Which definition of an ore is correct? In metallurgy, an ore is defined as that mineral from which metal can be obtained with economic viability.
I have also read that ores are rocks from which minerals or metals can be extracted profitably.
 If ores are rocks they cannot be minerals as rocks are aggregates of mineral and vice versa, so which of these is actually correct?

Some of the terms are pretty vague and it isn't worth getting hung up on a precise definition.

Of the terms you use, "rock" is the vaguest. I means just about any hard thing you find in nature (and even that last qualification isn't precise as some people might call something like concrete an artificial rock). You can't infer anything much from the term "rock". Sometimes geologists might classify particular rocks into useful categories such as sandstone, granite or limestone. But the term is vague enough that rock formations might have more than one type of rock in them.

Ore is a more specific term. It does have a meaning but that meaning can change over time. Generally an ore refers to whether a useful product can be economically extracted from something. So if a rock formation contains a small amount of gold then it would count as an ore if the cost of extracting it was less than the amount the resulting gold could be sold for. But, sometimes, a geologist might refer to a particular mineral as the ore from which the desired substance is extracted. For example, she might call galena an ore from which lead is extracted. But this ignores the fact that galena is found in formations that contain many other minerals and has to be separated from them to be useful. More precise usage would be to define the ore as the rock formation containing enough galena for economic extraction of lead.

A mineral is usually a very precise term referring to a pure substance. Rocks are usually made of mixtures of minerals. Granite, for example, is made from quartz, feldspar and mica. The separate crystals are usually large enough to see with the naked eye.

An ore is simply a rock formation (which might contain more than one type of rock) where one of the minerals in it is economically valuable.


Extraction of spilled methylamine from mud I was reading this question about Breaking Bad on the Movies & TV Stack Exchange site and I'm wondering:

If you spill dirt in methylamine, can you extract the methylamine back out?

My very basic level of understanding leads me to believe that if you heat the resulting mud, you should be able to dry out the dirt and return it to its normal state, and likely create some type of vapor.  But, I'm probably completely wrong.

So, is this possible?

The wording of the question (and confirmed in a comment) suggest a major problem with the scene in question. If mixing dirt with methylamine made mud, then it was either some solution of dissolved methylamine, or liquid methylamine, but not a solid methylamine salt like methylamine hydrochloride. The first scenario wouldn't make sense for the synthesis of methamphetamine, and the second case is the real problem: methylamine boils at -6oC.

Assuming they had a solution containing methylamine that had become contaminated with dirt, the recovery would actually be pretty easy. Because methylamine has such a low boiling point, simply basifying the solution during distillation would readily separate the methylamine from anything else that could be present in the contamination.


Why aren't soldering tips coated with platinum? The tips in a soldering iron tend to oxidize quite quickly (therefore becoming unuseable) if left heated and unused for more than a few seconds.

Why aren't they coated with platinum or some other material that sticks to solder and doesn't dissolve in it? The cost of the coating should pay for itself fairly quickly.

Sigh - as pointed out by @user83678 some 3+ years after first writing this, I answered for Pd, not Pt. That will teach me to pay attention. But, the general answer remains the same. There are enough interactions of platinum with either lead, tin, or silver (a common component in lead-free solders) such that a platinum tip does not make metallurgical sense.

Lets look first at Pt-Pb, with the CALPHAD parameters from Z.H. Long et al., J. Phase Equil. Diff 30 318-322 (2009):

While Pt does not have a high solubility in Pt, the presence of multiple intermetallics indicates problems: either Pb reacts with the Pt time, or Pt dissolves into the solder melt and forms an intermetallic there. Neither is desired.

For Pt-Sn, one gets (with CALPHAD parameters from V. Grolier and R. Schmid-Fetzer, J. Alloys and Compounds 450 264-271 (2008)):

Here the Pt happily sucks up lots of Sn, and then there are all those intermetallics.

For Pt-Ag (from I. Karakaya and W.T. Thompson, Bull. Alloys Phase Diagrams 8(4) 334-340 (1987)) one gets:

The fcc Ag and Pt have large solubilities for each other (but not complete solid solubility, somewhat surprisingly). Again, you get lots of interactions between a Pt tip and a lead-free solder containing Ag.

Aside from my comment that it is not hard to keep a soldering tip in tip-top shape, I will point out several materials issues.

For this purpose, I will stick to the interactions of Pd with classic Pb-Sn solder. A peek at the Pb-Pd phase diagram (ASM International, or G. Ghosh, Met. Trans. 30A, 5-18 (1999)) one readily sees that palladium has a large solubility of lead, more than 10% at room temperature. Since the solubility rises with temperature, cooling a Pb-saturated Pd tip from soldering temperature to room temperature would result in Pb precipitates forming, likely disrupting the mechanical stability of the Pd tip.

A glance at the Pd-Sn phase diagram (same paper referenced above) shows a fairly similar situation (not so surprising since Sn and Pb are in the same column, but that isn't a given in phase diagrams). Again, even at room temperature the solubility of Sn in Pd is in excess of 10%.

For both systems there is a binary phase, Pd$_{3}$Sn or Pd$_{3}$Pb, very close to the fcc palladium phase field, as well as a range of other phases stretching to higher Pb or Sn contents. The ternary diagrams included in the Ghosh paper show many phases present. This means that reactions between molten Pb-Sn solder and a platinum tip won't stop at just saturating the fcc Pd with Pb or Sn - the reaction will continue, making compounds across the phase diagram and rapidly ruining your fancy Pd tip.

For lead-free solders, note that Sn remains a primary component in many of them, so the same things apply.

I have included a (calculated from the CALPHAD parameters in Ghosh's paper) the Pb-Pd phase diagram.


Hazards of trivalent chromium and hexavalent chromium I am examining a material safety data sheet that reports on trivalent and hexavalent chromium in chemical films. It covers each type where it is used in a product individually.

Is there a practical difference in safety between the two, such that they should be reported separately?

Hexavalent chromium is a genotoxic carcinogen. Due to the similarity of the chromate(VI) and sulfate ions, it is transported around the body via the sulfate channels. Due to this it is a major health hazard. Trivalent chromium on the other hand is not generally regarded as a health hazard due to its poor solubility in water. With this information you should report both separately.


Metallic properties of iodine Why does iodine exhibit metallic properties at room temperature although being a non-metal? Are there any similar elements like this?

Iodine is a metalloid. Metalloid elements have one or more allotropes with properties intermediate between those of a metal and a nonmetal.

Metals are lustrous, ductile, and conductive of heat and electricity.

Nonmetals are matte, brittle, and insulators.

Metalloids may be semiconductors, like silicon, germanium, arsenic, and carbon (graphite). They may be lustrous, but brittle, like iodine and tellurium.


What is the quantum mechanical explanation of X-ray diffraction? One topic in crystallography that I've found a surprising dearth of information on is what the fundamental interaction behind the interaction of the X-ray and atom. Pretty much every book just treats them as classical waves instead of explaining the quantum mechanics behind it. By this I don't mean that interference that leads to diffraction, but what actually happens when the X-ray interacts with the electron cloud.

The only description I've found so far is in Glusker, Lewis and Rossi that talks about:

  "When X rays hit an atom, the rapidly oscillating electric field of the radiation sets the electrons of the atom into oscillation about their nuclei. This oscillation has the same frequency as that of the incident radiation. The result is that the electron acts as an oscillating dipole that serves as a source of secondary radiation with the frequency of the incident beam."

and goes on to note that this is coherent scattering and there is no wavelength shift.

Which doesn't make a ton of sense, isn't oscillating about the nuclei an old pre-quantum explanation of orbitals? Is this saying that this is similar to fluorescence/phosphorescence where the photo is absorbed, then another emitted? That is what most of the other grad students I've talked to have assumed, but if that was the case I'd expect a Stokes shift, the direction of emitted radiation to be random, and I don't see why you'd get a $180^\circ$ phase shift.

The other explanation I've heard is that you are actually sloshing electrons around within their orbitals, with the same frequency as the X-ray, similar to when you are dealing with optical properties, but I'm not sure what the mechanism for the interaction would be in this case.

The description given in your question pretty much explains things. I would put it in slightly different words which may or may not help, which is that the x-ray photon induces a dipole moment in the atom (a field induced dipole) which then radiates and so the x-ray is scattered. This is pretty much what happens in scattering  by visible photons and we must assume that the photon is distant in energy from any transitions. (In Compton scattering part of the x-ray energy is imparted to the electron, so the frequency of the scattered x-ray is changed).

The field induced dipole moment is proportional to the polarisability which for a single bound electron is
$$ \alpha = \frac{e^2}{m}\left ( \frac{1}{\omega_0^2-\omega^2} \right )$$

where $\omega_0$ is the resonance frequency of the bound electron and $\omega$ the x-ray frequency and so $\omega >> \omega_0$ which in the limit gives
$$\alpha \rightarrow -\frac{e^2}{m\omega^2} $$
Thus the calculation reduces to understanding the QM calculation for the polarisability. See for example Atkins & Friedmann 'Molecular Quantum Mechanics' Ch $12$.

(  By considering that the x-ray is a plane wave the intensity of scattered radiation relative to that incident is given by
$$ \frac{I}{I_0}= \frac{e^4}{m^2c^4R^2}P(\theta)$$

where R is the distance from the scattering centre to the detector, m the electron mass and $\theta$ the scattering angle and $P(\theta)=(1+\cos^2(\theta))/2$ is called the polarisation factor.  This formula is sometimes named after Thomson. The $m^{-2}$ shows why significant scattering occurs only from electrons and not from protons. The book by Flygare 'Molecular Structure and Dynamics' section 8.5 gives a detailed description of x-ray scattering in general and details such as calculating atomic scattering factors. )


Why do aqueous solutions always have to be electrically neutral? I was reviewing some analytical chemistry and stumbled upon a section that explained the imperfection of using a salt bridge.

It said that the using dissimilar ions is a problem because in, for example, the case of KCl the K+ and Cl- have different mobility and so you get regions that are rich in K+ and others rich in Cl-. This means that some specific parts of solution (even though it's small) to be not electrically neutral i.e. electrically charged.

So I imagine in theory at least it must be possible to somehow extract these charge-rich areas and put them into a beaker. So what is all this I've been learning about solutions having to be electrically neutral?

Because the electrostatic force is much stronger than you can imagine. Yes, you can get charge-rich areas in a cell. However, the amount of charge held here will be negligible. A rather small amount of charge can overcome the field near the plates and prevent more charge from approaching. The amount is quite small; it's not something that you would notice (and it won't give you a static shock1

1. Everyday static shocks are stronger by orders of magnitude because they deal with a lot of charge accumulated on the surface of an insulator, while here it is a relatively small amount of charge dispersed through the volume of a conductor


Does the Maxwell-Boltzmann distribution apply to gases only? The Maxwell-Boltzmann distribution can be used to determine the fraction of particles with sufficient energy to react. I know that the curve applies to gaseous reactants and would like to know whether solids and/or liquids are also described by a similar distribution.

In other words, can I use a Maxwell-Boltzmann distribution to interpret reaction rates at the molecular level in liquids and/or solids?

I'm going to go against the grain here:  the Maxwell distribution does describe the distribution of molecule speeds in any (3-D) matter, regardless of phase.

Suppose we have a system of $N$ molecules with masses $m_i$, positions $\vec{r}_i$, and velocities $\vec{v}_i$ (with $i = 1, ..., N$).  Assume that the total energy of this system is of the form
$$
E(\vec{r}_1, \dots, \vec{r}_N; \vec{p}_1, \dots, \vec{p}_N) = U(\vec{r}_1, \dots, \vec{r}_N) + \sum_{i = 1}^N \frac{\vec{p}_i^2}{2 m_i },
$$
i.e., a potential energy depending on the molecules' positions and a kinetic energy.  According to Boltzmann statistics, the probability of finding this system within a small volume of phase space $d^{3N} \vec{r} \, d^{3N} \vec{p}$ is
$$
\mathcal{P}(\vec{r}_1, \dots, \vec{r}_N; \vec{p}_1, \dots, \vec{p}_N) d^{3N} \vec{r} \, d^{3N} \vec{p} \propto e^{-E/kT} d^{3N} \vec{r} \, d^{3N} \vec{p} \\= e^{-U(\vec{r}_1, \dots, \vec{r}_N)/kT} \left[ \prod_{i=1}^N e^{-p_i^2/2m_i k T}\right]d^{3N} \vec{r} \, d^{3N} \vec{p}
$$
To find the probability distribution of finding molecule #1 with a particular momentum $\vec{p}_1$, we integrate over all other configuration variables (i.e., $\vec{r}_1$ through $\vec{r}_N$ and $\vec{p}_2$ through $\vec{p}_N$).  Because the $E$ is the sum of a contribution from the positions and a contribution from the momenta, the Boltzmann factors can be split between these integrals, with the result that
$$
\mathcal{P}(\vec{p}_1) \, d^3\vec{p}_1 \propto e^{-p_1^2/2 m_1 kT} d^3\vec{p}_1 \left[ \int e^{-U(\vec{r}_1, \dots, \vec{r}_N)/kT} d^{3N}\vec{r}\right] \left\{ \prod_{i = 2}^N \int e^{-p_i^2/2 m_i kT} d^3 \vec{p}_i \right\}
$$
These integrals are nasty (especially the one over all $N$ position vectors), but they're just constants with respect to $\vec{p}_1$, which means that the can be folded into the proportionality constant:
$$
\mathcal{P}(\vec{p}_1) \, d^3\vec{p}_1 \propto e^{-p_1^2/2 m_1 kT} d^3\vec{p}_1.
$$
A similar logic applies to every other molecule in my system.  In other words, the probability of finding a molecule with a momentum $\vec{p}$ doesn't depend at all on how they interact with each other, assuming that their interaction energy is dependent only on their collective positions.  Since these molecules obey the same momentum distribution, they must also have the same velocity distribution, and in particular they obey the Maxwell speed distribution.

So the answer to the question asked in your title, "Does the Maxwell-Boltzmann distribution apply to gases only?" is "no";  it applies to all phases of matter, in the sense that it describes the distribution of particle speeds and energies.  However, the question you ask at the bottom of your post, "Can I use a Maxwell-Boltzmann distribution to interpret reaction rates at the molecular level in liquids and/or solids?" may also be "no";  the connection between reaction rate and activation energy is not straightforward if the medium is dense, as was pointed out by @porphyrin in their answer.


Does the probability function for a molecule to have energy in a certain mode depend on the minimum required quantum energy level of that mode? I understand that the probability density for a particular velocity v+dv in one dimension for a molecule in a system is:
$$f(v) = \sqrt{\frac{m}{2\pi k_B T}} \cdot e^{\frac{-mv^2}{2k_B T}}$$
I'm noticing that this formula does not distinguish the type of freedom in which the concerning velocity is used. So basically this formula would say that a certain velocity in a translational degree of freedom has an equal probability as the same velocity in a rotational or vibrational degree of freedom, which is incorrect from what I know due to the different required minimum quantum energy for each type of mode (depending on the molecular structure) with respect to the energy that the molecule possesses.
Is there a formula that takes the type of mode into account for the probability?

I found a source that takes the frequency of a degree of freedom into account and calculates its minimum required energy according to hf = E but I don't know how to incorporate that formula into the probability function.

There's a couple of conceptual problems at the base of your question that need to be addressed.

1) Free translation of particles in space is not discrete - the distribution of quantum states is continuous, and therefore so is the distribution of energies or of velocities - but rotation or vibration are quantised, so they have discrete distributions, so the formula definitely can't be applied to them. You could apply it to molecules as classical objects, ignoring quantum behaviour, but what you are describing then is not a molecule; you won't get realistic results that you could compare to experimental results.

2) The equipartition theorem states that rotation or vibration degrees of freedom will have the same energy as translation degrees of freedom, but the same energy will not translate into the same velocity.

For instance, in a classical treatment, the energy distribution function per degree of freedom in equilibrium that we obtain from Maxwell-Boltzmann is

$$ f_\epsilon = \sqrt{ \cfrac{1}{\pi \epsilon_i k_B T} } \exp \left[ \cfrac{-\epsilon_i}{k_B T} \right] $$

Taking into account that classical rotational energy is

$$ E_{rot} = \frac{1}{2} I \omega^2 $$

we can derive the classical angular velocity distribution:

$$ f_\omega = \sqrt{\cfrac{I}{2 \pi k_B T}} \exp \left[ \cfrac{-I \omega_i^2}{k_B T} \right] $$

with the moment of inertia $I$ being dependent on the geometry of the molecule. For instance, in the simplest case of a diatomic molecule approximated as a rigid rotor, it is

$$ I = \mu d^2 $$

where $\mu$ is the reduced mass and $d$ is the interatomic distance.

However, as we said, this is not a correct description of an actual molecule, in which vibration and rotation movements are quantised. In a quantum treatment, the allowed values for rotational energy in a linear molecule are

$$ E_{rot} = J \left( J + 1 \right) \cfrac{\hbar^2}{2I} ~ \mathrm{with} ~ J=0,1,2... $$

note that rotational levels are degenerate (with unequal degeneracy $g_j = 2J+1$) and discrete, so we need to use a discrete partition function $Z_{rot}$:

$$ Z_{rot} = \sum_{J=0}^{\infty} \left( 2 J + 1 \right) \exp \left[ \cfrac{-J \left( J + 1 \right) \hbar^2 }{ 2 I k_B T} \right] $$

which in principle we can't calculate analytically. For molecules with a very large moment of inertia $I$ and/or at high enough temperatures $T$, the gap in energy between levels becomes very small and we can approximate this sum as a continuous integral over $J$ (the so-called high temperature limit):

$$ Z_{rot} \approx \int\limits_{J=0}^{\infty} \left( 2 J + 1 \right) \exp \left[ \cfrac{-J \left( J + 1 \right) \hbar^2 }{ 2 I k_B T} \right] dJ = \cfrac{ 2 I k_B T}{\hbar^2} $$

Under that assumption, the (discrete) angular momentum distribution of a rotational degree of freedom can be derived as

$$ F_J = \left( 2 J + 1 \right) \cfrac{ \pi \hbar^3}{ I k_B T} \exp \left[ \cfrac{-2 \pi \hbar^3 \left( J + 1 \right) }{ I k_B T} \right] $$

Note that we're using angular momentum and not angular velocity here because, while the former is a well-defined magnitude in quantum mechanics, it's not clear the later makes physical sense in a quantum treatment of molecules. We could naively calculate a semi-classical "angular velocity" distribution from that angular momentum distribution, but, again, we'd not be actually describing a molecule.

And a final conceptual problem:

3) Rotation and vibration (and electronic) motion is coupled; to give a simple example, in a diatomic molecule, as rotation angular momentum increases, interatomic distance (and therefore moment of inertia) changes; a simple approximation is to introduce a rotational term representing centrifugal distortion, but for higher precision both movements have to be considered at the same time, what makes for much more complex equations.

Keep also in mind that we've restricted ourselves to the simplest cases in rigid diatomic molecules; more complex systems have much more complex mathematics involved and often have to be treated empirically.


How can I determine if there are π-π interactions between an amide and an aromatic ring in a protein? In a crystal structure I've determined, a triazole ring on my ligand appears to be stacking with a tyrosine (top in picture):

However, there is also an amide, courtesy a glutamine, near it (bottom).  Is it likely this terminal amide is engaging in π-π interactions with the triazole or just weaker dispersion forces?  How would I quantify how significant the interaction is?

It seems that NickT was looking for an experimental solution. My post deals, however, with a computational solution.

  How would I quantify how significant the interaction is?

Determining the interaction energy between two defined monomers such as your aromatic triazole and amide is a rather straightforward process.  This process is referred to as the supramolecular approach.  I'll point you to a paper that analyzes the benzene dimer.  This method is strictly a computational one so a bit of knowledge in computational chemistry is necessary.

The Supramolecular Approach

The What

The supramolecular approach boils down to this.

$$E_{\mathrm{int}} =  E_{\mathrm{dimer}} - (E_{\mathrm{mon1}} + E_{\mathrm{mon2}})$$

Here we have some interaction energy ($E_{\mathrm{int}}$) determined from the difference of a dimer energy ($E_{\mathrm{dimer}}$) and the sum of the two monomers ($E_{\mathrm{monomer}}$).  If both monomers were equivalent (say, you were interested in the interaction energy of the benzene dimer where each monomer was a benzene ring), you could simplify the summation to two times the energy of one monomer ($2\times E_{\mathrm{mon}}$). In your particular case, you have two different monomers.

The whole idea is if I have two interacting molecules, I can determine the energy of each molecule individually (as if they were separated at infinite distance) as well as their complex.  So as you bring these molecules closer and closer together, the energy starts to go down (the interaction energy).

The How

NOTE: Your geometry is from a crystal structure.  Do NOT modify this geometry.  You will want to keep everything exactly as is.  This means you don't want to optimize your system.  You do not want to eyeball monomer placement.  Take everything from your known structure and be careful not to change it otherwise it can ruin this process.  We are modifying the structure by truncating and capping but intermolecular parameters must stay the same for whatever it is you are trying to model.  You will want to optimize your cap meaning, run an optimization on your capped-monomer but freeze everything but the thing you are using to cap.

Define your monomers.  You will need to determine what part of your 'dimer' system is important for describing this weak interaction.  I recommend keeping the aromatic ring and truncating the ring with something similar to what is being truncated.  You could cap your monomer with a hydrogen or a methyl group for example.  If the piece you've cut out is highly polarizable, cap with something with a similar property.  If your truncated piece is neutral in charge, cap with something that is neutral.  You get the idea.
Define your dimer.  Your dimer is simply a combination of your two defined monomers.
Determine the method you want to implement.  Wavefunction based or density functional theory (DFT) will work for this.  The former will be prohibitively expensive.
Define your basis set.  For aromatic systems your best bet would be to use Dunning-Hunzaga's correlation consistent family of basis sets.  I recommend using aug-cc-pVTZ for good results.  Whatever you decide, be sure your basis set includes polarization and diffuse functions.  The suggested basis set does this (augmented means diffuse on all atoms whereas pVTZ means 'polarized-valence triple zeta').
Determine the energies of your monomers and dimer.  You will want to run a single-point energy calculation on your monomers and your dimer. Optimize your monomers first but freeze all atoms except those you added to cap your monomer.
Determine the interaction energy.  Plug your energies into the equation given above and determine the interaction energy.  Convert to whatever units you wish to use (I prefer kJ/mol but most people use kcal/mol so you may want to use that).

I hope this helps.  Looks like an interesting project to say the least.

A quick note about the 'valence' basis set:

Typically core electrons do not participate in any sort of interesting chemistry (they are so much lower in energy than the valence electrons that they do not play a role in things like chemical bonding).  When you perform these energy computations, you will be invoking the frozen-core approximation (default on many quantum software packages but be sure to check the documentation first).  The pVTZ basis set expands the basis functions out to triple-zeta quality for valence electrons only.  The core electrons are given s-type functions and that is it.  If you ever find yourself in a situation where the core electrons are important to your system of interest, you would want to use the pCVXZ basis sets (where X = D,T,Q, etc.) which stands for polarized core-valence X zeta.  Here the core electrons are given the full set of X-zeta quality basis functions.  However this means you've just made your computation much more expensive.  The frozen-core approximation is typically used.


How to characterize gaseous ethylene that forms as byproduct? I'm running a reaction that has ethylene gas as a byproduct. I have observed gas evolution as my reaction proceeded and I have other ways of verifying the identity of the main product, but I thought it might be cool if I could also test for the presence of ethylene.
Are there any simple tests I can use for this purpose? Obviously it is not easy to detect since it's a fairly non-reactive gas, so it doesn't have to have 100% specificity.

If you want something simple and don't mind less than 100% certainty then you could bubble the gas through Bromine Water - in the presence of an alkene the bromine water would be decolourised from brown when reacting with the ethene.  Of course this does just indicate the presence of any alkene so isn't exactly specific! There is more about this on this website.


Why does a smaller bandgap correlate to increased absorption of visible light? I am currently involved in a project on metal organic frameworks in my AP chemistry class, and specifically their applications in $\ce{CO2}$ reduction into formate. All the explanations we can find on why a smaller bandgap correlates to higher absorption and quantum yields refers to PhD-level chemistry, e.g. the Density Functional Theory. Can someone provide a simplified version of this?

In your case, absorption takes an electron from a valence band and in to a conduction band. At energies below the narrowest gap between the valence and conduction band, the photon can't give enough energy to an electron to get it up from the valence to the conduction band - there is no absorption there.

Right at the band gap, you can now just get a small population of electrons from the valence to the conduction band, those right at the top of the valence band. Absorption starts to kick in.

As the photon energy increases, there is a larger and larger number of valence band states that can now be excited in to the conduction band.

Depending on details of the shapes of the conduction and valence bands (density of states of initial and final vs energy), and on the exact absorption process (if you need a phonon for momentum matching it is harder), you get various expressions relating the absorption to the band gap ranging from $\sqrt E$ to $E^{2}$ or higher powers.


Are `dTDP-4-dehydro-6-deoxy-alpha-D-galactose` and `dTDP-4-keto-6-deoxy-D-glucose` the same thing? Are dTDP-4-dehydro-6-deoxy-alpha-D-galactose and dTDP-4-keto-6-deoxy-D-glucose the same thing?
I am looking for the function of this enzyme in uniprot. It says

Catalyzes the synthesis of dTDP-4-amino-4,6-dideoxy-D-galactose (dTDP-Fuc4N) from dTDP-4-keto-6-deoxy-D-glucose (dTDP-D-Glc4O) and L-glutamate.
dTDP-4-amino-4,6-dideoxy-alpha-D-galactose + 2-oxoglutarate = dTDP-4-dehydro-6-deoxy-alpha-D-galactose + L-glutamate.

Note that glucose in the first line and galactose in the second line.
Or it is a typo there?

There are four stereocentres in hexoaldoses giving rise to a set of $2^4 = 16$ different molecules with the same connectivity; you can group these to eight pairs of enantiomers (perfect mirror images of each other). The enantiomers are labelled D and L depending on the orientation of the stereocentre which is farthest from the aldehyde group. In linear orientation, these are the following molecules:

(All images taken from Wikipedia; all D-isomers.)

In the order of reading these are allose, altrose, glucose, mannose, gulose, idose, galactose and talose. If you compare glucose (3rd in the first row) with galactose (3rd in the second row) you will realise that they differ only in the orientation of the hydroxy group in 4-position. The posh terminology is epimerism — D-glucose is the C4 epimer of D-galactose and vice-versa.

You asked about two labels where the sugar is preceded either by 4-dehydro-6-deoxy or 4-keto-6-deoxy. The agree in the 6-deoxy part which means that the oxygen on C6 is missing; you will find a methyl group $\ce{CH3}$ there instead. The labels 4-keto and 4-dehydro may also be confusing. The first says that a keto group is there instead of the hydroxy group, while the second says that two hydrogens have been removed from that position. The result is the same, instead of having $\ce{CHOH}$ as fourth carbon we have $\ce{C=O}$. This also deprives us of the stereocentre (a carbonyl group is never stereogenic).

Putting all of this together: Yes, the two are identical. If you start from the C4-epimers and remove the stereocentre at C4 it does not matter which you started from. Both galactose and glucose are common sugars so both choices are understandable (even though glucose is probably known even a tad more). I would, however, strongly object agains 4-keto-D-talose in the place of 4-keto-D-mannose (mannose and talose being C4-epimers) since mannose is much more known than talose.


Required temperature for the reduction of iron oxide using carbon When reducing iron oxide with carbon, must the substances be set alight or do they only need to be heated. Also, when the iron is formed will it immediately oxidise in the air?

The direct reduction with carbon is only possible if the iron oxide ($\ce{Fe2O3}$, $\ce{Fe3O4}$, or $\ce{FeO}$) is present as a liquid.

$$\ce{FeO$_n$ + $n$C -> Fe + $n$CO}$$

Accordingly, the required temperature is very high. Note the melting points of $\ce{Fe2O3}$ (1539 °C), $\ce{Fe3O4}$ (1597 °C), $\ce{FeO}$ (1377 °C), pure $\ce{Fe}$ (1538 °C), and the eutectic of iron containing 4.3 % carbon (about 1150 °C).

In the blast furnace process, solid iron oxides at lower temperatures are mainly reduced with carbon monoxide.

$$\ce{FeO$_n$ + $n$CO -> Fe + $n$CO2}$$

The required carbon monoxide is produced from the exothermic reaction of carbon with oxygen

$$\ce{2C + O2 -> 2CO}$$

and from the endothermic Boudouard rection of carbon with carbon dioxide.

$$\ce{C + CO2 <=> 2CO}$$

A temperature well above 700 °C is required to favour the formation of carbon monoxide. Hence, a part of the carbon has to be burned in order to generate the required heat and to produce carbon monoxide.

Hot fine-dispersed iron would be quickly oxidized in air (for example, steel wool can burn in air) whereas fused iron is only oxidized at the surface of the liquid. Note that the pig iron produced in the blast furnace process has a high carbon content, which significantly lowers the melting point.


Mixing of powders to make stock solutions For a solution such as:

Luria-Bertani (LB) Broth:

10 g tryptone
5 g yeast extract
10 g of NaCl ($M = 58.44\ \mathrm{g/mol}$)

q.s. to 1 l, pH to 7.2, autoclave.

I would like to make a bulk amount of premixed powder, i.e. combine 1 kg tryptone, 500 g yeast extract and 1 kg NaCl in a big bottle, shake it and keep it on shelf. Whenever I need to make LB now I just weight 25 g of this premix and add to 1 l of water.

I’m concerned about a couple things thought. First how thoroughly it gets mixed, and second any settling of different chemicals over time. Either would cause many different batches to be messed up.

I’m wondering if anyone has experience with this and ways to avoid those two problems. I’m interested in doing this for other solutions as well. Do manufacturers who make these premixed powders re pulverize them or process them in a certain way to make it work?

Once the powders are mixed, it is unlikely that they will separate unless the container is subjected to vibrations.

As for mixing, companies that need to mix batches of material use a mixer to make their batches. The second photo in the article is similar to what I have seen.


Thermal stability of interhalogen compounds I am striving over the question that what would be the stability order of the following interhalogen compounds : $\ce{ClF3, BrF3}$ and $\ce{IF3}$.

One probable answer I got is that stability of interhalogen compounds decreases as the size of central atom decreases due to electronegativity difference. Hence, the stability order should be $\ce{IF3} > \ce{BrF3} > \ce{ClF3}$.

Is the above mentioned logic correct? Can someone please elaborate?

More electronegativity difference generally gives you stronger bonds,  therefore higher thermal stability (i.e. you need to put in more energy to break them). Also, there are 2 lone pairs on the central atom and 3 on each F. Thus, interhalogen compounds of this type with larger central atom would experience less electron pair repulsion. So, I believe that the correct order is $\ce{IF3>BrF3>ClF3}$ as you suggested.


Why do metals appear lustrous? I came across a question asking me the reason for the lustrous appearance of many metals. The answer stated that it was due to the presence of free electrons in the metal. But I don't understand how this works. How do the free electrons affect the reflection of light from the metallic surface?

Thanks.

Light is an electromagnetic wave. A metal has a large cloud of relatively free electrons (electrons that are loosely bound to the metal surface). When a beam of light is incident on a metal surface, it polarizes the electron cloud, i.e. some regions on the metal become relative more "positive" while some regions relatively become more "negative". Thus, this induces a field which makes the electrons to start oscillating. This oscillation generates another electromagnetic wave which opposes the incident radiation, (An ideal metal will completely oppose the incident light radiations), and hence our incident light rays get reflected.


Better solvent for KCl than water? In my experiment, I used an atomizer to generate $\ce{KCl}$ aerosols. I saw that the nozzle that was dipped into the 10% wt $\ce{KCl}$ solution got partially clogged after an hour. I then had to disassemble the nozzle in order to clean it, which I did by rinsing it with water and drying it with a stream of air.
 Is there any way to avoid the clogging?

This seems like more of an engineering problem than "chemistry problem." I'm not sure how long you want the process to run, nor how critical the 10% salt concentration is.

I'd try adding a pure water bubbler in front of the salt solution to saturate the air being blown through with water.


Is it true that calcium carbonate is more soluble in NaCl than deionized water? I have two mixtures: $\ce{CaCO3}$ in NaCl and $\ce{CaCO3}$ in deionized water (DI). Is it possible to get more $\ce{Ca^2+}$ ions in NaCl solution than DI?

You claim to have a solid solution of Calcium carbonate in solid (or molten? you didn't specify temperature!) sodium chloride. I doubt it.
Because $\ce{CaCl2}$ is more soluble in water than $\ce{CaCO3}$, addition of $\ce{Cl^-}$ ions would be expected to increase the solubility of $\ce{CaCO3}$ in water. In other words, "yes, it is possible." One crude way to think about it is that the "average" environment for the $\ce{Ca^{2+}}$ ions is more "friendly" in a mix of $\ce{CO3^{2-}}$ and $\ce{Cl^-}$ ions than in a mixture of $\ce{CO3^{2-}}$ions alone. The effect is called the "Common Ion" effect. Look it up on Wikipedia.


SN1 or SN2, the kinetic isotope effect I know that you can't always trust Wikipedia, so I just want to check something regarding their kinetic isotope effect page.

The first example implies that $\sideset{^{12}}\ C$ and $\sideset{^{13}}\ C$ can be used to determine whether a reaction is $S_N1$ or $S_N2$:

  ...In the nucleophilic substitution reaction of methyl bromide with
  cyanide, the kinetic isotope effect of the methyl carbon, in this case
  defined as $k_{12}/k_{13}$, was found to be 1.082 ± 0.008 ... The observed
  methyl carbon kinetic isotope effect is indicative of an $S_N2$
  mechanism."

No explanation appears to be offered for this inference. I know that the reaction will not proceed via $S_N1$ because of the unstable carbocation but (how) does the KIE datum predict an $S_N2$ mechanism?

The rate constant depends on the energy difference between the ground state of the reactants and the transition state. The ratio of the rate constants due to the kinetic isotope effect therefore depends on the difference in vibrational zero point energy between the initial state and the transition state. Vibrational ZPE depends on bond stiffness and reduced mass. The change in reduced mass from $\ce{^{12}C}$ to $\ce{^{13}C}$ causes the kinetic isotope effect, and the bond stiffness determines the size of the KIE. Bond stiffness is to a first approximation proportional to bond order.

In both $S_N1$ and $S_N2$, the initial state has a single bond between the methyl $\ce{C}$ and $\ce{Br}$  (as well as some $\ce{C-H}$ bonds which stay the same throughout the reaction).
For $S_N2$, the transition state has partial bonds between the methyl $\ce{C}$ and both $\ce{Br}$ and the $\ce{C}$ of $\ce{CN}$.
In the $S_N1$ transition state, the methyl $\ce{C}$ has only a partial bond to $\ce{Br}$.

Since the change in bond order is greater between initial state and $S_N1$ transition state than between ground state and $S_N2$ transition state, $S_N1$ will have a greater kinetic isotope effect. Measuring the KIE of an unknown reaction and comparing it to the KIEs of reactions with known mechanisms will therefore distinguish between $S_N1$ and $S_N2$.


What is the name of the "primary" carbon in a benzene derivative? If we have a benzene with a single substituent, such as chlorobenzene, what do we call the carbon atom to which the group is attached? I'm familiar with the ortho- meta- para- nomenclature, but the last one escapes me.

It's ipso-.

Wikipedia page
IUPAC Gold Book definition of the ipso-attack


Why doesn't every pure enantiomer racemise? After having been acquainted with enantiomerism for nearly six months, I have one standing doubt.

Consider the structure of alanine. Suppose I have a pure mixture of exclusively 'S-alanine'(left in the picture) in a flask.

There are hundreds, if not, thousands of molecular collisions in the flask per second. Why can not a bump to one of the S-alanine change it to R-alanine? This way, the mixture should racemise itself if left standing for some time. But this does not happen. Why? Can someone please explain?

Because the activation energy for the enantiomeric H-Atom to jump to the other side of the carbon atom's sp3 tetraeder (by whatever mechanism) is so high that it is just not available.

You're right, statistically, every energy is available (with very bad odds). But very likely other processes, and more destructive ones, have a lower activation energy. Your compound will likely degrade and rot away long before it racemises. But even that, for alanine at room temperature, does not happen on any timescale that you can observe.


Does motor oil get more viscous when heated & what is the meaning of the SAE motor oil code? The motor oil in my 2004 Pontiac Vibe is 5W-30.

My (incorrect) understanding of this SAE nomenclature is that 5W means "Viscosity index of 5 in the winter" (or when really cold - like $\pu{-30 ^\circ F}$) and "Viscosity index of 30 at operating temperature" (or when the car has been running for a little while).

So how is this accomplished? I usually think of something like butter or coconut oil that solidifies at a lower temperatures and becomes fluid again as heated. To my understanding oil has a higher viscosity at lower temperatures (generally speaking).

Everyone keeps talking about "additives" that help accomplish this feat in "multi-grade" oil. What are these additives? Generally, how do they work? I'm no chemist and neither are these auto mechanics, so I figured this might be the right place to ask.

Since my understanding of the SAE motor oil code is incorrect, illumination on its exact meaning would be helpful.

A full blown chem answer and an "explain it like I'm 5" would be most appreciated :) I never studied chemistry past high school, so be easy on me!

The numbers in the code have nothing to do with viscosity index. However, your understanding of 'W' for winter is correct. The real code for your car is actually SAE 5W-30 where SAE stands for the Society of Automotive Engineers, which has established a multi-grade numerical code system for grading motor oils according to their viscosity characteristics. I said viscosity characteristics, not viscosity index. According to Wikipedia:

  The original viscosity grades were all mono-grades, e.g., a typical engine oil was a SAE 30. This is because as all oils thin when heated, so to get the right film thickness at operating temperatures, oil manufacturers needed to start with a thick oil. This meant that in cold weather it would be difficult to start the engine as the oil was too thick to crank. However, oil additive technology was introduced that allowed oils to thin more slowly (i.e. to retain a higher viscosity index); this allowed selection of a thinner oil to start with, e.g., "SAE 15W-30", a product that acts like an SAE 15 at cold temperatures (15W for winter) and like an SAE 30 at $\pu{100 ^\circ C}$ ($\pu{212 ^\circ F}$).

viscosity characteristics: Kinematic viscosity is graded by measuring the time it takes for a standard amount of oil to flow through a standard orifice at standard temperatures. The longer it takes, the higher the viscosity and thus the higher the SAE code. Larger numbers in the code are thicker.

The company called Anton Paar has analyzed these SAE coded motor oil for the parameters, Kinematic viscosity and oil density versus elevating temperatures. I have included some of graphs including SAE 5W-40, but failed to find the most common SAE 5W-30:

Finally, for your information, Viscosity Index is a dimensionless number that represents how the viscosity of a hydraulic fluid changes with temperature. The greater the viscosity index, the smaller the change in fluid viscosity for a given change in temperature, and vice versa. Synthetic oils usually have a higher viscosity index than do mineral oils. (Mobile Hydraulic Tips). The following graph illustrate how viscosity index influence the change of viscosity with temperature (Ref.1):

It is also nice to know how change of viscosity index by modifiers affect mono-grade oil to better multi-grade oil:

References:

Tiago Cousseau, “Film Thickness and Friction in Grease Lubricated Contacts: Application to Rolling Bearing Toque Loss,”  PhD Thesis; University of Porto (Universidade do Porto), Porto, Portugal, 2013.


Can someone walk me through this gas mixture question? Argon $(\ce{Ar})$ and helium $(\ce{He})$ are initially in separate compartments of a container at $\pu{25 °C}.$ The $\ce{Ar}$ in compartment A which has a volume $V_\ce{A}$ of $\pu{9.00 L}$ and a pressure of $\pu{2.00 bar}.$ The $\ce{He}$ in compartment B of unknown volume $V_\ce{B}$ has a pressure of $\pu{6.00 bar}.$ When the two compartments are connected and the gases allowed to mix, the total pressure of gas is $\pu{3.60 bar}.$ Assume both gases behave ideally.

  $$
\begin{array}{|l|l|}
\hline
\text{Chamber A} & \text{Chamber B} \\
V_\ce{A} = \pu{9.00 L} & V_\ce{B} = \pu{??? L} \\
p_\ce{A} = \pu{2.00 bar} & p_\ce{B} = \pu{6.00 bar} \\
\hline
\end{array}
$$

  [4 marks] Determine the volume of compartment B.

I've been stuck on this problem for a while.  Could someone walk me through the beginning steps? I should be good to go with some guidance.  Would I use Boyle's law in attempt to find the final volume once the partition is removed?

Solve for the amount of gas in A: $n_A=p_AV_A/RT=\pu{0.727 mol}$. You'll need to perform necessary unit conversions or use R in units that match the given T, p and V units.
Write the volume of chamber B in terms of knowns: $V_B=n_BRT/p_B=n_B \times\pu{4.13\times 10^3 m^3/mol}$.
Write the total volume when mixed: $V_{tot}=n_{tot}RT/p_{tot}=n_{tot} \times\pu{6.89\times 10^3 m^3/mol}$.
Solve this last expression for $n_B$ after inserting the equalities $V_{tot}=V_A+V_B$, $n_{tot}=n_A+n_B$  and the result of step 2:
$$\begin{align} V_A+V_B &=(n_A+n_B)\times \pu{6.89\times 10^{-3} m^3/mol}\\ &=V_A + n_B \times\pu{4.13\times 10^{-3} m^3/mol}\\\rightarrow n_B &= \frac{V_A - n_A\times \pu{6.89\times 10^{-3} m^3/mol}}{\pu{2.75\times 10^{-3} m^3/mol}} = \pu{1.45 mol}\end{align} $$
Compute the volume in chamber B: $V_B=\frac{n_{B}RT}{p_B}=\pu{6.00\times 10^{-3} m^3}$

It's usually a good idea to perform a check to see if we did things right. Compute back the final pressure: $p_{tot}=\frac{n_{tot}RT}{V_{tot}}=\pu{3.60 bar}$.

Answer: $V_B = \pu{6.00 L}$


Functional difference of Benedict's solution and Fehling's solution Chemically, Benedict's solution and Fehling's solution are very similar, with copper sulfate as a source of copper (II) ions, sodium carbonate, and sodium hydroxide respectively as alkalis, and sodium citrate and sodium potassium tartarate respectively as chelators. They are also both used to test aldehyde groups which can reduce the copper (II) ions into copper (I) ions, giving a precipitate of $\ce{Cu2O}$. Hence, they are often used in testing for the presence of aldehydes, and by extension, reducing sugars.

Are there any differences in function between the two reagents, perhaps in sensitivity, types of compounds that test positive, etc?

Fehling's solution consists of two parts, Fehling's A and Fehling's B. Fehling's A consists of aqueous copper(II) sulfate, while Fehling's B consists of aqueous sodium potassium tartrate ($\ce{NaK(C4H4O6)}$). Benedict's reagent consists of sodium carbonate, sodium citrate ($\ce{Na2(C6H6O7)}$), and copper(II) sulfate dissolved in water. It's pretty clear now, that the main difference between the two reagents is the complexing agent (citrate in Benedict's and tartrate in Fehling's).

Fehling's reagent was invented by Hermann von Fehling in 1849. You notice that this solution is given as two separate solutions, and is only mixed together while performing a test. This is because the complex formed when mixed ($\ce{[Cu(C4O6)2]}$) is kinetically favored, and that $\ce{Cu(OH)2}$ will eventually precipitate out due to being favored energetically. That's why the two solutions are stored, and only mixed during testing.

On the other hand, Benedict's reagent was discovered by Stanley Rossiter Benedict somewhere in between 1904 to 1936. Thus Benedict's reagent was developed after Fehling's was. The citrate complex in Benedict's reagent is much more stable that $\ce{Cu(OH)2}$, so Benedict's reagent doesn't precipitate on standing for long, unlike Fehling's. This is quite advantageous since you wouldn't have to mix two solutions together to perform the test. This is where Benedict's reagent is quite handy.

Both the reagents have a similar way of oxidizing aldehydes. The copper complex oxidized the aldehyde to a carboxylic acid, and while reducing itself from $\ce{Cu^2+}$ to $\ce{Cu+}$
 ($E^\circ = \pu{0.154 V}$). This reduction potential is sufficient to oxidize aldehydes to carboxylic acids. There is not much difference here though.

Neither of the reagents can oxidize aromatic aldehydes, due to the fact that they lack α-hydrogens, and thus oxidizing them requires stronger oxidizers (like Tollens’ reagent).

In short, Benedict's reagent is kind of like an "improved" version of Fehling's reagent, which is much more stable due to the difference in the complexing agent.


Reagents in Krapcho's decarboxylation Krapcho's decarboxylation is a reaction involving esters with a electron-withdrawing group $\beta$ to the carbonyl group and halide ions. It is typically conducted with $\ce {LiCl}$ and $\ce {DMSO}$ at high temperatures. However, in one of Krapcho's original papers, $\ce {NaCN}$ was also used alongside $\ce {DMSO}$.

What interests me is the first step of the reaction which involves the nucleophilic subsitution, likely by an $\ce {S_N2}$ pathway, of the alkyl group of the ester functionality. This is seemingly weird as the $\ce {C-O}$ bond usually does not break unless in acid catalysis. This is the case for the cleavage of ethers using $\ce {HX}$. I suppose the use of $\ce {LiCl}$ was to allow the $\ce {Li^+}$ to coordinate to the oxygen atom, allowing the $\ce {C-O}$ bond to cleave more easily. Why is it that $\ce {NaCN}$ can also do the job?

So as mentioned the Krapcho-Decarboxylation follows a SN2-mechanism and is therefore working best with a methylester.

The solvent choice for DMSO (polar aprotic) is due to low stabilisation of the chloride anion rendering it highly reactive for the nucleophilic attack. The resulting methyl chloride is gaseous at room temperature and leaves the reaction mixture (le Chatelier -> formation of product strongly favoured)
The leaving group of this substitution reaction decays into CO2 and the decarboxylated product. (Again: formation of a gas)

I think that the destabilisation of the cyanide in DMSO and the formation of stable acetonitrile can be considered as the driving force in the second reaction.


Removing water/moisture from car exhaust I know that car exhaust is almost entirely nitrogen, water and carbon dioxide. How would I go about removing the water? Would I have to condense it off? Any devices already exist for doing so?

Certainly devices exist to remove water vapour from gas streams. Many labs pass their nitrogen stream through a conc. sulfuric acid bubbler to ensure it is dry. Solid absorbents such as silica gel or Drierite can also be used. The question is what volume of exhaust gas do you want to dry? These absorbents need to be changed frequently to maintain the dryness of the gas stream. If you want to keep a lot of exhaust dry you'll need a lot of drying agent.


What's with the weird definition for the Calorie? I came across this definition for the calorie, which I found... strange:

"The (15°) calorie is the heat required to raise the temperature of one gram of water from 14.5°C to 15.5°C."

Subsequent searches on-line revealed that this was (one of) the standard definitions for the Calorie.

Now two issues came to my mind after I read that definition:

1) Why is the term 'heat' and not 'energy' used?

From what I recall from my Thermodynamics classes at school last year, heat is energy in transit. I'm under the impression 'heat' is more of a 'surface phenomenon' if you will; there's 'heat' at the junction between two surfaces, but no matter how much you 'heat' up a body it can never contain heat (strictly speaking).

2) What's so special about the 14.5°C and the 15.5°C range?

Seriously, why that particular range? Why not 0°C to 1°C, or even 1°C to 2°C? Heck even 14°C to 15°C would go down (mildly) well with me, but 14 .5 °C to 15 .5 °C? What's with the inordinate fondness for the 0.5°C? A 1°C difference is a 1°C difference in temperature, so why specify any range at all? Why not define the calorie as 'The energy required to raise the temperature of one gram of water by 1°C' ?

I didn't have this question before because I, for all practical purposes, only used to see the calorie as being roughly equal to 4.19 Joules.

Would anyone happen to have the answer to those two queries?

EDIT-

I've already seen this particular question, and I believe it is necessary to state that this does NOT answer my question. My question is distinct from it in the following respects:

1) I'm asking for a clarification regarding the use of the term 'heat' over energy.

2) I want to know why a generalization of the definition, replacing the arbitrary temperature range with a '1° difference'  wouldn't suffice.

3) I want to know why the range is demarcated by two fractions (14.5° and 15.5°) instead of integers.

A number of the Chemistry texts I've used or examined for adoption have indeed confused heat and thermal energy. They sometime refer to a body possessing heat. But Physics books are pretty unaminous that heat is the transfer of thermal energy and not a state property of matter.

I've never seen, in texts I've used, heat restricted strictly to the interface between the participants in the exchange. It refers to the total amount of thermal energy transferred between two bodies. It can also be used to describe how much energy was lost by one of the bodies in the exchange and how much was gained by the other. One way to reconcile this with your interface statement is that once the surface of a body warms due to contact with a hotter body, the higher temperature at the surface relative to the rest of the body will continue to cause heat to flow within the body.

As to the choice of temperature, it shouldn't near zero since this is where a phase change occurs.

Edit. In response to your edit, #2. Search for a table of "specific heat of water". Notice the specific heat is not constant as temperature varies.
I'll hazard a guess on the 14.5-15.5 part. Think 15 degrees. But, since an interval is required, let's begin a half degree before and end an equal amount after. As to why 15 degrees, hmmm. One answer to a recent question on the Physics forum says this is roughly room temperature. Pretty cold for me but maybe true when and where this was defined. (Sorry for no link, I'm not that fascile on iPad.)


Meaning of prefix p,p′-? I recently stumbled across the substance "p,p′-terphenyldicarboxylic acid" in a scientific journal when recently reading about metal–organic frameworks.

Could someone explain the prefix p,p'- to me?

I am guessing that this is the compound you are interested in:

Sigma labels this compound as p-Terphenyl-4,4′′-dicarboxylic acid (or 1,4-Bis(4-carboxyphenyl)benzene).

In your case, the p,p' indicates that the 2 carboxylic acids are in the para positions on 2 of the phenyl rings.


Is the reactivity of Benzene with an -N(R)₃⁺ group greater than its reactivity with an -NH₃⁺ group? Both are m-directors, and Electron Withdrawing groups, but which one is more reactive? At first sight, one might think clearly $\ce{-N(R)_3^+}$ is more reactive, but when you factor in that $\ce R$ is actually an o,p-director, hence activator, then I think $\ce{-N(R)_3^+}$ is indeed stronger than $\ce{ NH_3^+}$. I can't seem to find any verification on the internet for this.

$\ce{Ph-NR_3^+}$  is likely weaker than $\ce{Ph-NH_3^+}$ because the mechanism of deactivation is based on the amount of positive charge localized on the Nitrogen:  Alkylgroups will stabilized the positive charge through sigma donation.

That being said, the difference will be small.  Additionally, there are deprotonation side-reactions to consider for anilium ($\ce{Ph-NH_3^+}$) ions, which may compromise any discussion of deactivation.


Dependence of primary alcohol oxidation on presence of water and "strong" or "weak" oxidants It is known that the oxidation of primary alcohols can result in aldehydes or carboxylic acids. The product obtained from the reaction is described as dependent on reagents and conditions;[1]

  "Primary alcohols are oxidized to either aldehydes or carboxylic acids, depending on the reagents chosen and the conditions used."

A distinction is typically made between "weak" (eg. PCC) and "strong" (eg. chromic acid) oxidants[2], the former being capable of giving an aldehyde product while the latter carries out the oxidation to carboxylic acids. However, the accepted mechanism for the reaction shows that the oxidation always results on an aldehyde, which can then be further oxidized to a carboxylic acid through an hemiacetalic aldehyde hydrate forming in the presence of water.[1][3]

  "Most other commonly used oxidizing agents [...] oxidize primary alcohols directly to carboxylic acids. An aldehyde is involved as an intermediate in this reaction but can’t usually be isolated because it is further oxidized too rapidly."

This indicates that it is the presence of water in the reaction medium which determines the product of the oxidation. Thus, "weak" oxidants could still yield a carboxylic acid in aqueous solutions (in fact, they are always noted to be used in anhydrous conditions) and "strong" oxidants will result on an aldehyde when used with organic solvants. If so, I fail to understand the classification of "weak" and "strong".
Furthermore, some sources, like Wikipedia, seem to imply the existence of a direct oxidation from primary alcohol to carboxylic acid, but I've been unable to find such a mechanism in bibliography, always ending up with the hydrate intermediary method.
To summarize:

Is the distinction between "strong" and "weak" oxidants meaningful or does the presence of water always determine the reaction product?
Does a method for the direct oxidation of primary alcohols to carboxylic acids without an aldehyde hydrate exist?

McMurry, John. Organic Chemistry. Brooks Cole, 2011. 8th edition.
Alcohol Oxidation: “Strong” and “Weak” Oxidants
Reagent Friday: Chromic Acid, H2CrO4

After looking up more information and taking the comments of user55119 into account, I've decided to post an answer myself, hoping to clearly address the main question points.

The determinant factor of the products obtained from an oxidation reaction of primary alcohols is the presence of water in the reaction medium, as it allows the formation of an aldehyde hydrate which can be further oxidized. Thus, the distinction between "weak" and "strong" oxidants is not necessarily informative (and could even be misleading) and emphasis should be placed in the reaction mechanism alongside the reaction medium.
The oxidation of primary alcohols with Cr(VI) reagents takes place through the mechanism depicted in the question or slight variations of it (depending on the reagent chosen). Carboxylic acids are obtained only through an aldehyde intermediate.

Further reading:
McMurry, John. Organic Chemistry. 8th edition.
MasterOrganicChemistry


Shoulders of NMR peaks I want to ask a question about the shoulders of peaks when separating overlapping peaks.

I have the following spectrum circled with relevant shoulders:

I believe that the spectrum above shows an overlapping triplet and doublet, but I want to know what causes these shoulders in the NMR spectrum. I'm stuck to whether to decide there is further splitting of a small $\ce{J}$ value or to just assume it's a shoulder.

Below shows analysis of this region from Acta Pharm. 54 (2004) 163–176.

What is causing these shoulders? Are they part of the signal or just a result of poor preparation leading to artefacts?

A good question. The term 'shoulder' is really an informal description used to describe the appearance of parts of peaks in the NMR spectrum. When a peak tends to deviate from the standard Lorentzian, we might describe this as having a shoulder. Shoulders can come about from typically three origins:

Shimming anomaly. Fairly common, but the key to diagnosing this cause is that it will affect all peaks in the same way (within the natural linewidth of the peak). So, have a look at the TMS peak, or solvent peak, or that internal reference of ethyl acetate that your supervisor suggested was an impurity. They will all have exactly that same lineshape anomaly, with identical shoulders. Always check these small molecule peaks when you run your spectra to determine shim quality. As an aside on this spectrum, these 'shoulders' you are looking at aren't really shoulders - they are split peaks. If this lineshape was genuinely due to a shimming anomaly, that indicates a pretty serious z shim issue, and you've probably got some floating solid in the sample.
Overlapping peaks. NMR spectra are simply the superposition of all nuclei in the sample. If two singlet species have a non-identical chemical shift within their natural linewidths, then they will appear as a shouldered peak. Obviously, the relative populations will determine how they actually appear. This could account for this type of spectrum (but more than likely doesn't), and that is why you would do other experiments such as COSY or HSQC to verify this.
Splitting due to coupling Often it is difficult to fully resolve small couplings, especially those that come from aromatic systems or long range pathways, where the size of the splitting might be about 1Hz or so. This is where an understanding of the expected coupling pattern for the molecule is very helpful. So, take a deep breath in.....

For Salicylamide, as described in your paper, we would expect H6 to be a large doublet (ortho to H5, ~7-8Hz), but also show meta coupling to H4 (~1-3Hz), and so we expect this large doublet to exhibit small splittings. Indeed, that is what you are seeing. On a good shimming day, in better solvents than dmso, we might even expect to see para splitting (<1Hz - don't hold your breath though). For the H5, we expect, again, a large ortho doublet coupling to H6, but also a large ortho doublet coupling to H4 (again, ~7-8Hz). That gives us the reported triplet, which is in fact a dd. But, we also expect to see some meta coupling to H3 (again 1-3Hz). Again, we see this, and each of these of these should be split due to the small meta couplings. (And we'll leave alone the discussion about whether they have reported apparent splittings, as opposed to J couplings, otherwise J56=J65),

And breath out.
So what you are seeing is genuine splitting coming from small couplings. These are often not reported in experimental data, because it is quite difficult (or at least time consuming) to resolve all of these couplings. You can get a better understanding of these small splittings if you use resolution enhancement when you process your data. Use a window function such as a small negative line broadened exponential function, or a gaussian function, and you can often split these lines almost to the baseline.


What is Bisphenol A used for? I am trying to find out what BPA is used for in the industrial production of plastics. On sources such as Wikipedia I only found the information that it is used, but not why.

What are the advantages in using BPA in the production of plastics? Why is it needed?

Bisphenol-A is very inexpensive for its considerable formula weight - from benzene, propylene, air, acid; acetone (or other ketone or aldehyde) and acid.

1) Polycarbonate, as above.  If you don't like HCl, dimethyl carbonate will do.
2) Epoxy resin, by bis-etherifying with glycidyl chloride.  That is a huge market, including paints and coatings.
3) Bis-etherify with propargyl chloride.  That thermally B-stages (removes the shrinkage upon cure) then thermosets to an astounding thermally resistant super-crosslinked resin that adheres to any polar surface.
4) Bis-esterify with methacrylic acid to get into heavily cross-linked free radical or UV cures.
5) And if you really want something resistant, instead of acetone use fluorenone or a 2,7-disubstituted fluorenone.  Or be fiendishly clever and use 4,4'-dihydroxybenzophenone instead of acetone.  Now there is a tetrahedral tecton for synthesis!

Bisphenol-A is a claimed  endrocrine disruptor.  Eastman Kodak has immense production capacity for acetic anhydride (cellulose acetate photographic film).  It offshot into ketene dimer chemistry then mixed cis- and trans-2,2,4,4-tetramethyl-1,3-cyclobutanediol as a BPA replacement.  If you do not mind paying more and getting less, stuff as well as performance, the problem is solved.


what is the cross aldol product of ethanal and acetone(major product)? my teacher told that major product would be when acetone will form nucleophile because it will attack more positive carbon of ethanal(less +i than acetone) however i read in my book that since formation of carbanion is rds ethanal should be nucleophile (more stable due to less +i)

As pointed out by Waylander, the question essentially wants students to go for the most stable product, or thermodynamic product. This is sort of like a well-understood thing for aldol product determination questions. Why? Well... Simply because there is usually only 1 thermodynamic product while there are several possible kinetic products. It is therefore hard to argue that the kinetic product drawn is really the major product under kinetic control. Under kinetic control, you may just get a mixture of products with nothing formed in significantly higher quantity.

Based on this understanding, we can go on to determine the thermodynamic product. This reaction is a very straightforward aldol reaction due to there being only two possible reaction pathways since one reactant is a symmetrical ketone while the other is an aldehyde. So there really are only two possible products...

One possible product comes from aldehyde as a the electrophile and acetone behaving as the nucleophile. Upon dehydration, we obtain A, which is a ketone and has a di-substituted conjugated $\ce {C=C}$ double bond. The other possible product comes from aldehyde as the nucleophile and acetone as the electrophile. Upon dehydration, we obtain B, which is still an aldehyde but it has a tri-subustituted $\ce {C=C}$ bond.

Which one is the thermodynamic product? Usually, we will just look at the substitution of the conjugated double bond formed. Very clearly, the kinetic product must be A since the more nucleophilic enolate from acetone attacks the more electrophilic aldehyde. However, is B really more stable? No, it isn't. The explanation presented below is based on qualitative molecular orbital theory.

Well... The key differences between A and B are:

A, being a ketone, has extra hyperconjugative interactions between the $\ce {C-H \sigma}$ MOs and the $\ce {C=O \pi^*}$ MO, compared to the B
B instead has extra hyperconjugative interactions between the $\ce {C-H \sigma}$ MOs and the $\ce {C=C \pi^*}$ MO, compared to A

The former hyperconjugative interactions are stronger since the $\ce {C=O \pi^*}$ MO is lower in energy and thus more energetically-accessible to the $\ce {C-H \sigma}$ MO. This is due to the higher electronegativity of the oxygen atom. Thus, we would expect the thermodynamic stability of A to be higher than that of B.

To provide even more concrete substantiation for the higher stability of A over B, Karsten Theis has kindly provided a source of data which indicates the Gibbs free energy of formation of the two compounds. As stated on chemeo, here and here, A has $\Delta G_f = \ce {-57.48 kJ/mol}$ while B has $\Delta G_f = \ce {-36.63 kJ/mol}$. This very clearly indicates that A has higher stability than B.

Thus, A is the answer.


Resonance stabilized carbanion I am sure that the answer is (b) because I am familiar with the other rings and I think they are showing resonance (and it is a single choice question so it was easy to guess)

I wish to know why is (b) not showing resonance (because I think it can as far as I know) and if I am wrong about the answer then please correct me.

(b) does not show resonance because the second resonance structure (with negative charge at oxygen and double bond in the ring) violates Bredt's rule. The carbon $\ce{sp^3}$ orbital with the negative charge and the $\ce{p_z}$ orbital of the adjacent carbonyl C cannot overlap because they are not in plane, so the charge cannot be delocalized into the carbonyl group.


Why does period 3 of the periodic table contain 8 elements instead of 18? Period 1 of the periodic table contains 2 elements ($1s^1$ and $1s^2$).

Period 2 contains 8 elements ($2s^1$, $2s^2$, $2p^1$, $2p^3$, ..., $2p^6$).

By the same argument, period 3 might contain 18 elements ($3s^1$, $3s^2$, $3p^1$, $3p^3$, ..., $3p^6$, $3d^1$, $3d^2$, $3d^{10}$). Why does period 3 of the periodic table contain 8 elements instead of 18?

I think that it could be explained by means of atomic stability, but I need help to understand what happens.

Thank you in advance.

You could say that the period number tells you about the largest value of principal quantum number in which a electron is present.

Electrons are filled according to $n+l$ rule. It states as $n+l$ increases, the energy of the orbital increases. If $n+l$ is the same, then the bigger $n$ has larger energy.

So $\mathrm{4s}$ would be filled before $\mathrm{3d}$. Therefore elements containing $\mathrm{3d}$ electrons would have their $\mathrm{4s}$ orbital filled, and therefore, come in 4th period. There 3rd period contains elements having $\mathrm{3s}$ and $\mathrm{3p}$ electron as their last electron which are equal to 8.


cis- and trans-isomers in equilibrium Rotation of groups about double bonds are usually restricted due to large torsional strain and it is due to this reason that cis- and trans-isomers have their independent existence. However, certain structural modifications in the compound makes the rotation of groups about double bond possible and hence, for such compound, cis- and trans-isomer exist together in equilibrium.
Identify which of the following compounds has such structural feature and its cis and trans isomers exist together in equilibrium:
(A) $\ce{CH3-CH=CH-CHO}$
(B) $\ce{CH3-CH=CH-CH3}$
(C) $\ce{cyclohexyl-CH=CH-cyclohexyl}$
(D) $\ce{CH3-CH=NH}$

Could someone explain the answer step-by-step and the theory behind it?
I think that rotation of groups about a double bond can be possible due to conjugation of double bonds, so the answer should be (A). But I am afraid I haven't completely understood the logic behind it.

Option (D) is the answer. This may be predicted by redistribution of hybrid orbital theory or simply "Bent's Rule".
Clearly, during movement of the lone pair from one sp2 hybrid orbital to the other(to interconnect between the cis and trans isomers), there has to be an intermidiate transition state, during flipping when the lone pair resides in an unhybridized p-orbital. However the activation energy to achieve this intermidiate transition state is low for option (D).
Between N and H (as in the question), the electronegativity of N is higher than H. This pulls the shared electron pair in the N-H bond towards the N atom. In order to stabilize the high electron density, there is a redistribution of hybridization observed. The s character of the sp2 hybrid orbital of N (forming bond with the H ) increases ( eg from sp2 to sp1.8). This causes a corresponding decrease of s character of the other sp2 hybrid orbitals of N, such as the one containing the lone pair. The decrease in s character is associated with a a corresponding increase in p character of the lone pair containing orbital( eg from sp2 to sp2.3). Since it is easier to achieve transition from sp2.3 to unhybridized p orbital (having 100% p character), in the transition state it is easier to mediate between geometric isomer, possibly even existing in equilibrium.
For further reading see this answer:
Why are oxime geometrical isomers stable?


Why can't phosphorus trifluoride be used as a Lewis acid catalyst in electrophilic substitution of benzene? Which of the following cannot be used as Lewis acid in the electrophilic substitution of benzene?

  $\ce{BF3}$
  $\ce{SbCl3}$
  $\ce{FeCl3}$
  $\ce{PF3}$

My Attempt:

$\ce{BF3}$ and $\ce{FeCl3}$ are Lewis acids.
$\ce{PF3}$ though it has a filled s and p orbitals it can accept electrons from donors and expand its coordinate number up to six due to vacant d orbitals. It acts as a mild Lewis acid.
$\ce{SbCl3}$ has a completely filled s,p,d orbitals so it can no longer accept electrons, hence doesn't act as a Lewis acid.

My answer is option (2) but the answer provided is (4), where am I going wrong?

The best answer I can give is that there is enough space on $\ce{Sb}$ in $\ce{SbCl3}$ for an additional Lewis Base to interact with a nonbonding orbital on the $\ce{Sb}$ atom. While $\ce{P}$ can be hypervalent, the LUMO in this case would be $\ce{P-F}$ $\sigma^\ast$, which is really high in energy (meaning that it doesn't easily accept electron density from other species in solution).


Relation between freezing point depression and vapor pressure My lab manual defines the freezing point as "The temperature at which the vapor pressure of the substance in its liquid phase is equal to its vapor pressure in the solid phase." It goes on to explain why adding a non-volatile solute to a solvent, which lowers the vapor pressure compared to that of the pure liquid, lowers its freezing point.

What does the freezing point, which concerns liquid and solid phase, have to do with vapor pressure? If I do an experiment in a gas-tight syringe containing liquid and solid only, there will be no equilibrium with the vapor phase, but the freezing point will still change.

A freezing point is a temperature at which solid and liquid phases are in equilibrium with each other. In equilibrium the chemical potentials of each component in all phases present are equal. It means that if there are three phases in the system, solid, liquid and gas, then the chemical potential should be equal in all the three. As a consequence, the vapor pressure of a given component over the liquid phase should equal that over the solid phase. But if there are only solid and liquid phases, I see no need to involve gas. The basic derivation, which I was taught, of the freezing point depression of liquid solvent A (when the solute is insoluble in solid A) starts with $$\mu_\mathrm{A}(\mathrm{solid})=\mu_\mathrm{A}(\mathrm{solution})=\mu_\mathrm{A}(\mathrm{pure\,liquid})+RT\ln{x_\mathrm{A}}$$ - no gases here.

I might miss something here, but my guess is that your lab manual is just trying to be more obscure than it ought to be (no offence to its authors intended). While the expression for $\mu_\mathrm{A}(\mathrm{solution})$ itself can be derived from the Raoult's law, which requires gases, that's slightly different, although related, story. Provided that the expression for $\mu_\mathrm{A}(\mathrm{solution})$ is already known, I can see no sound reason for inviting the vapor pressures to explain cryoscopy.


Changing Temperature of a Galvanic/Voltaic Cell: Should I wait for the electrodes to also change in temperature? I'm conducting an experiment where I change the temperature of the sulfuric acid in a lead acid storage cell. My one question in controlling the variables is that should I let the electrodes also change in temperature, or should I keep them at room temperature and measure the cell potential? Does it even matter at all?

What I'm guessing is that a change in temperature affects the resistance of a metal, which would then affect the ability of the electrodes to transfer the electrons involved; therefore, the electrodes should be kept at room temperature.

If I should do one or the other, could someone explain to me why?

In general the higher the temperature of the electrolyte, the more power you can draw from the cell.

The reason for this is, because the ions of the electrolyte can move more freely when the temperature is higher and chemical reactions do in general happen quicker with increased temperature.

On the other hand side the hotter a metal gets, the bigger its resistance will become. This is because electrons can’t move as freely when the temperature is increased.

However this effect usually is of importance when temperatures are quite high, like in an light bulb.

If you just heat the electrodes for 30-40°C, I doubt you would be able to even measure the difference in resistance (even though it does exist).

So to sum it up, the output power of a cell in relation to the temperature will mainly be altered by the temperature of the electrolyte.


Effect of nitro group basicity order of o,m and p nitroanilines? Would someone please help me with this? The order of basicity of o,m and p nitroanilines is m>o>p. My book states that ortho isomer has -I effect, -M effect as well as ortho effect, and meta isomer has only -I effect while para isomer has -I and -M effect. These effects destabilize the conjugate acid.

My actual doubt is why exactly the nitro group has a negative mesomeric effect on ortho and para isomer, when actually it is a meta deactivator. Shouldn't it be having no effect at ortho and para positions, and destabilize the conjugate acid at meta position? I'm confused.

First we have to understand what affects basic nature of a compound.

Tendency to donate lone pair
Electronic effects— Inductive, resonance etc
Factors stabilizing conjugate acid (or acidity of conjugate acid
Nature of solvent

Tendency to donate lone pair is affected by the acidity of conjugate acid and electronic effects.

Grab a pen and paper and draw conjugate acids of

$$2-nitroaniline$$
$$3-nitroanliline$$
$$4-nitroaniline$$

And you know that electron withdrawing inductive (–I) effect increases acidity and electron withdrawing resonance (–M or –R) also increases acidity.

In conjugate acid of

$$2-nitroaniline$$
You will find there is only –I inductive effect and there is no –M effect of $-NO_2$ group because of SIR (In SIR - STERIC INHIBITION OF RESONANCE - the nitro $-NO_2$ group shows steric hindrance to $-NH_3^+$ group and the p–orbital of Nitrogen (in $-NH_3^+$ group) moves out of plane of p–orbital of Carbon (in benzene ring) which inhibits resonance or does not allow resonance to happen). So in conjugate acid of $2-nitroanliline$ there is only –I effect which increases acidity.
$$3-nitroanliline$$
You will find that there is no –M effect obviously!! But there is –I effect of nitro ($-NO_2$) group which increase the acidity of conjugate acid of $3-nitroanliline$ but it is less than what we have seen in $2-nitroanliline$ (due to distance factors). Therefore conjugate acid of $2-nitroanliline$ is more acidic than conjugate acid of $3-nitroanliline$
$$4-nitroanliline$$
You will find there are both –M and –I effect which increase acidity of conjugate acid of $4-nitroanliline$. So conjugate acid of $4-nitroanliline$ is more acidic than conjugate acid of $3-nitroanliline$ or $2-nitroanliline$

Conclusion
Order of acidity of conjugate acids
$$conj.\ acid\ of\ 3-nitroanliline<conj.\ acid\ of\ 2-nitroanliline<conj.\ acid\ of\ 4-nitroanliline$$

Since the order of basicity of 'orginal' compounds is just opposite of order of acidity of conjugate acids, so answer to your question is as follows
$$3-nitroanliline>2-nitroanliline>4-nitroanliline$$
Or m>o>p


What are the product(s) of red phosphorous when burned? A match box lighter strip typically consists of powdered glass and red phosphorous. When this part is burned, it's residue is like a sticky dust that, when applying friction, will release smoke or vapor. You can do it with you fingers and it looks really cool.

So what's the chemistry behind this?

The major product formed is -  Tetraphosphorus decaoxide ($\ce{P4O10}$)

Infact you could have googled it. There are many interesting videos on youtube.

Check this out -  Burning of Red Phosphorus


Is it possible for a substance to absorb visible light but still appear white? This is a UV-vis spectrum for zinc oxide

Your eyes are not sensitive to UV light so any absorption there will be invisible to you. The visible spectrum you give has fairly uniform absorbance across the visible. The whiteness you see is primarily due to scattering of white room lights or sunlight off the powdered zinc oxide. Light is scattered when there is a change in refractive index between two mediums, in this case air and zinc oxide. The effect is enhanced when the crystals are of a very irregular shape and so have many different scattering surfaces. Sugar and salt show the same effect when powdered  even though large crystals of each of them can be made that are clear and colourless. As the oxide has a more or less constant absorption no wavelength is very much more absorbed than any other, or alternatively, all wavelengths are equally scattered.

If you can find a solvent with similar refractive index to the oxide (and in which it does not dissolve) then the crystals will become far more transparent and effectively disappear. It is easier see this effect if you wet a piece of frosted glass with water as it will then become far more transparent. Alcohol or benzene would be better as they have refractive index closer to that of many types of glass. Zinc oxide has a value of 2 higher than any common solvent, such as diiodomethane approx 1.7.


What conditions to use to crystallise Equine Myoglobin? What are the crystallisation conditions for Equine Myoglobin?

I've tried repeating some from the PDB but no success.

Does the 1958 Nature paper by Kendrew include any useful details? (Sorry, I'm behind a paywall right now).


Effect of reduction potential on particle size of chemically precipitated silver particles Is there no mathematical relationship between redox potential and particle size saying that an $X$ potential should yield a $Y$ particle size? Right now its general that large redox potential results in smaller particles, but I want to know if, for example, for silver particles I use $\ce{AgNO3}$ with glucose reducing agent and say the redox potential is $\pu{0.33V}$, what the ballpark for the particle size should be based on that potential (assuming $\mathrm{pH}$ does not change)?

Here is an example of a study of fish oil nanoemulsions which examines, to quote:

  The influence of surfactant-to-oil-ratio on particle size and physical stability was evaluated.

Here is yet another study to quote:

  Although the role of surfactant in the synthesis of particles in the nanometer range has frequently been documented...The surface coverage per surfactant molecule could be related to the surface tension of the latex, thus providing a relationship between particle size and stability.

And, another more expansive work to quote:

  It was reported that the formation process of these NPs could be divided into three stages: growth, incubation and Oatwald ripening stages. The higher boiling point of 300 °C of paraffin affords a broader range of reaction temperature and makes it possible to effectively control the size of silver NPs by varying the heating temperature alone without changing the solvent. Moreover, the size of the colloidal silver NPs could be regulated not only by changing the heating temperature, or the ripening time, but also by adjusting the ratio of oleylamine to the silver precursor.

While chemical reduction methods are connected to reduction potential, the multi-stage aspect of the creation process for NPs suggests only a partial correlation, so not a 'mathematical relationship', but perhaps a statistical one.


What does oxidizing strength mean? What does oxidizing strength mean?

Does it mean the strength of a oxidation agent that is needed to oxidize the reduction agent, or does it just mean the tendency of a substance to lose electrons (oxidate)?

The strengths of oxidizing and reducing agents are indicated by their standard electrode potentials. Let's consider  standard potentials of some redox couples.

$ E^0(\ce{Li^+/Li})= -3.04V$,

$ E^0(\ce{K^+/K})= -2.92V$,

$ E^0(\ce{Cu^2+/Cu})= +0.34V$,

$ E^0(\ce{F_2/F^-})= +2.87V$

The values above are reduction potentials, so lithium at the top of the list has the most negative number, indicating that it is the strongest reducing agent. The strongest oxidizing agent is fluorine with the largest positive number for standard electrode potential.


Why left and right circularly polarized light have different index of refraction in chiral media? This is a followup question of (Dependence of the angle of rotation on the wavelength of plane polarized light).

Ron's answer tells about the difference between left and right circularly polarized light. But the explanation isn't in the answer.

So, what's the reason behind different index of refraction?

The answer to this is really a combination of the two answers given in the question you linked.

Namely, chiral molecules are unique in that they make a distinction between left and right circularly polarized light.

As porphyrin pointed out, the index of refraction is dependent on the ability of a dipole to be induced in a molecule by electromagnetic radiation. Linearly polarized light is itself achiral, thus any given wavelength of linearly polarized light has only one magnitude of dipole it will induce in a molecule, all else being the same. Circularly polarized light, on the other hand, is itself chiral. Thus, when we bring two chiral objects together, the molecule and the circularly polarized light, there is no reason at all to suspect the dipole induced in the molecule should be the same in both cases. The symmetry necessary to make such a claim simply is not present.

Thus, if you imagined tabulating the index of refraction for some molecule as a function of wavelength using circularly polarized light, you would need to have two columns. One would be for right-polarization and the other for left-polarization.

Also, because it is relevant to the answer, I will reiterate what Ron was saying in the comments. The photon is an inherently quantum object. It is a boson and thus has two possible values of spin angular momentum, $\pm\hbar$. These two values correspond to two allowable states the photon can be in, one which is polarized to the right and the other which is polarized to the left. That is, the electric field vector spins in a circle one way or the other. There also allowable solutions to the state of a photon which are some linear combination of these two polarization states. A 50/50 linear combination gives linearly polarized light. Thus, it seemed like you might have been picturing two photons whose polarizations were cancelling each other out. This is incorrect. Rather, you have to think of a single photon in a superposition of two polarization states.


calculate the number of shots of Captain Morgan’s rum I am facing a challenge, and this is my first question in this stack exchange. I have worked out the following problem but the answer I got (number of shots) is ridiculous. Could any of the chemist here please advise me where I went wrong?

Question: Assume that you are on vacation in Florida and enjoying the company of friends at a bar. Using the information in Table 80-1 in your text, calculate the number of shots of Captain Morgan’s rum required to take your blood ethanol concentration above twice the legal limit to drive. [Note: The legal limit in Florida is 80 mg/dL]. Assume that the shots are consumed in rapid succession (the wisdom of which you were later unable to explain). Show your complete calculation, including all inputs and assumptions.

My calculation:

Table reference

Let us assume that my binge-drinking has led to a blood alcohol concentration (BAC) of 162 mg/dL.

The formula for BAC is as follows:

`BAC = A/(R x W) – (0.15 x H)` where

A = weight of pure ethanol consumed (g)
R = Widmark’s rho factor (0.68 and 0.55 L/kg for men and women, respectively)
W = body weight (kg)
0.15 = average ethanol elimination rate in humans
H = drinking period (hr)

Inserting the values of 162 mg/dL for BAC, R = 0.55 L/kg, W =  50 kg, and H = 1 hr, the equation becomes:

162 = A/(0.55 x 50) – (0.15 x 1)
162 = A/27.5 – 0.15

Therefore,

A = 27.5(162 + 0.15)
    = 4459.125 g

According to table 80-1 in the book, one shot is 30 mL and contains 40% ethanol.
So one shot contains 12 mL ethanol.
Density of ethanol = 0.789 g/cm3
The volume of pure ethanol which I consumed = 4459.125/0.789                                                                       = 5651.62 mL
So, no. of shots consumed = 5651.62/12 = 471, over the course of 1 hour

Conculsion: 471 shots of Captain Morgan Rum is absurd for any person. So I think I am missing something in my calculation. I'm not able to figure out what it is.

As Zhe mentioned, you have inconsistent units in your calculation, so your numerical results are not physically meaningful.

Take for instance this expression:

162 = A/(0.55 x 50) – (0.15 x 1)

Here, you have different units for each term; from the values you supplied, the units are

162 mg/dL = A/(0.55 L/kg x 50 kg) - 0.15 g/(L•h) x 1 h

[Note that the units for the elimination rate weren't stated in the question, but a quick search of the literature gave studies measuring it at about 0.01-0.02 g/(dL•h), so I infer the 0.15 value refers to g/L]
If we use uniform units throughout, we get

1.62 g/L = A/27.5 L - 0.15 g/L

from where you get a more reasonable value for A:

A = (1.62 + 0.15) * 27.5 = 48.675 g

Each shot contains

V = 30 mL * 0.40 / 0.789 g/mL = 15.2 g

of ethanol, so you need to drink 48.675 / 15.2 = 3.2 shots of rhum over the course of an hour.

I'd suggest, however, that drinking shots in "rapid succession" implies a negligible drinking time (∆t ~ 0 h). In that case, A = 44.5 g and you need about 3 shots (2.93).


Meaning of depiction of atomic orbitals There was a section about atomic orbitals in my organic chemistry textbook that I did not quite understand. First the author explained a bit about the Schrödinger equation, which I understood to be (vaguely, they did not go to any detail, even omitting the formula all together) a equation that describes the likeliness of positions of an electron in 3d space. $\Psi$ (i.e. the Schrödinger equation) can be solved for different quantum numbers that respectively represent the different orbitals.

This made me wonder what a picture (like below) of an orbital actually depicts, as it are chances that are described by $\Psi$, not boundaries. My best guess was that such picture represents an area that has a high chance, say 90% or more, of containing the electron (how they determine the shape, I don't know, but my best guess again here is that they only take those parts of the real orbital that add a significant percentage to the total chance of the electron actually being in the depicted orbital).

Are all these assumptions right?

Without going into a long discussion about what the wave function is, I will try to briefly answer your questions:

1) For atoms, the (approximate) solution to $\Psi$ is the product of two functions:

the radial components (distance from the nucleus)
the angular/spherical components (behavior going around the nucleus, like latitude and longitude)

2) Since the radial components of each component extend quite far, decaying to zero only at infinity, it would be very difficult to make a plot of the 3D object on a 2D surface that does not cover up the features that would be nearer the nucleus (further from eye), without using varying degrees of transparency.

The spherical component is usually solved analytically using complex numbers (i.e. a+b*i) but they can be transformed into real values by linear combinations. So, instead of two "channels" for real/imaginary, we can use only one "channel" (real) for each function, 1s, 3dxy, etc.  Often, these are colored so that if the sign of the real function is positive, it is e.g. red, otherwise, e.g. blue.

A second feature of these types of graphics is called the "isosurface"--which means, plot only if the value of the function is equal to a chosen value.  Sometimes the chosen value is taken to be "90% of the corresponding electron density contribution from this atomic orbital is enclosed" (which, b.t.w. is $\Psi^* \Psi$ or simply $\Psi^2$.  Though $\Psi$ may be real/complex/negative/positive valued (hence need all sorts of colors and "channels" to represent it) the electron density is positive and real.  This is very important, as complex quantities are not observable, but are rather mathematical constructs (vehicles) for arriving at things that are actually observable, such as the electron density.

But practically speaking, in order to see certain features like the doughnut in $dz^2$, people play a little fast and loose with precisely what the value of the isosurface is.

The shading on the surfaces that you have provided could be misinterpreted, because the colors do not correspond in any way to the value of the function being plotted, but rather seem to be an aid to the volumetric rendering of an isosurface that has already been set to either the absolute value or the square of the atomic orbital.

Finally, note that in your center drawing, which adds all of the individual atomic orbitals, that you are beginning to form a sphere.  It had better, as atoms are, in fact, spherical!${}^{\dagger}$  The formal name of the functions are spherical harmonics, and the set of e.g. $p_x$, $p_y$, $p_z$ or the five $d$'s taken together form a spherical shape.

${}^{\dagger}$ assuming the nuclear quadrupole is zero


Why is tetracyanonickelate square planar? Why is tetracyanonickelate, $\ce{[Ni(CN)4]^{2-}}$ square planar?

$\ce{CN}$, for being a strong field, would not follow Hund's rule.
the central atom $\ce{Ni}$ would have a +2 charge so its configuration would change from $\ce{3d^8 4s^2}$ to $\ce{3d^8 4s^0}$
Using the statements above and using crystal field theory I get its hybridization as $\ce{dsp^3}$, which suggests that its shape must be trigonal bipyramidal while it is not.

This is the square planar splitting diagram

Many d8 metal complexes are usually square planar because a very stable electron configuration is achieved if you put the electrons on this splitting diagram -6 electrons stabilizing the compound and only two slightly destabilizing electrons-

Now you could ask, why then not d6 and d7 adopt this cinfiguration? Because for 6 or 7 electrons, an even more stable state is achieved with octahedral fields.

About hybridization... Forget about it when talking about coordination complexes. It's completely useless. It can't explain or predict anything that CFT/LFT can't


Birch reduction of 1,2-Cyclopentanedione and Diketones This answer given to this problem was that only one of the double bonds is reduced into an alcohol. I
think i understand how the double bond is reduced into an alcohol. An electron is transferred to the carbonyl carbon and the double bond cleaves homolytically, forming a carbanion (stabilised by resonance ) and the oxygen with an extra electron which then will gain another electron and both get hydrogens to form the alcohol.
First of all is my method right? I don't think so as both the double bonds would have been reduced into alcohols. Also i was unable to see any sources for the birch reduction of di-ketones, I was only able to find it for unsaturated ketones.

The course of the reaction depends on whether or not a proton source other than ammonia is present. In the absence of an alcohol (ethanol, tert-butanol) the α-dione 1 adds an electron from sodium to give a resonance stabilized radical anion 2. A second electron is added to afford enediolate 3. [This is the same species formed with sodium in the acyloin condensation on diethyl glutarate.]  Water workup provides the α-hydroxyketone 4. In the presence of alcohol, enolate 3 is protonated to form 5. [The reason I am reluctant to protonate the alkoxide of 5 is to avoid reductive elimination of hydroxide from 4 to leave the enolate of cyclopentanone. Further protonation/reduction may lead to the alkoxide of cyclopentanol.]  One-electron reduction of 5 yields ketyl 6, which via protonation and a second reduction leads to dialkoxide 8. Workup affords the diol 9.
The bottom line is if you want to form 4 by metal/ammonia reduction, avoid alcohol.