ability to stabalize a negative charge therefore more stable and that means stronger acid.
P + 4 oxygens > As +3 oxygens. Secondly, there are more terminal oxygens on the H3PO4 as well. All of the oxygen atoms on the H3AsO3 are bonded to the 3 H atoms.
IF they had an even number of electonegative atoms bonded to the central atom and equal (or even closely similar) number of terminal (not bonded to H) oxygens... then we can go ahead and compare their size since they are in the same column on the periodic table. When you have binary acids bonded to one other anion... go ahead and jump to electonegativity comparisons ( for comparisons in the same period) or size of anion comparisons (if in the same column or group.) first. With oxoacids like this first compare which one has more electronegative atoms .... then compare their electronegatively/ size.

The OP has in mind $\ce{H3AsO4}$, not $\ce{H3AsO3}$.


text:
Why doesn't pH = pKa1 in the buffer zone for this titration?
Below is an image from my textbook representing the titration curve for the titration of $5.00\ \mathrm{mL}$ of $0.010\ \mathrm M$ $\ce{H2S2O3}$ with $0.010\ \mathrm M$ $\ce{KOH}$, which has $\mathrm pK_\mathrm{a1} = 0.6$ and $\mathrm pK_\mathrm{a2} = 1.74$.
The Henderson–Hasselbalch equation implies that at the first half equivalence point, $\mathrm{pH} = \mathrm pK_\mathrm{a1}$, and at the second half equivalence point $\mathrm{pH} = \mathrm pK_\mathrm{a2}$, because there are equal changes of the acid and conjugate base.
This means that at the first buffer region b, I should expect the $\mathrm{pH} = \mathrm pK_\mathrm{a1}=0.6$ and at the second buffer region d, I should expect the $\mathrm{pH} = \mathrm pK_\mathrm{a2} = 1.74$. However, this is clearly not the case with the diagram.
What am I missing here? Is it because I have to account for the autoionization of water in the dilute solution? How would I do so?

How could you have pH=0.6 if concentration was only 0.01 to begin with?

@Mithoron So pKa doesn't always equal pH at the half equivalence point?

@KaienYang - Even if both protons ionized you could only get a acid concentration of 0.02 molar which is only a pH of  1.69. In other words the Henderson–Hasselbalch equation doesn't apply in this case.

@MaxW So it doesn't apply in this case because the original concentration of the acid is too high?

Sulfurous acid is pretty strong. It's pretty much fully deprotonated as soon as you add it to water.

@Zhe - $\ce{H2S2O3}$ is thiosulfuric acid not sulforous acid which is $\ce{H2SO3}$.

@MaxW Thanks for the correction, but a pKa value of 0.6 is still a fairly strong acid.

Your acid dissociation constants are $10^{-0.6}$ and $10^{-1.74}$ whereas your thiosulfate species have only a total molar concentration of $10^{-2.00}$. When the dissociation constants of an acidic solute are greater than the solute concentration the solute is rendered like a strong acid by dilution.  If you have 0.01 M of an acid with a $pK_a$ of 1.74, and you use the equilibrium relation to determine how much of the acid is dissociated, you find that in fact most of the acid is dissociated.
The proposed use of the H-H equation then does not work because this equation assumes that aqueous hydrogen or hydroxide ions are only a small part of the total solute concentration.  When the solute becomes like a strong acid (or strong base) as above, it floods the solution with aqueous hydrogen or hydroxide ions and the assumption needed for the H-H equation fails.


text:
pH throughout a titration

Estimate the $\ce{pH}$ of the solution that results when $\ce{75.0 mL}$ of $\ce{0.275 M HCl}$ is added to $\ce{50.0 mL}$ of $\ce{0.275 M Na2HPO4 (aq)}$.

My book gives the successive $\ce{Ka}$ values for phosphoric acid as: $\ce{Ka1 = 7.6*10^{-3}, Ka2 = 6.2*10^{-8}, and Ka3 = 2.1*10^{-13}}$.
I recognized that after the stoichiometric calculation, $\ce{6.875 mmol of H3PO4}$ and $\ce{6.875 mmol of H2PO4-}$ exist in the solution. I then used the Henderson-Hasselbalch equation to determine that the resulting $\ce{pH = -log(Ka1) = 2.12}$. I think ignoring the third deprotonation is justified because the successive $\ce{Ka}$ values are greater than 3 orders of magnitude from each other.
However, then answer in the textbook is $\ce{pH = 2.80}$. What am I doing incorrectly?

RE: What am I doing incorrectly? You used the Henderson-Hasselbalch equation to determine that the resulting pH and it doesn't apply. HINT recheck the millimoles.

@MaxW I apologize, there was a typo when I was phrasing the question. The number of milliliters of Na2HPO4 has been corrected and changed to 50 mL. Can you please help me solve it?

I think you are correct.

The molarity of the $\ce{HCl}$ and the $\ce{Na2HPO4}$ is the same. So 50 ml of $\ce{HCl}$ converts the $\ce{Na2HPO4}$ to $\ce{NaH2PO4}$. The remaining 25 ml of $\ce{HCl}$ essentially converts half the $\ce{NaH2PO4}$ to $\ce{H3PO4}$. So the solution is as if 6.875 millimoles of $\ce{NaH2PO4}$ were mixed with 6.875 millimoles of $\ce{H3PO4}$ in a total of 100 ml of solution.
The Henderson-Hasselbalch equation is:
$$\text{pH} = \text{pKa} + \log{\dfrac{\ce{[H2PO4^-]}}{\ce{[H3PO4]}}} = 2.1192 + \log{\dfrac{0.06875}{0.06875}} = 2.1192 \ce{ ->[Rounding] } 2.12$$
Let's check....
For the phosphate species let $C_\mathrm{P}$ be the concentration of all the phosphate species. The following equations give the fraction of each species as a function of $\ce{[H^+]}$. The equations were most easily calculated using a spreadsheet.
$$[C_\mathrm{P}] = \ce{[H3PO4] + [H2PO4^-] + [HPO4^{2-}] + [PO4^{3-}]}$$
$$\dfrac{\ce{[H3PO4]}}{[C_\mathrm{P}]} = \dfrac{\ce{[H^+]^3}}{\ce{[H^+]^3} + K_\mathrm{a1}\ce{[H^+]^2} + K_\mathrm{a1}K_\mathrm{a2}\ce{[H^+]} + K_\mathrm{a1}K_\mathrm{a2}K_\mathrm{a3}} = 0.49999$$
$$\dfrac{\ce{[H2PO4^-]}}{[C_\mathrm{P}]} = \dfrac{\ce{K_\mathrm{a1}[H^+]^2}}{\ce{[H^+]^3} + K_\mathrm{a1}\ce{[H^+]^2} + K_\mathrm{a1}K_\mathrm{a2}\ce{[H^+]} + K_\mathrm{a1}K_\mathrm{a2}K_\mathrm{a3}} = 0.50001 $$
$$\dfrac{\ce{[HPO4^{2-}]}}{[C_\mathrm{P}]} = \dfrac{\ce{K_\mathrm{a1}K_\mathrm{a2}[H^+]}}{\ce{[H^+]^3} + K_\mathrm{a1}\ce{[H^+]^2} + K_\mathrm{a1}K_\mathrm{a2}\ce{[H^+]} + K_\mathrm{a1}K_\mathrm{a2}K_\mathrm{a3}} = 4.0791\times10^{-06}$$
$$\dfrac{\ce{[PO4^{3-}]}}{[C_\mathrm{P}]} = \dfrac{K_\mathrm{a1}K_\mathrm{a2}K_\mathrm{a3}}{\ce{[H^+]^3} + K_\mathrm{a1}\ce{[H^+]^2} + K_\mathrm{a1}K_\mathrm{a2}\ce{[H^+]} + K_\mathrm{a1}K_\mathrm{a2}K_\mathrm{a3}} = 1.1272\times10^{-16} $$
Thus the assumption that the H_H equation can be used is valid.
The textbook answer is wrong.


text:
Specific heat problems
I am in need of help for my understanding in specific heat capacity. To be honest, I understand the lesson effectively, However, just a few days ago, I've encountered two similar yet different-in-approach problems. Here are the problems:
1.) A sample of gold was dropped to a glass of water. The water had a given initial temperature that was higher than the gold. Eventually, a final temperature for both elements were given. The problem was asking for the mass of the gold?
The solution showed to me that the heat of the gold (qgold) released is equal to the heat of water (qwater) absorbed.
qgold = -qwater
Now, I can see that the glass which served as the container of the water was not included in the equation. As far as I can get the logic of this, the second problem seemed to have a different approach but this time the container is involved in the heat transfer. Now this is the second problem:
2.) An unknown sample with a given mass and given initial temperature was dropped to a can with water. The can and the water had a similar initial temperature, but the two had a different mass. The final temperature of all elements present in this problem was given. The problem was asking for the specific heat of this unknown sample. The thing is, this is the part where I am a bit confused because the solution given to me was this:
q of unknown substance = q of the can + q of water
Now, what I really want to ask is this: When do I consider including the heat of the container of the system in this type of problem, as it is very situational in this case? My professor told me it's about material thing. However, I seem to not deem my professor's explanation to be sufficient, for I know a better explanation could be given.
I hope I would get a very comprehensive reply as soon as possible.
Thank you!

There are often some unstated simplifying assumptions in questions of this kind. For example, in both cases, no account is taken of heat transfer to the surroundings (the air?). Unless the system is isolated, one expects that eventually the system would reach the temperature of the surroundings. However, it is hardly worth stating this explicitly: our common experience is that we can neglect the transfer of heat to the surroundings, on the timescale of the experiment, because it happens comparatively slowly.
In the examples you give, some similar assumptions may apply. If you have a metal can, perhaps not very thick (low overall thermal capacity) and of high thermal conductivity, it's a reasonable assumption that it comes to thermal equilibrium with the contained liquid fairly quickly. On the other hand, for a glass, with lower thermal conductivity, it may be a reasonable approximation that no heat is transferred to it on the timescale of the experiment.
It is, as you say, situation-dependent: it depends on the material properties of the container. It would be most helpful if problems of this kind made it clear what you should assume. Alternatively you may be able to deduce what to assume, from the data provided (or not provided). Or indeed, perhaps it is expected that you can make the right assumptions, knowing something about the materials concerned.

So LonelyProf, you're saying that it all goes down to the kind of material being used to hold the water per se? If so, then why is that in a calorimeter, the coffee cup that is made of styrofoam, when computing for qcal, the equation is expressed in this way: qwarm water = -(qcoldwater + qcal). As you said, if the material has a very low thermal conductivity, we can neglect it. However, in spite of the calorimeter to be of low thermal conductive, it is included in the heat transfer equation. So, how can you explain this, though? Thank you very much.

Probably there are more issues than simply the thermal conductivity. I see that this has been discussed before, here: https://chemistry.stackexchange.com/questions/18889/metal-vs-styrofoam-cup-calorimeter.

Based on what I've read in the link you've provided, I see that there is a difference if you used metal as opposed to the styro ones. If that is the case, let's say that in the first problem I've provided, what if the glass is actually a styrocup, would the styro now be part of the equation as: qgold = qstyro + qwater?

I suppose so. It really depends on the specific properties of the container and the experiment. I don't think that I can usefully add any more to my original answer.

In the rush to get to a number for the answer the assumptions for a problem are usually cast aside. The pertinent assumptions really should be explicitly written down any  for problems like these.


text:
Synthesis of 1-methylcyclopentene from cyclopentene
I have to put a methyl group on a cyclopentene. My teacher made this solution (the picture). But my first guess was just to use $\ce{CH3Cl}$ and $\ce{AlCl3}$. Why can't you do that?

It doesn't appear to be a known reaction, as far as I can tell from SciFinder (possibly just because theres no point in doing it rather than it being impossible).  Your suggestion of a Friedel-Crafts type reaction is, however, unlikely to work as cyclopentene isn't aromatic. Your teachers route seems plausible, if not a little steppy

In Friedel–Crafts alkylation you form a strong electrophile to break the aromaticity of a benzene ring, resulting in a species that is also electrophilic, but not on the carbon atom. It is much more energetically favorable to reform the aromatic ring by losing a proton to $\ce{AlCl4-}$, evolving $\ce{HCl}$ gas.
The reaction of the electrophile with an alkene forms another carbon centered electrophile, however, which can rearrange and will go on to react with other alkenes or with $\ce{AlCl4-}$. There is also nothing to stop the reformed $\ce{AlCl3}$ from continuing to react with the newly formed chloroalkanes, so overall it would probably be a big mess. I found a resource here that discusses the reaction of $\ce{AlCl3}$ with alkenes:

For your reaction, though, I would propose something a little simpler and probably higher yielding like this:


text:
Organic molecules with 21.62% oxygen
I have been told to answer the following question:

Seven different compounds, all either alcohols or ethers,
  contained 21.62% oxygen. Find the structural formula for all of these compounds and name the alcohols.

So far I've tried using empirical formula calculations to work out a simple formula but this has yielded no results because I do not know the exact ratio of hydrogen to carbon in the formula nor do I know any masses which I could use to scale up the empirical formula into a molecular formula. It seems to me that there just isn't enough information to answer the question - after hours of searching I can't even find one, let alone seven. I'm sure the answer is stupidly easy but it's evaded me. Does anyone have any suggestions of things I might be able to try? This question is UK A-Level standard.

I am not sure if the precision that is implied in the question is correct. For example, butanol actually contains 21.58 % oxygen, which is suspiciously close to the given 21.62 %.

With this many significant digits, check the atomic weights used in this 21.62% figure.  Do they agree with your sources also to four significant digits?  If not you may be chasing your tail.

I also get butanol using empirical formulas. Note that UK A-Level uses one decimal place atomic masses.

"Alcohols or ethers," assuming no double bonds, suggests a structural formula $\ce{C_nH_{2n+2}O}$. Find the percentage of oxygen in such a compound as a function of $n$, set it equal to 21.62%, then solve for $n$. If a non-integral solution emerges, assume one double bond and repeat, etc. Then form seven isomers from the identified structural formula.

Disclaimer:
Here I take the molar masses of carbon, hydrogen and oxygen to be $\pu{12g, 1g, 16g}$ respectively, which is what would be expeccted of you at school.

For the sake of simplicity, let's assume you have $\pu{100g}$ of the said compound. Which would mean it contains $\pu{21.62g}$ of oxygen.
Employing a bit of basic math in the analysis of that fact, leads us to similar (but more useful) re-statement:

$\pu{74g}$ of the said compound will contain $\pu{16g}$ of oxygen (the number also being the same as the molar mass of oxygen).

Now our substance possess a general formula of $\ce{C_{x}H_{y}O_{z}}$.
So in $\pu{74g}$ of it (of which $\pu{16g}$ is oxygen), the molar masses of $\pu{C}$ ( times the number of moles of it, which is $x$ ) and $\pu{H}$ ( times the number of moles of it, which is $y$ ) must add up to $\pu{74-16=58}$. Which gives us the equation:
$$\ce{x(12) + y(1) = 58}$$
Now neither $x$ nor $y$ can carry the value $\pu{0}$ (else we won't be dealing with a hydrocarbon anymore). This gives us three mathematically (not necessarily physically) plausible options:

$\pu{x= 2; y= 34}$
$\pu{x= 3; y= 22}$
$\pu{x= 4; y= 10}$

Even if you've only just started studying organic chem. you should be able to tell that the first two options are absolutely ridiculous (that's way too many hydrogen atoms that can be accommodated around 2 or 3 carbon atoms). So we go with the third option: $\pu{x= 4; y= 10}$, which sounds promising.
Which gives us the formula  $\ce{C_{4}H_{10}O}$

You may now work out what alcohols or ethers that could give you ;-)


text:
Comparing the basicity of benzamide and acetamide
What if we were asked to compare the basicity of benzamide and acetamide?

Which one of them would be more basic? I read somewhere on the internet that benzamide must be more basic than acetamide due to cross conjugation (I am not aware of this term). Is this true? Please provide an explanation.

For starters, do you know which atom will get protonated? Then one needs to think what stabilisation resulting cation will have. Even if you get to solution with inductive and mesomeric effects which may satisfy someone who'd ask you it still may be not consistent with other effects and, more importantly,  experimental results, so it's more of a mind game.

Where do amides accept protons?
As Mithoron aptly pointed out in the comments, it is first important to establish which atom in the amide molecule receives the proton. Due to the strong delocalisation present in the amide, the nitrogen lone pair becomes very weakly basic and it is the lone pair on the carbonyl oxygen which is donated to form bonds with $\ce {H^+}$. Not only does the delocalisation diminish the basicity of the nitrogen lone pair, the donation of the lone pair from the nitrogen atom towards the carbonyl group effectively raises the energy of the lone pairs present on the oxygen atom.
This is more clearly illustrated in the molecular orbital picture, where we consider the interactions between the lone pair orbital on $\ce {N}$ and the $\ce {C=O}$ $\pi$ and $\pi^*$ MOs. The interaction with the $\pi$ MO helps to increase the energies of the electrons on the $\ce {O}$ atom while the interaction with the $\pi^*$ MO helps to lower the energy of the lone pair on the $\ce {N}$.
Let's look at some data
I personally find that a theoretical approach to this problem without looking at any data is non-trivial and may potentially give a false conclusion since many factors are at play. In fact, my initial thoughts without looking at any data was that benzamide would be more basic since the benzene ring can donate into the $\ce {C=O}$ group via resonance (in the same way as how a benzyl carbocation is stabilised by the donation of $\pi$ electrons from the ring) and this would certainly enhance the basicity of the carbonyl oxygen since the electron density at the oxygen atom would have increased significantly.
Cox et al. (1980) have determined via empirical measurements the $\ce {pk_a}$s of the conjugate acids of acetamide and benzamide to be $\ce {-0.73}$ and $\ce {-1.54}$ respectively, with an uncertainty of $\ce {0.02}$. This tells us that acetamide is actually more basic than benzamide. In the discussion of their results, the authors considered the factors which they thought underlied this basicity difference. These included the following:

Effect 1: inductive withdrawal due to the more electronegative $\ce {sp^2}$ hybridised carbon atoms of the aromatic ring, decreasing basicity of benzamide
Effect 2: resonance stabilisation of the positive charge in the conjugate acid of the benzamide molecule, increasing the basicity of benzamide
Effect 3: more negative entropy change of ionisation for benzamide, decreasing basicity of benzamide

The authors suggested that perhaps, effects 1 and 3 were more significant combined than effect 2, resulting in the observed difference. Effect 3 was not well-explained by the authors. I suppose effect 3 arises due to the difference in size of the methyl and phenyl substituents and the consequent difference in the extent of their solvation. The conjugate acid of benzamide may require more solvation and hence, a greater ordering of water molecules, giving rise to a more negative entropy change. This is my rationalisation of effect 3.
Further insights provided
In addition to determining the acidity constants of the conjugate acids, the authors also calculated the deviation from coplanarity of the the benzene ring of the various N-substituted benzamides. They observed that as the deviation from coplanarity increased due to the presence of large and bulky substituents on the $\ce {N}$ atom, the $\ce {pk_a}$ value of the N-substituted benzamides became more "acetamide-like". They then went on to say:

This shows that whatever the nature of the effect which causes benzamide to be less basic than acetamide, the degree of coplanarity in the molecule is important
  to it.

Perhaps, there is intramolecular hydrogen bonding?
I believe that the effect at play here that was not mentioned by Cox et al. may be the presence of intramolecular hydrogen bonding in the benzamide molecule. To highlight the significance of this effect, I will now discuss the computational findings of Muthuraja et al. (2017). The following image (the grahical abstract of their study) shows the subject of the computation study and it also shows the computed Mulliken charges on the different atoms. Notice that most of the aromatic $\ce {H}$s have much more signficant positive charge (i.e. in excess of $\ce {+0.1}$ compared to the other protons not bound to the aromatic ring. Additionally, note the large negative charge value on the carbonyl oxygens.

The authors posit high possibility of intramolecular hydrogen bonding between the carbonyl oxygen of the lactam with the neighbouring aromatic proton. They cite the following evidence justifying their claim:

Firstly, the $\ce {O...H}$ distance ($\ce {2.18}$ angstroms) is shorter than their van der Waal distance ($\ce {2.72}$ angstroms)
Secondly, the carbonyl stretching frequency is red-shifted from the normal stretching frequency

From this computation study, I simply wish to point out that there is a significant chance that intramolecular hydrogen bonding may be present in benzamide due to the proximity of the carbonyl oxygen to the aromatic proton. This interaction decreases the availability of the lone pair on the carbonyl oxygen for donation to acids. Thus, benzamide is less basic. Furthermore, this agrees with the additional insight provided by Cox et al. (1980): As deviation from coplanarity increases, there is less and less intramolecular $\ce {H}$ bonding and thus, greater and greater availability of the lone pairs on the $\ce {O}$ for donation to acids.

References

Cox, R. A.; Druet, L. M.; Klausner, A. E.; Modro, T. A.; Wan, P.; Yates, K. Protonation acidity constants for some benzamides, acetamides, and lactams. Can. J. Chem 1980, 59, 1568-1573.
Muthuraja, P.; Shanmugavadivua, T.; Joselin  Beaulab, T.; Bena Jothy, V.; Dhandapani, M. Influence of intramolecular hydrogen bonding interaction on the molecular properties of N-p-tolyl-5-oxo pyrrolidine-3-carboxylic acid: A theoretical and experimental study. Chem. Phys. Lett. 2017, 691, 114-121.


text:
Optical rotation in inversion of sucrose
In reading about the inversion of sucrose I came across the equation for the reaction rate constant below:
$$k = \left(\frac{2.303}{t}\right) \log\left[\frac{\alpha(0) - \alpha(\infty)}{\alpha(t) - \alpha(0)}\right] \tag{5}$$
What is the logic for difference of the rotations' sign for the logarithmic term? i.e conceptuatlly why is $\alpha(0)$ the minuend in the numerator and the subtrahend in the denominator?
I have seen other quantities like volume or pressure represented in this form but I don't understand the reason why reacting sugar would be presented this way.

In a first order reaction can we use anything directly proportional to concentration rather than concentration itself because these come as a ratio (and hence units cancel) as in the expression as above. Your particular case is described in the answer by MaxW below.

The above formula is wrong. The denominator should be $\alpha(t) - \alpha(\infty)$ instead of $\alpha(t) - \alpha(0)$.

Not quite sure what your asking, but I think I understand. If I misinterpreted the question please let me know.
The hydrolysis of sucrose to a mixture of the component simple sugars, fructose, and glucose, causes the direction of rotation to "invert" from right to left. This is because sucrose rotates to the right, but the combination of the two simpler sugars rotate to the left. So the rotation at time zero is for the concentration of sucrose, and the rotation expected when the reaction is complete at time infinity can be calculated from the concentration of the sucrose.

I still can't get the reasoning behind why the difference is directly proportional to the initial concentration and similarly the difference of rotation at  infinite time and any time t is directly proportional to  concentration at any other time

@harambe - Let's say that sucrose rotates $50^\circ$ to the right and the completely reacted product of fructose and glucose rotates $50^\circ$ to the left. There is thus a $100^\circ$ difference, so each 1% of the reaction will cause a $1^\circ$ change in rotation.


text:
Parallel first order reaction

I am not able to solve this question. Here's my working

Have I done this in the right way? Because my textbook gives answer as 4 moles

Ok, there are two reactions for the 2 moles of A
$$\ce{A ->[rate\space =\space k_1] 2B}$$
$$\ce{A ->[rate\space =\space k_2] 2C}$$
$$\dfrac{k_1}{k_2} = \dfrac{1}{2}$$
The question asks for the total number of moles of A+B+C when the reaction is 75% complete.

At that point 25% of A remains or 0.5 moles of A
It does matter what the ratio of B and C produced is since either reaction of A creates 2 moles of product.  So there is 2*1.5 = 3 moles of product (B+C).

Thus there is 0.5 + 3 = 3.5 moles of A+B+C in the solution when the reaction is 75% complete.
There are 4 moles when the reaction is 100% complete.
So the book answer is wrong.

I don't understand how either reaction produces 2 moles  of product ......how did you conclude this.Is my solution wrong because I am getting different concentrations of C and B

@GENESECT - It is just the stoichiometry of the reaction. 1 molecule of A makes either 2 molecules of B or 2 molecules of C. So 1 mole of A produces 2 moles of B+C.

So some part of A makes B and remaining part makes C .The sum of which is 2

But what have I done wrong in my solution.......I am also getting same answer but it's wrong as It seems from your comments

Yes. I could figure out how much of B and how much of C, but that is extra work and I am lazy. // Really -- the less stuff you calculate the less you're likely to do wrong. So solve the problem the simplest way possible.

@GENESECT - The way that you calculated isn't wrong, you just did extra work.

@MaxW Btw, the answer given in the book is 3.5 only. OP didn't check it correctly


text:
Why does freezing point decrease when vapour pressure decreases in dilute solutions?
I was studying colligative properties of dilute solutions, and one of the properties was freezing point depression. The confusion I had was that, if vapour pressure decreases, it should be easier to reach the freezing point (point where vapour pressure of solid and liquid are equal). So I thought the freezing point will increase, but freezing point actually decreases. Is there any logical explanation?

I thought of when there is increase in pressure there will always be increase in the density .For example when you add more pressure to your liquid it will tend to move to solid

Consider graphs of the vapor pressure of the liquid and of the solid, both as a function of temperature. Intuitively, both graphs increase with temperature, and intersect at the freezing point. By lowering the vapor pressure of the liquid, we lower the corresponding vapor pressure curve and hence the temperature at the point of intersection.

I think it of this way. There is a lot of intermolecular space between individual molecules of a solvent. When we decrease the temperature, the individual molecules come closer, lose their heat of fusion and eventually the intermolecular spaces goes on decreasing until they became a tightly packed solid.
Adding a solute hinders the process as the solute particles, which are different in size and geometry from the solvent particles, occupies these interstitial spaces and prevents solvent molecules from coming closer. Hence we need a much lower temperature than before to freeze my solution as at lower temperatures, the solvent-solute particle interaction gets relatively weakened.


text:
Heat transfer problem
For this problem, I'm wondering if the thermos should be involved in the heat transfer because right we have an equation in calorimetry: $q_\text{warm} =-(q_\text{cold} + q_\text{cal})$

A 125 g sample of cold water and a 283 g sample of hot water are mixed in an insulated thermos bottle and allowed to equilibrate. If the initial temperature of the cold water is 3.0 °C, and the initial temperature of the hot water is 91.0 °C, what will be the final temperature?

Since there is no information given about the thermos bottle, you can only make an approximation. In this case you might want to assume that the mixing and heat exchange of the two water samples is fast compared to the heat transfer to the thermos bottle, which is designed for good insulation, i.e. low heat transfer. Therefore, the heat transfer from the water to the thermos bottle may be neglected for the calculation.
Furthermore, you need to make an assumption about the pressure; however, the enthalpy of liquids is only slightly affected by differences of reasonable values of pressure. For simplicity’s sake, you might want to assume a constant pressure of $p=1\ \mathrm{bar}$.
Thus, the problem can be reduced to a simple enthalpy balance
$$H_\text{mixture}=H_\text{cold}+H_\text{hot}$$
The enthalpy for the cold sample can be obtained from the specific enthalpy of liquid water at $p=1\ \mathrm{bar}$ and $T_\text{cold}=3.0\ \mathrm{^\circ C}$, which can be looked up in steam tables as $h_\text{cold}=0.012704\ \mathrm{kJ\ g^{-1}}$.
$$\begin{align}H_\text{cold}&=h_\text{cold}m_\text{cold}\\[6pt]&=0.012704\ \mathrm{kJ\ g^{-1}}\times125\ \mathrm g\\[6pt]&=15.88\ \mathrm{kJ}\end{align}$$
Likewise, the enthalpy for the hot sample can be obtained from the specific enthalpy of liquid water at $p=1\ \mathrm{bar}$ and $T_\text{hot}=91.0\ \mathrm{^\circ C}$, which is $h_\text{hot}=0.38127\ \mathrm{kJ\ g^{-1}}$.
$$\begin{align}H_\text{hot}&=h_\text{got}m_\text{hot}\\[6pt]&=0.38127\ \mathrm{kJ\ g^{-1}}\times283\ \mathrm g\\[6pt]&=107.90\ \mathrm{kJ}\end{align}$$
The resulting total enthalpy of the mixture is
$$\begin{align}H_\text{mixture}&=H_\text{cold}+H_\text{hot}\\[6pt]&=15.88\ \mathrm{kJ}+107.90\ \mathrm{kJ}\\[6pt]&=123.78\ \mathrm{kJ}\end{align}$$
The total mass of the mixture is
$$\begin{align}m_\text{mixture}&=m_\text{cold}+m_\text{hot}\\[6pt]&=125\ \mathrm g+283\ \mathrm g\\[6pt]&=408\ \mathrm g\end{align}$$
Therefore, the specific enthalpy of the mixture is
$$\begin{align}h_\text{mixture}&=\frac{H_\text{mixture}}{m_\text{mixture}}\\[6pt]&=\frac{123.78\ \mathrm{kJ}}{408\ \mathrm g}\\[6pt]&=0.30338\ \mathrm{kJ\ g^{-1}}\end{align}$$
Looking up the temperature for $h_\text{mixture}=0.30338\ \mathrm{kJ\ g^{-1}}$ at $p=1\ \mathrm{bar}$ shows $T_\text{mixture}=72.4\ \mathrm{^\circ C}$.

By way of comparison, you might even want to simplify further and assume that the specific heat capacity $c_p$ of water is constant so that the temperature of the mixture is given by
$$\begin{align}T_\text{mixture}&=\frac{T_\text{cold}m_\text{cold}+T_\text{hot}m_\text{hot}}{m_\text{cold}+m_\text{hot}}\\[6pt]&=\frac{3.0\ \mathrm{^\circ C}\times125\ \mathrm g+91.0\ \mathrm{^\circ C}\times283\ \mathrm g}{125\ \mathrm g+283\ \mathrm g}\\[6pt]&=64\ \mathrm{^\circ C}\end{align}$$


text:
What is [U-14C]-glucose?
In Helen Tweeddale's article,[1] they labelled metabolites of E. coli with [U-$^{14}$C]glucose. What sort of glucose is it? I understand they used isotope 14, which is radioactive but what does the U stand for?

Helen Tweeddale, Lucinda Notley-McRobb, and Thomas Ferenci, J Bacteriol. 1998, 180(19), 5109–5116. PMID: 9748443 (pdf)

According to the IUPAC organic chemistry nomenclature 2013, section P-83 ISOTOPICALLY LABELED COMPOUNDS, it has to do with uniformly labeled compounds:

P-83.5 GENERAL AND UNIFORM LABELING
P-83.5.1 In the name of a selectively labeled compound in which all positions of the designated element are labeled, but not necessarily in the same isotopic ratio, the symbol ‘G’ is used in place of locants to indicate a ‘general’ labeling.
Examples:

Mixture of isotopically substituted compounds (selective labeling) $\ce{CH3-CH2-CH2-^{14}COOH}$, $\ce{CH3-CH2-^{14}CH2-COOH}$, $\ce{CH3-^{14}CH2-CH2-COOH}$, $\ce{^{14}CH3-CH2-CH2-^{14}COOH}$, etc., when added to $\ce{CH3-CH2-CH2-COOH}$, is designated as [G-14C]butanoic acid.
D-Glucose in which all six positions are labeled with 14C, but not necessarily uniformly, is designated as
  D-[G-14C]glucose.

P-83.5.2 In the name of a selectively labeled compound in which all positions of the
  designated element are labeled in the same isotopic ratio, the symbol ‘U’ is used in place of locants to denote ‘uniform’ labeling.
Examples:

Mixture of isotopically substituted compounds (uniform labeling) $\ce{CH3-CH2-CH2-^{14}COOH}$, $\ce{CH3-CH2-^{14}CH2-COOH}$, $\ce{CH3-^{14}CH2-CH2-COOH}$ and $\ce{^{14}CH3-CH2-CH2-COOH}$, in equal amounts, when added to $\ce{CH3-CH2-CH2-COOH}$, is designated as [U-14C]butanoic acid.
D-Glucose in which 14C is equally distributed among the six positions is designated as
  D-[U-14C]glucose.
Note: In the case of radioactive nuclides, ‘same isotopic ratio’ means ‘same specific radioactivity’.

(Please note that the D- stereodescriptor should be in smallcaps)


text:
Why is antimony pentafluoride a stronger Lewis acid than phosphorus pentafluoride or arsenic pentafluoride?
How can these relative reactivities as a Lewis acids be rationalized:
$$\ce{SbF5 > AsF5 > PF5}$$
One simple argument could be the size of the $\ce{Sb}$-center: Because it is bigger than for example a $\ce{P}$-center it can accept more ligands.
I couldn't find a more precise explanation or rationalisation. My knowledge of the MO diagram of hypercoordinative compounds is really basic, so maybe this could provide - energy differences of the LUMO's of the several compounds - a neat access?

I have not heard of a universally applicable Lewis acidity strength scale; so that statement is only partially correct. With those compounds, usually the fluoride affinity is compared. Antimony is (iirc) strongly in favour here because of the formation of oligomeric anions. Because of their size, the charge density is much lower than in other cases, i.e. the negative charge is dispersed better, making any counter cation extremely weakly coordinated and much better available for reaction.

Caution on the term hypervalent compounds. This term is deprecated and leads to more confusion, especially when looking at MO diagrams, since those usually involve d-orbitals. The involvement of those in bonding has been disproved. The preferable term is hyper-coordination, and the bonds tend to have a large ionic contribution.

@Martin-マーチン Thanks, I changed my "hypervalent" into "hypercoordinative"

might be of interest: https://pubs.acs.org/doi/abs/10.1021/ed073p701


text:
What is the composition of camel urine. Does it contain any particularly toxic substance?
I'd like to know the composition of camel urine. Furthermore, I'd like to know if it contains any known toxins.

According to this article at the Indian Journal for Pharmaceutical Sciences' website,

The chemical composition of camel urine has been reported to contain organic nitrogen, ammonia, urea, creatinine, creatine, hippuric acid and chloride.

hippuric acid?  cameluric, surely?


text:
How is an equivalence of a state and path function possible?
At constant pressure,∆H=q
Where ∆H=enthalpy
             q=energy transferred by heating
 ∆H doesn't depend on how we get from one state to another, but q does. How can they be equal in any case?
Is it because constant pressure implies just one Pathway from one point to another?

You should probably learn how the equality $\Delta H = q$ is derived, since it depends on more than just constant pressure.

I feel that this is not a duplicate because the proposed duplicate's answer does not address this question.

So, could somebody answer it then?

If you carried out an adiabatic  process in which you suddenly dropped the (external) pressure to a new lower value and then held the (external) pressure at that new value until the gas re-equilibrated at a new volume and temperature, would you consider that a constant pressure process?  Or does the external pressure have to be held constant at the original value for the entire expansion (say, while heat is added to the gas to increase its volume)?

It should be held constant for the entire expansion

Path functions are so called because they depend on the means by which a system goes from state to state. When we specify a path, the resulting “path function” no longer depends on the actual path taken, and can be considered a state function. For similar examples, consider $C_p$ and $C_V$. In other words:

How can they be equal in any case? Is it because constant pressure implies just one Pathway from one point to another?

Yes.

@user90596, I'm happy to help, but please note that no one on this site is obligated to answer your question, and in general posting comments like "please answer" is probably far less effective than editing your question statement with further thoughts you've had on the matter.


text:
Mechanism of substitution for disubstituted benzene

Here,  I count the individual effects of the two oppositely directing groups.  I assume that the deactivation of $\ce{NO_2}$ group at ortho and meta positions is higher.  While bromination,  I get the product as mentioned above. My query :
If OH group is more powerful than nitro group,  then why doesn't bromination occur at ortho position on between and para position ?
If I apply the same logic in the following compounds' bromination,  will the points marked red be the permissible position for substitution (separately)  ?

With two oppositely directing groups, I'd expect a complete mess.

@Ivan...  I'm in a mess...  I saw the first one in a website... Then I did the rest two myself.  However,  I didn't understand the symmetry in the third one...

What's to understand? One Cl directs with the same force as the other.


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How to safely clean tantalum jewelry
I have some tantalum jewelry that I would like to clean. I've tried just soap and water to no success, so I wanted to try one of the standard at-home cleaning methods, like this one that uses foil, hot water, baking soda, and salt.
I know that tantalum is very non-reactive, so I would expect this to be safe to the metal, but I want to confirm before I accidentally do any damage. Unfortunately because tantalum is such a non-standard metal for jewelry, I haven't been able to get a straight answer from the internet or even the jeweler for whether this will be okay.
Is using the foil and baking soda method safe for cleaning tantalum jewelry? Are there any common cleaning methods that would be safe for most metal jewelry but not for tantalum?

I don't think you can dissolve tantalum at home, either accidentally or on purpose.

@IvanNeretin That was the impression I got from reading around, and part of the reason we chose tantalum in the first place. I'm hoping the cleaning will help with the finish, otherwise I'll be suspicious of what metal was actually delivered...

Tantalum is pretty inert, impervious even to the human body, for which reason it's been used in prosthetic implanted devices.

Ultra Met uses tantalum for corrosion resistant coatings and has a list of corrosive substances that won't dissolve tantalum, comprising of hydrochloric, hydrobromic, and hydroiodic acids, nitric acid, sulfuric acid, acetic acid, hydrogen sulfide and even pirahna.
So unless you plan on using aqua regia, hydorfluoric acid, hot chlorine or other hot halogen gas, (none of which are safe for most metal jewelry) your ring should survive—at least the tantalum part will.

Note: the author of this answer has no affiliation nor commercial or personal interest in Ultra Met CPT.