Is it possible to manipulate some specific atoms of a liquid metal by dint of the magnetic field, electric field, etc? Let's deem, we have a liquid metal that is multicomponent, for example, type A, B, and C. Is it possible that we single out the type A and put them in a specific shape, for example, a circle?
Let me clarify it, I have a liquid metal, having three different elements, and I poured it in a square framework. I want to separate element C and then set them in several lines that are parallel with each other; so that, they play the role of reinforcement and the alloy of A and B is my matrix.

While Field Structured Composites are a thing, they use nano- to micro-scale particles in a polymer matrix, with particle alignment controlled/influenced by applying electric or magnetic fields during curing.
Trying to do this at the atomic level, as you ask about above, runs into several problems. These include:

Overcoming the desire of the liquid to mix, so you need your desired element to have a fairly well tuned positive heat of mixing to just balance the entropic contributions to the free energy. Or, in other words, just barely soluble.

Overcoming diffusion in the liquid would require strong interactions of your field with the specific element (and only that element). This runs up against the limits below.

In a metallic liquid, applying an electric field isn't going to specifically target any one atom type since the free electrons will shield the bulk from the field. Ions in a dielectric liquid would be a better choice (much like the field structure composites above), but that is just the reinvention of ion chromatography.

The magnetic moments of atoms are pretty darned small, so you won't get the ferromagnetic boost that can be achieved with particles. Applied magnetic fields will have to be enormous if they would even work at all.


How can iodine oxidise hydrogen peroxide? The following is mentioned in NCERT (government textbook in India used in all schools) under chemical properties of $\ce{H2O2}:$

(iv) Reducing action in basic medium
$$\ce{I2 + H2O2 + 2 OH- -> 2 I- + 2 H2O + O2}$$

I feel the reaction is not correct. Since $\ce{H2O2}$ is a moderate/strong oxidizing agent, I don't think it can be oxidized by iodine and the reverse reaction should be true since $\ce{O2}$ is a strong oxidising agent and $\ce{I-}$ is a reducing agent. Can someone please verify it?

Harsh Jain, It may be a correct reaction, but it is a very complicated one.

Let us see what happens when iodine is added to a large volume of an aqueous base in the absence of $\ce{H2O2}.$ It forms iodide and iodate ion. If iodine is in excess, triiodide ion is also formed.

The complication arises from adding hydrogen peroxide because it can act as an oxidizing agent as well as a reducing agent. Similarly, iodate ion and iodide ion can react with each other as well as $\ce{H2O2}.$ For example, iodine can react with $\ce{H2O2}$ and form iodide ions.
On the other hand, iodide ion can be oxidized by $\ce{H2O2}$ to free iodine.

I would suggest that you read a paper by Bray and Liebhafsky [1].
Here is the summary by the authors in their table:

References

Bray, W. C.; Liebhafsky, H. A. Reactions Involving Hydrogen Peroxide, Iodine and Iodate Ion. I. Introduction. J. Am. Chem. Soc. 1931, 53 (1), 38–44. DOI: 10.1021/ja01352a006.


Work Done Upon Molecules In which of the following processes is energy transferred into the substance by work ($w > 0$)?

  a) Expansion of a gas against the surroundings
  b) Expansion of a gas into a vacuum
  c) Vaporization of one mole of water at 50˚C in an open container
  d) Melting of 100 g of ice on a laboratory benchtop
  e) Combustion of methane in a constant volume container.

The correct answer is given as (d), but I needed to find a physicist to explain it to me.  This problem can being thought of entirely in terms of thermodynamic work.

So, we know that
$$w = P \Delta V.$$

From this, we see that no work is done in (b) and (e).  Also, since the gas expanded against its surroundings, it did work on its surroundings, which isn't what the question asked, thus (a) is not the correct answer.

Now, although (c) appears to be a heat transfer, and it is, there is a little more going on.  The heat that was given to the water is converted into work as the water vaporizes, this causes a volume expansion (the gas rising).  Intuitively, one can think of the surroundings as a tube with a movable piston.  When vaporizing, the water vapor will push against the piston, increasing the pressure and thus the piston will move, causing the water to do work.

The last choice is (d), the correct answer, but there is also an intuitive way to think about this.  As the ice is heated, it melts (this process is also, as we correctly deduced, a heat transfer), however, this heat also causes the surrounding atmosphere to do work on the ice.

Again, think of a tube with a movable piston with a cube of ice that is slowly melting in the presence of normal air.  As the ice melts, the volume will decrease, and thus the nitrogen and oxygen molecules in the air have more space to move about, decreasing the pressure, which causes the piston to move down, thus doing work upon the ice.

Hopefully this makes sense to everyone who has been either watching or contributing to this discussion.

Unfortunately, I don't know what the authors of this question had in mind. This came up as I was working through an old Chemistry GRE exam. It wouldn't have occurred to me to consider this problem through mechanical work. As a side note, only 18% of the people who answered this question got it right, which essentially means that people just guessed the answer.

This is a bad question.  But the answer is probably supposed to be (d) because:

Water is more dense than ice.
Thus when ice melts, the total volume of "substance" goes down.
This means that atmospheric pressure did PV work on the melting ice.

Of course, heat transfer would also be required to melt the ice but the question isn't asking about that.


Why do many organic halogen compounds smell or taste sweet? Is it just coincidence that many organic halogen compounds (especially chlorinated ones) tend to either taste or smell sweet? Examples include dichloromethane, vinyl chloride, chloroform, and sucralose. This chart also shows that many low-carbon haloalkanes have a sweet smell. If this isn't coincidence, what is the biochemical basis for this phenomenon?

The taste of halogenated organic compounds mainly depends on the halogen atoms involved. Study showed that halogenation of aliphatic series imparted sweetness and halogenation of aromatic compounds increased/decreased the intensity of sweetness/bitteness. This increase/decrease of intensity is due to the efficacy of halogen atoms which is directly related to electron withdrawing power (Hammett $\pu{\sigma}$ value). This is called AH-B theory of sweetness$\ce{^{[3]}}$. Effect of halogen substituent on aromatic compounds to impart bitterness was in opposite trend when compared to effect of halogen substituent on aliphatic compounds to impart sweetness. For example, halogenation of  aminonitrobenzene compounds impart sweetness$\ce{^{[1]}}$. Nitroaniline has a relative sweetness of 40 and substituting halogen at para position respect to nitro group increases sweetness (chlorine: $\pu{x 400}$, bromine:  $\pu{x 800}$, iodine:  $\pu{x 1250}$).

Commercially, chlorine played an important role in imparting sweetness in organic compounds. One instance is halogenation of saccharin. Substituting fluorine at para position imparts sweetness but a bitter aftertaste. Substituting chlorine imparts sweet but with accentuated bitterness. Substituting bromine enhances sweetness but it is extremely bitter. Substituting iodine loses is sweetness leaving a slight bitter aftertaste. Chlorination of other halogen compounds like sucrose and its derivatives, tryptophan, saccharides, lactose, maltose has also been studied where sweetness/bitterness increases by many folds$\ce{^{[2]}}$.
Notes and references

Taste by Lloyd M. Beidler
Taste Chemistry by R.S. Shallenberger
Robert Shallenberger and Terry Acree proposed the AH-B theory of
sweetness. Simply put, they proposed that to be sweet, a compound
must contain a hydrogen bond donor (AH) and a Lewis base (B)
separated by about 0.3 nanometres. According to this theory, the
AH-B unit of a sweetener binds with a corresponding AH-B unit on the
biological sweetness receptor to produce the sensation of sweetness. (Wikipedia)


Why is Libethenite (Cu2(OH)PO4) color green? It is not a complex compound and the copper atom is not oxidized so I'm wondering how does its color change. The book mentioned that at favorable conditions, adding water in Copper phosphate, which is blue, will result in the formation of Libethenite.

First of all, why do you think your copper is not oxidized? Each copper has a charge of +II while $\ce{OH^-}$ and $\ce{PO_4^3-}$ add up to -4.

For all further discussions we need to look deeper into the crystal structure. The compound crystallizes with two crystallographically destinguishable copper ions. One forms a $\ce{[Cu(PO_4)_4(OH)_2]^12-}$ octahedron while the other one is, at least to me only fivefold coordinated to form a trigonal bipyramid $\ce{[Cu(PO_4)_4(OH)]^11-}$.
Of course those are no simple $\ce{OH}$ units as there is further bridging and connections happening but for a simple view this should be okay. So for the color there are, among other effects, still $\ce{d-d}$ transitions happening.

EDIT:

To add something to the color change in the reaction itself. In your copper phosphate you have copper polyhedra that only feature phosphate as ligands, for the $\ce{[Cu_3(PO_4)_2]}$ these would be trigonal bipyramid and square plane.So you have the addition of $\ce{OH^-}$ to these polyhedra as well as some substitution. And introducing a different ligand will also change your splitting energy and therefore change the color.


Which of the following samples contains the largest number of atoms? Which of the following samples contains the largest number of atoms?
a.    2.0 moles of $\ce{H3PO4}$
b.    3.0 moles of $\ce{H2SO3}$
c.    4.0 moles of $\ce{HNO3}$
d.    6.0 moles of $\ce{HClO}$
e.    8.0 moles of $\ce{HBr}$

My logic would tell me that every mole has $6.022\times10^{23}$ atoms in it. Then you divide the amount of grams of each compound by the number of moles?
so for A I'd  use the following:
(Avg # *2) / 98 grams/per mole of $\ce{H3PO4}$   = $1.22\times10^{22}$ atoms
and so on, but I don't get the right answer :/
$(2\times(6.022\times10^{23})$ = number of atoms = $1.20\times10^{23}$
$\ce{H3PO4}$ = 98 grams/mole
The answer is C btw, I just don't know to set up the equation right to get to this answer.

Geeze, NaCl has 1 mole of Na and 1 mole of Cl per mole of NaCl. So that is 2 moles of atoms per mole of NaCl.

$\ce{K2SO4}$ has 2 moles of potassium, 1 mole of S and 4 moles of O per mole of $\ce{K2SO4}$. So that is 7 moles of atoms per mole $\ce{K2SO4}$.

You don't need to convert to atoms using Avogadro's number then back to moles. That complicates the problem needlessly. Just use the stoichiometry and you can do the arithmetic in your head.


Molecular electronic state symmetry I am trying to work out the ground state symmetry of electronic states. The paper states the GS symmetry is $^6A_1$ for the right hand side electron configuration and $^4B_1$ for the LHS in the D$_{2\mathrm{d}}$ point group. I understand that the symmetry is determined from direct products of the molecular orbital symmetries of the singly occupied orbitals. However I cannot get to the two symmetries listed. For the left electronic configuration it is $b_1 \times a_1 \times b_2 \times e \times e = [a_1+a_2+b_1+b_2]$ so I'm unsure of how to get $A_1$. Likewise the product e x e for the right electronic configuration produces the same results and again I'm unsure of how to arrive at $B_1$.

Each of the terms you generate has a different energy, and may have different spin multiplicity. The spin multiplicity $(2S+1)$ is given by the left superscript (6 and 4) . The quartet (right) it seems has the lowest energy with spin of a triplet (sym e) x doublet ($b_2) \to$  triplet + quartet.  To do this add spin vectors triplet $S_T=1/2+1/2$ , doublet $S_D= 1/2$ and  make series $S_T+S_D \to |S_T-S_D|$ in unit steps, (Clebsh-Gordon series) this gives total spin 3/2 & 1/2 or quartet($\uparrow\uparrow\uparrow$) + doublet($\uparrow\uparrow\downarrow$). The symmetry has next to be associated with the spin multiplicity and this is complicated. It is necessary to find the character ($\chi$) under, say, quartet multiplicity for each symmetry operation $R$ for each symmetry species $A_1,A_2 ..$  etc. $R^2$  means square the operation. In this case this is

$$\chi(quartet) = (1/6)\left([\chi(R)]^3-3\chi(R)\chi(R^2) +2\chi(R^3)\right)$$

The result has to be reduced into symmetric $\chi^+$ and antisymmetric components $\chi^-$  as $\chi^+=\left( [\chi(R)]^2+\chi(R^2)\right)/2 $ and   as $\chi^-=\left( [\chi(R)]^2-\chi(R^2)\right)/2 $ and the set of characters produced reduced to find the symmetry species.

Chapter 5 of 'Symmetry & Spectroscopy' by Harris & Bertolucci gives a detailed account of how to do this.


Quasi-static implies that T aprox equals Tsurr? A few days ago I re-visited my undergraduate thermodynamics notes and (as you may have seen from the amount of questions I have been posting) I have encountered many difficulties that I overlooked at that time.
And I think most of my problems come from how unclear is the concept of quasi-static process in books.
Many textbooks define a quasi-static process as:

  A process that has its macroscopic variables well defined and in the system it is at each instant of time in a state infinitesimally close to the state of equilibrium.

The first part of that definition is logical. But the second part, related to thermodynamical equilibrium confuses me.
What is clear is that if it is stated that a certain process is quasistatic, then the pressure of the system is infinitesimally similar to the external pressure, which allows us to easily calculate the volumetric work.

My concise doubt comes from the relationship of the external temperature with the internal temperature. We know that for a system to be in true equilibrium with the outside, both pressure and temperature must be the same (without considering material equilibrium).

If this is the case, T system should be infinitesimally equal to T of the environment which would completely eliminate the generation of entropy by transfer between finite temperatures.

Now, when one studies theoretical cycles, it is generally stated that they are internally reversible ( quasi-static internally and without internal friction) but irreversible at last because the unstopabble heat transfer between finite temperatures in many steps.

I see a contradiction here:
As I understand it, many define a quasistatic process as:

"An extremely slow process where internally the P and the T are equal, homogeneous". (Completely agree)
"The system  at each instant of time in a state infinitesimally close to the state of equilibrium". Dont' get it since many problems considers Pexternal=Psystem but T System different to T surroundigs.

To me, it should be called partially quasistatic process if Tsys does not equals to Tsurroundings.

Regards.

In my judgment, when they are talking about irreversible cycles, they are referring to idealized cases in which (a) the process is internally reversible with respect to the working fluid but (b) irreversible with regard to the region outside the working fluid where there is a finite thermal resistance (e.g., a cylinder wall with finite thermal conductivity) between an (external) ideal constant-temperature reservoir and the working fluid.  So the rate of heat transfer to the working fluid is very low (quasi static), and all the entropy generation takes place within the (higher) thermal resistance wall.  At least, this is how the irreversibility is often modeled.

To illustrate all this, see the following thread from Physics Stack Exchange:  https://physics.stackexchange.com/questions/443527/entropy-generation-in-an-endoreversible-engine


Identification of lithium and strontium salts I'm having trouble interpreting the results I got for my experiment.

As you should be well aware of, lithium and strontium both burn with crimson red when they are placed in a flame. Now, I can't decide which one it is.

Then I was told to do an experiment with the same salt where I had to add dilute hydrochloric acid and barium chloride. I had no visible change.

Does this last test help me with identifying between strontium and lithium? What element is it?

If your solution contains only $\ce{Li+}$ or $\ce{Sr^2+}$, you can differentiate them by adding a sulfate solution (for example, $\ce{Na2SO4}$ or diluted sulfuric acid). If strontium is present, a white, fine-crystalline precipitate of $\ce{SrSO4}$ will form. In the case of lithium, the solution will remain clear. This is because $\ce{SrSO4}$ is much less soluble in water than $\ce{Li2SO4}$ ($\pu{0.0132 g}$ versus $\pu{34.8 g}$ in $\pu{100 g}$ of $\ce{H2O}$ at $\pu{20 ^\circ C}$ (reference)).

Adding diluted $\ce{HCl_{aq}}$ and $\ce{BaCl2}$ to a solution is commonly used as a test for $\ce{SO4^2-}$ (if it is present, $\ce{BaSO4}$ precipitates). Maybe this could help you to distinguish between $\ce{Li+}$ and $\ce{Sr^2+}$ given the different solubilities of their sulfates. If your unknown salt readily dissolves in water and $\ce{SO4^2-}$ is found to be present in the solution, then it is likely that you have lithium sulfate. Strontium sulfate as unknown salt would not readily dissolve and therefore not give a positive test.

However, if your solution can contain more ions besides $\ce{Li+}$ or $\ce{Sr^2+}$, I would suggest to try the separation with ammonium carbonate which is already mentioned in the comments to your question.


Sigma bonds, pi bonds and the puzzling symmetry of methane My A-Level Chemistry textbook assures me that:
a sigma bond is formed when two s orbitals overlap.
a pi bond is formed when two p orbitals overlap.
all single bonds between carbon and any other atom are sigma bonds
(side note) double bonds between carbon atoms consist of a sigma bond plus a pi bond
(side-side note) triple bonds between carbon atoms consist of a sigma bond plus 2 pi bonds.
However, the electronic structure of carbon is [He]2s22p2, implying two s orbitals and two p orbitals. I deduce from this that e.g. methane (CH4) has 2 sigma bonds and 2 pi bonds.
I am pretty sure that my deduction is incorrect, as it contradicts the statement in bold above, but cannot see why.
(In addition, the book states that methane has a completely regular tetrahedral structure with all four C – H bonds exactly the same, at identical angles to each other.)
Edit: Although aspects of this question overlap(!) with Can a s orbital overlap with any p orbital to form a sigma bond?, I am most interested in why what appears to a mixture of s- and p- bonds ends up creating such a symmetrical molecule as CH4.

I am going to attempt to answer this, based on the link supplied by @Kartik and expanding on @Mitheron’s comment above.
As we know, covalent bonding is the process of “filling up” partially filled outer orbitals by sharing electrons between atoms. Carbon’s structure [He]2s22p2, or, more accurately, [He]2s22px12py1. When bonding occurs, however, two things happen:
1.  One of the 2s electrons gets “promoted” to a p orbital. This is an “excited” state in which the electrons have slightly more energy (though not enough to remove any of them). The configuration therefore becomes [He]2s12px12py12pz1, and voila, we now have 4 unpaired electrons available to bond with other atoms.
2.  The extra energy allows the orbitals to “hybridize”. In effect, the 2s and 2p orbitals morph into 4 identical orbitals of slightly higher average energy. These orbitals are called sp3 orbitals (one part s, three parts p). The electronic structure of the ready-to-bond atom is now [He] (sp3)4.
These four sp3 orbitals of course repel each other equally, and hence end up pointing towards the vertices of an imaginary tetrahedron. Hence the regular structure of CH4 and analogous compounds.
Returning to the question, the sigma bonds formed by carbon are neither from overlapping s or overlapping p orbitals, but from overlapping sp3 orbitals.
The thing that makes a bond a sigma bond is not the type of subshell from which the electrons “originated”. It is the fact that the orbital overlap directly, with the two atomic nuclei and the main concentration of the shared electron cloud in a straight line. Pi bonds, on the other hand, overlap “sideways”, with the main concentrations of the bonding electron cloud off to the sides of the bonded atoms.
Please chime in, people, if I have omitted or misunderstood anything.


Calculating Raman Intensities from Static Raman Activities Whenever I want to calculate a Raman spectrum, my quantum chemical programs give me Raman scattering activities. As the common Raman spectrum is dependent on the incident wavelength of the used laser this seems quite logic, because the program cannot guess which laser the experimentalists could have used.
At least in Gaussian 09, for some methods, there is the possibility to calculate the Raman scattering activities for a certain incident light frequency through cphf=rdfreq.

As far as I know, the latter would give me the dynamic Raman activities and whenever I don't enter the incident light frequency I will get static Raman activities.

To get the intensities out of the activities, I was given the following equation, which I was told only holds for the dynamic activies:
$$I_k \varpropto \frac{S_k}{\tilde{\nu}_k} \frac{\left(\tilde{\nu}_\text{in} - \tilde{\nu}_k\right)^4}{1-\exp\left(-\frac{h c \tilde{\nu}_k}{k_B T}\right)} $$

But what if the calculation method that I use is not able to produce dynamic Raman activities (i.e. because it's not implemented) and I am only able to get those static Raman activities? How do I get Raman intensities from them?

You have two options in that case.

You can try to understand the frequency dependent (or dynamic)  property in the formulation for Raman activity. This property is a bit complex and is known as polarizability. For Raman spectra,  certain invariants of this tensor are involved in the intensity formulation. Be careful with this.

You can then find the formulae for activity first (by refering to the Gaussian manual and cheking referred literature there). After having a formula, then calculate the components required (including polarizability) individually, and finally obtain the dynamic activity.

You can calculate static activity very accurately and scale it using dispersion relations. Dispersion relations are relations which relate a dynamic property to static ones. (I am not sure if Gaussian has dispersion relation calculation.) In some cases you can calculate dispersion coefficients also and use them for scaling. Important is that you need to verify whether static activity can be scaled in such a manner. (I know that polarizability can be scaled but, not near absorption regions.)


Why is Aluminium Chloride a Lewis Acid if it's an ionic compound? In my textbook it says in the reaction $\ce{AlCl3 + Cl- -> AlCl4-}$, Aluminium Chloride ($\ce{AlCl3}$) has six valence electrons and accepts a lone pair from $\ce{Cl-}$, but wouldn't that mean the aluminium atom covalently bonded with the three chlorine atoms to get six electrons (which is wrong as it's ionic)?

if it's an ionic compound

Actually, it is debatable. $\ce{Al^{3+}}$ is a very small cation and its formation is strongly unfavored. In fact, this cation, just like many others, may exist only in solvates, like $\ce{[Al(OH2)6]^{3+}}$, or unchemical environment.

In particular case of anhydrous $\ce{AlCl3}$ the molecule is a strong Lewis acid, strong enough to coordinate with lone pair of another similar molecule, so it forms either dimer $\ce{Al2Cl6}$ with two chlorine bridges, donating one pair to both Al ions, or polymer, with Al atoms exhibiting octahedral coordination. Unlike many ionic compounds, melted $\ce{AlCl3}$ has poor conductivity.

This, however, does no apply to $\ce{AlCl3.6H2O}$ which is, again, a ionic compound $\ce{[Al(OH2)6]^{3+} (Cl^{-})3}$ and does not exhibit any Lewis acidity (though it does exhibit strong Brønsted acidity, i.e. tendency for hydrolysis)

This said, however, I have to clarify: ionicity of compound does not prevent particular ion from being Lewis acid/base. In fact, cations often are excellent Lewis acids. The only thing needed for a particle to be a Lewis acid is presence of low  vacant orbital, and sometimes presence of such orbital is not derivable from octet rule. For example, $\ce{SnCl4}$ is Lewis acid, though in it $\ce{Sn}$ has full octet. The orbital in question is describable in terms of MO theory which I strongly encourage to study.

The most often considered Lewis acid is hydrogen cation. In fact, this Lewis acid is so strong, that it never occurs free, it is always bound to Lewis base (again, except unchemical environment). Another common Lewis acid is $\ce{Na^+}$. This cation is quite weak Lewis acid, but it is sterically accessible, and usually is paired with anion, which is usually a Lewis base. This said, despite mostly ionic nature of $\ce{NaCl}$, there is considerable covalent bonding in it. (of course, on the other hand there is such anion as $\ce{BPh4^-}$ which is not a lewis base of considerable strength, so its salts are almost purely ionic in nature)


How to write a reaction that you have never seen? I have been studying chemistry for a while but I feel a total noob about it, so please be clear when answering.

I was studying and I encountered this exercise:

  Write the reaction between copper and concentrated nitric acid.

Now, I know what copper is and also what nitric acid is, but how do I figure out that the reaction gives $\ce{Cu(NO3)2 + NO + H2O}$ ?

There is no general method for predicting the products of a chemical reaction, given only the starting materials and no other information.

You will learn certain trends, such as, for example, the increasing reactivity down the Group 1 metals with water, or dilute acids, but this often isn't enough to generalise to more complex situations.

You'll also learn about relative reactivity, that allows you to predict that, for example, although (because zinc is more reactive than copper) zinc reacts with copper sulphate to form zinc sulphate and copper, copper does not react with zinc sulphate in the same way.

Often, even after the products of a reaction are known, it can still take a good deal of research to establish exactly how it took place - how the starting materials interacted to form intermediates, for example, and how those intermediates interacted to form the products.  This is known as the reaction mechanism, and although many reaction mechanisms are known, plenty are not.

You should continue to study, and learn as much as you can about the principles of chemical reactivity, and you'll find that in more and more cases, there are patterns that let you predict what will happen in similar cases.  But there are often exceptions, that are only explainable once you have an in-depth understanding.

There is often no alternative to actually doing the reaction, and seeing if what is predicted actually happens.  This is why chemists do research!


Does hydroxytyrosol have the same neurotoxicity potential as dopamine? It has the catechol group, and it's the oxidation of the catechol group that makes dopamine produce reactive oxygen species (ROS).

Hydroxytyrosol isn't that different from dopamine (the distal N being replaced with an OH). I'm not sure if that really affects anything. Yet - it's often sold as an antioxidant.

Couldn't it have the same ROS potential that dopamine and other catecholamines have?

while hydroxytyrosol, ( lets call it HT because I can't keep spelling that ) is a catechol, it lacks the amine of dopamine. Dopamine gets metabolized to an aldehyde, this aldehyde can form schiff bases with amines on proteins. The catechol can also react with thiols on proteins. So in addition to dopamine's oxidative damage potential, it can crosslink proteins. HT doesn't have the amine, so it probably doesn't get metabolized in the same way.

Here's a paper with more information on dopamine metabolism and toxicity. Look up other papers by that group for more.


Competitive 'Wittig' vs 'Peterson' C=C bond formation If one considers the intermediate above, there are immediately two possibilities for subsequent reaction:

1. A Wittig-type olefination which would lead to a vinyl silane: This reaction happens to be unproductive as the vinyl silane can't do any further chemistry under the reaction conditions

2. A Peterson-type olefination: Which gives a vinyl phosphine (sp.), which may further be deprotonated to give another ylid (with concomitant formation of TMSOH) which will carry on to eventually give an allene.

My question is why is the Peterson faster than the Wittig. The immediate thing that springs to mind is the difference between acid and base catalysed Peterson reactions in which the base mediated version snaps shut before the C-C bond has a chance to rotate (i.e. kinetically very fast), but this isn't an explanation.

Potentially more promising is comparing the electronegativity/positivity of the elements, and noting that silicon is significantly more electropositive and hence a better electrophile however in the Peterson, the leaving group isn't such a driving force as the phosphine oxide from the Wittig.

Seems like the silicon atom is more oxophilic as compared to the phosphorus atom. This could in principal be explained by the higher strength of the $\ce{Si-O}$ bond, as indicated by the dissociation energies $D$ (see http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html):
$$\begin{array}{ccc}
\hline
\textbf{Bond} & \textbf{Dissociation energy / kJ mol}\mathbf{^{-1}} & \textbf{Bond length / pm} \\
\hline
\ce{Si-O} & 452 & 163 \\
\ce{P-O} & 335 & 163 \\ \hline
\end{array}$$
The $\ce{Si-O}$ bond is stronger thus given an explanation why the Peterson olefination is favoured.
It seems that the bond lengths $r_b$ are the same, though.

By the way, a nice discussion of oxophilicity was given recently in Inorg. Chem. 2016, 55 (18), 9461.


Why does an element's chemical properties rely only on its valence electrons and not on anything else? I understand that elements use their valence electrons for their reactions and whatnot, and that the whole idea of the electron-dot structure is all about valence electrons. But why can't the other electrons play a part in it?

Technically, other electrons could play a part in determining chemistry. However, there is a significant energy barrier between the p-orbitals of one period and the s-orbital of the following period.

Consider copper. The characteristic deep blue colour of $\ce{[Cu(NH3)4(H2O)2]^2+}$ derives from the absorption of photons whose energy corresponds to the difference of the $\mathrm{3d}_{z^2}$ and $\mathrm{3d}_{x^2 - y^2}$ orbitals. This is spin-allowed because copper(II) is a d⁹ system. The wavelength corresponds to approximately $610\:\mathrm{nm}$.

On the other hand, the energy between a $\ce{2p}$- and a $\ce{3s}$-orbital can be measured using x-ray transition — it corresponds to a photon of a wavelength of approximately $1\:\mathrm{nm}$. This is about six-hundred times the energy difference compared to inside the d-orbitals.

Of course, this is a slightly unfair comparison. The energies between $\ce{3d}$ and $\ce{4s}$ are noticeably larger than inside of the $\ce{3d}$ orbitals, and the energy difference between $\ce{4s}$ and $\ce{4p}$ should again be well larger — but they will remain significantly smaller than energy differences across shells. The former are typically visible or UV, the latter are typically x-ray.


When do reactions reach steady state? I was doing a multiple choice problem that asked about when the following nuclear decay sequence reaches steady state (When does the activity of $\ce{^99Mo}$ roughly equal the activity of $\ce{^{99m}Tc}$ ?)

The correct answer choice said after approximately 20 hours.

Let's denote $\ce{^99Mo}$ by $\ce{R}$ and $\ce{^{99m}Tc}$ by $\ce{I}$.

By definition, the time we're looking for is the time when $\frac{\ln 2}{65}N_R=\frac{\ln 2}{6}N_I$, which is when $\frac{N_R}{N_I}=10.83$
I don't have the formula for the exact amount of the intermediate vs. time memorized, but I don't think the problem required us to know the formula. I tried estimating the amounts of the various species present after 20 hours to see if steady state is actually reached then:

Let the initial ${N_R}=1$. I know that at $t=20$, $N_R=\frac{1}{2}^{20/65}=0.8079$. To estimate $N_I$, I did $N_I\approx(1-0.8079)*\frac{1}{2}^{20/6}=0.0191$. But using this approximation, it seems like the steady state condition is not met. Thus, I was wondering, what is a better way to estimate $N_I$ to get that the decay sequence is in steady state after 20 hours?

This type of consecutive reaction does not really come to steady state for very long, as seen in the figure, i.e there is only a small time over which $d[mTC]/dt \approx 0$.  The decay rate of mTC is faster than that of Mo and so mTc appears quickly and decays slowly. (The amount of mTC is multiplied by 10 in the figure is the amount formed is small].

The figure is drawn assuming  $k_1 = \ln(2)/65, \, k_2 =\ln(2)/6 $ and $\ce{[Mo]_0} = 1$ and that the other species are zero at $t=0$ The equations are
$\displaystyle \ce{[Mo]=[Mo]}_0e^{-k_1t}$ and
$$\displaystyle \ce{[mTc] = [Mo]}_0\frac{k_1}{k_1-k_2}\left( e^{-k_1t}-e^{-k_2t}  \right)$$ The amount of Tc is found as $\ce {[Tc] = [Mo]_0-[Mo]-[mTc]}$

The maximum amount of mTC, which is assumed to be 'steady state'  is found by differentiating.

The time at maximum is $\displaystyle t_m= \ln(k_2/k_1)/(k_2-k_1)$ which is 22.7 hours so the 'steady state' is established just before this and lasts a little past this time. How much in either way is somewhat subjective unless you use a definition of say $\pm 5$ % of the maximum etc.


How does epoxy stick to stuff? According to Wikipedia what gives epoxy its holding power is apparently "ionic" interactions between the epoxy and the surface.

But I thought that epoxy was overall neutral when cured. How can it make ionic bonds with the surface it is bonded to? Or are there unprotonated alkoxides left in the epoxy which then make the bonds with the surface?

To sum it up, the adhesive property of an epoxy is the result of inter-molecular forces and the formation of mechanical bonding.

Adhesives cure when the small resin molecules (mers) join together to form extremely large molecules known as polymers.
Some adhesives are used only with specific substrates because of the cure time needed to allow for good bonding (both with the substrate and for polymer formation.)
I had trouble finding information saying that all epoxies use ionic bonding, and it  seems to me that most epoxies will actually just utilize intermolecular forces such as

Ion-induced dipole forces
Ion-dipole forces
Van der Waals forces
Hydrogen bonding

An important first step in applying an epoxy is the wetting of the surface and then the polymerization of the epoxy. The ability to wet is related to viscosity and the inter-molecular forces. If there is a low viscosity and strong attraction between the substrate and the adhesive then we can witness the formation of mechanical bonding. Below is an explanation:

  In order for an adhesive to bond  two surfaces, there must be several types of interaction between the adhesive and both substrates. The first type of interaction is that the adhesive must wet the substrate, meaning that the adhesive must spread itself out into a film that covers the substrate surface. In order for this to happen, the adhesive must have a low enough viscosity so that it will flow. Because viscosity is temperature dependent, the application of a cold adhesive to a substrate, or the application of an adhesive to a cold substrate, may result in poor wetting. Another factor that affects wetting is the relative strengths of cohesive forces (between like molecules, such as two adhesive molecules) and those of adhesive forces (between unlike molecules, such as an adhesive molecule and a substrate molecule). If the cohesive forces among adhesive molecules are weaker than the adhesive forces between the adhesive molecules and the substrate surface, then the adhesive molecules will spread out over the substrate and wet its surface. An adhesive that has a relatively low viscosity and is able to wet the substrate surface will flow into any tiny cracks or pores on the substrate surface, thus promoting what is known as mechanical bonding. Mechanical bonding increases the strength of an adhesive bond and, as a result, a forced separation of the two substrate surfaces is more apt to tear the substrate surfaces.

  Mechanical bonding is one of several ways that an adhesive bonds substrates. All surfaces, except those that are highly polished, have pores. If the adhesive flows into these pores and then polymerizes, a mechanical bond is formed. It is similar to placing a wick into liquid candle wax. Once the wax solidifies the wick can not be easily removed. A mechanical bond has formed.

Quotation and information from this answer were sourced here.


What is the reaction mechanism between boric acid and 2-ethyl-1-hexanol Here is a photo from an article, Hydrometallurgy 2016, 160, 129–136

This indicates the reaction mechanism as esterification. However, the written reaction equation ($\ce{H3BO3 + nROH -> H3BO3.nROH}$) suggests the mechanism as solvation.
Shouldn't the reaction be $\ce{H3BO3 + ROH -> H2RBO3 + H2O}$ or something like that based on the scheme?

The research paper sighted in the question (Ref.1) is about the extraction of boron (as boric acid: $\ce{H3BO3}$) from salt lake brine using 2-ethylhexanol/kerosene as organic phase. It is true that the given diagram in this paper (as shown in the question) is showing an esterification, but the authors refer it to two references, one of them is Ref.2. However, the association used to calculate the distribution of boric acid between aqueous and organic phases in Ref.1 is not the esterification. Accordingly to the paper, the association reaction is as follows:
$$\ce{H3BO3 + nROH <=> H3BO3.nROH}$$
Since there are no water molecule elimination during the reaction, this is not an esterification. It is a solvolysis reaction as OP correctly recognized.
Extraction of boron from existing sources is a challenging problem (Ref.3). Boron can be extracted as neutral ester $(\ce{B(OR)3})$, as boric acid complexes $(\ce{(H3BO3)_x.(ROH)_y})$ or as tetrahydroxy borate $(\ce{R+B(OH)4-})$. However, the extraction of boric acid by monohydric alcohols is caused by purely physical actions and the amount of extracted boric acid is roughly proportional to the concentration of an extractant (Ref.3). Accordingly, it is safe to say the extraction process is based on complexation of boric acid with intended alcohol.

References:

Ran Zhang, Yingming Xie, Jianfeng Song, Lixin Xing, Dingfeng Kong, Xue-Mei Li, Tao He, "Extraction of boron from salt lake brine using 2-ethylhexanol," Hydrometallurgy 2016, 160,  129–136 (DOI: https://doi.org/10.1016/j.hydromet.2016.01.001)(PDF).
Jianhua LÜ, Jidong LIU, Yujie SUN, Chunli LI, "Kinetics of Forward Extraction of Boric Acid from Salt Lake Brine by 2-Ethyl-1,3-hexanediol in Toluene Using Single Drop Technique," Chinese Journal of Chemical Engineering 2014, 22(5), 496-502 (DOI: https://doi.org/10.1016/S1004-9541(14)60059-8).
Adam Balinski, Volker Recksiek, Norman Kelly, "Solvent Extraction of Boric Acid: Comparison of Five Different Monohydric Alcohols and Equilibrium Modeling with Numerical Methods," Processes 2021, 9(2),  398 (15 pages) (DOI: https://doi.org/10.3390/pr9020398).


How to determine whether a given compound is paramagnetic or diamagnetic? Well i am very confused whether we can determine the magnetic property of compounds..

If it us asked for elements we can say by looking at electronic configuration...

For ex-
how is N2O diamagnetic?
How is NO paramagnetic?
How is NO2 paramagnetic?
What about N2O5

Paramagnetism is a consequence of having one or more un-paired electrons in the outer electronic configuration. e.g. O$_2$, and NO. An an odd number of electrons can clearly produce a paramagnetic ion, radical or molecule. However, molecules with an even number of electrons can also be paramagnetic, the most famous being dioxygen (O$_2$). The experimental observation is that oxygen is attracted to a magnet.(This is best observed with liquid oxygen.) Nitrogen (N$_2$) is not paramagnetic but diamagnetic.  Whether a molecule is paramagnetic or not is only explained using molecular orbital theory. In the case of O$_2$ it is found that the lowest energy state is one that has one electron in each of two orbitals. An alternative would be to expend energy to pair up these two electrons into one orbital. As this requires energy it does not occur. High and low spin transition metal complexes show the same effect of pairing vs. placing electrons into different orbitals.


Understanding molecule infrared absorption My chemistry course requires us to know which molecules absorb infrared, and under which vibrational modes that they do. My professor provides us the following rule, which is very confusing for me and I still don't know how to start.

  Rules for IR absorption: (1) The frequency of the IR radiation
  matches the frequency of the vibrational mode. AND (2) There is a net
  change in the symmetry of electrical charges during the vibrational
  mode.

So, say we have a molecule such as CF2Cl2. This absorbs infrared under all vibrational modes: symmetric stretching, asymmetric stretching, and bending. How do you get to that answer?

Thank you very much.

All forms of  radiation match the difference in two energy levels, electronic, vibrational, rotational & nmr etc. For  IR transitions we refer to molecular vibrations. The molecule must have a permanent dipole or produce a transient dipole as a result of vibrating in one of its normal modes.

So homonuclear diatomics have no IR spectrum, they of course have energy levels, but the radiation cannot 'couple' to these and so be absorbed. Heteronuclear molecule have a dipole and so also have IR spectra. The classical idea is that the electrons oscillate as the nuclei move in a vibration, this oscillating charge can interact with electromagnetic radiation of the correct energy and cause a transition from a lower energy level to a higher one.

Molecules such as benzene vibrate in many ($3N-6$) normal modes, some such as asymmetric stretches produce transient dipoles and have IR transitions. Others do not and so have no transitions. What determines the are called selection rules.

The only secure way is to look at the symmetry of the molecules using a point group table, its very easy when you know how, but a bit daunting at first sight. Have a look at this answer Understanding group theory easily and quickly for more details.

(The molecule you mention belongs to the $\ce{C_{2v}}$ point group, see molecule-viewer.com for a 3D model with symmetry elements.)


How do I make a dysprosium chloride solution from dysprosium oxide? I know that dysprosium oxide is insoluble in water.

I was told that I need to add concentrated $\ce{HCl}$ to the compound and then evaporate the $\ce{HCl}$ off. From looking at the reaction equation, I can see that I will get dysprosium chloride and water as products.

Once I'm able to dissolve the $\ce{Dy2O3}$, how is it that I'm supposed to evaporate off the excess $\ce{HCl}$? Do I just let it sit, or do I need to gently heat the solution? How can I tell when all the excess $\ce{HCl}$ is gone and I can start adding water to obtain the concentration of dysprosium chloride that I need?

DyCl3 is often prepared by the "ammonium chloride route," starting from either Dy2O3 or hydrated chloride or oxychloride.[2][3] or DyCl3·6H2O.[4] These methods produce (NH4)2[DyCl5]:

    10 NH4Cl + Dy2O3 → 2 (NH4)2[DyCl5] + 6 NH3 + 3 H2O

    DyCl3·6H2O + 2 NH4Cl → (NH4)2[DyCl5] + 6 H2O

The pentachloride decomposes thermally according to the following equation:

    (NH4)2[DyCl5] → 2 NH4Cl + DyCl3

The thermolysis reaction proceeds via the intermediacy of (NH4)[Dy2Cl7].

Treating Dy2O3 with aqueous HCl produces hydrated chloride (DyCl3·6H2O). This salt cannot be rendered anhydrous by heating. Instead one obtains an oxychloride.

source: wikipedia


Malic acid 3-step synthesis I was quite puzzled by the answer for this question in my Carboxylic Acids and Derivatives tutorial question set. It is question 6 (a), on the synthesis of malic acid, as shown in the image below.

The idea for the synthesis is to convert the $\ce {Br}$ to $\ce {CN}$ and the carbonyl group to a cyanohydrin, then convert the $\ce {CN}$ to $\ce {COOH}$. I have no disputes about the last step, which is to heat with aqueous hydrochloric acid, or some other mineral acid. However, what I proposed was to perform the two initial conversions in one step, which is to heat the compound with $\ce {KCN (aq)}$ and ethanol. Would that not allow for both the nucleophilic addition, as well as the substitution, to take place, all in one step?

The proposed answer for steps 1 and 2 was to first heat with reflux, with ethanolic $\ce {KCN}$, then react with $\ce {HCN (aq)}$, with trace $\ce {KCN}$, or $\ce {NaOH}$ as catalyst, under temperature of 10-20 degrees C. Is it really necessary to perform the initial two conversions in two separate steps?

I have clarified with my chemistry tutor and it appears that performing the substitution and the addition as two separate steps with two separate sets of conditions is absolutely necessary. The most important reason would be the temperature, which is also the most stark difference between the two different set of conditions.

Due to entropy considerations, addition should ideally be carried out at low temperatures. Thus, if we would like to have the addition of $\ce {HCN}$ to the substrate, it should be done at much lower temperatures as the reaction has a negative entropy change. However, the substition of $\ce {Br}$ by $\ce {CN}$ does not need such low temperlatures and should be done at higher temperatures to provide enough energy of activitation to break the carbon-bromide bond.


Stereochemistry of cyclic hemiketal formation $\hspace{5cm}$

I know that the acid protonates the carbonyl oxygen and this opens a pathway to an intramolecular cyclization reaction because the $\ce{-OH}$ group will attack the carbonyl carbon and a six-membered ring is formed.

I know that the carbonyl part is flat, and that the $\ce{-OH}$ can either attack from either the re or the si face. I'm not concerned about that. What happens to the methyl group on the wedge? Does that change depending on which side the $\ce{-OH}$ attacks the carbonyl? I am having a hard time visualizing what might happen to it.

No, the stereochemistry of the methyl group is not dependent on which face the $\ce{-OH}$ attacks from. This is apparent in the products of the cyclization of aldopyranoses; only two isomers (called anomers) are formed, which are classified as either $\alpha$ or $\beta$:

Depending on which position we are considering the stereochemistry of, a wedge in the straight chain could appear as either a wedge or a dash in the cyclic form. For the molecule you are considering:

$\hspace{.7cm}$

To visualize this, convert the skeletal structure into its Fischer projection:

$\hspace{4.5cm}$

$\hspace{2cm}$


Question about the setting of DFT (Density Function Theory)-NEB (Nudged Elastic Band) calculation I'm attempting to calculate Mg ion vacancy migration inside Pyrope unit cell. System has total 159 atoms, beginning and end geometry is optimized. For intermediate image, I'm using one coordinate that I acquired from crashed NEB case. This is not perfect intermediate geometry, but I thought this should be enough one to guide NEB.

1) Is there any rule of thumb to decide "num_of_images" based on moving distance of target atom?
In my script, I wish to move one Mg atom from one position to next vacancy which is about 3.5 angstrom away. In this case, how many images for NEB would proper? For example, if I choose 1 image per 0.5 angstrom, then 8 images should be chosen including the beginning and the end. But I'm not sure if 1 image per 0.5 angstrom would be the proper value. How do you decide "number of images" per given distance?

2) How many core number should be chosen for NEB?
I have some sense of core number decision based on my experience, but this is for relaxation. For NEB, I'm not sure how much core number is proper for given atom number or given number of images setting. I wish to learn if there are rule of thumb to decide core number for given NEB conditions.

3) How to decide intermediate image?
First, I tried my best guess of Mg ion location between two vacancies as an intermediate image. Then, I'm using one from one of the crashed NEB simulation result. Is this good approach?

4) Can freezing atom helps convergence? (freezing = fixing x, y, z coordinate of atom)
Is this better to freeze (put 1 1 1 beside the atom coordinate) the target atom for the beginning, intermediate, and end geometry? Or surrounding atoms around target atom? For now I'm not inducing any freeze to any atom in my NEB setting, but I'm curious if freezing helps convergence of NEB calculation.

Thank you

There is not a general rule for choosing the right displacement between images. For very big systems, I sometimes launch two NEB calculations: one with shallow parameters (eg: steps of 0.5 angstrom on the reaction coordinate), and later I use the almost converged geometries near the TS for another NEB calculation, with finer (less than 0.1 angstrom) steps. Of course the nature of the system might help you choose: sometimes things behave funny, and you have to choose a strategy for each system. Anticipating your last question, freezing or constraining sometime helps: while it doesn't improve convergence, it can speed up the calculation (by telling the program that some movements are forbidden, so that the NEB algorithm doesn't need to try them before discarding them). The intermediate image should not be mandatory for a NEB calculation (if not given, it is generated by interpolation and optimized at every cycle), but it can speed up a calculation, if provided. It can be taken, for instance, from a geometrical scan, or from an approximate TS geometry.
As for the core number, if you are talking about processors, then generally speaking as much as possibile is a good choice (edit: provided that the software you are using is capable of parallelizing efficiently the slowest step of the calculation: your choice could be the same you use for geometrical optimizations at the same level of theory)


How can Iodine bond with 5 Fluorines in Iodine Pentafluoride? According to this lewis diagram, Iodine has 12 valence electrons? How is the possible? Only thing I can think of is that I'm mistaken in counting the covalent bonds as two valence electrons each. Do they only count as one in this scenario?

Explanations of bonding and hybridization in non-metals that involve d-orbitals have generally fallen out of favor for two reasons:

Usually d-orbitals are considerably higher in energy than the s- and p-orbitals they would interact with leading to poor mixing, and
d-orbitals are very diffuse leading to poor overlap with other orbitals and any resulting bonds would be very weak

A reasonable hybridization alternative involves what is termed hypercoordinated bonding; where 3 center, 4 electron bonds are involved.  Such bonding involves p-orbitals with overlap geometry and molecular orbitals as pictured in the following diagram.

In the case of $\ce{IF5}$, the molecule has the following geometry.

Using the hypercoordinated bonding approach, one would say that the 4 identical "equatorial" fluorines (they're not quite equatorial, we'll return to that later) and 2 p-orbitals on iodine would participate in this type of bonding.  Each of the two iodine p-orbitals contains two electrons from iodine and bonds to two fluorines which each contribute one electron.  We wind up with two linear hypercoordinate bonds at right angles to each other that lie in a plane.  Each of these bonds is comprised of 3 atoms ($\ce{F-I-F}$) and contains 4 electrons as described in the MO diagram above.  The only orbitals remaining on iodine are an s- and a p-orbital.  These two orbitals mix in traditional fashion (like the carbon in acetylene) to produce two $\ce{sp}$ orbitals.  these two orbitals are arranged in a linear fashion (again, like acetylene) and are perpendicular to the plane containing the two 3 atom - 4 electron hypercoordinate bonds.  One of these $\ce{sp}$ orbitals contains an iodine lone pair of electrons and the other forms a bond to fluorine using one iodine electron and one fluorine electron.

Because of the asymmetry introduced by having a lone pair on one side of the molecule and a fluorine on the other side, our 4 "equatorial" $\ce{I-F}$ bonds distort from planarity tilting away from the electrostatically repulsive lone pair as illustrated in our figure (above) showing the geometry of $\ce{IF5}$.

In this bonding scenario the central iodine is sharing 4 electrons in each of the two hypercoordiante bonds (total=8) and two electrons in the apical $\ce{I-F}$ bond; plus there is a lone pair on the iodine.  Therefor the iodine is neutral (7-[(8+2)/2]-2 = 0), but is it surrounded by an octet of electrons?  The answer is, yes.  Look again at the MO diagram for the hypercoordinate 3 center - 4 electron bond.  Notice that the nonbonding orbital ($\Psi_2$) contains 2 electrons, but these electrons are centered entirely on the fluorine atoms, not the central iodine.  Therefore, around iodine we have 2 (lone pair) plus 2 (apical $\ce{C-F}$ $\ce{sp}$ hybridized bond) plus 2 times the two $\Psi_1$ electrons in our hypercoordinate bonds, yielding a total of 8 electrons around the iodine.


Is there a pH at which all proteins are negatively charged? How can it be determined? I know that depending on the amino-acid-composition of the peptide the distance to the isoelectric point determines the charge. Is there a universal point at which all proteins are definitely negatively charged?

That depends highly on what you mean by "all." Since acid-base chemistry is equilibrium based, there is in principle always some of the less preferred species around even though it might be quite small. At the very least, there is a nonzero probability of finding the disfavored species, in your case the neutral species.


Does the natural abundance of radioactive elements change? Given that the half-lives differ between isotopes of a particular radioactive element, wouldn't the natural abundance change (fairly rapidly)?

For example, if we isolated a bunch of thorium, wouldn't the relative amount of the various isotopes of thorium in the sample differ over time, with the more slowly decaying isotope eventually making up the largest proportion?

Yes, you are right. The abundances of (radioactive and stable) nuclides change due to radioactive decay.

Regarding this, most environmental radionuclides can be divided into four categories:

Cosmogenic radionuclides are constantly produced by cosmic irradiation of the atmosphere. Therefore, their average concentrations in the atmosphere stay roughly constant even if their half-lives are short.

Primordial radionuclides have lifetimes comparable to the age of the earth. These nuclides have been formed at the time of the formation of the earth, the solar system, or even before. Their abundance is slowly decreasing due to radioactive decay. Therefore, the radiation exposure was higher some billion years ago when life evolved on earth than today. Furthermore, due to different half-lives, the relative abundance of these nuclides (e.g. the natural $\ce{^{235}U/^{238}U}$ isotopic ratio) also changes.

Radionuclides that are part of the natural decay chains of $\ce{^{238}U}$, $\ce{^{235}U}$ and $\ce{^{232}Th}$ are constantly reproduced by the radioactive decay of their mother nuclides. Therefore, their abundance is only slowly decreasing with the long half-life of $\ce{^{238}U}$, $\ce{^{235}U}$, or $\ce{^{232}Th}$ even if their own half-lives are short (this situation is called secular equilibrium).

Anthropogenic radionuclides are produced by modern technology. Since the discovery of nuclear fission, the activity concentrations of various anthropogenic radionuclides in the environment were increasing until the limited nuclear-test-ban treaty banned atmospheric tests of nuclear weapons. After that, the activity concentrations of most anthropogenic radionuclides decreased again.

If you isolate and chemically purify a radioactive sample, you remove any other mother or daughter nuclides and thus you disturb any decay chain or radioactive equilibrium. Thereby you start a radioactive clock.

For example, a solid rock sample may contain $\ce{^{226}Ra}$ $\left(t_{1/2}=1600\ \mathrm{a}\right)$ and its daughter nuclide $\ce{^{222}Rn}$ $\left(t_{1/2}=3.83\ \mathrm{d}\right)$ in a secular equilibrium. If you dissolve the sample and release the $\ce{^{222}Rn}$ (radon is a noble gas), the released $\ce{^{222}Rn}$ decays with its half-life of 3.83 days. If you purify the $\ce{^{226}Ra}$ and put it in a closed container, initially the container does not contain any $\ce{^{222}Rn}$, but new $\ce{^{222}Rn}$ is produced by radioactive decay of $\ce{^{226}Ra}$.

A similar thing occurs when natural minerals solidify. Radionuclides become trapped and decay. As they decay, decay products accumulate in the closed mineral.

By measuring the amount of mother and daughter nuclide, it is possible to calculate how long this sample has existed. This is the basis for nuclear dating.

If you isolate a sample that contains multiple radionuclides of the same chemical element, you can observe changing isotopic ratios – as you have expected in your question.

Your example, freshly purified thorium from a natural sample, mainly contains six different thorium nuclides:

$\ce{^{234}Th}$ $\left(t_{1/2}=24.10\ \mathrm{d}\right)$
$\ce{^{232}Th}$ $\left(t_{1/2}=1.405\cdot10^{10}\ \mathrm{a}\right)$
$\ce{^{231}Th}$ $\left(t_{1/2}=25.5\ \mathrm{h}\right)$
$\ce{^{230}Th}$ $\left(t_{1/2}=7.54\cdot10^{4}\ \mathrm{a}\right)$
$\ce{^{228}Th}$ $\left(t_{1/2}=1.91\ \mathrm{a}\right)$
$\ce{^{227}Th}$ $\left(t_{1/2}=18.7\ \mathrm{d}\right)$

The individual half-lives differ significantly; thus, the different thorium nuclides decay at different rates. The isotopic ratios change accordingly.
However, the calculation is slightly complicated since $\ce{^{234}Th}$ indirectly decays to $\ce{^{230}Th}$, $\ce{^{231}Th}$ indirectly decays to $\ce{^{227}Th}$, and $\ce{^{232}Th}$ indirectly decays to $\ce{^{228}Th}$.


What concentration of small molecule binding is considered weak or strong? In the chemistry literature, I often see a stated concentration (nanomolar, micromolar, millimolar) followed by qualitative judgment on the potency of the small molecule for protein inhibition. What concentration range is considered weak, and what range might a small molecule be strong enough to be lead optimized?

The threshold for activity varies based on the target, but hit-level activity is typically 1 µM or lower. Targets are normally enzymes or receptors, so the term activity refers to an IC50 or EC50 value.

Smaller molecules have fewer potential sites for intermolecular binding than larger molecules. Therefore, small molecules rarely bind as strongly as larger compounds. For example, a hit from a typical combinatorial library may show activity (KD, IC50, Ki, EC50) at concentrations of 1 µM or lower. In contrast, a hit in a fragment library may be selected with activities of around 1 mM, which is a 1,000- fold difference in activity (a larger IC50 value implies weaker enzyme inhibitor binding).

  In 1999, Teague distinguished between lead-like and drug-like hits,
  and combinatorial library collections. 20 Lead-like hits are
  characterized as having lower molecular weights (<350), activity (>0.1
  µM), and clog P values (<3). The lower values give lead-like compounds
  room to grow into an optimized, high-affinity drug that still
  satisfies Lipinski’s rules. Drug-like hits have higher molecular
  weights (>350) and clog P values (>3) but still modest affinity (~0.1
  µM). The definition of lead-like has since been used as a preliminary
  filter for selecting more promising hits from a screen.

The cut-off for the required activity of a hit is somewhat arbitrary. The threshold may also be based on the performance of the entire library. For example, the discovery group may count all compounds that are two or three standard deviations more active than the average of the full library.

While hits are often selected at an activity level of 1 µM (KD, IC50, etc.), structural changes may provide a lead with activity at concentrations of 0.1 µM (100 nM). Eventually, potency will typically be improved down to the 1–10 nM level during the lead optimization stage.

References:

Medicinal Chemistry E. Stevens


(stereochemistry)epoxide from alkene picture above describes two ways of forming epoxides from alkene.
(a) peroxy acid
(b) Br2 and H2O / NaOH
it says R groups stay in the same position.

and I want to ask some about (b).

is it possible R groups that were on the same side before reaction occurs can be opposite side of where the double bond used to be after product is formed?
I've drawn this in the picture below.

The problem with NK Yu's diagram is that the bond rotation is unnecessary to obtain the correct conformation for epoxide formation. The epoxidation and the bromination/base give the same product and both processes are stereospecific. (Z)-Butene (1) gives the cis-oxirane 2 while (E)-butene affords the trans-oxirane (not shown). Peracid epoxidation adds oxygen to either face of the π-system of (Z)-butene to provide meso-oxirane 2. Bromination of 1 affords an intermediate meso-bromonium ion 3 by the same type of addition as was seen in the peroxidation. When hydroxide effects an SN2 (blue arrow) inversion of stereochemistry occurs (R --> S), the S,S-bromohydrin 3 is formed. With equal probability the path of the red arrow leads to the R,R-enantiomer 4. Deprotonation of the racemic bromohydrin leads to a second inversion of stereochemistry to form meso-oxirane 2, i.e., R,R --> R,S and S,S --> R,S. The commonality here is that direct, peracid epoxidation involves no inversions of stereochemistry while the bromination/base route involves two inversions of stereochemistry. The bottom line is that an even number of inversions, zero being considered even, produces the same stereochemistry.


Is there any difference between negative electron gain enthalpy and electron affinity? Electron affinity is the amount of energy "released" during the addition of an electron in the valence shell of an isolated gaseous atom. The sign convention is opposite to that of thermodynamics convention i.e. positive means release of energy.

For many non-metals, the first electron affinity is exothermic, i.e. positive while all second electron affinities are endothermic, i.e. negative.

As we already know that electron affinity is energy released (the term "affinity" implies an attraction for the electron), then why we consider endothermic cases using this term?

Well, energy is not always "released" when an electron is added to an atom in the ground state.
Noble gases for example, have a stable configuration and a high amount of energy is required to force the electron into a higher energy state and a less stable configuration, thus making the electron affinity negative.


Reduction of butadiene with Na/NH3 - stereoselectivity Why is the major product of the reduction of butadiene with sodium in liquid ammonia cis-2-butene (60:40) when it is thermodynamically less stable than trans-2-butene?

At first ... I don't like books that don't give proper solutions to their questions$^{*0}$. This doesn't help anyone ... or at least not me.

Old sources

I started my search on Reaxys and there are two old papers that I found when searching for $$\text{buta-1,3-diene} \xrightarrow{\ce{Na}} \text{but-2-ene}$$
They are:

K. Ziegler, F. Häffner, H. Grimm, Liebigs Ann. Chem. 1937, 528, 105
W. Hückel, H. Bretscheinder, Liebigs Ann. Chem. 1939, 540, 157

In both (german$^{*1}$) papers they work at very low temperatures around $\mathrm{-70\ldots-60~^\circ C}$. Both say that they get mainly the Z-isomer and discuss somehow why this happens. They think about their reaction as follows: $$\ce{dien + 2 M + 2R2NH -> dien.H2 + 2R2N.M}$$
They think that the butadiene is somehow pre-oriented towards the sodium surface from which the preference for the cis-isomer follows.

Old but not soooo old sources

Through the paper that M.A.Ramezani mentioned (W. Reeve, D. R. Kuroda, J. Org. Chem. 1980, 45 (12), 2305–2307), I found another paper (N. L. Bauld, J. Am. Chem. Soc. 1962, 84 (22), 4347–4348) that clarifies some of the mysteries. It is about the stereochemistry of metal addition to conjugated systems, esp. dienes. One of the first things that is shown is the following table.

  $$\small\begin{array}{ccc}
\hline
\text{Diene} & \text{Temp., }\mathrm{^\circ C} & \text{% cis-Olefin}\\
\hline
\text{Butadiene} & -33 & 13\\
\text{Butadiene} & -78 & 50\\
\text{1,3-Pentadiene} & -33,\,-78 & 68\\
\hline
\end{array}$$

It contains the cis-yields from v.p.c. analyses for two different temperatures for the reduction of butadiene with sodium in liquid ammonia. As one can see, there is less Z-but-2-ene produced when the temperature is higher$^{*2}$ (13 %) and more when the temperature is very low (50 %).

What the author also mentions is, that there is in every case more Z-product produced than one would expect, because there exists only about 3-7 % of Z-butadiene at room temperature (which must be even lower at lower temperatures). So somehow the metal is included, wich is described by Bauld with a cyclic complex of sodium and the radical anion that is supported by the experiments of Reeve and Kuroda.

As there are many possibilities how the sodium ion "binds" to E- and Z-form$^{*3}$ and there is also the possibility that there are two (or maybe none?) sodium ions involved, this is far off from the scope of this answer.

Besides that, I roughly calculated the structures from the mechanism that is schown in Scheme 8 from the other link by M.A.Ramezani --which, as I found out in this moment, is also from Prof. N. L. Bauld--. Below are the corresponding $\pi$-MO's$^{*4}$ -- you can imagine the mechanism with involved sodium ion(s) to be similar but somehow with the above cyclic intermediate.

TL,DR

The question in Flemings book lacks some informations. The reaction in liquid ammonia does not only depend on temperature (which was not stated!) but also on the rests that are applied to the diene. For Buta-1,3-diene in $\ce{Na/NH3}$, the Z-isomer is favored when cool temperatures are applied and therefore yield the kinetic product of the reaction. This follows from a cyclic radical anion intermediate with the sodium ion. But when the reaction temperature is rised, the thermodynamically preferred E-product is favored, as is shown by Bauld (see the table).

To get to know more about what happens when different methyl and ethyl groups are added, read the paper by Reeve & Kuroda.

$^{*0}$ Dare you, Szabo and Ostlund ಠ_ಠ
$^{*1}$ Old german papers are a joy to read ... usually the older the better.
$^{*2}$ $\mathrm{-33~^\circ C}$ is the boiling point of ammonia
$^{*3}$ In my case it seems to be purely vdW-interactions but it is gas-phase, this is a small basis set and even if the functional fits there was no dispersion correction.
$^{*4}$ be aware ... those are Kohn-Sham orbitals


How is it possible to dissolve zinc oxide safely for skin use, preferably without acid? Zinc oxide comes as a powder that is insoluble in water but soluble in most acids.

I once received a clear liquid containing zinc oxide from an Italian pharmacy as a deodorant.

What could they have used to make the $\ce{ZnO}$ soluble and safe for use on the skin?

As I previously stated in a comment, I think that the zinc oxide is not actually dissolved. That would just leave ions which you would hardly consider zinc oxide any more. $$\ce{ZnO + 2HCl <=> Zn^2+ (aq) + 2Cl- (aq) + H2O}$$

Instead I think that there are nanoparticles in use, that appear to be dissolved. In deodorant or other skin products it is likely to have quite a low concentration.

An admittedly short search turned up a patent by Marc Paye, in which is stated:[1]

  [...]
  (f) 0.01% to 1%, more preferably 0.03% to 0.5% of Nanoparticulates of zinc oxide which functions as an anti-irritant system for anionic surfactants,
  [...]

Or from another patent of the same company:[2]

  [...]
  (c) 0.5-10% (particularly 0.5-7.0, and more particularly 0.5-5.0%) of a small particle size zinc oxide, particularly a micronized zinc oxide or a nanoparticle size zinc oxide having a particle size in the range of 20 nanometers-200 microns;
  [...]

I also found, but have not completely read an article that confirmed my theory.[3]

  As an ingredient in dry deodorants to reduce wetness under the arm, ZnO can be used between 0.05 and 10% by weight with average particle size in the range of 0.02–200 microns. ZnO may be used to provide a pH range desirable for deodorants designed for use on sensitive skins [174].

At least in the last article you should also find some information about how to synthesise these particles. Spoiler: It's nothing you would try at home.

Paye, M.; Zinc oxide containing surfactant solution. US6774096 B1, 2004.
Hall-Puzio, P.A.; Gale, A.E.V.; Brahms, J.C.F.; Deodorant with small particle zinc oxide. US 6358499 B2, 2002.
Moezzi, A.; McDonagh, A. M.; Cortie, M.B.; Chem. Eng. J., 2012, 185–186, 1-22..


Miscibility of pairwise miscible liquids If you have a mixture of 3 (or more) liquids which are "pairwise miscible", but which don't react with each other, does that guarantee that the mixture will only have one layer?

By "pairwise miscible", I mean that for every combination of two substances in the mixture, they are miscible with each other, i.e. soluble in all proportions.

For example, ethanol/water/acetone.

So you want to know whether there is a ternary system with an "island of immiscibility". Well, it turns out that people have thought of this before; such systems have a name ("type 0 systems"), and this paper mentions water/DMSO/THF as an example.


Explanation of calculating equilibrium constant Hydrogen iodide is a colourless gas that will decompose into colourless hydrogen gas and purple iodine gas according to the following endothermic reaction.

  $$\ce{2 HI (g) <=> H2 (g) + I2 (g)}$$

  A $\pu{1.0 L}$ glass container was filled with $\pu{0.60 mol}$ of hydrogen iodide gas. When equilibrium was established, there were $\pu{0.25 mol}$ of iodine gas present in the container. Calculate the equilibrium constant for this reaction.
  $$
\begin{array}{|c|c|c|c|}\hline
&\ce{HI}&\ce{H2}&\ce{I2}\\\hline
\text{Initial}&0.60&0&0\\\hline
\text{Used/made}&0.50&0.25&0.25\\\hline
\text{Equilibrium}&0.10&0.25&0.25\\\hline
\end{array}
$$
  $$K=\frac{[\ce{H2}][\ce{I2}]}{[\ce{HI}]^2}=\frac{0.25\times0.25}{0.10^2}=6.25$$

This is what I understand:

Product always goes on the top, reactant on bottom
Balancing coefficients eg $\ce{2HI}$ become powers $\ce{-> HI^2}$
Initially, 0.6 moles of hydrogen iodide gas was placed, then at equilibrium 0.25 moles of iodine gas.
If there was 0.25 moles of iodine gas, there has to be 0.25 moles of hydrogen gas aswell thus the top part of the fraction becomes $0.25 \times 0.25$
I have no idea where the 0.1 moles come from …

I don't want to 'understand' these as much as I just want to be able to do them.

Your reasoning is correct. Initially, you saw that there was no iodine in the vessel. After reaching equilibrium, you saw 0.25 mol. Where did this come from?

Answer: Hydrogen iodide broke down to form this iodine and hydrogen. How much hydrogen iodide? Well this is a super simple moles problem.

0.25 moles of iodine was formed, and the mole ratio for $\ce{HI:I_2}$ is $2:1$. For 0.25 moles of iodine to be formed, 0.5 moles of hydrogen iodide must have been used (just look at your balanced equation).

If 0.5 moles of hydrogen iodide was used, how much hydrogen iodide is left? 0.60 moles minus 0.50 moles to leave 0.10 mol.

Then, as you did correctly, you plug it into the $K_c$ formula.

While you have stated that you just want to know how to do it (which you should be able to now), it is useful to know what's going on. Basically, the $K_c$ is a measure of where the equilibrium is. If your $K_c$ value is high, there will be a higher concentration of the products at equilibrium.


Why are actinides not commonly included in rare earth metals? According to the German Wikipedia, the rare earth metals include all elements of the third side group except actinium, and all lanthanides.

  Zu den Metallen der Seltenen Erden gehören die chemischen Elemente der 3. Nebengruppe des Periodensystems (mit Ausnahme des Actiniums) und die Lanthanoide – insgesamt also 17 Elemente.

https://de.wikipedia.org/wiki/Metalle_der_Seltenen_Erden

Since actinides (like lanthanides) should be part of the third side group, why aren't they considered part of the rare earth metals?

It comes down to the history of their discovery, natural occurrence and applications. It's a fascinating story that you can read about in here:

http://www.vanderkrogt.net/elements/rareearths.php

The rare earths (i.e. lanthanides) were discovered over a few decades in the late 1800s. They were laboriously separated from "rare earths" - the name used back then to describe the rare minerals (silicates, oxides, phosphates, etc) that included these elements. These are all naturally occurring rocks. You can literally pick up from the earth a mineral that has rare earths in it. They are also stable (apart from Pm). Therefore, they are used in industry. There are rare earth mines, rare earth magnets, rare earth lasers. You get the stuff from the earth.

Actinides, were mostly discovered in the early to mid 1900s, after rare earths were a "thing". Other than U and Th, you do not find them in nature. You do not mine them from the earth, and they do not exist in "earths". Their radioactivity precludes their use in industrial applications other than a few niche uses. Although their chemical properties are similar to the lanthanides, there are many differences. They are simply not rare earths. Almost, but not quite.

I'll also add that the definition of "rare earths" is a loose one. The broad chemical definition includes the lanthanides La–Lu, Y and Sc. As far as geologists care, Sc is not a rare earth element, and Y is only "honorary". Industry and mining usually consider the lanthanide oxides as REO (rare earth oxides) and use REY (rare earths + yttrium) when they are talking also about Y, because it's not obvious to them that it is a rare earth.


Why do we need incubation time during the glucose test by chronoamperometry? Also, is there any difference between with and without incubation time in the glucose test? For example, the voltage results acquired will be different?

Well, Although I'm not a chemical student, I think I could share my understand of the Chronoamperometry by the parameters that could affect the result in my project.

According to my test result, the incubation time will affect the magnitude of the current response at the chemical cell(also for the trigger time). This is related to the property of the glucose test strip(Truetrack in my design) and the circuit design. And I assume that this is because of the distribution of the glucose in the test strip, specially the amount of the glucose close to the surface of the electrode.

I hope if there is a chemical guy that can explain the detailed principle of Chronoamperometry, very detailed stuff. Because I have finished my project and thus the theoretical staff used for the lecture is not enough to attract my attention.
Also, if someone has interest in this, I can share my experiment result and discuss with him personally.
Here is my email: matthew97829782@gmail.com


How much water has boiled off when precipitation begins? Here's a challenge problem I created! None of my students could get it; can you?

$25$ grams of table salt (NaCl) is dissolved in $150$ grams of water. The temperature of the solution is raised to its boiling point until sodium chloride begins to precipitate.

How much water has boiled off when precipitation begins? Also, what would be the temperature of the boiling solution when precipitation begins?

One expects 25 g of $\ce{NaCl}$ to saturate 62.5 g of water at 100 °C. However, 25 g of $\ce{NaCl}$, 58.443 g/mol, is 0.4278 mol in 62.5 ml of water at 100 °C. Colligative property boiling point elevation is +0.52 (°C kg)/mol. 0.8556 mol of particles in 62.5 ml of water is 13.69 mol/kg or 7.1 °C elevated.

$\ce{NaCl}$ solubility/temp increases from about 36 g/100 g at 0 °C to 40 g/100 g at 100 °C or 0.04 g/°C. This gives first approximation 0.2847 g more solubility. "25.28" g $\ce{NaCl}$ in 62.5 g water at 107 °C is then 25 g $\ce{NaCl}$ in 61.8 g water. 88.2 g water boiled off (with some decimal trim for Zeno’s paradox).

Repeat the problem with ammonium cyanate.


Percent Dissociated from Titration Curve Question 818 references the titration curve. Answer is A because $\ce{H+}$ conc $= 10^{-4}$. This is conc of dissociated acid. The conc of the undissociated acid is the original concentration minus this: $0.1 - 0.0001$, which is about $0.1$. So then its $\frac{0.0001}{0.1}\times 100 = 0.1\%$.

Where in the world are they getting the original concentration? how did they get $0.1$ as the concentration of undissociated acid?

I'm not sure I follow your logic.
For a monobasic acid S $(\ce{HA})$ dissociation degree $α$ is

$$α = \frac{[\ce{H+}]}{c_\mathrm{a}},$$

where $c_\mathrm{a}$ is the initial concentration of the acid which you are determining via titration with the defined volume of a strong base $V_\mathrm{b}$:

$$c_\mathrm{a}V_\mathrm{a} = c_\mathrm{b}V_\mathrm{b} \implies c_\mathrm{a} = \frac{c_\mathrm{b}V_\mathrm{b}}{V_\mathrm{a}}$$

Finally, taking $V_\mathrm{a} = V_\mathrm{b} = \pu{50 mL}$ (from the figure's caption and equilibrium point) and, as you already assumed from pH, $[\ce{H+}] \approx \pu{1e-4 mol L-1}$:

$$α = \frac{[\ce{H+}]V_\mathrm{a}}{c_\mathrm{b}V_\mathrm{b}} = \frac{\pu{1e-4 mol L-1}\cdot\pu{50 mL}}{\pu{0.1 mol L-1}\cdot\pu{50 mL}} = \pu{1e-3}~\text{or}~0.1\%$$


Trustworthiness of IC50 values My two questions: Is replication of recording $\text{IC}_{50}$ values prior to (primary?) publication less common than performing a reaction several times to determine a more reliable record? Provided the very small quantities used in these tests (vide infra), should a synthetic chemist rather stay skeptic if a newly isolated material is described as highly potent?

In a literature seminar, the story/history of maoecrystal V, a highly congested polycyclic natural product was presented:

(source first part of an overview)

Initially, both the molecular architecture as well as the the $\text{IC}_{50}$ value ($\pu{0.02 ug/mL}$ at a molar mass of $\pu{330 g/mol}$, $\pu{60 nmol}$) in comparison to cis-platin ($\pu{0.99 ug/mL}$ at a similar molar mass of $\pu{300.01 g/mol}$) against cervical HeLa cancer cells reported by Li et al. in  Org. Lett. (doi 10.1021/ol0481535) in 2004 triggered the interest of several chemists to synthesize this material.

About a year ago, the Baran group published again efforts towards this compound (doi 10.1021/jacs.6b06623, ACS AuthorChoice / open access), displaying again tremendous efforts in the lab.  With larger quantities of the material in hand, however, the initially claimed activity against cancer cells (not only against HeLa) could not be confirmed.

The supporting information of the last paper sheds some light on the frugal scale the $\text{IC}_{50}$ values were recorded:

  Synthetic (–)-maoecrystal V (1) was dissolved in 100% DMSO at a     concentration of 1 mM. [...]  A day after cell seeding, 10 μL of prediluted compound was added to the cells (1:16 dilution).

Which would refer to low umol quantities of the material deployed per test.  Still, the question arises why the initially reported activity differs so greatly from the one eventually determined.  (Speculation: perhaps fewer groups would have joined the challenge to prepare the material if interested merely in its architecture.)

Addition

For the interaction of Maoecrystal V and other, already known, agents for example facing HeLa cells, page S52 of the SI of the 2016 publication shows this plot:

Note that Actinomycin D is considerably heavier ($\pu{1255 g/mol}$) than Maoecyrstal V ($\pu{330 g/mol}$).

I have worked in natural product isolation a little. Although I have never witnessed tests with HeLa cells, I can provide some insight into the general screening procedure.

In the lab I worked in which dealt with isolation of myxobacterial natural products, a novel strain was first tested as a crude extract in a preliminary screening via serial dilution assay. Tests were performed against a set of standard screening organisms including S. aureus, Candida albicans, E. coli and a few others I forgot.

If an isolated showed enough inhibitory activity at this stage, a semipreparative HPLC was performed to locate the peak(s) which caused said activity. The HPLC fractions were collected in a 96-well plate, the solvent removed by nitrogen flow and then the well plate was infected with the organism of interest. Hopefully, one or two wells would remain with out bacterial or fungal growth and hopefully these would not be in the fatty acid realm at the end of the 96 wells.

After that, a slightly larger separation was performed with the goal of gaining a few milligrams of the active compound (after confirming via HR-MS and dereplication that it was not already known) to perform another serial dilution assay to identify the minimal inhibitory concentration (MIC) — the isolationists’ equivalent to an $\mathrm{LD_{50}}$. This was typically performed around three times.

After that, the paper was drafted and submitted to peer review.

(Okay, there are a few steps I left out, but they contain other, unrelated stuff like making sure the myxobacterium was properly classified, elucidating the structure, hoping for crystals and maybe even performing some low-scale fermentation of the relevant myxobacteria to gain enough sample for proper structure elucidation. Ths last step could also preceed the MIC value determination.)


Carbonyl oxygen as nucleophile Aldehydes and ketones have electrophilic centres at their carbonyl C, since the carbonyl O polarises the bond and attracts electron density towards itself on account of its high electronegativity. But despite this electron density, the oxygen atom hardly ever acts as a nucleophilic centre. If at all, it is the alpha C, which in enols and enolates acts as a nucleophile. I only know of the Wittig reaction in which the O also reacts with the reagent (the P atom in this case). Is there any reason for the low reactivity of the oxygen? It is joined to C with a pi bond, which in any case is easy to break on attack by a reagent.

The carbonyl oxygen can act as a nucleophile but it is strongly dependent on the conditions of the reaction.

Enolates can react as a nucleophile through either the $\alpha$-carbon or the oxygen.

(source)

The ratio of carbon to oxygen substituted products depends on the balance of orbital interactions and electrostatic interactions between the reactants. As you can see from the diagram above, the HOMO of the enolate has the greatest contribution from the $\alpha$-carbon and so reactions which are controlled by orbital energy factors take place through there. Conversely, the greatest negative charge is on the oxygen due to its greater electronegativity and so if the reaction is controlled by electrostatic factors then it will take place at the oxygen.

Changing the electrophile:
Alkylation is a common reaction of enolates and provides a good insight into the factors affecting the selectivity. If methyl iodide is used as the alkylating agent in a polar aprotic solvent then the reaction produces mostly the carbon alkylated product. Switching to methyl bromide or methyl chloride increases the amount of oxygen alkylated product and if methyl tosylate is used then the reaction produces mainly the oxygen alkylated product. This is because the increasing electronegativity of the substituent increases the positive charge on the methyl carbon, meaning that electrostatic interactions are more important. Additionally, the $\ce{C-X}~\sigma^*$ orbital (the LUMO of the electrophile) will be lower in energy in methyl iodide than in methyl tosylate due to the poorer energy match between the $\ce{C}$ and $\ce{I}$ atomic orbitals compared to $\ce{C}$ and $\ce{O}$. This brings the electrophile LUMO closer in orbital to the enolate HOMO, making the orbital interactions more important.

This thinking can be extended to other electrophiles. Many hard Lewis acids such as $\ce{R2BCl}$ and $\ce{(CH3)3SiCl}$ (credit to @Jan for pointing this out) will give almost exclusively oxygen substituted products. This is because the central atoms are small and highly charged, and so have strong electrostatic interactions with the oxygen end of the enolate. This can be viewed as an application of 'hard soft acid base theory', where the silicon and boron based electrophiles are considered hard acids and the carbonyl oxygen is a hard base. By comparison, neutral carbon electrophiles are soft acids and so exhibit less affinity for the oxygen.

Changing the solvent:
Changing the solvent is one of the easiest ways to influence the outcome of the reaction, particularly in the case of alkylation. In polar aprotic solvents, the metal cation is strongly solvated but the enolate is weakly solvated due to the lack of acidic hydrogens on the solvent to form hydrogen bonds. In polar protic solvents, the oxygen end of the enolate is strongly solvated by hydrogen bonding which hinders reaction at the oxygen, strongly favouring C-alkylation, even with an alkyl tosylate.

Some more reading can be found here, here, here and here.


Which one is the best notation for sodium acetate I have seen two different notations for sodium acetate. The first one is:

$$\text{NaCH}_3\text{COO}$$

The second one is:

$$\text{CH}_3\text{COONa}$$

Now I'm confused, which one is the best to use?

There is nothing wrong with either formula. And you can use even more:

$\ce{NaC2H3O2}$
$\ce{C2H3NaO2}$

It really depends on which point you want to bring across.

$$\ce{NaCH3COO}$$

This formula, being analogous to formulae like $\ce{NaCl}$ stresses the inorganic salt view more. It shows that there is a cation ($\ce{Na+}$) and an anion ($\ce{CH3COO-}$) which form a salt crystal together much like the anion $\ce{Cl-}$ would do with the same cation. Inorganic nomenclature prefers cations to be written first.

$$\ce{NaC2H3O2}$$

Is basically the same except for saying ‘I don’t care how those seven atoms combine to form the anion, all I care for is what’s in there.’

$$\ce{CH3COONa}$$

This one stresses a more organic-chemical point of view where it’s relevant where the cation is actually bound to within the molecule. Oftentimes, organic chemists would even write structural formulae with a bond between oxygen and sodium as if it were hydrogen. The reasoning behind this is ‘I don’t care if the structure ends up being a salt, all I need to know is that the sodium somehow connects to the carboxyl group.’

Note that you can combine the first’s and the third’s idea to give $\ce{NaOOCCH3}$ which works, but is a lot less used than either of them.

$$\ce{C2H3NaO2}$$

This is the Beilstein-type lookup formula. You just know the elemental composition of your compound and now want to look it up — you can’t know whether it’s sodium acetate or sodium hydroxyethanalate (if that name is even correct). Or maybe you just don’t care. The reasoning behind this is $\ce{C}$ first, $\ce{H}$ second, the remaining elements following in alphabetic order.

Finally, there are also shortened formulae. The most common (the one I actually use most) would be $\ce{NaOAc}$. Here, $\ce{Ac}$ is an abbreviation for $\ce{H3C-C=O}$ where whatever follows or precedes is bound to the carbonyl carbon. Note that this formula is not considered standard, and would need to be included in a list of abbreviations or defined otherwise if you decide to use it.


Residual catalyst leaching from PETE I am trying to look for healthy options for 5 gallon water jugs. The problem is all the plastic 5 gallon water jugs on the market are either made of #1 PETE (BPA free) or #7 (which contains BPA). It seems #1 PETE is healthier option but it turns out #1 PETE is known to leach a bit of antimony oxide ($\ce{Sb2O3}$ catalyst, used in its synthesis) into water.

So my question is if I bathe the jug in water, so that the antimony oxide can leach into the water and wash the jug one more time, can all the antimony be removed that way?

Let's ignore the safety aspect here, and focus on the 'if i bathe the jug in water so that all the antimony can leach into the water and wash the jug one more time, can all the antimony be removed that way'. In short, no.

There have been a number of studies on antimony in PET bottles. They look at various factors that influence the rate of leaching of Sb from PET bottles. These factors include temperature, UV, colour of bottle, pH, bottle volume and time. There is evidence that leaching is still occurring well beyond 6 months. One study suggests that leaching levels out after about 800 days. Another study calculates that the total levels of Sb in the plastic are very unlikely to be completely leached out.

Factors that have significant influence of amount of Sb leached include:

time stored (major influence)
temperature (higher temps have major influence)
volume of bottle (smaller volume bottles have higher SA/vol ratio and increase leaching)
type/origin of PET (significant differences between the colour of bottles used in one study)

I'm not aware of a study that attempts to find the conditions whereby Sb no longer leaches into the water, but effectively you would need to bathe the the bottle at high temperature for several days at a time, repeatedly, over months. That's just not going to happen.

Some useful references for you include:

doi.org/10.1016/j.watres.2007.07.048
doi.org/10.1016/j.scitotenv.2009.04.025
doi.org/10.1021/es061511+


Are the following compounds aromatic? I know Huckel's rule states that an aromatic species must have $4n+2$ π-electrons. Is the last molecule also aromatic?

Molecule 1, the cyclopentadienyl cation, is not aromatic. It only has 4 π electrons and does not fit the $4n+2$ rule.

Molecule 2, the cycloheptatrienyl cation (or tropylium cation), is aromatic; it has 6 π electrons and fits the $4n+2$ rule with $n=1$.

Molecule 3, azulene, is generally considered to be aromatic. It fits the $4n+2$ rule with $n=2$, and is often referred to as a "nonbenzenoid" aromatic compound. It can also be viewed as a fusion of two other aromatic units, the tropylium cation and the cyclopentadienyl anion.


Configuration change at chiral carbon in SN2 reaction Consider $\ce{A}=\ce{CH(CH3)(C2H5)(C3H7)}$ in R configuration. Suppose somehow in a $\mathrm{S_N2}$ reaction, $\ce{Br-}$ attacks and replaces $\ce{-H}$ (this compound we call $\ce{C}$). Now we expect change of configuration (Walden inversion), i.e. it to become S. If the structure was inverted without replacing $\ce{-H}$, we would have got the S-stereoisomer of $\ce{A}$, call this $\ce{B}$. Replace the $\ce{-H}$ in $\ce{B}$ with $\ce{Br}$, we obviously get $\ce{C}$. Now since $\ce{H}<\ce{C}<\ce{Br}$ in the Cahn-Ingold-Prelog rules, $\ce{C}$ is R. But $\ce{C}$ should have been S.

Where is my logic wrong?

EDIT: As noted by A.K. below, there is no problem in this case, and $\ce{C}$ is in-fact S. But everything depends on the atomic number of $\ce{Br}$. Is there a general proof of correctness for the scheme?

I am going to use ‘Pseudo-Fischer’ structures in this post, i.e. anything vertical should be considered behind the PC screen plane and anything horizontal in front of it.

Your original molecule and the hypothetic $\mathrm{S_N2}$-type inversion is here:

   Pr            Pr
   |             |
 H-C--Me ---> Me-C--Br
   |             |
   Et            Et

  (A)           (C)

If you determine both compounds’ absolute configuration, naturally we will have (R)-A but look closely at C and realise that it is also (R)-C. This is because of one of two fallacies students often have with CIP priorities (I’m presenting the other one here just for completeness):

  A chemical inversion of a specific stereocentre does not have to mean an inversion of the stereodescriptor.

  Just because a stereocentre does not participate in a reaction does not mean its stereodescriptor remains identical.

In our case, we can easily arrive at that conclusion by the two-step process you outlined. First, invert the original structure A to give B, then replace (under retention) the hydrogen in B to give C.

   Pr            Pr            Pr
   |             |             |
 H-C--Me ---> Me-C--H  ---> Me-C--Br
   |             |             |
   Et            Et            Et

  (A)           (B)           (C)

We already established that A is (R)-A. Going to be, this is a strict inversion and only in this case of a strict inversion will the fallacy of bullet point one actually hold true; thus, it is (S)-B. Finally, we replace the atom of lowest priority with one that will automatically become the highest priority. To explain what happens requires further elaboration.

First, assume without loss of generalisation that whichever group of B had second priority (in our case: ethyl) is pointing upwards in a Pseudo-Fischer structure while whichever group originally had third priority in B (here: methyl) is pointing downwards. This gives us depiction 1.

   Et
   |
Pr-C--H
   |
   Me

  (1)

As we have the highest priority on the left above and the lowest priority on the right above, it is easy to see that this compound has an (S) configuration. If the hydrogen and propyl groups had been switched, we would have had (R). Now, perform the replacement in a manner such that the former fourth priority becomes first. The original third priority is now fourth, second became third and first became second. This is shown in depiction 2.

   Et
   |
Pr-C--Br
   |
   Me

  (2)

Now, since our lowest priority is pointing downwards, to assign a stereodescriptor in 2 we must look from above. The first and second priorities are above the PC screen plane and on the left and right-hand sides. The third priority is behind the plane and centred. It is rather easy to see that in the case of 2 as drawn, with the right-hand group having a higher priority than the left-hand group, the stereodescriptor is (R). If they were reversed, it would be (S). We see that replacing the fourth priority with something of highest priority will invert the stereochemistry.

What happens if you replace the lowest priority with something of intermediate priority (i.e. something that will come in second or third)? The third case is easy; you are just switching two groups. Switching two groups will always give you a formal inversion. What if our replacement group takes second priority? I refer back to depiction 1 but this time replace hydrogen with $\ce{BEt} = \ce{(pin)B-C2H4}$ giving depiction 3.

   Et
   |
Pr-C--BEt
   |
   Me

  (3)

Again, we must look from above. The highest priority is on the left-hand sinde in 3, the second is on the right-hand side, the third is in the centre and behind. Our journey is counterclockwise or (S). Had the original configuration been inverted it would be identical. Thus, replacing a fourth priority with something of second priority retains the stereochemistry.

This is a one step analysis on a per-case basis. A.K. showed the usage of the switching method (two groups switch at a time) to its fullest; thereby, these independently established rules can also be derived as was shown.