when using glass to glass stoppers, is it best to have them dry or use lubricant? I have some glass flasks that have glass stoppers. It’s basically the typical glass stopper with the frosted tip which fits into a glass opening. Basically think of a typical glass volumetric flask. i attached a pic of my flasks with the stopper. they are not threaded connections. i believe they slight self lock. like i put the stopper in and push down and turn to get it tight.
So my question is, if I am travelling/moving theses flasks around in a situation where they may tip over or be upside down for some time with fluid inside them , is it better to have the glass to glass contact between the glass stopper and the glass opening fully dry? Or is it better to add some oil to the glass stopper?
If I do glass to glass and they both are dry, it’s usually pretty sealed however I believe I have always been told to use some type of oil as it helps it seal better but I tbink I am against using oil becusse I feel like the stopper has a higher chance of coming off or slightly moving causing a leak becusse of the slipperiness of the oil.
*the connection is not threaded, it is just flat like a typical glass volumetric flask”
I of course will be using parafilm/tape on top so I am just wondering which method is best so no liquid leaks out. There will be no gas build up, its just liquid like water or alcohol sometimes but i dont care if they slightly evaporate. i just dont want leaks
Basically I am looking for the best method to reduce leaks
Thanks

Sometimes dry glass-to-glass sealing of the stopper leads to seizing of the stopper and can become very hard to remove. Therefore vacuum grease is often applied lightly. However, if your material in the flask can dissolve the vacuum grease, the material can become contaminated by the vacuum grease. If your material is volatile, you will need the vacuum grease to ensure an airtight seal. Another method you can use is to use teflon tape between the stopper and the ground glass joint to make a seal that will not seize. But, vacuum grease will give you the best seal.


Synthesis of a bicyclo[2.2.2]octane from benzoic acid I think that it is a Diels Alder reaction, but I don't recognize the start materials!
First I make the tertiary alcohol from carboxylic acid with organo-lithium reagent; after I try to make a Diels Alder reaction, but I don't find the right conjugated diene and dienophile.

Supposed that the synthesis does involve a Diels-Alder reaction, it is nowhere written that it is the last step in the sequence.

Take another look at your product and imagine a $\ce{C=C}$ double bond somewhere. Does it now look more like a Diels-Alder product? The $\ce{C=C}$ can be turned to $\ce{C-C}$ using... (You figure that out).

Work your path back from here and try to identify diene and dienophile.

Don't look until you've figured out the diene ;)

   Convert acid to ester, 2 x $\ce{MeMgBr}$, Diels-Alder reaction with vinylmethyl ether, hydrogenate double bond. You didn't cheat, did you? ;)


How to detect lead paint on a bicycle? Recently I started renovating a bicycle. The frame has been painted probably over 30 years ago, and despite holding well, big patches of rust are showing. So it's time to give the bike a new paint job. And before that, I need to sand the frame and remove the current paint.

Sadly it came to my attention that some bikes around that age have lead paint. Toxic, obviously.

I'm by no means a pro chemist. Far from that actually: if you show me a bunch of formulas, I might understand it, but I'm not skilled for manipulations neither properly equipped.

Provided that I can scratch a bit of that paint with sandpaper, what's the easiest way to test it for presence of lead? Apparently some lead test kits exist, but they're incredibly hard to find here.

Two chemicals are used in lead test kits: sodium sulfide or a rhodizonate salt. Sodium sulfide should be easy to obtain (or synthesize), and is useful for light-colored paint, as it produces a gray or black color change. Potassium rhodizonate can be synthesized from inositol, but you might be better off getting sodium rhodizonate, e.g. from Sigma-Aldrich for ~US$35 for 5 grams.


Reaction of methacrylic acid with BH3 If methacrylic acid (2-methylprop-2-enoic acid) is reacted with first $\ce{BH_3/THF}$ and second  $\ce{H_2O_2/HO^{-}}$ then $\ce{OH}$ group is attached at less subtituted carbon.

But if it is treated with $\ce{CH_3COOH}$ in the second step, what is the end product?

It's simply a protonolysis:

Except in the picture D is used instead of H. Hydroboration based reactions always proceed via this boron-ate complex (negatively charged) then migration and loss of leaving group. Search google for carbonylation, amination, cyanidation using boron.


Determining central atom in NCl3 Less electronegative atom is the central atom. $\ce{N}$ is more electronegative than $\ce{Cl}$ as $\ce{Cl}$ has larger atomic radius. So why in $\ce{NCl3}$, $\ce{N}$ is the central atom?

Cl is slightly more electronegative according to the tables I've seen: http://en.wikipedia.org/wiki/Electronegativity

But I wouldn't say it's an absolute rule that the less electronegative atom is central.  For example, $\ce{NH3}$ has the N central and $\ce{Cl_2O}$ has the O central.


How to check whether intramolecular hydrogen bonding takes place in cyclic compounds? How can we confirm or have a clear idea whether hydrogen bonding is taking place in substituents of cyclic compounds? For example, it takes place in ortho-nitrophenol, but it doesn't take place in ortho-flurophenol despite the fact that the nitro group and the fluorine are present at the same position and both are highly electronegative.

From a structure diagram alone it is hard to determine these interactions. The reason for this is that important structural parameters are stretched or shortened in order to flatten the molecule for a 2D drawing.

In a first order approximation a molecular modelling kit based on balls and sticks could already be very helpful. It'll let you approximate the 3D structure and get a 'feeling' (also known as chemical intuition) about where to expect close contacts, or what bonds are strained, etc.; just to play around with it will give you a better visualisation of the shapes of molecules in general.

More advanced is using computers. While this might seem intimidating at first, it is actually something you can do at home. I've previously written extensively about how to do these identifications in this Q&A: How to identify hydrogen bonds and other non-covalent interactions from structure considerations? There are quite a few programs out there that let you build ball and stick models of compounds using tabulated bond lengths. That is probably already enough to identify likely positions for these interactions. Many of these molecular editor programs can run optimisations on a quick but low level, too. My personal favourite is Molden (very lightweight), but Avogadro does the job equally well, and probably has a nicer user interface. (Geoff, if you read this, we are all still hoping for a major update.)

Obviously you can increase the level at which you are running the calculations; for more complex structures you may need to. A lot of this can still be done with a home computer. You can start out with semi-empirical models, of which I recommend the fantastic open-source xtb, go on to ab-initio wave function methods (esp. HF), climb Jacob's ladder (AIP Conference Proceedings 2001, or see these slides by Perdew via helper.ipam.ucla.edu), and finally make your way to the various post-HF methods. (the last one you probably don't want to do on a laptop computer.) In this context I would like to point you towards Orca.

For the two molecules you mentioned, I've done the analysis with xtb and Multiwfn.

With these two molecules it is still quite simple, because they are flat (mostly at least; they are of course not static). Here is a model of the 3D structures with the hydrogen bond measured. For ortho-fluorophenol it is 2.25 Å and for ortho-nitrophenol it is 1.73 Å. You can see a clear difference already.

And here are the non-covalent interactions from Multiwfn analysis (Option 20, 2, visualised with ChemCraft at contour value 0.2 and in range ±0.05):

The red coloured blobs correspond to repulsive interactions, while the green and blue coloured values correspond to attractive interactions. As you can see, there is no hydrogen bonding in ortho-fluorophenol, but there is some in o-nitrophenol.


Why are NaCl and NaBr not soluble, whereas NaI is soluble in acetone (in the Finkelstein reaction)? My book states that:

  Finkelstein reaction is particularly useful for preparing iodoalkanes. The iodoalkanes are obtained by heating chloro or bromoalkanes with a concentrated solution of sodium iodide in dry acetone.
  $$\ce{R-X~+~NaI->[\ce{acetone,~reflux}]R-I~+NaX}$$
  $(\ce{X=Cl,~Br;~R=alkyl ~group})$
    Sodium chloride and sodium bromide being less soluble in acetone get precipitated from the solution and can be removed by filtration. This also prevents the backward reaction to occur according to Le Chatelier's principle. The reaction gives the best results with primary halides.

My question:

Why are $\ce{NaCl}$ and $\ce{NaBr}$ not soluble, whereas $\ce{NaI}$ is soluble in acetone?
All we can observe is that, sodium iodide has greater molecular weight than sodium chloride or sodium bromide. But, how do this account for solubility of sodium iodide?

The solubility of a salt in a solvent is for a good part determined by the difference of stability in the dissolved state (solvated ions) vs. the solid state (ions in crystal lattice). The different solubility of the sodium halides in acetone can be explained by the crystal structures.

With iodide, the solvate $\ce{NaI \cdot 3(CH3)2CO}$ can be formed, in which each $\ce{Na+}$ is coordinated by 6 acetone ligands via oxygen lone pairs, and the $\ce{I-}$ ions fill in the gaps between these octahedral units, being surrounded by the methyl groups of the acetones. $\ce{I-}$ has a larger ionic radius compared to $\ce{Cl-}$ and $\ce{Br-}$ and thus fills the gap completely, thereby minimizing empty space in the lattice and its distance to $\ce{Na+}$, which is, however, still larger than in solvent-free sodium iodide. This larger distance between the ions decreases the lattice energy, which in turn makes the compound readily soluble in acetone.

$\ce{Cl-}$ and $\ce{Br-}$ are too small to fit in these gaps, and as the cation-anion distance becomes too large, the only way to minimize the lattice energy is the formation of a solvent-free structure $\ce{NaX}$ (X = $\ce{Cl-}$, $\ce{Br-}$). When the lattice energy is larger than the energy released by solvation of the ions, namely that of $\ce{Na+}$ by complexation with acetone, the compound becomes insoluble and precipitates from the solution (source).


Chemical reaction and shift in magnetic susceptibility sign I'm trying to find an easy experiment involving a chemical reaction in which two paramagnetic reactants A and B react to give a product which is diamagnetic (or the reverse). The idea is to show that A is paramagnetic, B is paramagnetic, but A+B --> C is diamagnetic.

Or, another experiment: an experiment in which a change in temperature changes paramagnetic to diamagnetic (or the reverse). I've seen such an experiment but cannot remember the details. All I can say is that there was a color change and, at the same moment, a paramagnetic/diamagnetic shift

Seems to me any reaction involving the recombination of two radicals leads to a closed-shell diamagnetic species (which oxygen is definitely not!). I can't think of any other examples immediately. You could make singlet oxygen, which is diamagnetic, from paramagnetic triplet oxygen by irradiation, but then there is only one reactant.

If you heat a paramagnetic substance enough so that the thermal energy can overcome the energy that is gained by the alignment of your magnetic moments, I believe the substance will not exhibit paramagnetism in a magnetic field above that temperature. So that seems doable.

Edit: Actually, I found an example: ozone is diamagnetic and can be made from paramagnetic oxygen and an oxygen radical. So that could be a possible reaction where you have two paramagnetic reagents combining into a diamagnetic product.


What defines matter's phase at room temperature? So, a simple question: what makes neon a gas; what makes water liquid; what makes plutonium-238 a solid at room temperature? Of course, the phase is determined by the excitation of the molecules, but what determines that excitation?

It's all in the intermolecular forces (and the air pressure). Intermolecular forces are the forces of attraction and repulsion between molecules to neighboring particles. To understand how they determine physical properties like enthalpies of vaporization, fusion, etc..., you need to think about what states of matter are and how they affect the molecules. When you heat up water to a temperature (average kinetic energy) of 90 °C, the energy within the molecules isn't strong enough to overcome the intermolecular bonds (and the atmospheric pressure above the sample) and become a gas. However, it does have enough energy to overcome the bonds to a certain extent and thus can remain in a liquid state instead of in a solid state. This principle applies to any substance. In order for there to be a change in state of matter, the energy within the molecules has to overcome the intermolecular bonds and atmospheric pressure.

As mentioned several times above, another factor is the atmospheric pressure. A common experiment is to put water in a vacuum and watch it boil, something many first-time-experimenters wouldn't expect. The reason water is able to vaporize at room temperature is because it doesn't have the atmosphere to prevent it from vaporizing and therefore, in order to reach an equilibrium, the water is able to vaporize with much less energy than normally required at STP (standard temperature/pressure).

As a direct response to your question about neon, the reason that the noble gases are usually gases unless at extreme temperatures has to do with the weak intermolecular forces they have (van der Waals forces, or more specifically, london dispersion forces). Because the noble gases don't usually bond intramolecularly with anything else, including themselves, and have full energy levels, they have extremely weak intermolecular forces allowing them to almost always be in the gas phase.

I've listed some helpful links below if you want to take your research to the next step.

http://en.wikipedia.org/wiki/Intermolecular_force
http://wps.prenhall.com/esm_hillpetrucci_genchem_4/16/4219/1080299.cw/index.html
http://www.youtube.com/watch?v=A30lguyG0WM
http://chemed.chem.purdue.edu/genchem/topicreview/bp/intermol/intermol.html
http://en.wikipedia.org/wiki/Van_der_Waals_force


Why does p mean −log(x)? $\mathrm{pH} = -\log[\ce{H+}]$ while $\mathrm{pOH} = -\log[\ce{OH-}]$

Why does $\mathrm p$ represent $-\log$ of something? Is it due to some historical reason or due to some scientific reason?

According to the international standard ISO 80000 Quantities and units – Part 9: Physical chemistry and molecular physics, the symbol $\mathrm p$ is interpreted as an operator $(\mathrm px=-\lg x)$. It is used in particular (but not only) in the definition of the quantity $\mathrm{pH}$:

$$\mathrm{pH}=\mathrm pa_{\ce{H+}}=-\lg\left( a_{\ce{H+}}\right)=-\lg\left(m_{\ce{H+}}\gamma_{\mathrm m,\ce{H+}}/m^\circ\right).$$

where $a_{\ce{H+}}$ is the activity of $\ce{H+}$ in solution and $\gamma_{\mathrm m,\ce{H+}}$ is the activity coefficient of $\ce{H+}$ on the molality basis at molality $m_{\ce{H+}}$. The standard molality $m^\circ$ is chosen to be equal to $1\ \mathrm{mol\cdot kg^{-1}}$.

The definition of $\mathrm p$ (and $\mathrm{pH}$) given by ISO is actually quoted from IUPAC Quantities, Units and Symbols in Physical Chemistry (Green Book).

In its current form, this definition can be traced back to a recommendation given in
  R. G. Bates and E. A. Guggenheim
  Report on the standardization of pH and related terminology
  Pure Appl. Chem., 1960, Vol. 1, No. 1, pp. 163–168
  http://dx.doi.org/10.1351/pac196001010163
which reads

  There already exists international agreement that $\mathrm{pH}$ should be written and printed on line in roman type. We recommend that, with the unique exception of $\mathrm{pH}$, the operator $\mathrm p$ (printed in roman) should denote $-\log_{10}$.
  For example
  $\mathrm p m_\mathrm H$ means $-\log_{10}m_\mathrm H$
  (…)
  where $m$ denotes molality (…)


What is the balanced chemical equation for a redox reaction between a lithium-cobalt battery and a graphite electrode? I'm doing a project for my last year of secondary school on lithium cobalt batteries and got confused about the equation.

This is the only equation I can find (source: www.science.org.au):

\begin{align}
\ce{LiC6 &→ $x$ Li+ + $x$ e- + C6}\tag{anode}\\
\ce{Li_{(1−x)}CoO2 + $x$ Li+ + $x$ e- &→ LiCoO2}\tag{cathode}\\
\hline
\ce{C6 + LiCoO2 &→ Li_xC6 + Li_{(1−x)}CoO2}\\
\end{align}

But I don't understand it at all! I figured it must be balanced, but the '$x$' part of the equation is beyond my level of understanding. Is there a simpler way of writing this formula for someone at my level? If not, can someone please explain what this equation means?

Below we have a simpler way of the $\ce{Li-Co}$ cell reactions.

Cathode half-reaction:

$$\ce{CoO2 + Li+ + e- <=> LiCoO2}$$

Anode half-reaction:

$$\ce{LiC6 <=> C6 + Li+ + e-}$$

Net-cell reaction:

$$\ce{LiC6 + CoO2 <=> C6 + LiCoO2}$$

Over-discharging supersaturates lithium cobalt oxide, leading to the production of lithium oxide.

$$\ce{Li+ + e- + LiCoO2 -> Li2O + CoO}$$

Similarly overcharging (up to $\pu{5.2 V}$) leads to following reaction:

$$\ce{LiCoO2 -> Li+ + CoO2 + e-}$$


Colliding particles and activation energy If two particles collide and they are under activation energy, why do they JUST “bounce apart”? Isn’t the activation energy steadily decreasing if those particles continuously collide?

I always viewed the activation energy as a "barrier" that prevents collisions from producing a reaction, if sufficient energy is not met. I thought that if particles collide and they do not have enough energy to cause a reaction, the "barrier" gets weakened. Therefore, I found it surprising that this "barrier" doesn't get weakened, and the activation energy isn't reduced.

This is the nature of the quantum world. An action, at some level, either takes place or does not... there is no half-way state. Another example is the photoelectric effect. Planck and Einstein explained the requirement for at least a minimum energy of a photon before it can raise an electron to a higher energy level. A million photons just under that energy will not* raise an electron to a higher level, but just a single photon of the required energy (or higher) can bump up the electron.

Caveat: There is also the uncertainty principle, allowing electrons to tunnel to a higher level, or for multi-photon absorption to cause ionization, but they're not as  common as single-photon absorption and ionization.

If you're bothered by this non-intuitive behavior, you're not alone. Bohr, Bose, Boltzmann, Einstein, and others were forced to this theory through experimental results and the horrors of the ultraviolet catastrophe. Those who fail to appreciate the ultraviolet catastrophe shall be made, energetically, to walk the Planck.


What is the role of hydrochloric acid in a copper electroplating bath? Considering water as a given, the third most important ingredient in a copper electroplating solution, as a short survey of related recipe webpages will show you, is hydrochloric acid (HCl). Here's a sample link: http://www.thinktink.com/faqs/cupltfaq/cupltf01.htm

I can guess that the role of sulfuric acid (the second most important ingredient; the first being copper sulfate) is to provide you with sufficient anions to allow copper cations to easily enter the solution at the anode. But in almost all recipes that I have seen, a small amount of hydrochloric acid is added. I wanted to know how useful it is and what is the opportunity cost associated with it.

As already mentioned by @GM it indeed is used to take out $\ce{Cu^{2+}}$, but it is also used to take out $\ce{Cu+}$.

I know this because I once used a mixture of $\ce{HCl}$ and $\ce{H2O2}$ to etch copper boards, in that process both $\ce{Cu^{2+}}$ and $\ce{Cu+}$ were dissolved by the formation of $\ce{CuCl2}$ and $\ce{CuCl}$ with $\ce{Cl-}$ from $\ce{HCl}$.


How to calculate the pH of an acetic acid/ calcium acetate buffer? Problem:

  A buffer was made by mixing $\pu{500 mL}$ of $\pu{1 M}$ acetic acid and $\pu{500 mL}$ of $\pu{0.5 M}$ calcium acetate.

  What are the resulting concentrations of acetic acid (henceforth called $\ce{AcH}$), $\ce{Ca^2+}$ and acetate (henceforth called $\ce{Ac-}$, and what is the resulting $\mathrm{pH}$? $K_\mathrm{a}$(acetic acid) = $1.75\times10^{-5}.$

My attempted solution:

This problem gives me two equilibrium reactions to take into account.

\begin{align}
\ce{AcH + H2O &<=> Ac- + H3O+}\tag{1}\\
\ce{Ac- + H2O &<=> AcH + OH- }\tag{2}\\
\end{align}

In (2), I assume that calcium acetate completely dissociates in water. I also assume that the concentration of $\ce{OH- }$ from the auto ionization of water is negligible.

First I calculate the concentrations of $\ce{AcH}$, $\ce{Ac- }$ and $\ce{H3O+}$ in (1), ignoring what I assume to be minuscule amounts of $\ce{H+}$ from the auto ionization of water. To do this, I use the equation for the acid dissociation constant:

$$K_\mathrm{a}=\frac{[\ce{Ac- }][\ce{H3O+}]}{[\ce{AcH}]}\tag{3}$$

Initial values for $\ce{AcH}$, $\ce{Ac- }$ and $\ce{H3O+}$ are set to 1, 0 and 0 M, respectively. Equilibrium concentrations will then be $1-x$, $x$ and $x$ M.

Plugging this into (3), I get $x = 4.17\times10^{-3}.$

I have now found (or can at least easily calculate) the concentrations of the compounds in (1) at equilibrium.

On to (2), I know that $K_\mathrm{a}K_\mathrm{b}=K_\mathrm{w}$, and I have the values for $K_\mathrm{a}$ and $K_\mathrm{w}$, which gives me the value $5.71\times10^{-10}$ for $K_\mathrm{b}.$

Initial concentrations in (2) for $\ce{Ac- }$, $\ce{AcH}$ and $\ce{OH- }$ are 1 (because 0.5 moles of calcium acetate gives $2\times0.5=1$ moles of acetate), 0 and 0 M, respectively. Equilibrium concentrations will then be $1-x$, $x$ and $x$. Plugging this into (3) (though using the equation for the base dissociation constant instead), I get $x = 2.39\times10^{-5}.$

I have now found all the values I need to calculate the final concentrations and pH when adding the two solutions to each other, and up till this point I am fairly sure that all my calculations are correct.

Questions:

How do I write the chemical equation for the mixing of the two solutions? I tried the following; $\ce{AcH + Ac- <=> Ac- + AcH},$ but this doesn’t seem to help at all.
From my younger days, I seem to remember a much simpler and quicker method of doing these calculations than what I've shown here; does such a method exist? Please note that if it does exist I’d like to learn about it, but I would also like answers based on the method I've used.
Let’s say you mix one weak acid with one weak base, and they aren't conjugate pairs. How would the method differ from the one in this problem? I am specifically wondering about whether both the dissociation constants must be provided to be able to solve, and also how to write the chemical equation, but other observations are more than welcome as well.

In general: $\ce{acid + H2O <=> base + H_3O^+}$
It is called the Henderson-Hasselbalch equation
$\ce{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{\ce{[base]}}{\ce{[acid]}}$
I believe the equilibrium constant for a such a neutralisation would look like $K_n = K_\mathrm{a}K_\mathrm{b}\frac{1}{K_\mathrm{w}}$


Using RDKit to determine "nonring, nonterminal atoms" I am a non-chemist using RDKit to determine the number of "nonring, nonterminal atoms" in a molecule.
I can't find anything in the RDKit documentation that helps with "nonterminal". Can it be used for that?
I can using RDKit's AtomRings function to identify atoms in rings, but perhaps there is a more direct way?

I think you want to use SMARTS patterns to perform searches. These are something like "regular expression" patterns for molecular graphs.
At a rough guess, your query converts to [*R0!D1]:

* matches any atom (i.e., you don't care if it's a carbon or an oxygen)
R0 matches an atom in a ring with size 0 (i.e., not a ring atom)
D1 matches an atom with explicitly one bond (so !D1 matches atoms with more than one bond).

It's often helpful to use SMARTSview to test patterns.
More on RDKit's SMARTS support can be found in the documentation


What is the reasoning behind the assumptions of Transition State Theory? We were told in a first year introductory Thermodynamics and Kinetics lecture, and also from Wikipedia, that there are 3 main assumptions for the derivations used in Transition State Theory, but I'm not sure how or why they have been assumed?

Reactants are in constant equilibrium with the transition state structure.
The energy of the particles follow a Boltzmann distribution.
Once reactants become the transition state, the transition state structure does not collapse back to the reactants.

I'm concerned with 1. and 3. particularly, as it seems as though they contradict each other, because how can the reactants be in equilibrium with the TS, if the TS structure cannot revert back into reactants? How do we then utilise these assumptions, is it so that we can write
$$\ce{A + B <=> AB^‡ -> P}?$$

This is deeper discussed in Rate Constant Units and Eyring Equation, but I am going to post a very short take-home message here.
First, the three assumptions you have cited are not complete. They completely gloss over the most important one, i.e. nuclear and electron motion can be separated, energies (translation, vibration, rotation) can be treated additively.
Having gotten that out of the way, reactants are in constant thermal equilibrium with the transition state structure. This actually means that the reactant structure(s) is in equilibrium with all possible structures that can be reached from it. It is hence more correct to say that the reactants are in thermal equilibrium with all possible transition state structures.
This also basically ensures, that all these structures can be treated within a single, combined Boltzmann statistic, where the energy is distributed according to all the states of the assembly/molecule.
The next assumption is that the transition state will be crossed only once (point of no return), which means that when an assembly/molecule reaches this (high) energy state, it will form the product. This basically sets the limit to the Boltzmann statistic as all states following the transition state are not included anymore.
Putting both together we arrive at
\begin{align}
  \ce{A + B <=> &AB^\ddagger -> P}\\
  \ce{P <=> &AB^\ddagger -> A + B}.
\end{align}
However, you see that you can look at it from both channels, but that'll become to involved here. Please read the other post first.
All these assumption basically lead up to that the transition vector can be treated classical as translation.
In conclusion, the equilibrium part is necessary to write the combined Boltzmann statistic; the transition state collapses into products part is necessary to only include everything up to and including the TS, but nothing more into the statistic.

Also related: References for Transition State Theory and Quantum Transtion State Theory


Have we ever created artificial metals? I really don't know anything about chemistry, but I was wondering about the following vague question: has any substance been created that does not contain any metallic elements, but is metallic?  In fact, the only nonelemental metals I know of are alloys, so I may as well ask whether we have ever created any metallic compounds, with or without metallic constituents (without preferred).  Both Wikipedia and the tag wiki here claim that metallic compounds do exist but don't give examples.

My vague understanding of what makes an element metallic is that it has something to do with the sharing of electron clouds between atoms, and this question is motivated by wondering whether this kind of quantum effect can be induced from nothing in an appropriately constructed molecule.

Metals have several general properties in common (to some degree):

Luster - metals are shiny
Electrical conductivity - metals conduct electricity
Thermal conductivity - metals conduct heat
Ductility - metals can be drawn into wires
Malleability - metals can be beaten into shapes
Fusibility - metals can be melted and forged

All of these properties derive from the "metallic bond", a delocalized sharing of electrons throughout the metal lattice, which is the key necessity for displaying macroscopic metallic properties. All metals do not behave the same way for these properties. For example, gold is more malleable than iron.

Metalloids are those elements with one or more allotropes that have metallic characteristics. Few if any metalloids display all of the metallic characteristics (otherwise they would be metals!). For example, graphite is lustrous and electrically conductive, but not thermally conductive, malleable, ductile, or fusible. The metalloid with the most metallic behavior is tellurium, which is fusible and ductile.

As for compounds, many are like the metalloids - they have some properties of metals but not all. For example, silicon carbide is lustrous, but that's about it. Otherwise it is hard, brittle, and an insulator.

Polyacetylene and derivative polymers are sometimes call "synthetic metals" because of their electric conductivity and sometimes luster. However, again, they are thermally insulating. They also tend to burn instead of fuse.

Metalloid elements and compounds are not quite metallic because their electron delocalization is in only one (polyacetylene) or two (graphite) dimensions, not all three.

  Update!

Based on Snipergirl's comment, it appears that we are closer than I originally thought to preparing true synthetic metals. This highlight at phys-org suggests that supramolecular lattices of redox-coupled conjugated organic ions can exhibit metallic behavior at low temperature. The full article is at Nature Communications and has been made open access. These compounds appear to be close to achieving delocalization in all three dimensions. The behavior of this organic material is not surprising: metals and metalloids become even more metallic at cryogenic temperatures. The induced superconductivity allows for such cool things as magnetic levitation and MRI.


Size of dispersed phase particles in colloid? I am really confused about what the size of particles of dispersed phase in a colloid is: one of my textbook says its between $1 - 100nm$, another says its between $1-1000nm$, and Wikipedia says its between $2-500nm$. Is there any reliable source giving the exact information?

Its between $1-1000nm$ as stated on wikipedia in a paragraph name IUPAC definition.

  Colloid: Short synonym for colloidal system.
  Note: Quotation from refs.[3][4]
  Colloidal: State of subdivision such that the molecules or polymolecular particles dispersed in a medium have at least one dimension between approximately 1 nm and 1 μm, or that in a system discontinuities are found at distances of that order.[5]
  Note: Quotation from refs. [3],[4]

Edit: Since it is mention under IUPAC definition that size lies between $1nm-1000nm$, therefore this must be the correct size.


Is there a relationship between bond strength and polarity? I believe that a more polar bond would be stronger, since it would be similar to an ionic bond.

However, this seems to be wrong when applied to the study of acidity of some compounds.

In terms of acidity, ionic(polar) bonds dissociate more easily than covalent(polar) because the electrons are shared less equally in the ionic bond leading to easier breaking apart of the molecule. An acid for with an ionic bond will dissociate more easily leading to a stronger pH than an acid with a covalent bond.


How are non-carbon stereogenic centers named (S/R)? The R/S naming system for stereocenters relies on the Cahn–Ingold–Prelog priority rules to rank the four substituents of a stereocenter; the R or S name is then attributed depending on the spatial orientation of the ranked substituents. While this is mostly used for asymmetric carbon atoms, it holds true of other stereocenters with four substituents.

However, nitrogen and phosphorus atoms can also act as stereocenters if the inversion of their configuration is blocked (e.g. because they're in a cyclic molecule) or slowed enough (e.g. because of bulky substituents). Apparently, these stereocenters are also labelled in R/S fashion. Examples I found include Tröger's base:

(above is the (5S,11S)-enantiomer, below is the (5R,11R)-enantiomer),
and DIPAMP:

Having said that, I can't find any reference as to how the R/S naming scheme is applied to these tetrahedral stereocenters which don't have four substituents, but a lone electron pair and three substituents instead. I assume that the same scheme can be followed if one considers the lone pair as the lowest-ranked substituent, but I can't find any canonical reference as to that.

So, how are such stereocenters named? What authoritative reference can I find on this topic?

but I can't find any canonical reference as to that.

Look harder ;-)

It seems to be that the lone pair has lowest priority.

OK, I can't find any authoritative reference to that (I don't have access to the original paper), but I've got plenty of others:

Alright, IUPAC says:

  Except for hydrogen, ligancy, if not already four, is made up to
  four by adding "phantom atoms" which have atomic number zero
  (0) and are thus always last in order of preference.

  This has
  various uses but perhaps the most interesting is where nitrogen
  occurs in a rigid skeleton, as for example in a-isosparteine (14);
  here the phantom atom can be placed where the nitrogen lone pair
  of electrons is; then N-i appears as shown alongside the formula,
  and the chiraiity (R) is the consequence; the same applies to N-i6.
  Phantom atoms are similarly used when assigning chirality symbols to chiral sulfoxides (see example to Rule E—4.9).

When ligancy is less than four, we obviously have some lone pairs hanging around (except in Hydrogen--which has been excluded, and some other cases like carbocations--which are planar anyway)

So the lone pair becomes a "phantom atom" and has 0 priority.

Now I wonder what happens to these rules if you have two lone pairs, one of them muonic ;-)

Check the first revision of this post for my original, non-authoritative references


How does sodium in ethanol reduce carbonyl compounds? According to my teacher, Na in ethanol reduces carbonyl compounds (e.g. aldehydes and ketones) to alcohols. However, I can't see where the hydrogen needed for reduction comes from. How does the reduction take place? (or is my teacher wrong?)

Sodium is a powerful reducing agent. It wants to donate an electron to things that are electron deficient.

The mechanism of the reduction of aldehydes and ketones by sodium would be similar to that of the Bouveault-Blanc reduction. Ethanol is the source of the protons (at least early in the reaction). The dots in the reactions below signify radicals (unpaired electrons), which are represented to the best mchem and MathJax's ability (which is not great).

  Step 1 - one electron transfer from sodium to carbonyl group:

$$\ce{Na. + RCH=O->Na+ + RCH.-O^-}$$

  Step 2 - abstraction of a proton from ethanol:

$$\ce{CH3CH2OH + RCH.-O^- <=> CH3CH2O- + RCH.-OH}$$
This step is an equilibrium step, and is driven toward products by the second electron transfer (which consumes the $\ce{RCH.-OH}$.

  Step 3 - one electron transfer from a second sodium atom to the organic radical:

$$\ce{Na. +RCH.-OH -> Na+ + RCH:^{-}{}-OH}$$

  Step 4 - second proton transfer from ethanol:

$$\ce{CH3CH2OH + RCH:^{-}{}-OH -> CH3CH2O- + RCH2OH}$$

This reaction, while technically an acid-base equilibrium, essentially goes to completion because of the vast disparity in acidity of $\ce{C-H}$ and $\ce{O-H}$.

The net reaction is:

$$\ce{2Na + RCH=O + 2CH3CH2OH -> RCH2OH + 2CH3CH2ONa}$$
$%edit$


How can Hydrogen NMR be used to distinguish between these two isomers? I am having trouble explaining how and why the spectra of these two compounds would differ:

I am sure it has something to do with how the protons will couple. In the first isomer (left), the protons are anti-periplanar whereas in the second (right), I think they are staggered? So the first may be able to couple stronger than the second?

In the absence of any other configuration anchoring factors, I would not expect the conformation of the two structures to be reasonably stable at room temperature to be able to distinguish both by NMR. You may well need lower temperatures to freeze chair inversion and create a set of distinct, non-averaged peaks.

This issue aside, it is very easy to distinguish the two once you know you have a stable conformation. The key is the Karplus relationship which allows you to link a vicinal coupling constant to the dihedral angle between the two hydrogens. In general, a $^3J_\ce{HH}$ coupling constant will be very high ($^3J_\ce{HH} > \pu{9Hz}$) if the dihedral angle is either $0^\circ$ or $180^\circ$ — typically, if the hydrogens are anti-configured. On the other hand, the coupling constant becomes minimal for values around $90^\circ$ ($^3J_\ce{HH} < \pu{2Hz}$). For a $60^\circ$ gauche arrangement, a coupling constant of $^3J_\ce{HH} \approx \pu{3 Hz}$ is expected.

So the coupling between $\ce{H{1}}$ and $\ce{H{2}}$ would be around $\pu{9Hz}$ or $\pu{10Hz}$ in the first case but only around $\pu{3Hz}$ in the second — if the configuration is long-living on the NMR timescale.


Is (high) pressure required to form (quartz) crystals? I've been reading a lot of sources about the hydrothermal synthesis of quartz using autoclaves. I want to try doing this myself, but before I can I need to know the minimum pressure required.

If you need a refresher, hydrothermal synthesis uses steam as a solvent, and creates a supersaturated solution that places the quartz molecules onto a seed to grow a pure crystal.

I've gotten many different numbers in terms of how much pressure [from the steam] is required for the process, ranging from 7500 psi to 30000 psi. I want to work with as low of a pressure as possible.

So then my question is: is high pressure necessary for quartz crystal formation? My understanding is that the pressure in the case of hydrothermal synthesis is created from having large amounts of steam in order for more material to be dissolved in said steam, so the crystal will form faster. So is this the case, in which having a lower pressure will work, or is pressure required for the crystal to form properly? As in, does the pressure actually affect the crystal structure in some significant way?

Pressure does not affect the crystal structure (unless it gets insanely great, so that stishovite formation kicks in; that's not going to happen in an autoclave, though). Pressure, however, does affect the temperature (indirectly), and hence solubility, and hence the rate of crystal formation. There is no lower threshold, that's why you keep getting wildly different estimates for the "required" pressure. It is just that the higher you get, the faster your crystals grow. They would grow at normal atmospheric pressure as well, just very slow (so that only your great-grandchildren may possibly be able to see them).

I suggest that you aim at something less extreme. Calcite, maybe?


Is there intramolecular hydrogen bonding in salicylaldehyde? Is there intramolecular hydrogen bonding between the hydroxyl and carbonyl groups?

I think personally is that it does but what is concerning me is that I have got no adequate answer or response to my suggestion, so if anyone could help on this one, that would be great.

It can indeed hydrogen bond, this can (to a certain extent) be observed by NMR. Due to the internal hydrogen bonding, the phenolic proton is surprisingly consistent, and shows no dependence on concentration.

  Image reproduced from Basic one- and two-dimensional NMR spectroscopy, Wiley, 2010.


Heterogenous equilibrium nernst equation In the nernst equation, when a gas - ion electrode is used eg when we write the equation for standard reduction potential for a particular electrode : we get a term of activity of h2 gas and h+ ion. We respectively put 1 atm and 1 molar. However how can we use partial pressure and molarity in the same expression for Q? Shouldn't it be one or the other?

You can mix activity with fugacity in the expression of equilibrium coefficient or reaction quotient. In fact that is quite common when the reaction involves both solution and gas phase reactants/products.

The definition of the equilibrium constant actually involves dividing the standard pressure $p^\unicode{x29b5}=1\,\rm atm$ or the standard concentration $c^\unicode{x29b5}=1\,\rm mol/L$. In your case, the equilibrium constant will be

$$K^\unicode{x29b5}=\frac{[a(\ce{H+})/c^\unicode{x29b5}]^2}{[f(\ce{H2})/p^\unicode{x29b5}]}$$

which is dimensionless, and is related to the change of Gibbs free energy by $\Delta G=\Delta G^\unicode{x29b5}-RT\ln K^\unicode{x29b5}$.


Experimental melting-point of Arm-and-Hammer sodium bicarbonate is remarkably high(?) Online literature reports the melting point of $\ce{NaHCO3}$ as ~50 deg. C.

Using a standard Mel-Temp II melting-point apparatus and a standard alcohol thermometer, I didn't observe any melting of any sample lower than 150 deg. C. I don't think hydration of the salt can even explain this. Is there some other material in commercial baking soda that would alter the melting point this much?

As a raw calibration, I found that some Bayer Aspirin powder (possibly containing cellulose or other binding agent) had a melting point of 120 deg. C., a good deal below the reported ~135 deg. C.

Does anyone know what may be present in commercial Arm-and-Hammer baking soda besides sodium bicarbonate? The box reports 100% purity but I have trouble believing it now.

I've heard rumors that some baking-soda products contain aluminum and this has raised some alarm about ingestion hazards. Is this valid?

See the MSDS for Arm and Hammer Baking Soda here, from the manufacturer (Church & Dwight Co.) (it will link to a PDF) in which it is stated:

CONDITIONS TO AVOID: Temperatures above 65°C (150°F).
INCOMPATIBILITY WITH OTHER MATERIALS: Reacts with acids to yield carbon dioxide. May also yield free caustic in presence of lime dust (CaO) and moisture (i.e., water, perspiration). Dangerous reaction with monoammonium phosphate or a sodium-potassium alloy may occur.
HAZARDOUS DECOMPOSITION PRODUCTS: Heating above 100°C may cause dangerous levels of carbon dioxide gas to be present in confined spaces. Yields sodium oxide if exposed to temperatures above 850°C. Avoid inhalation, eye and skin contact with sodium oxide.

The ingredients are listed as 100% sodium bicarbonate with CAS number 144-55-8.
Your choice of thermometer is questionable in the present case. See this Wikipedia entry which reliably reports that the upper limit on temperature measurement from an alcohol-based thermometer is 78 $^{\circ}$C. A mercury thermometer has a higher upper temperature range (see this Wikipedia article) and is preferable to the alcohol thermometer. Alternatively, use a gallium thermometer or a digital model with an operating temperature range that suits your needs.
You ask:

I've heard rumors that some baking-soda products contain aluminum and this has raised some alarm about ingestion hazards. Is this valid?

Probably not in the case of the specific brand in question here, based on the MSDS.


Colour in vanadyl sulfate What is the reason that vanadyl sulfate shows blue colour, even in its anhydrous form? If it's d-d transition, can you explain how it works in this compound?

Vanadyl sulfate or $\ce{VOSO4}$ has $\ce{VO^2+}$ ion. It exists as a blue hydrated crystalline salt, known as acid vanadyl sulfate, $\ce{VOSO4 \cdot H2SO4 \cdot xH2O}$. If you're to heat it at $\pu{533 K}$ with concentrated sulfuric acid, you obtain anhydrous vanadyl sulfate, which is rather a greyish-green crystalline powder.¹

The ion $\ce{VO^2+}$ is more accurately written as a complex $\ce{[VO2(H2O)4]+}$.² Inductive effects of ligands and different symmetries of orbitals causes the d-orbitals to split up and thus they become non-degenerate.

The electrons possibly absorb and emit electromagnetic radiation in the visible spectrum. A rough estimate leads me to conclude that this complex would've been absorbing radiations towards the higher-wavelength (possibly red) region of the visible-spectrum thus emitting radiations corresponding to a blueish tone of light.

The colour is possibly due to d-d transition as the central metal ion ($\ce{V+4}$) has one electron in its d-orbital.

References:

B.Sc. Chemistry–II (UGC), R.L. Madan
Chemguide – Vanadium


Which of the following can be a source of error in Spectrophotometry? Mark the wrong answer a) Differences in solvent
b) Acid base equilibrium (differences of pH)
c) Parasitic or stray radiation
d) Monochromatic radiation
I have some doubts about pH in spectrophotometry. I'm not sure if it is correct but here is what I thought:
a)Differences in solvent. It can be a source of error because disociation, reaction or asociation of the analyte with solvent can generate absorption products different of the analyte.
b)Acid base equilibrium. I don't know why pH is a variable to take into account in spectrophotometry. I thought maybe the case of the determination of phenolphthalein that  depends of pH medium. Spectrophotometry doesn´t work when phenolphtalein is in acidic medium. So pH is a multiplicative interference in this case.
c)Parasitic radiation. It is a source of error because  it is radiation that doesn't come from the analyte and it generates signal.
d)Monochromatic radiation. I think that is the wrong answer because it is what we need to avoid big changes in the absorbance as a function of the wavelength.

It is a very open ended question, but first three choices will cause errors. Regarding your pH issue, you are right, it is multiplicative error. Take the example of phenolphthalein, as you say that its color (=absorption spectrum) is different in acidic or basic medium. So pH change can shift the lambda max of analytes, hence the cause of error.
Phenolphthalein still absorbs light in the acidic medium. I cannot find phenolphthalein spectra, but let us take phenol red indicator, you can see drastic changes in absorbances at the lambda max as a function of pH.
If by some random chance, you were measuring at the isosbestic point, then pH will not affect absorbance but in all other cases pH will change the A values drastically.


Why is propanoic acid weaker than butanoic acid? I was reading about electronic effects in organic compounds, and I found the explanation that carboxylic acids are weakening as their number of carbon atoms increase.

But I found a table of $K_\mathrm{a}$ for different acids, and it writes that $K_\mathrm{a}$ for propanoic acid is $1.34\times 10^{-5}$ while butanoic acid's acidic constant is $1.48\times 10^{-5}$. This means that butanoic acid is stronger than propanoic acid, but isn't this a contradiction to the inductive effect?

As a trend is usually a linear fit through experimental values it does neither mean that there are no minor variations between the values nor that there is a constant rise between values but only that one can see a difference between earlier and later values of that measured property.

And besides all the answers on the actual influence of the longer alkyl chain on the acidity of the carboxylic group there are no big differences between the acidities for those small alkanoic acids between chain lengths of two to ten carbons.

However, there is a small overall trend $\left(+0.02~\mathrm{pK_a}\right)$ which can be seen in the following graph.

So one is a little bit stronger than the other but it is not very much.

One could start to argue why exactly those two differ "so much" in strength but looking at those nine values above and the "noise" around the trend I guess that makes no sense.


What is the difference between resonance and π conjugation I have seen electron delocalization mentioned frequently in 2 separate contexts: resonance (ozone, carbonate / nitrate...) and π-conjugation systems (aromatic compounds).
While the 2 initially sound the same, 2 different points lead me to believe otherwise:

π conjugation requires p- orbitals to be in the same plane, whereas many compounds said to have resonance ($\ce{ClO4-}$, $\ce{SO4^{2-}}$) are not planar so resonance can't be a subset of π conjugation
All molecules with (significant) resonance have multiple equivalent stable-lewis structures (nitrate has 3, sulfate has 4...) whereas many molecules with π-conjugation do not (haxa-1,3-diene has only one "most stable" lewis structure) so π-conjugation cant be a subset of resonance

NOTE ON POINT 2: I am aware that hexa-1,3-diene has minor contributing structures, but so do all possible molecules. However, the only molecules for which resonance hybrids are used are those with multiple major contributing lewis structures.
The only intersection between the 2 concepts I find is benzene; it has 3 equivalent major contributing structures implying resonance and 6 overlapping p-orbitals lying in the same plane, implying π conjugation
I feel like I'm missing some part of the bigger picture. What's actually happening?

You might be confused but not missing anything big or fundamental.
A pi-conjugated system is one in which delocalisation of pi-electrons can take place. It is all.
When effective, pi-conjugation is a subset, or a type, of delocalisation than can be described by resonance.
However:
to be strict, conjugation refers to bonds, and specifically to an alternating single and double bonds structure. It is just a positional nomenclature.
It turns out that in cases and in different extent the structure with such conjugated alternating single and double bonds does not represent the real molecule right because delocalisation takes place.
While the terms are linked, a not very accurate use of them leads to their treatment as if they are synonyms. That is what I have more or less done above, but just using a more nuanced term as pi-conjugation, thereby referring to electrons rather than bonds.
This is as to say that a sequence of alternating single and double bonds (as in at least one limiting form) is conjugated by definition. The conjugation (would be better saying delocalisation, here) can eventually/effectively be so true to the point of erasing the original meaning, i.e. in fact there are even not localized conjugated bonds.
Whit all the above in mind, a perfect organic metal with equalised bonds would be seen as "the king" of conjugated polymers.
Finally, within the context and dictionary of  VB the above delocalisation is described by resonance, This again lead to the use of delocalisation and resonance as interchangeable terms.


What is the mechanism for this reaction? The closest reaction I can find is Lossen rearrangement but in that hydroxyamic acid is first converted to its derivative but how will that be possible here?

Moreover, even if I make an amine in this step I cannot proceed further, how will the amine and alcohol react in the next step? Any hints would be appreciated.

In your question, you have mentioned:

The closest reaction I can find is Lossen rearrangement but in that hydroxyamic acid is first converted to its derivative but how will that be possible here?

You have recognized the reaction correctly but the Lossen rearrangement is the conversion of a hydroxamate ester to an isocyanate (only the first step of given mechanism). Although the original Lossen rearrangement needed hydroxamate ester $\left(\ce{R(C=O)NH-O(C=O)R'}\right)$ to rearrange,  recent research has shown that the rearrangement is possible with free hydroxamic acids $\left(\ce{R(C=O)NH-OH}\right)$ as well (Ref.1):

Without using activating agents, a variety of free aromatichydroxamic acids could be rearranged to aromatic amines in the presence of base alone:

Source of the image: Ref.1
According to $\bf{(a)}$, free aromatichydroxamic acid $\left(\bf{1}; \ce{R(C=O)NH-OH}\right)$ react with a base to give  the corresponding carboxylate $\left(\ce{R(C=O)-O-}\right)$ as intermediate and hydroxyamine $\left(\ce{NH2-OH}\right)$ byproduct. The intermediate $\left(\ce{R(C=O)-O-}\right)$ reacts with another aromatichydroxamic acid $(\bf{1})$ to give necessary hydroxamate ester $\bf{A}$, which undergoes Lossen rearrangement in basic conditions to give the isocyanate $\bf{B}$. Mechanism for aromatic amine from isocyanate is depicted in $\bf{(b)}$.
For instance, the base-mediated rearrangement of free 4-methylphenylhydroxamic acid $\bf{(1a)}$ to  4-methylaniline $\bf{(2a)}$ is given as follows:

A mixture of N-hydroxy-4-methylbenzamide $\bf{(1a)}$ $(\pu{0.363 g}, \  \pu{2.4 mmol}),$ $\ce{K2CO3}$ $(\pu{0.332 g},$  $\pu{2.4 mmol})$, and DMSO $(\pu{2 mL})$ was heated to $\pu{90 ^\circ C}$ and stirred at that temperature for $\pu{2 h}$. The mixture was cooled to rt, and then treated with $\pu{2 M} \ \ce{ HCl}$ $(ca. \pu{3 mL})$. After the mixture became the clear solution, $\pu{2 M} \ \ce{NaOH}$ $(ca. \pu{3 mL})$ was added and extracted with $\ce{Et2O}$ $(\pu{15 mL} \times 3)$. The combined organic layers were dried over anhydrous $\ce{Na2SO4}$, filtered and evaporated under reduced pressure. The residue was purified by silica gel column chromatography $(\ce{hexane/Et2O}, 1:1)$ to yield the pure 4-methylaniline $(\pu{0.253 g}, \ 98\%)$ as a white crystalline solid.

The effects of the amount of base, reaction time, and presence of water are summarized in following diagram (Ref.1):

Thus, as shown in above diagram, presence of water may have direct effect on the conversion of the substrate to corresponding amine (even in DMSO). Further, Ref.1 clearly shows that this reaction requires polar aprotic solvent to proceed through Lossen rearrangement to give thiocyanate intermediate, which eventually converted to corresponding amine in the presence of aqueous base. For instance, if $\bf{(1a)}$ is reacted with $\ce{K2CO3}$ in methanol instead of in DMSO or DMF at for $\pu{90 ^\circ C}$ for $\pu{2 h}$, the yield was found to be only $10\%$ while it was $98\%$ in DMSO and $92\%$ in DMF under the same conditions. The authors reported $86\%$ of the starting compound has been recovered unreacted (none of it recovered in DMSO or DMF).
Note that the first reaction given in the question is in aqueous medium. Thus, according to the results found in Ref.1, the resulting product in the given reaction would be a trace of corresponding amine $\left(\ce{R-NH2}\right)$ and most of substrate would be recovered unreacted. Consequently, there won't be a reactive isocyanate $\left(\ce{R-N=C=O}\right)$, thus the reaction would not proceed to give the anticipated product in methanol as indicated.
Suppose this first reaction has been done in $\ce{OH-/DMSO}$ in place of $\ce{OH-/H2O}$. That would give you the anticipated isocyanate intermediate $\left(\ce{R-N=C=O}\right)$. It is well known that isocyanates reacts very differently in the presence of water, alcohol, and amine. In the presence of water, it would give you corresponding amine with evolution of $\ce{CO2}$. However, in the presence of alcohol or amine, it would give you corresponding carbamate or urea derivative, respectively$:

Since the given second reaction is in methanol (an alcohol), the product would be a methyl carbamate as shown in the scheme above (Ref.2). For example, polyurethane is prepared from diisocyantes:

References:

Yujiro Hoshino, Moriaki Okuno, Eri Kawamura, Kiyoshi Hondab and Seiichi Inoue, "Base-mediated rearrangement of free aromatichydroxamic acids (ArCO–NHOH) to anilines," Chem. Commun. 2009, (17), 2281-2283 (DOI: https://doi.org/10.1039/B822806J).
Greet Raspoet, Minh Tho Nguyen, Michelle McGarraghy, and Anthony Frank Hegarty, "The Alcoholysis Reaction of Isocyanates Giving Urethanes:  Evidence for a Multimolecular Mechanism," J. Org. Chem. 1998, 63(20), 6878–6885 (DOI: https://doi.org/10.1021/jo9806411).


Does ultrasonic bath cleaning damage laboratory volumetric glassware calibration? As an analytical chemist, I'm always ensuring my glassware is thoroughly clean before I start an analysis. However, my supervisor always states that using ultrasonic bath to clean volumetric glassware damages its calibration, rendering it imprecise for analytical usage. Does that hold true, or is it okay to use an ultrasonic cleaner?

According to technical information for volumetric measurement provided by Brand, it is acceptable to use an ultrasonic bath.

  Both glass and plastic labware may be cleaned in an ultrasonic bath. However, direct contact with the sonic membranes must be avoided.

Nevertheless,

  For gentle treatment of labware, clean immediately after use – at low temperatures, with brief soaking times, and at low alkalinity. Glass volumetric instruments should not be exposed to prolonged immersion times in alkaline media above 70 °C, as such treatment causes volume changes through glass corrosion, and destruction of graduations.


Is it possible for Hydrogen to lose its electron? I know the basics of Chemistry and one thing I've always wondered if it is possible for Hydrogen to give up it's one electron? I know Hydrogen is eager to share its election through covalent bonds, but is there a scenario where Hydrogen will give up its only electron?

Hydrogen can lose an electron meaning it can be in the +1 oxidation state. However, just like any other cation or anion it never occurs free in condensed matter, it always is in contact with solvent and/or anions. Moreover, because of extremely small size of proton, it is an extremely powerful Lewis acid. Consequently, in common conditions proton would react with first electron pair it comes in contact with, up to and including inert gas electron pairs and covalent bond pairs.

On the other hand, hydrogen ions are quite easy to generate in electric discharge and/or under extreme heating. In fact, producing and confinement of super-hot plasma, consisting of hydrogen ions and electrons, is an area of active research for several decades and, well, producing and confinement of relatively cold plasma is not a problem. Confinement of several billions Kelvin hot plasma, however, is still a problem.


Can d-orbitals participate in bonding? My textbook says:

It is appropriate to remark that the trick of rationalizing apparent violations of the octet rule by invoking the participation of d-orbitals in the bonding scheme is a matter of some controversy. Some scientists feel that the d-orbitals lie so high in energy that they should not be treated as valence orbitals, but as excited orbitals which cannot confer any appreciable stability to chemical bonds. Others feel that while d-orbitals are of high energy in free atoms, their energy decreases as other atoms approach to make bonds.
This author has not been convinced of the absolute validity of either side of this argument, and interprets the controversy merely as more evidence that the description of chemical bonds in terms of a few atomic orbitals is a highly approximate procedure. The most convincing reason for invoking d-orbitals in chemical bonding is that it is a simple idea which works, if it is used carefully and not over-interpreted.

Since this book is quite an old edition, I wanted to know whether there are any newer developments in this direction. Which of the two views is generally acceptable?

Yes it works, according to result they have with organometallic chemistry. You may not know $e_g$ and $t_{2g}$ orbitals. Have a look at this diagram of orbitals :

Its diagram for an octaedric complex, where M is the metal. Positions of $e_g$ and $t_{2g}$ depends of what L is.


How many empty orbitals do elements have? A student was contemplating a question regarding elements with "no empty valence orbitals." An argument was made that elements all have infinite number of valence orbitals.

  Every atom basically has an infinite number of shells. The thing is that almost all of those shells are empty (they don’t have electrons in them).

https://van.physics.illinois.edu/qa/listing.php?id=1224

So it is facile to say that, say, the fluoride anion has no empty valence orbitals on the basis of it having 8 valence electrons and the $2n^2$ rule (where n = 2)?

In my view there is an important difference between empty orbitals and empty valence orbital.  There are an infinite number of orbitals, a finite number of filled orbitals, but only one valence orbital.

The definition of a valence electron is usually something like, "The electrons in the outermost shell are the valence electrons--the electrons on an atom that can be gained or lost in a chemical reaction."  Therefore, a valence orbital would be the orbital holding those electrons.  If the valence orbital is full, it is full; there are no other valence orbitals.

Based on this understanding, it would be correct to say "the fluoride anion has no empty valence orbitals."


dissociation of aluminum sulfate in water I had a question as to what happens when aluminum sulfate is placed into water. After conducting research online, I got varied answers: some say that it dissociates into Al$^{3+}$ and SO$_4$$^{2-}$ ions, while others claim that aluminum sulfate will react with water to form aluminum hydroxide and sulfuric acid.

I even conducted a mini-lab by placing aluminum sulfate in water and sometimes there was precipitate (assumed to be aluminum hydroxide) and sometimes there was none.

What is really happening?

Aluminum sulfate has special properites and uses. It is true that aluminum sulfate is soluble in water, and it has specific solubility values for certain temperatures:

$31.2\ \mathrm g$ per $100\ \mathrm{mL}$ at $0\ \mathrm{^\circ C}$
$36.4\ \mathrm g$ per $100\ \mathrm{mL}$ at $20\ \mathrm{^\circ C}$
$89.0\ \mathrm g$ per $100\ \mathrm{mL}$ at $100\ \mathrm{^\circ C}$

You stated you used $12\ \mathrm g$ in $100\ \mathrm{mL}$, so it would seem to completely dissolve. However, it this situation, since it isn't being dissolved in very large amounts of water, it doesn't dissociate and form aluminum hydroxide. Aluminum sulfate can be used in water purification in small amounts, because it causes impurities to group up into larger particles and then settle to the bottom of the container (or be filtered out) more easily. The solid you see sometimes are the impurities of the water that have been grouped up and look like precipitate. If you use more pure water, you will see less of these impure solids, or possibly nothing if the water is pure enough (such as if it's distilled water).


Does aluminium foil have an odour/gas off? I have used some aluminium foil to cover some windows and I notice a light smell in my small flat. I could be mistaken and it could something as I've been moving things around but this leads me to wonder does aluminium foil have an odour/gas off?

I notice it sticks to things as well so might this be an added solvent or what property of the foil gives it this characteristic?

Here is related material I once posted:

Also, commentary in Bretherick, Volume 1, page 23-24, to quote:
A student mixed aluminium foil and drain cleaner in a soft drink bottle, which started emitting gas. Another student carried the bottle outside and was claimed to have been overcome by the toxic fumes [1]. Most drain cleaners are alkalis, so that aluminium will dissolve to produce hydrogen. The bleach that it is is suggested may have been present will produce no toxic fumes in alkali, and one would be surprised to find arsenic or antimony compounds present. If the collapse was not purely hysterical, the remaining, though remote, possibility would be phosphine. The soft drink the bottle had contained was one of the many perhaps best described as impure dilute phosphoric acid [2]. "
A related comment attributing the formation of gases like arsine and stibine from nascent hydrogen per this source.
Education on nascent hydrogen interaction is important so as to avoid toxic gases that may be inadvertently and unexpectedly created in significant amounts.

Now, in the context of extensive employment of Al foil in your home, there is a potential danger that an acid/base is attacking the foil. This could release active (nascent) hydrogen atoms.
The latter could interact, for example, with compounds in physical contact, like paint (containing perhaps zinc sulfide). This may, hypothetically, result in the release of odorous hydrogen sulfide. This is a problematic gas, but even more deadly gases could arise also (per source above).
As such, I would advise removing the Aluminum foil (or, at least areas in contact with paints,...).


Why is ionization energy of indium less than gallium? In group 13 we observe an irregular trend in ionization energy: B > Tl > Ga > Al > In.

Gallium has a filled 3d subshell, but indium has a filled 4d and 3d subshell. Thus it should have more poor shielding along with more ionization energy, but that is not the case why?

The theoretical values of ionisation potential are pretty close to each other.in fact ionisation potential of gallium is almost equal to that of aluminium.Now due to MORE shielding in indium but lack of lanthanoid contraction the affect balances out.As the atomic radii and shielding affect simultaneously takes place from your viewpoint you can claim that the increase in the atomic radii predominates over the zeff.thus indium will have lesser ionisation energy.


Mechanism for the chlorination of Fluorene with NCS? From literature, it was described that fluorene can react with NCS and conc. HCl to produce the 2,7-diCl derivative. Please help me understand the mechanism.

According to this paper here

it has been described recently that at very low pH ($\mathrm{pH <  3}$)  $\ce{Cl2}$  functions  as  an  active  chlorinating  agent

In the authors' view the system works by $\ce{NCS}$ oxidising chloride to chlorine which then provides $\ce{Cl+}$ as the chlorinating agent.


Why do higher fluorine oxyanions not exist? I asked this question from my teacher and he said that's because fluorine has a high electronegativity and does not form dative bond with oxygen. That's why only hypofluorite $\ce{FO-}$ exists and not $\ce{FO_n-}$ ($n = 2,3,4$).

But then I was wondering why $\ce{BF4-}$ exists. And according to this Wikipedia article, fluorine does form dative bond.

so why the anions in the title do not exist?

Thanks for your help!

There are two factors at play here:

Fluorine is more electronegative than oxygen. In oxyanions, the central atom carries a substantial positive charge due to the oxygen atom pulling away electron density. In oxyfluorides, oxygen actually donates electron density to fluorine and oxygen would logically be the central atom. There are theoretical studies into $\ce{OF3+}$ but no successful syntheses yet.
The fluorine atom is tiny, even by period 2 standards. There's simply not enough space around the proximity of the atom to fit three or more other atoms.


Why don't the electrons of opposite spins collide in an orbital? As I was reading through my textbook, I got to know that in an orbital, there are 2 electrons of opposite spins. My question is: Shouldn't they collide whilst they are rotating?

In reality, electrons in a molecule are colliding all the time insofar as they are experiencing a force (the Coulomb force) due to the presence of other electrons, and are correlating with each other as a result. Orbital models like the Hartree-Fock mean field approximation mostly neglect these interactions, so the answer to your question is they don't because the model doesn't consider such dynamics. But many sophisticated methods for calculating the effects of this correlation exist. E.g. "Full Configuration Interaction" and "Density Functional Theory" are two such methods.

So keep in mind that orbital models are only approximate. It is fine for many purposes, but inadequate for many too. The true ground state wavefunction of a molecule cannot be represented by a single determinant, and therefore cannot be thought of strictly as a collection of electrons occupying orbitals.


How to reproduce the Gaussian 09 SDD basis set in Gamess-US? I'm trying to migrate away from Gaussian 09, using Gamess-US instead for molecular modelling of some heavy alkaline earth metal complexes, using density functional theory (DFT). In some of my previous Gaussian inputs I used the SDD (Stuttgart/Dresden) pseudopotential with the metallic cation ($\ce{Sr^2+}$). Now I intend to reproduce this calculation in Gamess-US.

After a couple of days figuring out how to use mixed basis sets in Gamess, I searched the Basis Set Exchange for this Stuttgart/Dresden ECP specification. Then I got confused, because there's no basis listed with that exact name. The closest I've found available for strontium was Stuttgart RLC and Stuttgart RSC 1997.

Which of these (if any) is the best match to the SDD pseudopotential available in Gaussian?

SDD is a basis set description, which uses Stuttgart/Dresden ECPs. They are readily available for download at the BSE, but also via Univ. Köln. And they are pretty easy to pick out. See the Pseudo keyword for what is Gaussian 16 is using currently, which should be the same for 09. At least in the BSE there hasn't been an update to the S/D ECPs since their initial upload. See the related notes included at BSE and pick the appropriate size. The more problematic part is finding the BS for the orbitals that are actually described. I believe this is the correct choice, i.e. Stuttgart 1997 RSC. Please review the literature specified on the Gaussian homepage about the implementation of SDD (should be about 20 to 30 papers), and then you can cross-reference them to the references saved in the BSE library.

The work is from the earlier days of computational chemistry, so it is quite fragmented. I believe the ECPs are still popular and used in the correlation consistent Dunning basis sets. The orbital parts I wouldn't be too sure about. There are plenty of other choices out there and it would be wise to review the literature to see what is the state of the art for strontium.

Why not use the move away from Gaussian also as a move to a less fragmented, different level of theory? If you want to compare the data to previous calculations, you have to recalculate them anyway, why not choose a method which is a bit more consistent than the SDD basis. And depending on what you are calculating, you might also want to look at different software suits, e.g. Orca, Psi 4, Turbomole. Gamess certainly has its strengths, basis set implementation and handling is unfortunately not among those.


Alkenes v/s Alkynes in two different reactions Alkynes are less reactive towards electrophylic addition than alkenes. However, in case of catalytic hydrogenation, alkynes are said to be more reactive. Why? Also, if I want to add H+ as an electrophile, then will alkenes be still more reactive than alkynes?

Does the reason lie in the mechanism of both the reactions (that was obvious, I guess)? If so, please explain, I cannot seem to justify for the later one.

The first stage of electophilic addition is addition of a cation to alkene/alkyne with formation of a carbocations. Vinyl carbocations are much less stable then alkyl cations, so the path is strongly hindered for alkynes.

The first stage of catalitic hydrogenation is addtion of hydride-ion on the metal-surface to unsaturated hydrocarbon, forming a carbanion bound to metal surface. Vinyl-anion are more stable than alkyl-anyons for the same reasons vinyl-cations are less stable than alklyl-cation. Acetylenes also are much less hindered sterically than alkenes and adsorb on a metal surface much easier.

Please, note, than nearby substituents may affect these properties dramatically. Deactivation of alkenes in electrophilic addition by strong $\pi$-electron-withdrawing groups is well known.


What is the value of the gas constant derived from the kinetic model of gases? I used this resource (http://digitalcommons.unl.edu/cgi/viewcontent.cgi?article=1151&context=physicskatz) to derive the ideal gas law from the kinetic model of gases:

$$PV=n(2/3)N_oT$$
with T being equal to $(1/2)mv_{average}^2$.

And now I'm super confused. Does this mean that the universal gas constant R is simply equal to (2/3)*avogadro's number?

$T$ here does not refer to temperature but rather the average kinetic energy of a molecule.

I can see why there is some confusion. Sadly we only have so many symbols and kinetic energy also happens to be commonly denoted with $T$.

To relate the average kinetic energy to the temperature, you need to bring in equipartition theory

$$\frac{1}{2}m\langle v^2\rangle = \frac{3}{2}kT$$

(assuming three translational degrees of freedom only). If you substitute that into your above equation

$$pV = \frac{2}{3}nN_\mathrm{A}\cdot m\langle v^2\rangle$$

then you will find that $pV = n(N_\mathrm{A}k_\mathrm{B})T \Longrightarrow R = N_\mathrm{A}k_\mathrm{B}$, hopefully a familiar expression. The question of how the value of $R$ is derived then becomes a question of how equipartition is derived - and that deserves another question itself.


The definition of formal Charge Ok, most people including me know that formal charge is just a book-keeping tool. I know how to calculate it, apply it to Lewis structures etc etc. But I am confused in its very basic definition which says - "Formal charge is the charge on an atom if the electronegativity differences are ignored" . If the electronegativity difference is ignored, then how can an atom have charge ? I know this is not a place to ask such a basic question , but still I am very confused about this miniscule definition.

A good description appears in the book, "Chemistry A Molecular Approach".

In the molecule of hydrogen fluoride, we know that it has a dipole moment, and fluoride is slightly negative. Ignoring this information (electronegativity difference) and we share the bonding electrons equally, the formal charge can be calculated, and both of them are determined to be 0.

In reality, we know that fluoride has the electron from the hydrogen most of the time (due to electronegativity difference). We say the hydrogen has a partial positive charge and the fluoride has a partial negative charge.

From this, it is apparent that the formal charge is different from the charge on the atoms in a molecule.


In practice, do polar molecules actually align in an electric field? We have all seen the following diagram (or similar) in our first chemistry class, depicting polar molecules aligned in an electric field. Is this just one of the half-truths of beginner chemistry or is this what really happens? I realize there will still be kinetic motion (spin, etc), but even statistically, is it possible for such a fluid/gas to have a net directional orientation throughout?

I have often seen the diagram depicted with water, but as the top answer on this question explains:

  "...the ions will rush to the respective electrodes and form the so-called electric double layer. Its thickness, known as Debye length, is usually quite small. As for the rest of the solution, it will be effectively shielded from the electric field."

Perhaps this would be the same issue with hydrogen fluoride?
Is it possible the Debye length would be longer for a polar covalent molecule like Acetonitrile or Chloromethane?

If there are in fact polar molecules that exhibit this field-orientated behavior on a macro scale, what would be a way that it could be demonstrated?

These are two entirely different and unrelated phenomena, to the point that you'd better unsee that other question and my answer to it, if you can. Ions are very much not like molecules. Ions are charged; molecules are not. Molecules don't form the double layer, nor do they even know what is Debye length. The electric field (though somewhat diminished, by a factor of dielectric constant) penetrates the whole volume of liquid, so all molecules feel it the same way; none is shielded.

Yes, polar molecules get oriented in the electric field, albeit to a very tiny degree. Think of it this way: the molecule still bounces and rotates around, but it spends a little bit longer in the field-aligned orientation than in any other. To orient all molecules in a truly stiff way, like soldiers in formation, the energy of alignment has to overtake the energy of thermal motion, which requires enormous electric fields and is ultimately impossible, because a breakdown would occur way sooner than you get there.


How does Dichlorofluoromethane get into the stratosphere? I was wondering why FCKWs can destroy the ozone layer.
It seems like these gases have a high molecular weight in comparison to $\ce{N2}$ or $\ce{O2}$.
How can these gasses come up to the outer layer of our atmosphere? In particular, how long (on average) would it take to get to such an altitude (through the diffusion of such a particle)?
I mean, they are basically several times heavier than all the other components of our atmosphere combined. Hence, it shouldn't be very energetically stable for them to float on the outer layers of space on top of other lighter particles.

This is one of the more frequently asked questions about ozone depletion. I will try to give a short summary.

First of all, it is important to know that Chlorofluorocarbons (CFCs) are extremely stable at room temperature and pressure. It is therefore only natural, that in the lower atmosphere these gases are enriched, therefore they have plenty of time getting in higher altitudes via diffusion.

The earths atmosphere is in constant motion, you can experience that on a macroscopic level via different winds. The sun heats up the earth's surface, causing all sorts of liquids to be evaporated. While evaporating water, these small molecules may form little conglomerates, taking other molecules up to higher heights. This will happen to dust, so it will also happen to CFCs. There will be local areas (or volumes, layers) of the air in different heights of the atmosphere that have different densities, concentrations and composition. As a result they will also have different air pressures. You can experience this on a plane at higher altitudes. Assuming that all reactions tend to react towards equilibrium, there will be forces acting between these areas. In a consequence there will be also upward forces acting on the CFC containing gas mixture. Slowly these molecules will move upwards to the stratosphere.

As you can see, this process is heavily dependent on many different factors and may well require years. I think it is actually very hard to find any accurate numbers for this process. My research on the subject turned up nothing about that. In most of the cases it was just stated, that these gasses are there.

Some further reading:

Why does free chlorine in the stratosphere lose its ozone-depleting potential after about 100,000 reactions?
A collection of nature articles. Some are free access.
Ozonewatch @ NASA
http://www.ozzyozone.org/ (for children, in english, french, spanish)
A hole in the sky @ Science in school (different languages available)


Why can air accommodate more water when the air has a higher temperature? The amount of water(g) that air can accommodate depends on temperature. The absolute humidity then is defined as the percentage of water to air and decreases when the temperature decreases. The relative humidity is the ratio between the current and the maximum humidity, which increases with decreasing temperature. The temperature at which the relative humidity reaches 100% is called the dew point.

Anyway, why can air keep more water when it is warmer?

For a substance to be a gas, the kinetic energy of the molecules that make it up must be higher than the electrostatic potential energies that hold it together. When we talk about air "holding" water we are really talking about the amount of water that would be in the vapor phase relative to the amount that would be liquid at equilibrium.

Since temperature is a measure of the average kinetic energy of molecules, as the temperature increases, the fraction of water that can exist as vapor increases. Since air is already a gas under any conditions where water vapor can exist, it looks like the air "holds" more water as temperature increases.


Do Lewis acids and bases turn change colour of litmus Well, the question is short. Can lewis acids and bases change the colour of a litmus paper? I can't seem to find the answer online (at least according to my high school level chemistry search history).
The intuitive answer I feel is that they shouldn't be able to, as lewis acids and bases are defined only by their ability to donate or accept electrons. On the other hand, Arrenhius and Broensted-Lowry acids seem to be able to change the colour of litmus, possibly due to the presence of H+ or OH-
As a bonus question, is there any indicators that can be used to test the presence of lewis acids that change their colour after donating a lone pair of electrons?

You should keep in mind that the concept of acid-bases is not restricted to water. Lewis acid definition is applicable to gaseous, aqueous and non-aqueous systems. Consequently, all Bronsted-Lowry acids are Lewis acids. The idea applies to bases. Note that the converse is not true just like an apple is fruit but not all fruits are apples. The idea of litmus test is rather restricted to aqueous systems. So in short, the all relatively strong Lewis acids and bases in water can affect the litmus.
If you put a blue litmus is very dry HCl (gas), it would still turn it red because paper is never dry. It always has moisture absorbed in it.


What is the difference between an element and an atom? First, I would like to quote sentences from a book introducing elements and atoms:

An element is a fundamental (pure) form of matter that cannot be broken down to a simpler form.
Elements are made up of particles called atoms. An atom is the smallest unit of any element [...]

The first quote states that element cannot be broken down to a simpler form, but the second quote says elements are formed by atoms, which means element can be split into a simpler form (atoms)?
The difference between atom and element confuses me.

The wording in your quote is a bit unclear, I agree. A very nice definition is given by the Jefferson Lab page What is the simplest way of explaining what atoms, elements, compounds and mixtures are?

Atoms

  are the smallest bits of ordinary matter

Whereas an element is

  a substance that is made entirely from one type of atom

Atoms themselves can be 'broken down' into smaller sub-atomic particles (protons, neutrons and electrons) - the amounts of these (particularly the number of protons) define each element.

For example, in their un-ionised form, the element carbon is made entirely from carbon atoms, which have 6 protons and 6 electrons each, and cadmium atoms, which have 48 protons and 48 electrons, make up the element cadmium.

The amount of neutrons can vary with atoms of a particular element, forming isotopes, which (from the link):

  For example, carbon-12, carbon-13 and carbon-14 are three isotopes of the element carbon with mass numbers 12, 13 and 14 respectively. The atomic number of carbon is 6, which means that every carbon atom has 6 protons, so that the neutron numbers of these isotopes are 6, 7 and 8 respectively.

Different arrangements of atoms of single elements can form allotropes, such as for carbon, with diamond and graphite (among others - shown below). They both are composed of atoms of the element carbon, but in different configurations.

Below are diagrams of the allotropes of carbon featured in the article Work Function Engineering of Graphene (Garg et al. 2014)


Why do halogen-metal exchanges happen? I'm trying to understand organometallic raegents but am stuck at halogen-metal exchanges.

For example if you have $\ce{PhBr + BuLi}$ you get $\ce{PhLi + BuBr}$, I don't understand why $\ce{PhBu + BrLi}$ wouldn't form.

How Did Wittig Come to Discover Lithium-Halogen-Exchange?

Georg Wittig, discoverer of the Wittig reaction, was studying the effect of ring strain on the strength of two neighboring carbons; by adding enough strain he expected to coax C-C bonds into forming diradicals:

Unfortunately tetraphenylbenzocyclobutane rearranges to triphenyldihydroanthrancene instead of forming a diyl. This and other “failures” did not deter Wittig; they elected to continue by replacing phenyl groups with anisyls. To their surprise, reactions with anisyls gave unexpected products. They focused their efforts on a simple derivatization with benzil to characterize the major product. They were trying to synthesize $p$-anisyldiphenylcarbinol, but instead what happened is:

Continuing along they discovered phenyllithium would perform the same reaction. As part of their investigation they varied other substituents on the aryl halide and discovered an even more peculiar product which violated their chemical intuition. Wittig was the first to speculate a lithium-bromide exchange reaction in 1938, below is the reaction he and coworkers carried out in diethyl ether giving a high yield (95%), the preferred product led them to infer it went through an aryllithium intermediate:

This same reaction occured with iodine, bromine, and chlorine. Yet, fluorine gave products via an entirely different mechanism. This led them to postulate an aryne intermediate. They then set off in a new direction, by trying to weaken carbon-hydrogen bonds with a neighboring quaternary ammonium, this led them to discover N-ylides. It was found this more readily occurred with phosphonium, where they discovered P-ylides. Now you can see the rough outline of the path required to discover the Wittig reaction.

How Does Lithium-Halogen Exchange Work?

This initial discovery by Wittig and then Gilman encouraged further efforts because it allowed the formation of a number of previously inaccessible organometallics. In the above reaction, the Grignard reagent cannot simply replace phenyllithiun (some Mg cases have been reported for other halides, but Li is the stalwart metal).

Today this reversible lithium-halogen exchange reaction can be formulated as,
$$\ce{RLi + R’X <=>[{K}]  R’Li + RX}.$$

As an equilibrium process, we know that the more stable lithium species is preferred. Interestingly when $\ce{R’}$ = aryl (particularly bromides and iodides) this is best written as, $$\ce{RLi + R’X <=>> R’Li + RX}.$$

When $\ce{R’} = $ allylic or benzylic, Wurtz coupling is favored. $\ce{R’} = $ alkyl tend to give unsatisfactory results. In general, the equilibrium favors those species which most stabilize the carbonanion intermediate. Depending on the specific halide, side-product formation is a problem: $\beta$-elimination, $\alpha$-metalation, or Wurtz-type reactions can occur.

Gilman discovered that alkyllithiums, for example t-butyllithium, were more adaptable in their reactivity than aryllithiums, which explains their widespread success in the modern chemical literature.

How the reaction proceeds remains an unresolved problem, however, there is evidence to suggest the following mechanisms are at play:

(1) intermediates proceeding through an “ate” complex

Perhaps unsurprising to you by now, Wittig also hypothesized the existence of an "ate" complex intermediate:

This is in other words, a nucleophilic attack of the halide by the alkyllithium with a carbonanion leaving group.

(2) radical mechanisms

A one electron transfer from the nucleophile to halide thereby producing a radical intermediate that then can give rise to a number of products. There are data from EPR and radical caging experiments to suggest radicals at least form.

(3) mixtures of (1) and (2)

As with many systems, the truth is both mechanisms compete with each other; of course which pathway, if any, dominates will depend on the solvent, temperature, substrate, and certainly the lithium species used.

tert-Butyllithium requires 2 equivalents, resulting in a non-equilibrium process where: $$\ce{RX + (Me)3CLi -> RLi + (Me)3CX}$$ $$\ce{(Me)3CX + (Me)3CLi -> (Me)3CH ^ + (Me)2C=CH2 ^ + LiX}.$$


What barriers prevent plastic from becoming a fiber optic, data transmission medium? This is a question that has bothered me for a while.  In the past many years, no company has been able to develop a material that can transmit light nearly as well as fiber optic glass, or at least not as cheaply.

After doing a bit of research, I found that although there is fiber optic plastic, the quality is considerably poor when compared to the standard glass medium.  I would assume that developing a highly-transparent, high-quality plastic that is cheap to manufacture would be considered the holy grail of data communication (not to mention, quite profitable.)

Why is it that in the many, many years that fiber optic data transmission has been around, no one has been able to develop such a form of plastic?  In fact, I've never even heard of such research making headlines on the more common tech news sites.

In fact, I've never even heard of such research making headlines on
  the more common tech news sites.

Tasks like that are called deveopment, not research, normally.
And no industrial company will talk much about such work
if its not sucessful.
In fact there are such organic fibers for minor purposes (eg the Laserjets
II and III used such a fiber to send a synchronising signal from the
laser scanning unit down to the control board, for about 10 inches.
30 years ago one could buy such fibres for experiments at radio shack)
If You read some literature on optical fibres, You will learn that
low scattering is the grail. And plastics scatter a lot, for
very basic reasons of polymer structure. Try to look throug a plate of
acrylic glass "edge on", and You will understand.


Definition of enthalpy of vaporization The amount of heat required to vapourize one mole of a liquid at
constant temperature and under standard pressure (1 bar) is called its
standard enthalpy of vapourization.
The enthalpy of vaporization is often quoted for the normal boiling
temperature of the substance.

Suppose, some amount of water is kept inside a perfect thermally conducting container at 1 bar pressure and $\mathrm 50~^\circ \mathrm{C}$ within a thermal reservoir of $\mathrm 50~^\circ \mathrm{C}$ also. Due to evaporation, slowly liquid will convert to vapours.
After one mole of water evaporates, if we measure the heat energy received by the container from the reservoir, would it be equal to standard enthalpy of vapourization of water at $\mathrm 50~^\circ \mathrm{C}$?

Assuming that the reservoir does not absorb any heat and there is no loss of heat from the setup by:-

Radiation of heat from the reservoir,
Convection currents in the air/surrounding fluid (if any),
The entire water is at the same temperature so that heat absorbed is not used for increase in temperature,
Conduction loss from the reservoir to the ground/support it is on and air(negligible but thought I should mention)

Fulfilling these conditions ensures that the energy is only used by the water for vaporization.
So, yes, these conditions are fulfilled. But, that is never the case when experiments are conducted in labs.
Edit: The above conditions are more than just an isolated system.


Agar gel vs aqueous solution Can someone tell me why - in experiments to observe corrosion of iron nails - we often use a preparation of agar gel (with $\ce{NaCl}$, phenolphthalein, etc.) instead of an aqueous solution?

If you are trying to observe corrosion in a specific area, agar is a convenient way to prevent convection currents from obscuring what caused the corrosion and for localizing the indicator.


How is latent heat of vapourisation utilized in changing state? If latent heat of vapourisation is utilized in overcoming intermolecular forces of attraction between the participle of a material then how is it still hidden there? If latent heat is used in overcoming forces of attraction between the particles of a material then why doesn't it increase the kinetic energy of the particles of that material?

It is raising the energy level of the system by breaking the bonds that hold it together as a liquid. Once the bonds are all broken within a certain volume, the liquid will vaporize.

The energy seems to 'disappear' or 'hide' because we expect to see a change in the properties of the chemical as energy is added, but the breaking and reforming of intermolecular bonds is happening constantly within a liquid.

When the liquid has enough energy, random bond-less distributions will occur within the liquid that are large enough to overcome the pressure of the liquid and its container. This is how water boils. If enough of the water molecules become unbound, they will form a bubble. Inside the bubble, the water vapor will be exactly equal to the pressure of it's environment.

This gives us our definition of boiling point: vapor pressure equals atmospheric pressure. The water or other liquid can't stay in the liquid state because there's not enough surrounding pressure to hold it together.


Does the hydrolysis speed increase when more H+ ions are in the solution? In hydrolysis of polysaccharides, does the hydrolysis speed increase when more $\ce{H+}$ ions are present in the solution?

So if I were to use $\ce{H2SO4}$ instead of $\ce{HCl}$ (same volumes) then $\ce{H2SO4}$ will be the most effective?

The glycosidic bond of a polysaccharide chain is essentially an acetal moiety, and hydrolysis of acetals proceeds under acid-catalyzed conditions. Hence, the addition of acid, within some ambit of concentration, will accelerate the hydrolysis reaction. The optimum proportion of acid, however, is probably something that needs to be determined experimentally on a case-by-case basis. It isn't necessarily true that the rate of the reaction will continuously increase, without bound, as the concentration of catalyst increases.

The mechanism proceeds through a series of proton transfers and SN1 reactions. As far as I'm aware, the rate-determining step is typically the actual cleavage of the glycosidic bond after an initial protonation of the oxygen of the leaving group. Given that, I would expect increases in acid concentration to have marginal effect beyond some threshold value. That's just a surmise on my part, however.

I did find one study that examined the kinetics of starch hydrolysis, which found the optimum pH to be 3 (with values of 2-5 being tested).


How to determine amount of CO₂ in soda water I've got a problem in determining the amount of $\ce{CO2}$ in soda water. Basically, I have boiled some purified water to remove the dissolved gas.  By using a SodaStream machine, I made carbonated water.

After that, I have tried to determine the amount of carbon dioxide through titration with phenolphthalein and also $c(\ce{NaOH})=0.1~\mathrm{mol/L}$ sodium hydroxide.

The problem is I am not sure if this is the correct method to determine the amount of dissolved $\ce{CO2}$ in water.

First of all, after degasing by heating you have to be sure that no $\ce{CaCO_3}$ is formed, so you can minimise this problem by transfering the water in another flask.

More accurate would be to start directly with demineralised water which you can buy cheap at filling stations.

Secondly, you shouldn't wait to much while titrating, otherwise the water is taking up $\ce{CO_2}$ from the air.

You should use a bulb which doesn't contain much air (which contains $\ce{CO_2}$ and can interact as well. You can use a round flask and directly connect it with a stopper.

Third, Phenolphtalein is a good indicator because you titrate a weak acid with a strong base.

Make sure, you don't use too much Phenolphtalein.

Last but not least be sure that the $\ce{CO_2}$ from flask doesn't go away in the air or into the NaoH of the burette.

To cool will help, because $\ce{CO_2}$ is more soluble in cold water.

If you want to be a super-purist, do a potentiometric measurement and plot the point of change using a simple voltmeter.