Unitary Transformation of the HF equations I am working on a code to preform a HF procedure and I am a bit confused about the unitary transformation of the basis. My instructions are as follows.

specify the basis and geometry (done)
Evaluate and store overlap, Kinetic energy, nuclear attraction, and electron repulsion integrals (done)
diagonalize $ \mathbf{S}$ and find $\mathbf{S}^{-1/2}$ which is the unitary transformation matrix.
Use the core hamiltonian to find the initial Fock matrix:
$$ \mathbf{S^{-1/2}}^{T} \mathbf{H}~\mathbf{S^{-1/2}}\equiv\mathbf{F}_0$$
Diagonalize the Fock matrix to find the mo coefficients:
$$\mathbf{F}_0 \mathbf{C}^{'}_0=\mathbf{C}^{'}_0\epsilon_0$$
Back transform the coefficients.
$$\mathbf{C^{'}_0}~\mathbf{S^{-1/2}}=\mathbf{C}_0$$
compute the density, use the density to evaluate new fock matrix, find new coefficients repeat until convergeance.

Now step 7 kinds glazes over a lot of the work but I understand that part, so It isn't important to the question.

I am confused by the $\mathbf{S}^{-1/2}$ matrix.  Is this literally one over the square root of each element of $\mathbf{S}$? Or is there some decomposition I should preform that I just do not understand.  I can carry out the whole procedure but without orthonormalization my results are nonsense.

You are going to want to check out Szabo and Ostlund page 143, they explain all the nitty-gritty for this matrix. The math itself is on page 22 and 23. But if you don't have a copy...

The $\mathbf{S}^{-1/2}$ comes from the need to orthogonalize your basis so that you get your equation (5) instead of the $\mathbf{FC}=\epsilon\mathbf{SC}$. It's one method of several to achieve this orthogonalization, and it's called symmetric orthogonalization.

So how do you compute this function of a Hermitian matrix? Well, functions on diagonal matrices are super easy; you just apply the function to each diagonal element. For non-diagonal matrices, this is not true. More to your question, you can't just take one over the square root of each element of $\mathbf{S}$. But you can if it's diagonal. And you can make any Hermitian matrix diagonal via a unitary transformation.

The trick, then, is to diagonalize $\mathbf{S}$ first, apply the inverse square root to each eigenvalue, then transform back using the eigenvectors. So you're going to want to

Solve $\mathbf{SU} = \mathbf{sU}$. Where $\mathbf{s}$ are the eigenvalues of $\mathbf{S}$, and $\mathbf{U}$ are the eigenvectors.
For each $s$ in $\mathbf{s}$ do $s^{-1/2}$.
Transform back to original basis by $\mathbf{S}^{-1/2} = \mathbf{Us}^{-1/2}\mathbf{U}^{\dagger}$


Electron Ionization and the Franck-Condon Diagram: vibrationally excited and vibrationally ground states I am currently studying Mass Spectrometry: A Textbook, third edition, by Jürgen H. Gross. On page 41, the author says the following:

  As explained by the Franck-Condon diagram, hardly any molecular ions will be generated in their vibrational ground state. Instead, the majority of the ions created by EI is vibrationally excited and many of them are well above the dissociation energy level, the source of this energy being the 70 eV electrons.

On page 37, the author presented the following Franck-Condon diagram:

In this diagram, the energy is plotted on the ordinate, and the bond length is plotted on the abscissa.

My understanding is that the vertical transitions represent the electron ionization of the neutral $M$.

Following this diagram, on page 38 the author said the following:

  The counterpart of the vertical ionization is a process where ionization of the neutral in its vibrational ground state would yield the radical ion also in its vibrational ground state, i.e., the $(0 \leftarrow 0)$ transition. This is termed adiabatic ionization and should be represented by a diagonal line in the diagram. The difference $IE_{\text{vert}} - IE_{\text{as}}$ can lead to errors in ionization energies in the order of 0.1 - 0.7 eV.

So, If i'm correctly interpreting the description below the Franck-Condon diagram, the only way for the molecular ions to be generated in their vibrationally ground state is if the bond lengths of the molecular ion in its ground state somehow end up being larger than the bond lengths of the molecular ion in its vibrationally excited state?

If my interpretation so far is correct, then two related questions immediately follow:

Why is it unusual (as it seems to be implicitly implied) that the bond lengths of the molecular ion in its ground state somehow end up being larger than the bond lengths of the molecular ion in its vibrationally excited state?
How does a situation arise where the bond lengths between such molecular ions differ so significantly that some end up in a vibrationally excited state and others in a vibrationally ground state? (If I had to give an educated guess, I would say that this is just due to random or non-primary factors.)

I'm studying this from a non-chemistry background, so I'd appreciate it if people would please take the time to explain this in a more novice-friendly way.

Why is it unusual (as it seems to be implicitly implied) that the bond lengths of the molecular ion in its ground state somehow end up being larger than the bond lengths of the molecular ion in its vibrationally excited state?

It's not so much about whether the bond length in $\ce{M^.+}$ is larger or smaller than that in $\ce{M}$; it's more about whether it's different. If you think pictorially about the vibration of a diatomic molecule or a spring, it makes sense that as you increase the energy (i.e. go to higher vibrational states), the amplitude of the vibration increases. The only way for a molecule to access bond lengths that are substantially different from its equilibrium bond length is therefore to be in a vibrational excited state.

Ionisation of $\ce{M}$ effectively always takes place from the vibrational ground state, so the initial bond length will always be very close to the equilibrium bond length of $\ce{M}$. If ionisation is almost instantaneous, such that the bond length cannot change during the ionisation, then $\ce{M^.+}$ is going to be generated with a bond length that is far away from its own equilibrium. This corresponds to an excited vibrational state. Whether it's larger or smaller than equilibrium doesn't matter; it just matters that it's different.

Of course, to say that this is the case presupposes that the equilibrium bond lengths of $\ce{M}$ and $\ce{M^.+}$ are substantially different. But this appears to be the assumption that the author makes. For more insight into this, @atbm's comments are spot on: the change in the equilibrium bond length will depend on the bonding character of the electron being removed. If a bonding electron is removed, then the bonds holding atoms together are weakened, and consequently $\ce{M^.+}$ will have a larger equilibrium bond length than $\ce{M}$; similar considerations hold for nonbonding and antibonding electrons.

  How does a situation arise where the bond lengths between such molecular ions differ so significantly that some end up in a vibrationally excited state and others in a vibrationally ground state? (If I had to give an educated guess, I would say that this is just due to random or non-primary factors.)

I suspect that the FC principle is no longer relevant to the adiabatic ionisation process. Adiabatic processes are associated with "slow" transitions: see e.g. https://en.wikipedia.org/wiki/Adiabatic_theorem, whereas the FC principle is to do with "instantaneous" processes where there is no possibility of the bond length changing during the ionisation.

This is not my area of expertise, but I will go out on a limb and say that the adiabatic ionisation process is largely hypothetical (at least in the present context). Wikipedia writes that

  However, due to experimental limitations, the adiabatic ionization energy is often difficult to determine, whereas the vertical detachment energy is easily identifiable and measurable.

I am not sure what "error" the author is talking about, but my interpretation is that the adiabatic ionisation energy is the "true" or "actual" ionisation energy of the molecule; however, because physical ionisation does not take place in an adiabatic fashion, the measured ionisation energy differs from the "true" ionisation energy by ca. 0.1–0.7 eV.

So I would not interpret this as being two different physical processes that actually occur in the mass spectrometer; rather, it seems to be more of a thought experiment to me. Of course, I am open to being corrected on this.


What is the correct IUPAC name of this compund? I thought C=C is the "functional group" here and Br is a substituent so the numbering should start from the top left C, making the name 4-Bromo-3-methylbut-2-ene, but my book says 1-Bromo-2-methylbut-2-ene. Which is correct?

According to P-14.4 in Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 (Blue Book):

P-14.4 Numbering
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
[...]
(e) saturation/unsaturation: [...]
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;

In this compound, unsaturation, in the form of the double bond, has the priority for low locants. Since both numbering schemes (shown below) assign the locant '2' to the double bond, criterion (e) is not sufficient for the purposes of deciding between the two options.

However, criterion (f) does allow for a choice to be made. The detachable prefixes in this name are 'bromo' and 'methyl'. Since the locant set (1,2) is lower than the locant set (3,4) (note that both sets are ordered in increasing numerical order as stipulated), the preferred IUPAC name (PIN) will use the locant set (1,2).
Note, however, that your double bond also contains configurational information. According to the Cahn–Ingold–Prelog rules (see P-92 of the Blue Book), the double bond in your molecule as drawn is (E)-configured. Stereodescriptors such as 'E' are cited together with the locant describing the position of the stereogenic unit (see P-91.3).
Hence, the PIN of the molecule as depicted in your question is (2E)-1-bromo-2-methylbut-2-ene.

Note that contrary to existing answers, there is no inherent preference between "bromo" and "methyl" for low locants, unless the application of criterion (f) does not lead to a difference. In that case, criterion (g) must be invoked. Since 'bromo' is earlier than 'methyl' in alphanumerical order, it will be the prefix cited first in the name, and will therefore have priority for a lower locant. For example:


How dependent are computed charges using the quantum theory of atoms in molecules on the used level of theory? The quantum theory of atoms in molecules is based on the topology of the electron density. This mathematical analysis allows to find critical points and hence has a unambiguous way of separating a molecule into atoms. Each atom can be assigned a basin, and integrating the electron density in it yields a partial charge of that atom (taking the charge of the nucleus into consideration).

In principle this analysis can be carried out from observed electron densities. It is also a more and more common tool to analyse calculated electron densities. Since it is only possible to solve the Schrödinger equation approximately, I also expect a dependency of the calculated charges on the level of theory. It is well known, that the Hartree-Fock method only accounts for Pauli correlation, while other methods also account for coulomb correlation. I would expect, that the higher the level of theory the more accurate the calculated charges are. Hence the following questions:

How dependent are computed charges using the quantum theory of atoms in molecules on the used level of theory?
Do calculated charges converge to a specific value with higher level of theory?
Are charges from calculated electron densities comparable to charges from measured electron densities?
(Optional) Are there more reliable methods to obtain partial charges?

I have been doing quite a lot of reading recently about the theory of atoms in molecules and think that I have semi-satisfactory answers to your questions, so I will share them here and perhaps others will add complementary answers as there actually seems to be quite an extensive literature on this subject, but there is still a lot of debate over these topics.

First, a quick note on how the charge density, $\rho$, can be used in the theory of atoms in molecules (AIM) to determine chemically intuitive properties. This is usually done by the definition of an index. For instance, the first index one might think to define is the electron number index, $N(\Omega)$
, where $\Omega$ is the space defined by the critical points surrounding a single nucleus. Essentially AIM claims that $\Omega$ and the very idea of an atom are interchangeable, so $N(\Omega)$ is conceptually identical to answering how many electrons are associated with a partiular atom. Hence, the AIM definition of the charge on an atom follows as,
$$
Q_{\Omega_i}=Z_{\Omega_i}-N(\Omega_i)
$$
where $Z_{\Omega_i}$ is the charge of the nucleus of atom $i$ in the basin $\Omega_i$.

Not surprisingly, $N(\Omega_i)$ is defined as[1],
$$
N(\Omega_i)=\int_{\Omega}\rho(\textbf{r})d\tau
$$
which is clearly just the charge density in a particular basin. This is interchangeable with a number density of electrons by division by the electric charge $e$.

These indices are always defined in terms of the charge density. The reason these are called indices, rather than something more exact, is because we are really computing a quantity which dimensionally satisfies the property we are interested in, and then interpreting it as the thing itself. The ambiguity comes from the idea of a basin, $\Omega$, which is mathematically well-defined but is not guaranteed to be physically exact for any reason.

  How dependent are computed charges using the quantum theory of atoms in molecules on the used level of theory?

Ref. [2] is a review of the AIM theory with a specific emphasis on the electron localization functions (ELFs) to describe aromaticity. More on these ELFs in a moment. They present the calculation of the index $N(\Omega)$ for several diatomic molecules using DFT, HF, and CISD calculations. The total charge in each of the basins stays more or less the same for all of these methods. The value of $N(\Omega)$ generally only changes in the first or second decimal place for the calculations they present (so tenths to hundredths of an electron). The trend when Coulomb correlation is included is that $N(\Omega)$ decreases for what would traditionally be considered the more electronegative atom. That is to say that coulomb correlation makes the bonds more covalent and less ionic. For mostly ionic systems such as $\ce{LiF}$, the addition of coulomb correlation has a small effect because the electrons are already mostly localized in the basins [2]. All of this is to say that the actual charges computed based on $N(\Omega)$ seem to be relatively insensitive to coulomb correlation. This makes sense based on the fact that the primary features of the spatial distribution of $\rho$ will be dictated by the requirement of antisymmetrization of the wavefunction which is, of course, included in a HF description.

That covers the question of charges based on $N(\Omega)$, but this does not actually give us a picture of how electrons are shared between atoms. In other words, how localized and how delocalized are the electrons? In order to answer this question, ref. [3] provides definitons of both localization and delocalization indices, which are unique because the localization index describes localization of electrons to a single basin, while the delocalization index describes the sharing of electrons with either a specific basin, or with all other basin in a molecule.

While $N(\Omega)$ is not greatly affected by coulomb correlation, the localization index, $\lambda(A)$, and the delocalization index, $\delta(A,B)$ are. For instance, $\delta(A,B)$ which is a measure of the sharing of electrons between the atoms $A$ and $B$, changes by almost $1.0$ when going from the HF description of $\ce{N2}$ to the CISD description of $\ce{N2}$. In general, for homodiatomics, coulomb correlation has the effect of increasing the density in individual basins. And, consistent with the statement above about charge, the localization index for polar diatomics tends to increase the charge on the less electronegative element when a correlated wavefunction is used.

I would highly recommend reading ref. [3] as a place to gain some intuition about the behavior of these different indices.

  Do calculated charges converge to a specific value with higher level of theory?

I have not been able to find any explicit discussion of this, but of course any method which approaches the exact wavefunction, and hence the exact density, must have the same limiting value for any index defined in terms of the density. This is really the power of the AIM model. It is completely quantum mechanical in the sense that the exact wavefunction will yield the exact basins. Whether or not these basins mean what AIM wants them to mean, however, is what people debate.

Another related point is how the various indices are affected by the particular basis set used. I have seen a couple papers which mention this, and note that the results really do not depend much on the particular basis set chosen[1]. The reason which is speculated for this behavior is that the partitioning used in AIM is in real space rather than in the Hilbert space to which the basis set belongs. Unsurprisingly, however, the indices are sensitive to the basis set when the density itself is very sensitive to the basis set. Open-shell molecules seem to be more complicated on this front [2].

  Are charges from calculated electron densities comparable to charges from measured electron densities?

The primary purpose of ref. [1] is to compare the value of properties determined from AIM and from a density determined by x-ray diffraction of p-nitroaniline. This molecule was chosen, I believe, because it has a large dipole moment.

One complication in using the experimental density is that it is not as simple as just measuring the density and then having the whole scalar field in front of you. Rather, from what I understand, you measure the diffraction data and then use certain theoretical models to translate this diffraction data into a charge density. Thus, there is ambiguity in what charge density you actually measure. Ref. [1] uses the UMM and KRMM multipole formalisms which are discussed in this wikipedia article. The point of bringing this up is that the answer for the charges actually depends on which method you use to construct the experimental density.

Ref. [1] indicates that the KRMM multipole method which uses the AIM analysis of charges agrees much more closely with an AIM analysis of the charges from a theoretical density found from periodic-HF and periodic-DFT calculations (this is all solid phase). The p-HF and moreso the p-DFT charges agree very well with experimental charges which are calculated from KRMM using the AIM method. See Table 2 of [1].

The dipole moments also agree very well, but only after various refinements are made to both the partitioning of the experimental density as well as the use of the KRMM method. See Table 4 of Ref. [1].

One point which was mentioned in this book[4] (the relevant chapter is available on Google Books, ch. 3) is that the location of bond critical points, saddle points on the density between two atoms, can be very different for theoretical and experimental densities. Thus, properties derived from the basins can, in principle, be very different for the densities. It seems, however, that despite this problem of locating the bond critical point accurately, properties which are derived from the theoretical and experimental basins are not very different. This is because in the region of the saddle point, the density is usually quite flat, thus shifting the bond critical point does not have much effect on how much density is in each basin.

  Are there more reliable methods to obtain partial charges?

Only briefly, I will point out that various population analyses are always possible to determine charges on atoms. The answer to this really depends on what you mean by reliable. In a sense, AIM is the only reliable method of determining charges on atoms in molecules because it is the only method I am aware of which bothers to define what an atom in a molecule actually is. Additionally, it is reliable in the sense that it exists in a completely self-contained mathematical framework. Population analyses, however, can be very dependent on the particular basis set used, and also are not invariant to unitary transformations of the orbitals. In this sense they are very unreliable.

Mulliken charges are a common form of finding charges from a population analysis, but these are known to very basis set dependent. A better scheme, I think, is the CHELPG scheme which fits the molecular electrostatic potential of a system and hence conserves the dipole moment when it assigns charges to atoms. This is a very nice property for a population analysis. Nonetheless, it will be sensitive to method and basis set depending on how much the electrostatic potential changes upon changing either of these.

I think that generally the AIM charges and charges from population analyses will not be too different, but one notable exception is given in ref. [4] on page 162 where it is noted that the AIM charge on boron in $\ce{(H3BNH3)2}$ is positive, $q(\Omega_{\ce{B}})=2.15$, while the Mulliken population analysis gives a negative charge of $-0.26$. So that's a very stark difference. I guess I would trust AIM, but I don't really know.

As a concluding point, everything I have discussed above calculates properties from indices based on $\rho$, but there is no reason to limit the definition of indices to the density. Rather, some authors have defined indices, for physically motivated reasons, based on the gradient of the density, $\nabla\rho$ and the Laplacian of the density, $\nabla^2\rho$.

An interesting application of one of these based on the gradient is for some alkanediols where there is no bond critical point between the two $\ce{O-H}$ groups despite the obvious presence of an internal hydrogen bond [5].

Ref. [5] is a really cool paper. Definitely give it a read.

[1] Volkov, A., Gatti, C., Abramov, Y., & Coppens, P. (2000). Evaluation of net atomic charges and atomic and molecular electrostatic moments through topological analysis of the experimental charge density. Acta Crystallographica Section A: Foundations of Crystallography, 56(3), 252-258.

[2] Poater, J., Duran, M., Sola, M., & Silvi, B. (2005). Theoretical evaluation of electron delocalization in aromatic molecules by means of atoms in molecules (AIM) and electron localization function (ELF) topological approaches. Chemical reviews, 105(10), 3911-3947.

[3] Fradera, X., Austen, M. A., & Bader, R. F. (1999). The Lewis model and beyond. The Journal of Physical Chemistry A, 103(2), 304-314.

[4] Popelier, P. L. A., Aicken, F. M., & O’Brien, S. E. (2000). Atoms in molecules. Chemical Modelling: Applications and Theory, 1, 143À198.

[5] Lane, J. R., Contreras-García, J., Piquemal, J. P., Miller, B. J., & Kjaergaard, H. G. (2013). Are bond critical points really critical for hydrogen bonding?. Journal of chemical theory and computation, 9(8), 3263-3266.


Why do these reactions occur? Identify the products $[A]$ to $[D]$:

  $\ce{CH3COONa}+\ce{NaNH2} + \triangle \rightarrow [A]$

  $[A] +\ce{CO_2} \rightarrow [B]$

  $[B] + \ce{H2O} \rightarrow [C]$

  $[C] +\ce{P2O5 } \rightarrow [D]$

  $[A],[B],[C],[D]$ are the main product in each step.

According to my book's answer

$[A]=\ce{NaCH2COONa}$

$[B]=\ce{NaOOC-COONa}$

$[C]= \ce{(CH2COOH)}_2$

$[D]= \ce{C_3O_2}$

I can intuitively understand that product $[A]$ is correct.However I never learnt the next three reactions.Can someone explain/give a source from which I can learn why the next three reactions occur ?

Thanks.

Abstraction of $\ce{H+}$ by strong base $\ce{NH2^-}$
$$\ce{CH3-COO- + NH2^- -> ^-CH2-COO- + NH3}$$

Carboxylation to form malonate
$$\ce{^-CH2-COO- + CO2 -> ^-OOC-CH2-COO-}$$

Free malonic acid
$$\ce{^-OOC-CH2-COO- + 2 H+ -> HOOC-CH2-COOH}$$

Dehydration of malonic acid with $\ce{P4O10}$ yields carbon suboxide (unbalanced)
$$\ce{HOOC-CH2-COOH + P4O10-> O=C=C=C=O + H3PO4}$$


Indigo and the Environment: is it that bad? TL;DR: In what aspects are the products of the indigo vat dyeing process (sodium sulphate, sulphite and thiosulphate ions) harmful to the environment or to living beings?

I was recently looking into the wastewater created in the process of vat dying using indigo. I found out that since indigo is quite insoluble ($\pu{990\mu g}$ at $\pu{25^{\circ}C}$) in water for reasons of intra-molecular hydrogen bonding, a reducing agent like sodium dithionite ($\ce{Na2S2O4}$) is required, which gives leucoindigo by reducing indigo in an alkaline medium.

A related passage from a distantly related paper:

  For this process, sodium dithionite ($\ce{Na2S2O4}$) has been considered as a best reducing agent in traditional and industrial dying but the reducing agent releases sodium sulphate, sulphite and thiosulphate ions into the media. Those ions not only cause environmental concerns as major pollutants of the modern textile industry waste waters but they present also no negligible risks of toxicity against living organisms.

A few papers suggest that the "wastewater" generated by the process is not suitable for disposal as such. But I still find no clarity, despite prolonged efforts at scouring journals, in why these exact ions cause the suggested adverse effects to the environment. I would be grateful if someone could please share any relevant knowledge.

These ions appear pretty harmless to me..

References:

The role of natural indigo dye in alleviation of genotoxicity of sodium dithionite as a reducing agent, Bektas I, Karaman S, Diraz E, Celik M, Cytotechnology, 2016 Dec; 68(6): 2245–2255, DOI: 10.1007/s10616-016-0018-7

BONUS: Are there any other harmful substances or environmental hazards in this nice indigo broth?

For this process, sodium dithionite (Na2S2O4) has been considered as a best reducing agent in traditional and industrial dying but the reducing agent releases sodium sulphate, sulphite and thiosulphate ions into the media. Those ions not only cause environmental concerns as major pollutants of the modern textile industry waste waters but they present also no negligible risks of toxicity against living organisms.

It is a very convoluted sentence with so many negations. My simple definition of environmentally dangerous stuff is that anything which is not found in nature is bad for the environment in excessive amounts. Sodium thiosulfate and sulfite are strong reducing agents, in excess amounts, they will slowly deplete the oxygen levels in water. Read the concept of "Chemical Oxygen Demand" in natural waters. These ions will heavily affect the COD, which is a measure of water quality. Sodium sulfate should be fine. You have to check the half-life of these ions (thiosulfate and sulfite) in natural waters.

Untreated textile waste is worse in the sense because it is colored. If large amounts of untreated textile waste, such as those of indigo, are discarded into natural streams they might block the light reaching the depths.


Peptide bond formation involving side chains of charged amino acids Peptide bonds are formed as such:

Aspartic acid, glutamic acid and lysine all contain either one extra $\ce{-COOH}$ or $\ce{-NH2}$ group in them. Why does that extra group not participate in peptide bond formation? For reference, the structure are given below:

It should be possible for the $\ce{-COOH}$ group attached to $\ce{C}$3 in aspartic acid and $\ce{C}$4 in Glutamic acid to form a peptide bond with $\ce{-NH2}$ of another amino acid. Do these bonds occur in any peptide linkage? If they do, what is the nomenclature associated with them and how do their chemical properties differ?

Equilibrium in aqueous solution

In aqueous solution (such as our digestive tract), the equilibrium of the reaction as written is on the side of the reactants. Digestive enzymes catalyze the hydrolysis of peptide bonds, showing that they are not thermodynamically stable in water. However, in the absence of catalysts at neutral $\mathrm{pH}$, the peptide bond is very stable against hydrolysis, explaining how it is possible for proteins to fulfill their biological role.

Peptide formation in the cell

In the cell, amino acids form a mixed phosphoric acid anhydride by reacting with ATP, which then reacts with hydroxyl groups of tRNA to form an ester. This ester reacts with the growing peptide chain, catalyzed by ribosomes using mRNA as a template to make a specific polypeptide rather than a random one. All steps are enzyme-catalyzed, ensuring that only the main chain carboxylate reacts while the side chain carboxylate does not.

Peptide solid state synthesis

In a peptide synthesizer, amino acid building blocks start out as esters and amides with suitable protecting groups on the side chains. Only when the entire polypeptide has been synthesized are the side chain protecting groups removed. Below is a schematic of the process:

Source: https://www.biosyn.com/images/userfiles/image/general-scheme-solid-phase.jpg

  Why does that extra group not participate in peptide bond formation?

It depends on the context but it does not react because

it is not activated (carboxylic acid in water does not form amide)
it is protected
enzymes are selective for specific functional groups on a molecule

  It should be possible for the $\ce{-COOH}$ group attached to $\ce{C}$3 in aspartic acid and $\ce{C}$4 in Glutamic acid to form a peptide bond with $\ce{−NH2}$ of another amino acid. Do these bonds occur in any peptide linkage? If they do, what is the nomenclature associated with them and how do their chemical properties differ?

For side chains with amino groups, there are cases of peptide formation, for example in ubiquitination reactions where the protein ubiquitin is attached to a lysine via its carboxy-terminus. Other side chains are glycosylated, methylated, or attached to lipids. Together, these reactions are described as post-translational modifications. There is a report of glutamate methylation (https://www.ncbi.nlm.nih.gov/pubmed/16707700).

Some cyclic peptide antibiotics form peptide bonds with the glutamate side chain called gamma peptides (instead of alpha peptides). Here is an example:

Source: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1176672/?page=7


Is NADPH sensitive and unstable in UV light? According to the MSDS for NADPH Tetrasodium Salt, it says to store it in a place away from heat and light. Does this include UV light or just visible light? I was told that a pure NADPH stock solution (with pH 7 phosphate buffer as solvent) could be re-used after absorbing UV light at 340 nm, but based on the MSDS it seems that shouldn't be the case. So my question is: could I reuse a sample of pure NADPH stock solution that was exposed to UV for my enzyme activity assay?

datasheets.scbt.com/sc-202725.pdf

https://www.interprise.com.br/produtos/download/90289.pdf

NADPH isn't destroyed by absorbing 340 nm light; if it were, then spectrophotometry wouldn't be a good technique to monitor its concentration.  The primary stability concern with NADPH and NADH is acid-catalyzed decomposition.  A great 1986 article summarizing this phenomenon is available here.

Some highlights from the article:

At 30 °C, the pseudo-first order rate constant for NADPH degradation in aqueous solution goes from 0.5 min$^{-1}$ at pH ~3 to $10^{-5}$ min$^{-1}$ at pH 10.  At pH 7 the rate constant is $10^{-3}$ min$^{-1}$.
The rate is elevated slightly by higher concentrations of phosphate buffer, about 10x higher for 1 molar (!) phosphate vs. extrapolation to infinite-dilution  phosphate.

A secondary pathway for NAD(P)H decomposition is oxidation by atmospheric air, but it is usually pretty minor compared to acid degradation.

So I don't think there will be a big problem with light destroying your NADPH.  You might be able to re-use it, but it depends how long you let it sit in solution, and at what temperature.


Is there are any universal methods or software for reaction rate modelling? In software like DWSIM are used databanks which contains enthalpy and entropy at 298 K and more other information about chemicals. This software can create a steady-state system, where showed state of matter, temperature of reaction, etc.

But I can't find any software for calculation of reaction rate. I know, that chemical kinetics is using experimentally measured time of reaction and concentration of matter. So what should we do, if we want to know an estimated time of reaction?

Most modern process simulators, including DWSIM, feature reactor models which can calculate and display reaction rates. For PFR and CSTR models you must also consider the reactor volume to estimate the reaction time.


How is volatility useful in the production of acids? Sulfuric acid because of its low volatility can be used to manufacture more volatile acids from their corresponding salts.

How does volatility affect the production of acids? Isn't it that sulfuric acid being stronger than the salt of weak acid replaces the anion part of salt so the weak acid is produced or does just the low volatility of sulfuric acid matter? Will it be feasible to produce a highly volatile acid even stronger than $\ce{H2SO4}$ by exploiting the low volatile nature of $\ce{H2SO4}$?  What is the real process involved?

Yes, this is indeed the case. The reasoning behind it is using chemical equilibria to their fullest.

If you have a Brønsted acid and a Brønsted base in the same vessel, you will always have an equilibrium of the following kind:

$$\ce{HA + B- <=> A- + HB}$$

It depends on the nature of the acid and the base — i.e. their $\mathrm{p}K_\mathrm{a}/\mathrm{p}K_\mathrm{b}$ values — which side of the equation is favoured in a closed system (i.e. solution). For example, dissolving $\ce{NaHSO4}$ and $\ce{NaSH}$ in the same sample of water will result in the following set of equations:

$$\ce{H2SO4 + Na2S <=>> NaHSO4 + NaSH <=>> Na2SO4 + H2S}$$

Because the $\mathrm{p}K_\mathrm{a}$ values of both compare as follows:

$$\begin{array}{ccc}\hline
\text{acid} & \mathrm{p}K_\mathrm{a1} & \mathrm{p}K_\mathrm{a2}\\
\hline
\ce{H2SO4} & \approx -3 & 1.9\\
\ce{H2S} & 7.0 & 12.9\\
\hline\end{array}$$

We already have a way of producing $\ce{H2S}$ by means of protonating sulfide ions in solution here. However, $\mathrm{p}K_\mathrm{a1}\left (\ce{HCl}\right ) \approx -6$, which is more acidic than sulphuric acid, so adding $\ce{NaCl}$ to $\ce{H2SO4}$ should result in a net reaction of nearly nothing.

This is where vapour pressure comes in as a second factor. $\ce{HCl}$ is a gas under standard temperature and pressure while $\ce{H2SO4}$ is a liquid. ($\ce{H2S}$ is also a gas. If we added liquid sulphuric acid to sodium sulfide, $\ce{H2S}$ gas would evolve, but that’s what the $\mathrm{p}K_\mathrm{a}$ value predicts anyway.) If we have an open vessel — i.e. gases can diffuse away — then any $\ce{HCl}$ produced in equilibrium will dissociate away in the long run. This is due to a second equilibrium that we could write as follows:

$$\ce{HCl (g) <=> HCl (aq)}$$

The ‘equilibrium constant’ of this physical process is more or less what we call vapour pressure, and for hydrogen chloride both sides are significantly more balanced than for sulphuric acid. Therefore, we should describe the entire system in the following way:

$$\ce{NaCl + H2SO4 <=> NaHSO4 + HCl (diss) <=>> NaHSO4 + HCl (g) ^}$$

Where $\ce{(diss)}$ is a shorthand for dissolved in sulphuric acid, our reaction medium.

Drawing away the hydrogen chloride gas in the rightmost equation, for example by reducing pressure or inducing an air flow, will shift all equilibria to the right since that product is constantly being removed. And thus we can liberate $\ce{HCl}$ gas with sulphuric acid, even though the $\mathrm{p}K_\mathrm{a}$ of hydrogen chloride is lower than that of sulphuric acid.


How to determine optical activity of a compound without asymmetric carbons theoretically? I know that a compound doesn't need to have a chiral carbon to be optically active. So how can I find out whether a compound drawn on paper is chiral without access to any experiments? e.g. 1,2,2-tribromopropane

The question is expected to be solved using only pen and paper

A substance will be optically active if it is not identical to its mirror image. Check out this great answer to see why. So the dullest and brute-force-est way to find out whether a molecule is chiral is to draw it, to mirror it, and then to attempt to overlay image and mirror image using only rotation. (Remember to draw molecules three-dimensionally!)

But there are a few shortcuts. You can look at the molecule as a whole and check out what symmetry features it has — determine its point group. If a molecule has a centre of symmetry (i.e. inversion results in the identical molecule), a plane of symmetry or improper rotation (which is a rotation followed by mirroring along a plane orthogonal to the rotation axis) it is achiral. If you cannot find any of these symmetry elements, the molecule is chiral.


Methods: Does failing at intermolecular properties mean the same for intramolecular properties? Let us consider the following:

1.) Method X fails at describing intermolecular non-covalent interactions.

2.) Method X excels at describing molecules of type Y.

3.) Method X ___?___ at describing molecules of type Y_1 containing intramolecular non-covalent interactions.

Is there an existing general consensus on number 3? I haven't found an example in the literature yet which would highlight a clear answer to this.

EDIT for clarification:

Example 1: An non-covalent intermolecular interaction would be the interaction of two stacked benzene molecules.

Example 2: A molecule of type Y could be like an amino acid. This molecule has well defined, strong covalent bonds.

Example 3: An organic molecule with a non-covalent interaction occurring between two functional groups.

So... regarding #3, would we expect the method to do well at characterizing a molecule in Example 3, would we expect the method to fail (as it did in Example 1) or is there no general consensus on the matter which is essentially the same as saying, "Just have to try it and see!"?

I think the distinction that you are trying to draw is (as you have hinted) better expressed in terms of covalent and noncovalent interactions, rather than inter- versus intramolecular interactions. A technique that fails to correctly describe noncovalent interactions will fail whether or not these interactions are between two moieties on the same molecule, or on different molecules. The usefulness of a technique that fails noncovalent interactions is contingent on whether or not those interactions are significant to the particular properties of a molecule that you are investigating.


How to cancel once-mentioned element in thermochemical reactions If I have three thermochemical reactions in a series(in order to find the enthalpy change) how can I cancel an element having not included in the reaction that's being asked about?

and only mentioned in one of the reactions ( either products or reactants)

e.g: $$\ce{NO + O3 \rightarrow NO2 + O2}$$
     $$\ce{O3 \rightarrow 3/2 O2}$$
     $$\ce{O2 \rightarrow 2O}$$

The reaction that is being asked: $$\ce{2NO2 \rightarrow 2NO + O2}$$

We can make use of this formula: $$\Delta H = \Sigma ~\Delta H_f~(\text{products}) - \Sigma ~\Delta H_f~(\text{reactants}) $$

Take $\ce{O3 \rightarrow \frac{3}{2} O2}$ into consideration. Ozone molecule is the less stable form of O. If you know that how much the $\Delta H$ for that reaction is, you can in fact predict the $\Delta H_f$ for $\ce{O3}$. If you're interested, wikipedia has documented some of the enthalpies of formation. Notice that the most thermodynamically stable form of the element is considered to have $\Delta H_f = 0$.

Now that the $\Delta H_f$ for ozone is determined, you can apply the formula above for the first reaction: ($x$ is $\Delta H$) $$x = (\Delta H_f \ce{(NO2)} + 0) - (\Delta H_f (\ce{NO}) + \Delta H_f (\ce{O3}))$$

So, from the equation above we can conclude that $$x + (\Delta H_f (\ce{NO}) + \Delta H_f (\ce{O3})) = (\Delta H_f \ce{(NO2)}) $$

Now write the master formula for the last reaction: $$(2 * \Delta H (\ce{NO}) + 0) - 2 * \Delta H_f (\ce{NO2})$$

What did we conclude from the last reaction? $x + (\Delta H_f (\ce{NO}) + \Delta H_f (\ce{O3})) = (\Delta H_f \ce{(NO2)}) $ Assuming that $x$ is at hand, do you really need to know the $\Delta H_f$ for $\ce{NO}$ & $\ce{NO2}$?


What is the reaction mechanism of nitrogen leaving from diazo compounds? When a diazo compound is formed, and then a nucleophile is added to the mixture. Will nitrogen gas leave first or will the nucleophile have to "do" a substitution reaction? What I mean is will the carbocation be formed due to nitrogen leaving or will it be an SN2 reaction? Or will it depend on the carbon that the diazo group is attached to, (like primary, secondary, etc.) depending on the stability of carbocation formed? In short, I mean to ask if it's SN1 or SN2.

Is it a thumb rule that when an unstable diazo compound is formed, that the nitrogen leaves without any nucleophile intervention?

The reaction is $\ce{S_N1}$

  Among the evidence for the $\ce{S_N1}$ mechanism with aryl cations as
  intermediates, is the following:

  The reaction rate is first order in diazonium salt and independent of the concentration of Y.
  When high concentrations of halide salts are added, the product is an aryl halide but the rate is independent of the concentration of the
  added salts.
  The effects of ring substituents on the rate are consistent with a unimolecular rate-determining cleavage.
  When reactions were run with substrate deuterated in the ortho position, isotope effects of $\sim 1.22$ were obtained. It is
  difficult to account for such high secondary isotope effects in any
  other way except that an incipient phenyl cation is stabilized by
  hyperconjugation, which is reduced when hydrogen is replaced by
  deuterium.

  That the first step is reversible cleavage was demonstrated by the observation that when  $\ce{Ar^{15}N+ #N}$ was the reaction species,
  recovered starting material contained not only $\ce{Ar^{15}N+ #N}$ but
  also $\ce{ArN^{+} #N}$This could arise only If the nitrogen breaks
  away from the ring and then returns. Additional evidence was obtained
  by treating $\ce{PhN+ #N^{15}}$ with unlabeled $\ce{N2}$ at various
  pressures. At $\pu{300 atm}$, the recovered product had lost $3 \%$ of
  the labeled nitrogen, indicating that $\ce{PhN2+}$ was exchanging with
  atmospheric $\ce{N2}$

Source: March's Advanced Organic Chemistry: Aromatic Substitution: Nucleophilic and Organometallic


Why are larger alicyclic rings more flexible? My lecturer said that the increasing flexibility of an alicyclic ring is an inherent property. I'm not entirely sure what they mean by flexibility. Number of feasible conformations? How far the bond angles can be away from non-ideal?

Is there an actual explanation for the increasing flexibility?

It's geometry and you can get a petty good intuition by building simple models of the molecules

Saturated carbon-carbon bonds have two key properties of relevance. The angle between a pair of bonds on a carbon prefer to point to the corners of a perfect tetrahedron (~109.5 ° between each bond) if there are no spatial or geometric constraints. Also, C-C bonds can rotate freely unless something else in the molecule prevents that rotation.

In small cyclic aliphatic molecules (cyclopropane to cyclopentane) there is no easy way to satisfy the tetrahedral preference. No combination of bond rotations can give C-C-C bond angles of 109.5 °. If you build skeletal models of the compounds with tetrahedral carbon atoms this is easy to see: cyclopropane is rigid and flat; cyclobutane can twist slightly away from planarity but can't get those bond angles close to the unstrained angle, cyclopropane has slightly more freedom but is still pretty rigid.

But, from cyclohexane up there are plenty of extra freedoms. Cyclohexane can twist into a "boat" or "chair" shape and those give essentially unstrained bond angles at the carbons (the angles would be 120° in a flat hexagon, but when the bonds can also twist this does not constrain anything as enough combinations of twists are available to give an unstrained result in contrast to any of the smaller cyclic compounds). Plus, the twists cost little energy so they can interconvert easily given the amount of energy available at room temperature. Again, this is easy to see with physical models.

Higher cyclic compounds have even more freedom because there are more bonds to twist to give the ideal bond angle at carbon. So the increasing flexibility is a product of the fact that bond rotation is easy to do in the absence of geometric constraints. The more bonds you have to twist, the larger the number of conformations of the molecule available to give unstrained angles.


Arranging the phenol derivative in decreasing order of acidity I am facing difficulty in the following problem.

What is the decreasing order of acidity of the following phenol derivatives?

In my view benzene-1,2-diol 1 should be most acidic because $-\ce{OH}$ is a strong $-I$ group thereby it will stabilise the negative charge on oxygen formed after removal of a proton.
$\ce{OCH3}$ is a $+R$ and $-I$ group. Near the carbanion $-I$ effect should be more effective. Hence the second most acidic should be 2-methoxyphenol 2. However the answer key gives it as 4-methoxyphenol 3.

The acidity order should be 1 > 2 > 3 as they have pKa values of 9.48, 9.98, and 10.21 respectively.

+R effect of $\ce{-OCH3}$ is more significant than +R effect of $\ce{-OH}$

In case of 1, -I effect works at -ortho position and also +R effect works at -ortho position.

In case of 2, -I effect works at -ortho position and also +R effect works at -ortho position. Net effect is +R effect as +R effect is dominant than -I effect.

In case of 3, -I effect works at -para position and also +R effect works at -para position. Net effect is +R effect as +R effect is dominant than -I effect. But this +R effect is more dominant than in case 2 because -I effect is more dominant at -ortho position than at -para position but+R effect of equally effective at both -ortho and -para positions.

As a result of which, acidity is reduced to greater extent in the third compound compared to the second compound.


Pinacol pinacolone rearrangement I was reading the topic alcohol in that I saw a reaction Pinacol pinacolone rearrangement .

But I am not able to find the sterochemistry of the reaction .

Can anyone please tell me the stereochemistry

The migrating group attacks trans to the leaving OH group:


Inductive vs resonance effects and the acidity of phenol A lot of people rationalize the acidity of phenol by saying that resonance is responsible for much of phenol's acidity as opposed to aliphatic alcohols.

However, this image suggests that in fact the inductive effect is responsible for a great deal of phenol's additional acidity. As we can see in the picture immediately above, it seems that the sp$^2$ carbon neighboring to the oxygen is pulling most of the weight here; that the resonance contributors with charges on carbons are minor contributors. The additional resonance contributors for phenol seem to be doing almost nothing here.

So, who's right? Resonance>inductive effect in phenol or the other way around?

EDIT: The argument I have summarized above is presented here at this website entitled "I Judge People By Their Grammar and Knowledge of Phenol"

  While the chemistry in the paper was good, as The Chem Blog has noted,
  the authors’ lack of attention to detail was borderline disrespectful.
  I expected more from one of the most storied labs in synthetic organic
  chemistry.

  [...]

  It turns out that an inductive effect—not a resonance effect—is the
  predominant reason for the increased acidity of phenol relative to
  aliphatic alcohols.

  [...]

  For those interested, these data come from Evans’ Chem 206 lecture
  notes (Lecture 20, restricted access), where the point is hammered
  home in glorious detail. Professor Evans’ PowerPoint slides should be
  framed and displayed in the Smithsonian.

Addressing jerepierre's point

$\ce{Phenol + acetone~enolate \leftrightharpoons phenoxide + acetone ~enol}$

So, from the pKa values we know that the above equilibrium lies mostly to the right (according to jerepierre). Given this, we can conclude that the reactants are less stable in the acid-base sense than the products; acetone enolate is less stable than phenoxide.

Phenol is also less stable than acetone enol.

But I don't see how this contradict's the blog article. It does contradict ron's point about being unable to use the provided pKa values to compare the relative stabilities of the acids.

The acidity of $\ce{A-H}$ is a measure of where the following equilibrium lies.

$$\ce{A-H <=> A^- + H+}$$

This means that we are comparing the relative stabilities of the products and reactants ($\ce{\Delta G=-RTlnK}$).

In the present example (assuming your $\mathrm{p}K_\mathrm{a}$ for propen-2-ol is really being measured in water, note too that in water the relative concentration of this enol of acetone is ~ $10^{-8}$, how can you measure the $\mathrm{p}K_\mathrm{a}$ of something so dilute?) we are comparing the relative stability between [phenol and phenoxide] to the relative stability between [propen-2-ol and the corresponding anion].  Just because the two systems have the same difference in relative stabilities does not mean that phenol and propen-2-ol have the same stability.  In other words, just because the two systems have similar $\mathrm{p}K_\mathrm{a}$'s this cannot be interpreted to mean that the additional resonance structures in phenol and phenoxide are not significant contributors to both their stability and description.

EDIT: Jerepierre's comment as worked up by Dissenter (Addressing jerepierre's point)

in his edit Dissenter wrote,

  Given this, we can conclude that the reactants are less stable in the acid-base sense than the products

I agree

Dissenter further posits,

  acetone enolate is less stable than phenoxide. Phenol is also less stable than acetone enol.

That's where I disagree.  We know that the reactants are less stable than the products, but I don't see how we know which reactant(s) is less stable than which of the products.

Let's say the reactants are $10\,\mathrm{kcal/mol}$ (made up number) less stable than the products.  Is phenol $5\,\mathrm{kcal/mol}$ less stable than phenoxide and acetone enolate $5\,\mathrm{kcal/mol}$ less stable than acetone enol?  Or is it $10$ and $0$ or $0$ and $10$, or $15$ and $-5$?  How can you break down the overall reactant/product difference of $10\,\mathrm{kcal/mol}$ further?


What is the dipole moment direction of an aliphatic amine, i.e., R-NH2? Since the electron donating property of $\ce{N}$ and $\ce{C}$ is different, there is partial charge transfer between $\ce{N}$ and $\ce{C}$ atoms in aliphatic amines, namely $\ce{R-NH2}$. What is the charge transfer direction? Can the following be true? And how large is this dipole moment?

More generally, how can one determine both the direction and quantity of the charge transfer dipole moment given two common organic groups? Or what is the order of electron donating property of common groups?

I really do not see an easy and straight forward way of determining the dipole moment's direction and quantity. It is very much dependent on how the molecule is currently shaped, i.e. rotations of the alkyl chain and also the electron cloud. Of course you could go ahead and calculate the dipole moment from tabulated values of certain bonds (for example), but this can be really tedious and it might now be always right.

You can probably find tabulated values for most common compounds online. Try http://cccbdb.nist.gov/ or maybe just google will do. If you are lucky enough to have access, the handbook of chemistry and physics has a large number of values. There is an online verison, but you can often find it in libraries, too.

For the meantime, I will try to illustrate something. In the series $\ce{R-NH2}$ the simplest one is $\ce{R~=~H}$, ammonia. Here the dipole moment is straight forward isn't it? I believe it is a little more complicated than that. Unfortunately you cannot just take the charges of the atoms in the molecule, since they are not point charges. Have a look at the total electron density, you will notice, that the isosurface is more extended on top of the Nitrogen, which is expectable, since it is the area where the lone pair is supposed to be. This also coincides with the highest occupied molecular orbital (HOMO). In this special case, the dipole moment is almost a direct correlation of that. (If you would add up all dipole moments of all occupied molecular orbitals, you will also obtain the total dipole moment.)
Below you see an isosurface of the total electron density in blue on the left, the isosurface of the homo in the middle and natural charges with the resulting dipole moment on the right. The values are taken from a DF-BP86/def2-TZVPP calculation with Gaussian09.

What we can learn from this is, that the lone pair has a huge effect on the dipole moment.

Let's go a bit further and look at $\ce{R~=~CH3}$, methyl amine. It can be expected, that the dipole moment is somewhat smaller than in ammonia, due to the methyl group. What you can also observe, that it is dependent on how the methyl group is rotated. Of course in room temperature, the group is freely rotatable and it will result in a mean (pulsating) value for the dipole moment. However, what you can observe here is, that the direction of the dipole moment is aligned with the direction of the lone pair. As you go further down the series this generally will hold, However, the magnitude will be decreasing since your "quite unpolar" moiety is becoming bigger.

For ethyl amine, $\ce{R~=~C2H5}$, the trend holds. But to not extend this answer too much, I only pick one representative amongst all possible conformations.

And for propyl amine, $\ce{R~=~(CH2)2CH3}$, we can observe it analogously, with decreasing magnitude of the dipole moment.

As the last example I have chosen heptanyl amine, $\ce{R~=~(CH2)6CH3}$, to example the trend. I removed some of the charges to not overcrowd the image. If you have a close look at the HOMO, you will notice, that it only significantly extends to the first carbon. The situation for any aliphatic amine can be expected to be very similar.

(I have no idea, why the vector for the dipole got bigger and bigger, while the magnitude was decreasing.)

TL;DR:
The dipole moment in aliphatic amines is about 1 Debye and the direction is through the nitrogen directed to the lone pair.


At what temperature do complex hydrocarbons become ashes? When heating a metal tube with polypropelene in it, i found that i still had brittle black foam and flakes in it even though it had been in a blue flame at around 500'C

What happens to complex hydrocarbons/plastics when they turn into black foam at high temperatures?
At what temperature do tars and heavy hydrocarbons completely lose all their Carbon and only ashes are left?

What happens depends whether it is being heated in the presence or absence of oxygen.  For example, in the manufacture of coke (mostly carbon), coal is heated without oxygen and from Coke Production For Blast Furnace Ironmaking:

  The coal-to-coke transformation takes place as follows: The heat is transferred from the heated brick walls into the coal charge. From about 375°C to 475°C, the coal decomposes to form plastic layers near each wall. At about 475°C to 600°C, there is a marked evolution of tar, and aromatic hydrocarbon compounds, followed by resolidification of the plastic mass into semi-coke. At 600°C to 1100°C, the coke stabilization phase begins.

Without oxygen, the carbon will remain upto its boiling point of 3825 °C.

With oxygen, carbon as coke with start combustion at its autoignition temperature of about 700°C


Displacement of leaving groups in organic reactions When I initially learned about substitution reactions, there would be times where a poor leaving group had to be changed into a better one in order to proceed with a substitution reaction. For instance, changing a poor leaving group (-OH) to a better one, such as water:

After the "good" leaving group is made, it is displaced via an SN2 reaction. The displacement of the water required a nucleophile. However, I recently came across this mechanism (dehydration):

In this case, the water leaves by itself. My question is:

If I ever come across an instance where water (a good leaving group), must leave, how should I assess whether it can leave BY ITSELF or not?

In that kind of reactions, the full explanation would require advanced quantic calculation to determine the energy of all the present entities.

In chemistry, we usually prefer giving a half-quantitative explanation to explain the formation of the observed products.

In your example A —> B + C + D, A is a secondary alcohol and it CAN be transformed to a carbocation leading to B... but you can see it is far from perfect since B represents only 50% of the yield. The result would be much more clear-cut with a primary alcohol (unstable carbocation) or a tertiary one (very stable carbocation). I think you are expected to explain all that.


Environmental Alkylating Agents I am looking for the names of some environmental alkylating agents (in context of DNA damages) found for example in the fume produced by industry. Most of the papers which I have found is concentrated on cigarette smoke, chemotherapy ect. I found something about methyl chloride, that is produced by plants and fungi, but also industrially. However there is no more details about that industrial part and in general the amount of methyl chloride in the air is rather low. Are there more relevant examples? Also I came across the information that alkylating agents may be found in food, but again there was no details.

A much more significant example of this is methyl bromide, which has historically been used as a soil fumigant. Its use has been slowed phased out due to its toxicity and ozone depletion abilities.


Logical reasoning behind the formula for molar conductance of an electrolyte I am studying some introductory material on electrochemistry. My book and all standard websites I've found define the molar conductance of an electrolyte as "conductivity of an electrolyte solution divided by the molar concentration of the electrolyte."

I am unable to figure out how this definition was arrived at. Can anyone provide a logical, beginner-level, step-wise approach to how this definition was reached?

For most physical properties of substances, it's useful to know how those properties depend on other parameters.  For example, the various coefficients of thermal expansion (CTEs) indicate how rapidly the dimension(s) of a substance change when the temperature is changed by a given amount. A common CTE used when working with materials is the linear CTE:

$$
\alpha_\mathrm L = {1\over L} {dL \over dT}
$$

When thermal expansion occurs, if the temperature change is relatively small the amount of thermal expansion is roughly proportional to the original length $L$. So, the definition of $\alpha_\mathrm L$ divides the length dependence $dL/dT$ by $L$ to yield a parameter that is approximately constant within a particular temperature range.

A similar rationale underlies the definition of the molar conductance (also called the 'molar conductivity') as:

$$
\Lambda_\mathrm m = {\kappa \over c}
$$

The conductivity $\kappa$ of an electrolyte is a strong function of the concentration $c$ of the electrolyte of interest. However, for electrolytes containing only strong acids and/or bases (or salts of these), $\kappa$ is nearly a linear function of $c$ over a very wide concentration range, and thus $\Lambda_\mathrm m$ for these electrolytes is nearly constant.

Thus, for species that show this strong linearity, defining $\mathbf{\Lambda_m}\equiv\kappa / c$ is useful because a single $\mathbf{\Lambda_m}$ value can be employed to calculate the absolute conductivity $\mathbf\kappa$ across a wide range of electrolyte concentrations $c$, as:

$$\kappa = \Lambda_\mathrm m c$$


Rearrangement considerations in feasibility of SN1 in given reactants I have been asked to compare the feasibility of Unimolecular Nucleophilic Substitution ($\mathrm{S_N1}$) in the following question.

I have received instruction that the reaction will be feasible only in the first compound (which results in a 3° stable carbocation) since 1-chloro-1-methylcyclopropane is angle strained and the resulting carbocation in this case will be unstable.

However, if we consider the possibility of hydride rearrangement in the second molecule, after formation of the carbocation, the following rearrangement should stabilise the carbocation formed, through the delocalisation of sigma electron pair as is seen in the case of cyclopropyl methyl carbocations.

Where exactly am I going wrong?

The proposed rearrangement would produce a primary carbocation which, relatively lacking in hyperconjugation, is itself unstable.  The proposed rearrangement therefore is not really plausible.
The "instruction" that the highly strained 1-methyl-cyclopropyl cation cannot be formed may be wrong anyway.  This paper reports a substituted cyclopropyl cation as an intermediate in a reaction between a more complex chlorocyclopropane and an alcohol.


Ordering the activity of metals So I'm stuck on this question because I'm having trouble breaking it down. So I know the answer already but like most stuck students I'm not sure how to get to it.

The problem is that I am not sure how to break this question down.

The activity series summarizes the experimental behavior of metals and their cations. A more active metal will displace the cation of a less reactive metal from solution.

Let's consider the case of zinc and copper. Zinc is higher on the activity series than copper. Zinc will displace copper ions from solution:

$$\ce{Zn(s) + Cu^2+ (aq) -> Zn^2+ (aq) + Cu(s)}$$

If we place a strip of zinc metal into a solution containing $\ce{Cu^2+}$ ions, the some of the zinc will dissolve into the solution, while some of the copper will plate onto the zinc.

What happens if we put a copper strip into a solution of $\ce{Zn^2+}$? Very little. Copper is less active than zinc. Copper metal will not displace zinc ions from solution.

$$\ce{Cu^2+ (aq) + Zn(s) -> No\ reaction}$$

If you did not know which was more active, copper or zinc, these two experiments would give you the information you need. Zinc displaces copper from solution. Copper does not displace zinc. Zinc is more active than copper $(\ce{Zn>Cu})$.

How can you apply this logic to your question?


Cause of m/z = 56 peak in 4-methyl-1-pentanol This is the mass spectrum for 4-methylpentan-1-ol:

From its structural formula, what could be the fragmentation which results in a peak at $m/z = 56?$

I was considering a few possibilities; could it be $\ce{C3H4O+}?$

Unfortunately I did not find a high resolution spectrum of this compound, which would have quickly answered your question.

Nevertheless, ionized alcohols and even more primary alcohols have a main fragmentation pathway which is the loss of water to yield an ionized alkene. In this case, the corresponding alkene would be 4-methyl-1-pentene, at m/z 84. The peak is present although rather small in the EI spectrum. From there, it is quite obvious, when one looks at the fragmentation spectrum of 4-methyl-1-pentene (NIST Webbook spectrum), that it shares a lot of similarities with the 4-methylpentan-1-ol EI spectrum. (See for instance peaks at m/z 69, 56, 43, 42, 41.) From there, it is quite likely that the m/z 56 peak is ionized isobutene ($\ce{C_4H_8}^{\bullet +}$) arising from a retro-ene type rearrangement.


How do I calculate the total volume when i mix 1000 kg of glucose with 5000 L of water I want to know how to calculate the resulting volume when I mix 1000 kg of glucose into 5000 L of water.

How does the calculation change if I'm mixing NaCl into the water, instead of glucose? Or how does the calculation change if im mixing several things into water (CaCl, Nacl, ect.)?

Sorry if this is a basic question, I have searched alot for a way to calculate this, but could not find answers that clearly explained how to calculate.

You could calculate it only if

you know the density of the substance being dissolved and of the solvent
there is no interaction between the 2 compounds, causing the deviation from the Raoult law and the shift of the final density.

If both assumptions are met, the final volume is
$$V_{\rm mix}=\frac{m_{\rm 1}}{\rho_{\rm 1}}+\frac{m_{\rm 1}}{\rho_{\rm 1}}$$

where m and $\rho$ are respective masses and densities.

On the other hand, if the mentioned deviation is in place, the calculation will be wrong.

E.g. if you mix $\pu{1 L}$ of ethanol with $\pu{1 L}$ of water, then after cooling down ( as the mixture would heat itself), the result will not be $\pu{2 L}$, but $\pu{1.9 L}$

In such cases, it can be determined experimentally, or from tabulated data of solution densities.

Or, the calculation above can be usually used as a good estimation, especially if there are no significant temperature changes during dissolution.


Comparing the stability of geminal diols Arrange the given gemdiols in decreasing order of their stability:

I thought that high angle strain would correspond to low stability, and therefore I compared the stability of the rings, which yields the answer as C > B > A, but, the answer given in my textbook is the opposite i.e., A > B > C.
On googling, I found out that, in this case, when the corresponding ketone hydrolyses to form the gemdiol then the one which gets greater relief of angle strain will have higher stability. But why do we have to check that? And I don't  know the reliability of the source. Moreover, what is wrong in my reasoning?

Your reasoning is correct, but I think you'd mistakenly compared the ring strain of the rings, while I would have compared the ring strain in the corresponding ketones.
Note that, you are asked about the stability of gem-diols, which could convert into the corresponding ketone. The point is, the ring isn't going to change, so the stability of rings don't matter.
Now, as you know the corresponding cyclic ketones would be $\ce{sp^2}$ hybridized. So, the ring angle should be $120^\circ$ in order to minimize the ring strain in ketone. The most closest is in C and least is in A.
Hence, the order.


Lewis structure for magnesium diboride What is the Lewis structure for $\ce{MgB2}$? I just read that it is a superconducting material (up to a relatively high critical temperature, 39 K) and am wondering if this has to do with its electronic structure.

In $\ce{MgB2}$, $\ce{Mg}$ has 2 valence electrons and each $\ce{B}$ has 3 for a total of 8. I know that $\ce{B}$ is "satisfied" with 6 valence electrons, but I don't see how to get 8 around $\ce{Mg}$ with 6 on each $\ce{B}$.

It's not really a molecule, so you don't really have a Lewis structure consisting of one magnesium and two boron atoms.  For a lattice structure see Wikipedia.  The lattice structure is hexagonal with alternating magnesium and boron layers, the boron layers having twice as many atoms per unit area as the magnesium layers.

The referenced picture is given here  and consists of green balls (magnesium) plus pink balls (boron), the latter covalently linked with each other.  If we model the magnesium as $2+$ ions (yes, the electronegativity difference is only 0.8, but de la Mora at al. indicates that the magnesium-boron bonding is predominantly ionic), then each boron atom has one negative charge and so becomes isoelectronic with carbon.  Then each of the hexagonal, covalently bonded boron layers is isoelectronic and isostructural with a graphite layer.


Why would substitution of H with D alter chemical shift? The $\ce{^{119}Sn}$ chemical shift of the dimeric tin hydride shown below (R=terphenyl) changes on addition of $\ce{D2}$ from a 1:2:1 triplet at $\pu{657.9 ppm}$ (due to coupling between $\ce{Sn}$ and bridging $\ce{H}$) to a doublet at $\pu{650.6 ppm}$. I assume that's because of reversible dissociation of one of the bridging $\ce{H}$, which allows formation of a mixed species where both $\ce{H}$ and $\ce{D}$ are present (the doublet being due to coupling with the remaining $\ce{H}$). However, I'm not quite certain why that alters the $\ce{^{119}Sn}$ chemical shift (I know the $\ce{Sn-D}$ bond will be shorter/stronger due to the lower ZPE). I would be interested in any thoughts/comments.

OP's main question: Why would substitution of $\ce{H}$ with $\ce{D}$ alter chemical shift?

OP has correctly indicated that the $\ce{Sn-D}$ bond will be shorter/stronger than $\ce{Sn-H}$ due to the lower ZPE, which may cause the chemical shift difference. That is the major reason for isotope effect in chemical kinetics mainly due to bond strength (shorter the bond distance, stronger the bond). However, here, how change of isotope effected the chemical shift is caused by increasing the electron density of the adjacent atom ($\ce{Sn}$). Since $\ce{Sn-D}$ distance is shorter than $\ce{Sn-H}$, the sharing electron pair in $\ce{Sn-D}$ bond is closer to $\ce{Sn}$ tan that in $\ce{Sn-H}$ bond. Thus, it is safe to say that $\ce{Sn}$ of $\ce{Sn-D}$ is shielded more than that of $\ce{Sn-H}$. Therefore, we can expect up-field shift (red-shift) in relevant $\ce{Sn}$ chemical shift. That's what you have experienced $(\pu{650.6 ppm} - \pu{657.9 ppm} = -\pu{7.3 ppm})$ when you have changed the $\ce{H}$ with $\ce{D}$ of the dimer. This phenomena is common among deuterated organic compounds (Ref.1):
$$
\begin{array}{llrrr}
\hline
\text{Hydrated cpd} & \text{deuterated cpd}  & \delta \ \left(\ce{^{13}C-H}\right)/\pu{ppm} &  \delta \ \left(\ce{^{13}C-D}\right)/\pu{ppm} & \Delta \delta/\pu{ppm}\\
\hline
\ce{CHCl3} & \ce{CDCl3} & 77.36  & 77.16 \pm 0.06 & -0.20  \\
\ce{CH3C#N} & \ce{CD3C#N} & 1.79  & 1.32 \pm 0.02 & -0.47  \\
\ce{(CH3)2C=O} & \ce{(CD3)2C=O} & 30.60 & 29.84 \pm 0.01 & -0.76 \\
\ce{C6H6} & \ce{C6D6} & 128.62 & 128.06 \pm 0.02 & -0.56  \\
\ce{CH3-OH} & \ce{CD3-OH} & 49.86 & 49.00 \pm 0.01 & -0.86 \\
\hline
\end{array}
$$
These chemical shifts are also depend on the other factors such as temperature and solvent. For comparison purposes, all chemical shifts of hydrated compounds have been recorded in their deuterated versions (e.g., $\ce{CHCl3}$ in $\ce{CDCl3}$) at a constant temperature (Ref.1).
Note: I also realized that $^1\!J_{\ce{Sn-D}}$ constant must be zero or negligible according to your observation. I tried to find the value of $^1\!J_{\ce{Sn-D}}$, but failed. Note that in deuterated organic solvents, $^1\!J_{\ce{C-D}}$ varies between $\pu{20 Hz}$ and $\pu{30 Hz}$ (Ref.1).
References:

Hugo E. Gottlieb, Vadim Kotlyar, Abraham Nudelman, "NMR Chemical Shifts of Common Laboratory Solvents as Trace Impurities," J. Org. Chem. 1997, 62(21), 7512-7515 (DOI: https://doi.org/10.1021/jo971176v).


Material that is flexible when cold but becomes brittle when hot I'm looking for a material that I can use to make a mesh netting.  It should be flexible and have reasonable tensile strength when cold but brittle when hot (>$\pu{400^\circ F}$).  I want to use it to contain charcoal when cold but disintegrate when charcoal is burning and some force is applied to it.

Any ideas?

It's indeed not common, as the brittle/ductile transition is usually the other way around. You can, however, find counter-examples (in a limited range of temperature) by searching for “inverse brittle-to-ductile transition”.


Melting Point Trend For Non Metals? I noticed that the melting point for non metals decreases across a period (i.e. melting point of nitrogen is higher than oxygen, oxygen's is higher than fluorine, so on). It is significantly more noticeable for chlorine and argon. Why is this? Furthermore, I a question came up in my homework asking sulfur (S8) had a higher melting point than phosphorus (P4). Their explanation was that sulfur is a larger molecule, but I don't see how this affects melting point. Is there a better explanation?

Note - this is the graph I was referring to for melting point: http://antoine.frostburg.edu/chem/senese/101/periodic/images/meltingpoint.gif

as one moves along the period, the atomic radii decrease. You see, this decrease leads to an increased net Z-effective and hence, the electrons come very close. In the case of Nitrogen, which exists as a diatomic molecule - this shrinking in size is less than that for oxygen and fluorine.
    the melting point is determined by the intermolecular interactions as well as the geometry of molecules. greater the number of bonds to break, higher is the melting point. For oxygen and especially Fluorine, there can be a forced compression, but yet again - these small atoms repel each other because of their high electron density. This repulsion lowers the melting and boiling points.

As for noble gases like Argon, they do not need to form any sorts of bond. The only kind of forces acing between two atoms are very weak London forces. These London forces are responsible for the freezing of Helium, at temperature as low as -273K, very close to absolute zero!!

I hope that helps!!


Intensive and Extensive Properties--Resistance Is resistance an intensive quantity or an extensive one? And why?
Let me present 2 scenarios.

Resistance of a wire:
The resistance of a wire depends on area and length. This implies it depends on the "quantity" of the wire taken. Does this mean it is extensive?
Resistance of a cell:
As we know,
resistance of a cell is $R = \dot G/\sigma$, where $\dot G$ is the cell constant and the property of the cell, and $\sigma$ is the conductivity. Now, this won't depend on the amount of electrolyte taken in the cell. Does this imply resistance is an intensive quantity?

So, my final question is, whether the resistance is an intensive or an extensive quantity and why?

To the extent that it's appropriate to treat it as a thermodynamic quantity, given that it's generally a description of a specific, macroscopic system (and thus violates the statistical-mechanical large-numbers assumption of thermo), resistance is extensive, always.

Resistance of a wire: The resistance of a wire depends on area and length. This implies it depends on the "quantity" of the wire taken. Does this mean it is extensive?

Yep! You're spot on on this.

Resistance of a cell: As we know, Resistance of a cell is $R=\dot G/\sigma$ where $\dot G$ is the cell constant and the property of the cell, and $\sigma$ is the conductivity. Now, this won't depend on the amount of electrolyte taken in the cell. Does this imply resistance is an intensive quantity?

Actually, this resistance measure will depend on the amount of electrolyte in the cell. Per here, $\dot G$ is a function of the dimensions of the active portion of the cell, in particular the sizes and shapes of the electrodes, and how far they are from one another. These geometrical characteristics then affect how much electrolyte must be used in order to fill the cell. Thus, the resistance $R=\dot G/\sigma$ is still extensive.


Liquid nitrogen superheating with a layer of oil? edit: Thanks Ivan, also further research.

Pure water can be superheated to a few degrees above 100 degrees. However I remember reading reading that water can be superheated to even higher temperatures by covering it in a thin layer of oil that more effectively prevented the water from boiling. When disturbed the water can then suddenly boil and vaporise at a very fast rate. Apparently a microwave is an effective way to heat the water for this purpose.

However this can easily become very dangerous.

As liquid nitrogen boils at -196 degrees C, would using liquid nitrogen instead, with a thin layer of oil allow the same suppression of boiling, superheating, and sudden vaporisation to be observed but at much safer temperatures, as well as keeping the liquid nitrogen in it's liquid state at higher temperatures?

See also: physics.stackexchange.com/questions/29392/why-droplets-of-water-under-oil-explode-when-heated

Thank you Nicolau Saker Neto.

Superheating is possible up to 300°C for water, yet disappointingly only -163°C for liquid nitrogen.

Source: http://www.nist.gov/data/PDFfiles/jpcrd276.pdf


Effect of impurities on Enthalpy of vapourization How does the Enthalpy of vapourization change when impurities are added? I know that the boiling point of a substance increases on addition of impurities, but how does the Enthalpy of vapourization change? Is less heat needed to convert the liquid from a solution to a gas?

Related: This question deals with a similar problem, but for Enthalpy of fusion instead of Enthalpy of vapourization.

Adding impurities(which mostly have low heat capacity, like salt) decreases the enthalpy of vaporization. This is because the impurities lower the heat capacity of the solution as a whole, making the enthalpy of vaporization lesser than the original solvent.

This is why even though the boiling point of a solution increases when impurities are added it is heated up faster(boils faster) than the one without impurities. For example, saltwater (salt is a non-volatile solute) boils faster than pure water because adding salt lowered the heat capacity of water as a result of which the enthalpy of vaporization is lesser.

Sources:

https://www.studocu.com/en/document/western-governors-university/natural-science-lab/mandatory-assignments/task-1-boiling-salt-water/5667142/view
https://www.thoughtco.com/adding-salt-to-boiling-water-607427

The addition of impurities lowers the heat capacity since the solvent molecules bind to the solute particles in a specific manner and hence require less heat to vaporize than they'd require when they're randomly moving.

Source:

https://socratic.org/questions/how-does-salt-change-the-specific-heat-capacity-of-water#143950


Sensitivity of Carbon NMR Compared to Proton NMR I am trying to figure out exactly how much less sensitive carbon NMR is from proton NMR. In a lecture, my professor said that carbon NMR is 1/5800th as sensitive as proton NMR for the following reasons:

Abundance of Carbon-13 is 1.1% so sensitivity is decreased by a factor of (0.011)
The precession frequency (difference in energy between the alpha and beta states) is about 1/4 of the precession frequency of a proton. Because this is equivalent to electrical output on the oscillator coil, this will be 1/4 as powerful (0.25)
The magnetogyric ratio of Carbon-13 is 1/4 that of Hydrogen-1 (0.25)
The dipole generated by Carbon-13 is 1/4 that of Hydrogen-1 (0.25)

Thus: 0.011 * 0.25 * 0.25 * .25 = 1/5800 (roughly)

My only objection is that the magnetogyric ratio is proportional to the magnetic moment and the spin number. Thus, I would not expect the 1/4 factor to be counted twice since it is assumed based on the definition of magnetogyric ratio. For this reason, I would expect the sensitivity of carbon NMR to only by 1/1600th of proton NMR.

i.e. 0.011 (abundance) * 0.25 (lower precession frequency) * 0.25 (magnetogyric ratio / dipole moment) = 1/1600

I am not sure who is right. I cannot find any sources that will detail the exact sensitivy differences between proton and carbon NMR and why it is so

The receptivity ratio is ca. $5800:1$, as listed in the Bruker tables (under the column Receptivity > Natural rel. $\ce{^13C}$). Receptivity is a product of the natural abundance and an "inherent NMR sensitivity", which scales as $\gamma^3 I(I+1)$. For proton and carbon, $I = 1/2$, so the $I(I+1)$ term cancels out and we are left with

$$\begin{align}
\frac{\text{receptivity }\ce{^1H}}{\text{receptivity }\ce{^13C}} &= \frac{\text{abundance }\ce{^1H}}{\text{abundance }\ce{^13C}}\cdot \left(\frac{\gamma_\ce{H}}{\gamma_{C}}\right)^3 \\
&= \frac{100\%}{1.1\%} \cdot (4)^3 \\
&\approx 5800.
\end{align}$$

Why $\gamma^3$? One must bear in mind that at the end of the day, the signal that is detected is not the magnetisation directly. The precessing magnetisation that is excited in an NMR experiment induces a voltage in the coil that surrounds the sample. This voltage is the actual quantity that is detected.

By Faraday's Law, the voltage generated is proportional to the rate of precession, i.e. the Larmor frequency $\omega_0$, as well as the actual magnitude of the magnetisation $M$. The Larmor frequency is itself proportional to $\gamma$ (it is defined by $\omega_0 = -\gamma B_0$), so that accounts for one out of three.

Any magnetisation that we can measure must be derived from the equilibrium magnetisation $M_0$. Since the equilibrium magnetisation points along the $z$-axis, $M_0$ is the expectation value of $z$-magnetisation $\langle M_z \rangle$ at equilibrium. $\langle M_z \rangle$ itself can be related to the populations of the spin-up and spin-down states, conventionally denoted as $n_\alpha$ and $n_\beta$:*

$$\begin{align}
\langle M_z \rangle &= \gamma \langle I_z \rangle \\
&= \gamma \left[ \left(\frac{\hbar}{2}\right)n_\alpha + \left(-\frac{\hbar}{2}\right)n_\beta \right] \\
&= \frac{\hbar\gamma}{2}(n_\alpha - n_\beta)
\end{align}$$

So that accounts for one more occurrence of $\gamma$. The third and final one comes from the population difference $(n_\alpha - n_\beta)$. According to the Boltzmann distribution, we have

$$\begin{align}
\frac{n_\alpha}{n_\beta} &= \exp\left(\frac{\Delta E}{k_\mathrm{B}T}\right) \\
&\approx 1 + \frac{\Delta E}{k_\mathrm{B}T}
\end{align}$$

where $\Delta E = E_\beta - E_\alpha$ is the (positive) energy difference between the $\alpha$ and $\beta$ states. In the second step we have expanded the exponential as a Taylor series and truncated at first-order, making the (in this context, very valid) assumption that $\Delta E \ll k_\mathrm{B}T$. So now:

$$\begin{align}
n_\alpha - n_\beta &= \left(1 + \frac{\Delta E}{k_\mathrm{B}T}\right) n_\beta - n_\beta \\
&= \frac{\Delta E}{k_\mathrm{B}T} \cdot n_\beta
\end{align}$$

The energy of the spin states in an external magnetic field (the Zeeman splitting) will depend on the gyromagnetic ratio:

$$\Delta E = \gamma\hbar B$$

So, working backwards: a nuclide with a smaller gyromagnetic ratio, when placed in an external magnetic field,

has a smaller difference between the population of spin-up and spin-down states (by a factor of $\gamma$);
which leads to a smaller equilibrium magnetisation (by a factor of $\gamma$);
which, when excited, generates a smaller voltage in the detection coil (by a factor of $\gamma$).

Finally, it should be noted that this does not take into account any noise in the spectrum; it is just a measure of the signal that is available. As far as I can tell, the (root-mean-square) noise also scales with the frequency, and this leads to a theoretical signal-to-noise ratio that scales as either $\gamma^{3/2}$ or $\gamma^{11/4}$ depending on the analysis made (essentially, the exponent is slightly reduced from the nominal value of 3). I don't have the ability to explain this properly, but I can provide references that are probably helpful:

Abragam, A. The Principles of Nuclear Magnetism; Clarendon Press: 1961, especially pp 78 onwards.
Hill, H. D. W.; Richards, R. E. Limits of measurement in magnetic resonance. J. Phys. E: Sci. Instrum. 1968, 1 (10), 977–983
Hoult, D. I.; Richards, R. E. The signal-to-noise ratio of the nuclear magnetic resonance experiment. J. Magn. Reson. 1976, 24 (1), 71–85

* Populations should not be interpreted as the actual number of particles in a specific eigenstate, but rather the ensemble average of the coefficient corresponding to the specific eigenstate in the wavefunction. In other words, if each spin has a state $|\psi\rangle = c_\alpha|\alpha\rangle + c_\beta|\beta\rangle$, then we have $n_\alpha = \overline{c_\alpha^* c_\alpha}$ where the bar indicates an average over all spins in the ensemble. For more information please consult a suitable resource on density matrices.


Why does FeCl3 react with carboxylic acid, but not with sulphonic acid in this question? I was solving a question where I was asked to identify the reagents given in the question which could react with a compound provided.
The first compound was methanoic acid. One of the reagents given was $\ce{FeCl_3}$. According to the answer key, these 2 could react. On googling the reaction, I found that methanoic acid loses its acidic proton and is attacked by $\ce{FeCl_3}$ and the proton released then removes chlorine to form the complex. It all made sense because $\ce{CH_3COOH}$ was an acid.
But the next compound was 3-carboxy-4-nitro-benzene sulfonic acid. And I wanted to check its reactivity with $\ce{FeCl3}$. So, I felt that the compound being acidic due to the sulfonic group would work similarly as the first compound and thus react with $\ce{FeCl_3}$. However on verifying with the answer key, I found that it would not react with $\ce{FeCl_3}$. But I couldn't think of a reason why.

It all made sense because $\ce{CH_3COOH}$ was an acid

$\ce{CH_3COOH}$ being an acid facilitates the complex formation as the chlorido ligands can be easily removed as $\ce{HCl}$. But, that's not the only reason why the red colored complex $\ce{[Fe(OAc)_3]}$ was formed.
You have to understand that the stability of the complexes is also determined by how well the ligand interacts with the central metal ion.Here, the ligand forming the coordination complex is $\ce{OAc^-}$ and this is associated with enough negative charge density to stabilize the complex formed.

So, I felt that the compound being acidic due to the sulfonic group would work similarly as the first compound and thus react with $\ce{FeCl_3}$. However on verifying with the answer key, I found that it would not react with $\ce{FeCl_3}$

This is because although as you have said has a sulfonic acid group, it is not the acidic strength of this ( or acidic strength of $\ce{CH_3COOH}$ in the first question) that drives the complex formation.It mainly depends upon the strength of the ligand and effective nuclear charge of the central metal ion.The ligand in this case, is associated with lot of electron withdrawing groups like $\ce{-NO_2}$ and that makes this a poor ligand.Also, the steric hindrance due to the bulky groups also has a role to play for the complex formation not to take place.
NOTE:
All my arguments made regarding stability of complexes is as per the Crystal Field Theory which assumes the interaction between metal ion and ligand to be purely ionic.That's why, I was comparing the charge density of the ligands in the 2 cases.


What is meant by complete outer shell? Why do the noble gases have zero valency? Does having 8 or 2 electron in the outmost shell mean its outmost shell is full and its valency is zero?

I know that the 3rd and 4th shell can contain 18 and 32 electrons.
Then how can Argon's (2,8,8) outmost shell be full, though it does not contain the highest number of electron (18) in it's 3rd shell?

It is the same in the cases of other noble gases too? Why?

When we refer to "outer shells" we are talking about the highest n-level, but also limiting ourselves to the s- and p-orbitals. The reason for this has to do with how the effective nuclear charge "felt" by the valence electrons changes as you move through successively higher energy configurations.

In summary, $s$- and $p$- electrons are screened less effectively by inner shells, and so for a given n-value, the $s$ and $p$ orbitals of the next n fill before the current n (they have lower energy).

In your example using argon, this means that the 3d shell actually has no electrons - the condensed configuration is:

$$
\ce{Ar: \space [Ne]}3s^23p^6
$$

For the next noble gas, krypton, the condensed electron configuration is:

$$
\ce{Kr: \space [Ar]}4s^23d^{10}4p^6
$$

Note that it has $3d$ electrons at a higher energy than what argon had, but since the value of n roughly corresponds to the distance of the orbital from the nucleus, the $3d$ orbitals are physically closer to the nucleus than the $4s$ and $4p$ orbitals. In other words, for single atoms, the lowest energy electron configuration always fills the s- and p- orbitals first, and so when we are looking  for the valence or "outermost" electrons, that is where we need to look - we sort of ignore the $d$- and $f$-orbital electrons for these kinds of problems.


Calculation of equilibrium constant in a gaseous system A ⇌ B + C I am supposed to determine whether the following two statements are true or false:

For the reaction $\ce{A (g) <=> B (g) + C(g)}$, $K_p = \pu{1 atm}$. If we start with equal moles of all gases at $\pu{9 atm}$ of initial pressure, then at equilibrium the partial pressure of $\ce{A}$ increases.

The reaction quotient $Q_p > K_p$, hence the equilibrium shifts in the backward direction.

I discussed it with my friend, who gave me a weird explanation abut how to find the reaction quotient $Q$ at the start. He claimed that it would be equal to the total pressure at the starting point, $\pu{9 atm}$. But I believe that my friend is wrong, according to the following calculation:
Let us assume that $x~\mathrm{mol}$ of each gas is present. Hence the mole fraction of each gas is $x/(3x)$ = $1/3$. The partial pressure of $\ce{A}$ is thus $p_\ce{A} = (1/3) \times 9 = 3$ . Similarly we get $p_\ce{B} = 3$ and $p_\ce{C} = 3$, which gives
$$Q_p = (p_\ce{C} \times p_\ce{B})/p_\ce{A} = (3 \times 3)/3 = 3$$
Regardless of whether $Q$ is 3 or 9, it is still larger than $K$, so the eventual answer is not changed. However, I wanted to know what the correct value of $Q$ was.

Your analysis is correct, except that you don't account for how the the units cancel out in your calculation.

The initial reaction quotient, $Q_P$, is indeed 3, as you have calculated.  And since $K_P$ is 1, the reaction will shift to the left, increasing $P_A$.  Hence Statement-1 is nearly correct.  I say "nearly" because $K_P$ is dimensionless, so it should be 1, not 1 atm (each partial pressure should be divided by whatever you define as the standard pressure, 1 atm or 1 bar -- that's what you've left out of your calculation; you should try this yourself as an exercise).

Statement-2 is not quite correct, because it says the "equilibrium shifts in the backward direction."  Assuming you are not changing the temperature, the equilibrium does not shift (it's a constant: $K_P$ = 1).  Rather, the reaction  (and thus $Q_P$) shifts in the backwards direction until equilibrium is attained.


C-C bond dissociation energy Have the various $\ce{C-C}$ bonds in straight chain hydrocarbons, have the same dissociation energies? Or does it depend on the location of the carbon in the chain?
In other words, are the terminal bonds weaker or stronger than the internal bonds?

The terminal bonds are stronger than the internal bonds.

In simple hydrocarbons where strain or electronegative substituents are absent, the bond strengths will correlate with the stability of the resultant radicals that would be formed when the bond breaks.

Because radicals are electron deficient species (they are one electron short of an octet) they are stabilized by electron donating substituents (alkyl groups for example).  Hence, a tertiary radical is more stable than a secondary radical which is more stable than a primary radical which is more stable than a methyl radical.

Using this information we can see that
$$\ce{R-CH2. + .CH2-R}$$
is more stable than
$$\ce{CH3. + .CH2-R}$$
Hence the internal bond will have a lower bond dissociation energy and break more readily.


Usage of lowest locant rule How should we name this compound by IUPAC system?
Is it

2,3-dimethylcyclohexene
 (or)
 1,6-dimethylcyclohexene?

In the first case locant position adds to 2+3 = 5, while in second it is 1+6 = 7. However, in second case, first substituent gets lower position, i.e., 1 , while in first case it gets 2.

According to latest IUPAC recommendations, lowest locant rule is the valid rule followed for organic nomenclature and presently, the lowest sum rule is no longer followed.

So, as you have discussed in the question, the name that follows the lowest locant rule is the proper name of the compound.

So, it is 1,6-dimethylcyclohexene.


How to solidify sand I'm doing a research project where we analyzing the structure of sand after a certain process. To get a better idea of the internal structure of the sand, we use different colors to visualize the movement of the particles. However we can't actually observe that structure since sand is not a solid substance. Therefore we need a way to solidify the sand without changing the structure.

We figured we would need some sort of liquid which we can pour into the sand container (the container is 2x1x0.2 meters) so that it will solidify. In other words, how can we solidify sand?

Using a transparent container is not really an option since we are curious about the structure in the center which we cannot see... unless we use transparent sand ;)

Note: we are using 'regular' beach sand.

EDIT:
I have read through all the answers and they are great! I can't accept any answers just yet since we need to figure out which method we will use, but it looks promising! We started with 0 and now we have some great ideas going with your help!

Based on your comments, it sounds like you have some sort of colored sand and you want to immobilize it. My suggestion would be some sort of polymer matrix.

There are multiple possibilities, but one of the easiest is polydimethylsiloxane PDMS which is optically clear and pretty easy to work with (i.e., safe). There's the "base" and a curing agent which creates cross linking.

Probably the easiest how-to comes from U. Wisconsin.

If done right (i.e., no air bubbles) you should be able to pour in the PDMS, cure, and have a polymerized matrix around the sand. Then you could cut it up or do whatever needed to see different regions.


Base-promoted rearrangement of β-ketoester Propose a reasonable mechanism for the following transformation:

I guess an enolate probably forms first, but then what does it attack to give something like the product in the question?

This problem has probably been on everyone's mechanistic chemistry exam at some time or another. Does the alkyl (allyl) just "jump" from one side of the carbonyl group to the other? In most examples, the allyl group is a saturated alkyl group. Under such circumstances I am in full agreement with @orthocresol as to the mechanism of the process (1 --> 9). The reaction is not catalytic in methoxide and requires sufficiently concentrated alkali to assure sufficient ester enolate, lest the reaction stops at the diester 4. Notice that the red hydrogens in β-ketoester 1 retain their position in the product 9.

While @Carlson Miller suggested the formation of an enolate, his proposal brings up an interesting possibility in this particular example. The ensuing mechanisms are improbable in protic media but imagine running the reaction with a kinetic base ((NaH, (TMS)2NNa, LDA) in an aprotic solvent is intriguing. There are two possible mechanisms for the reaction. First, enolate 10 can fragment to a ketene-ester enolate that must lose methoxide and generate diketene 11. At this point the reaction is stoichiometric in diketene and methoxide. These two species lead to enolate 12. Bimolecular hydrogen exchange affords the more stable enolate 8. Although I am not particularly enamored with this option, it is nonetheless possible in spite of the limited amount of base relative to the protic reaction. Incidently, any product of intramolecular ketene dimerization would be susceptible to opening by methoxide. This mechanistic pathway, as with the protic reaction, retains the red hydrogens in their original location.

However, the more likely pathway for enolate 10 is via an oxy-Cope-like rearrangement. Although enolate formation is the driving force for the oxy-Cope-like rearrangement (13 --> 14), the rearrangement of enolate 10 gives the more stable enolate of the β-ketoester 8. In this case, the red vinylic hydrogens of ketoester 1 are now located on the methylene group of the allyl residue. Indeed, the allyl group does "jump" in the oxy-Cope.
Enolate 16 is generally formed by alkylation of the dianion of an acetoacetic ester with an allyl halide. Is anyone in SE-land aware of the transformation 15 --> 16?