What happens to boiled (BMIM)PF6? [BMIM]PF6, or 1-butyl-3-methylimidazolium hexafluorophosphate, is commonly used as an ionic liquid. As has been shown by the accepted answer to this question, sodium chloride that has been made to boil actually exists either as a monomer (Molecule of NaCl), or a dimer, rather than ions, in the gas phase. What happens to 1-butyl-3-methylimidazolium hexafluorophosphate, or any other ionic liquid, that has been boiled? Would it, like NaCl, also exist as a monomer? What would happen to an ionic liquid that has been placed into a vacuum chamber from which air had been removed?

Ionic liquids are not magical liquids. They are just typical organic compounds with certain properties. These compounds generated a storm in the tea cup in the late 1990s and 2000s, but they are as toxic as other solvents. Certainly they have niche applications especially as a gas chromatography stationary phase, where they can be used at high temperature (up to 350 centigrade!).
Now you can guess based on the GC application, that ionic liquids cannot be distilled or vaporized. They start to decompose by complex path ways above these temperatures. We can easily see that in a gas chromatograph. Surprisingly, not much has been studied as what happens to ILs at high temperatures.
Only one recent study exists Dicationic ionic liquid thermal decomposition pathways. There are complex decomposition products in the absence of oxygen. Like any organic compound, they will burn in the presence of air. NaCl is another story. It is not an organic compound. In short do not compare NaCl with ionic liquids.


Stability of iron(III) complexes Amongst the following the most stable complex is:
A) $\ce{[Fe(H2O)6]^3+}$
B) $\ce{[Fe(NH3)6]^3+}$
C) $\ce{[Fe(C2O4)3]^3-}$
D) $\ce{[FeCl6]^3-}$

I know that the answer should be $\ce{[Fe(C2O4)3]^3-},$ but I could not find a satisfactory explanation.

This is known as the chelate effect. Complexes with bi- or polydentate ligands — chelating ligands to be exact — are observed to be more stable than their nonchelating counterparts.
To understand this, we need to compare apples with apples, so your example is not the best one. (However, based on the trend we can still give a prediction; more on that later.) Instead, I will base my argument on comparing $\ce{[Fe(NH3)6]^3+}$ and $\ce{[Fe(en)3]^3+}$, where $\ce{en}$ is ethylenediamine $\ce{H2N-CH2-CH2-NH2}$.
Consider the following equilibrium:
$$\ce{[Fe(NH3)6]^3+ + 3 en <=> [Fe(en)3]^3+ + 6 NH3}\tag{1}$$
To determine which side is favoured (‘more stable’), we need to take a look at the equilibrium constant $K$. This is defined as
$$K = \frac{[\ce{[Fe(en)3]^3+}][\ce{NH3}]^6}{[\ce{[Fe(NH3)6]^3+}][\ce{en}]^3}\tag{2}$$
But to get a hint on its numeric value without measuring it, we need to apply the Gibbs-Helmholtz equation:
$$\Delta_\mathrm{r}G^⦵ = - RT \ln K\tag{3}$$
Equation (3) tells us that $\Delta_\mathrm{r}G^⦵$ will be negative (exergonic) if $\ln K$ is positive ($R$ and $T$ are positive by definition) — so if $K >1$, i.e. the product side is more favoured than the reactant side. So far no new information. We need to apply another equation, the definition of the Gibbs free energy:
$$\Delta_\mathrm{r}G^⦵ = \Delta_\mathrm{r}H^⦵ - T\Delta_\mathrm{r}S^⦵ \tag{4}$$
For the enthalpic term $\Delta_\mathrm{r}H^⦵$, we need to ask ourselves which bonds are broken and which are reformed. We break six $\ce{Fe\bond{<-}N}$ bonds and form six $\ce{Fe\bond{<-}N}$ bonds over the course of the reaction. Thus we can assume:
$$\Delta_\mathrm{r}H^⦵ \approx 0\tag{5}$$
Therefore:
$$\Delta_\mathrm{r}G^⦵ = - T\Delta_\mathrm{r}S^⦵ \tag{4'}$$
The entropic term $\Delta_\mathrm{r}S^⦵$ is most influenced by the number of particles in solution. The reactant side has four free particles in solution while the product side has seven. Since $7>4$, we can assume the product side’s entropy to be larger. Thus giving:
$$\begin{align}\Delta_\mathrm{r}S^⦵ > 0\tag{6}\\
- T \Delta_\mathrm{r}S^⦵ < 0\tag{7}\\
\Delta_\mathrm{r}G^⦵ < 0\tag{4''}\end{align}$$
Therefore, the reaction as written in equation (1) is exergonic, the products are more stable then the reactants.
Applying this to the problem, we see that most of the ligands are similar enough to not warrant any other stronger effects. Thus, the complex with three bidentate, chelating ligands is most stable.


Electrolysis using electrodes of metals whose ions are in solution So when we perform electrolysis of say an aqueous solution of silver nitrate with silver electrodes, silver ions having a greater reduction potential than water, accepts electrons and deposits at the cathode.At the anode however, silver just leaves the electrode and enters into its solution.
The anode must just be a metal rod whose sea of electrons is much poorer than what it must be.Why does the silver get oxidized and leaves the rod as an ion? Water could get oxidized instead and that appears to be a more logical expectation.
Secondly why does this not happen if we do the same reaction using platinum electrodes? What is preventing the metal from leaving the anode as an ion,moving to the cathode causing gradual thinning of one electrode and thickening of another?

The reason why silver dissolves (or water is oxidized) is well known. The reason why platinum does not dissolve is more elusive. Thanks to Maurice's comment for pointing this out.
Let's assume your solution is 1M in AgNO3 and 1M in acid ($H^+$). The potential required to oxidize silver is 0.80 V; the voltage required to split water (oxidize it) is 1.23 V, but there are complications of overvoltage, mostly with hydrogen, but also with oxygen. To sum up: it is easier to grab an electron from silver than to pump another hydrogen ion into solution. AHA! How about reducing the acid concentration, or not adding any at all? (Of course, the reason for adding the acid was to make the silver dissolve easier.) If the pH is increased so that hydrogen ion concentration is reduced to give a lower oxidation potential than silver, then water will be oxidized, hydrogen ion will be produced, and eventually, silver will become more easily oxidized. This electrolysis wil go thru a phase where hydrogen and silver are codeposited, probably making a porous, weak deposit of silver.
For platinum to yield 2 electrons requires 1.188 V. Although this is less than required for oxidizing water, overvoltage effects due to oxygen and hydrogen are apparently important. Although platinum can be electroplated with a platinum anode, apparently the process fails to work well; older methods of platinum plating were kept refreshed with additions of platinum salts rather than relying on dissolution of the anode (Ref 1). More modern methods use a pulsed electropotential, but their success seems to be from a solving difficulties at the cathode rather than getting the anode to dissolve (Ref 2). They deposit a monolayer of platinum attached to a monolayer of hydrogen, then oxidize off the hydrogen and repeat. The electrode potentials strongly suggest that increasing the cell potential, if you had a platinum anode, would result in dissolving platinum metal, but the catalytic reactivity of platinum with respect to H2O2 suggests that maybe the oxidation reaction occurs in the water layer, and the platinum acts merely as an electronic conductor.
Ref 1. https://www.technology.matthey.com/article/25/1/32-41/
Ref 2. https://www.chemistryworld.com/news/platinum-plating-at-the-flick-of-a-switch-/5720.article


Polarity of xenon fluoride The structure of xenon fluoride is a capped octahedron. The lone pair is stereochemically active, i.e. it will rotate about.

The dipole moments of the axial fluorides will get cancelled, and so will those of the four equatorial fluorides will also cancel each other. Thus, overall, the molecule should be nonpolar.

But what effect will the stereochemically active lone pair have on the polarity? Will the polarity keep on changing?

First of all, the point group for $\ce{XeF6}$ is disputed. The latest I saw was $C_\mathrm{3v}$, not $O_\mathrm h$ but the $O_\mathrm h$ is so energetically close to its $C_\mathrm{3v}$ that it is likely in a perpetual state of transition between the two. As for polarity, due to this it is hard to think it would have a permanent dipole. One would assume it would be easy to induce one, though.

$\ce{XeF6}$ is not the typical molecule the different models were made for, so YMMV.


Vapour density dependence on temperature Does vapour density change with temperature or is it solely dependent on molecular mass
as per the formula VD=molar mass/2

For ideal gas vapour behaviour, the vapour density is the ratio of its density and of the density of hydrogen at the same conditions
$$\frac { \frac {pM}{RT}}{ \frac {pM_{\ce{H2}}}{RT}} \simeq \frac M2 $$.
There for assumption of vapour ideal gas begaviour, "vapour density" does not depend either on temperature either on pressure.

I see a clash of definitions, or rather their terminology for the relative definition of vapour density versus the absolute density definition with the unit kg/m3 in sense of "density of vapour" or substance density.
It is easy not to be aware of "vapour density" is not "density of vapour". In fact, lack of context gives the reader the confusing choice if vapour is an adjective to density or a part of the term.


marijuana testing process Does anyone know a bit about how costly and accurate are the current THC measurement techniques?

Please speculate what drives cost chemicals, consumables or instrument time.

As a starting point, THC testing is advertised online for the following prices:

--Green Leaf Lab: $75

--Pharm Labs : $75

These companies both use HPLC. Without looking anything up, I'd guess that: The machine is around $100,000. It measures a sample every 30 minutes. (There might be modern marvels of engineering that run several samples in parallel and run way faster - idk.) Solvents and consumables are cheap. Maintaining the HPLC is expensive. You need a trained technician to prepare samples, run the analysis, and troubleshoot. That's expensive. You also need someone to deal with customers, which can be time-consuming as well.

You are not the only person thinking about THC testing. Google finds a handful of articles on what a business opportunity it is, and equipment manufacturers are ready to sell you stuff.

Good luck!


Why are the vinylic bonds in alkenes slightly polar? when searching in the internet about why are alkenes more polar than alkanes (but still they are non-polar), they are always saying that the double bond is more polarizable , i understood the whole story about how the sp2 attract electrons from the sp3 and forming london forces, but they always said that vinylic bonds are slightly polar but i never understood why.

Maybe because of resonance:

$H_{2}C=CH_{2}$    $\leftrightarrow$    $H_{2}C^{-}-C^{+}H_{2}$


Stability of tertiary carbocation Our teacher told us that a carbocation connected to 3 cyclopropane rings is among the most stable carbocations. But I am unable to understand how, as resonance (the most important factor governing stability) is absent. How did other factors bring so much stability?

Your teacher is correct, that

  a carbocation connected to 3 cyclopropane rings is among the most stable carbocations.

This because of various electrical effects, such as:

The Inductive Effect
Resonance
The Mesomeric Effect
Electromeric Effect
Hyperconjugation

Here's a link for you, to study them: Click Here.
Please make sure that you learn this correctly and from standard resources, in order to avoid confusion.

There may be a little confusion between mesomeric effect and resonance effect, so here's another link just to avoid confusion. : refer to the answer by Maria Dar

Please, do ask your teacher to help you understand this topic better :-)


Un-denaturing industrial alcohol I have a question regarding the denaturation of industrial ethanol with methanol to discourage its consumption.

The boiling point of methanol is 64.7 °C, while the boiling point of ethanol is 78.37 °C. Wouldn't it be trivial then, to separate the methanol by simply distilling industrial ethanol at 64.7 °C ?

Is there some technical reason as to why this wouldn't work? Because if it does work, then why isn't it a widespread practice among criminal elements and such.

(I'm no chemist, but here's my take on it)

Let's see: I’m an alcoholic, and I really hate having to dole out large amounts of cash to purchase your usual "drinking"-ethanol.

So I get this amazing idea to separate the ethanol by distillation (methanol has a lower boiling point than methanol, so what I'll really be doing is distilling out the methanol, leaving behind ethanol and water). Pretty smart, right (for someone who's still inebriated)? Thus invigorated, I sally forth... and return with a dozen gallons  of industrial (denatured) ethanol. However, as I slowly get sober, I begin to see two major obstacles to my plan:

It’s going to be very, very difficult to maintain a constant temperature of exactly 64.7 °C (boiling point of methanol). Now, even though ethanol boils at around 78 °C, trying to maintain the denatured alcohol at under (say) 70 or 75°C isn’t really an option either. Remember, what I’m trying to do here is criminal (where I'm from at least), plus I don’t really have a Nurdrage-worthy laboratory. If say, I were to simply buy a large metal vat with a distillation apparatus clamped on the top and light a fire below, at such a large scale (a dozen gallons in my case) and with the non-professional equipment I use, it’ll be impossible to ensure that the vessel is uniformly heated at a specific temperature. So what’ll happen is, I’ll end up distilling out a mixture of ethanol and methanol, and the ethanol that remains will also have significant amounts of methanol in it (10ml of methanol's enough to make you go blind… just to give you an idea about what I mean by “significant”). Regardless, I wouldn't want to engage in something that involves producing methanol vapors (be it in a proper laboratory or otherwise)

Say I did manage to get hold of the required laboratory apparatus (I, uh, happen to know this cancer-stricken, bald, high-school Chemistry teacher who was willing to lend me them), and I were to distill the alcohol in small batches to ensure I evenly heat the solution at the desired temperature, but the whole process would be so time-consuming and labor intensive that it would probably just be better if I just went and bought "drinking"-ethanol in the first place.
(Should have mentioned this first, but since you were talking about methanol, ah well…) To begin with, I probably wouldn't know what the manufacturers actually used to denature the alcohol. They could have used methanol, pyridine, or some other foul-tasting, foul-smelling additive for all I know, or maybe even mixed all of them together. So trying to get ethanol out of that mess, is going to be a HUGE pain in the neck.

Also you might want to read up on denatured alcohol here.

I wasn't able to find anything on ethanol-methanol azeotropes in this Wiki page, so if anyone knows where I can find something on it, gimme a shout! :-)


How to calculate the new concentration of a solution after adding more solute and convert it to ppm? My question is the following:

  $2~\mathrm{g}$ table salt is added to $0.5~\mathrm{m^3}$ water whose salt concentration is $10~\mathrm{mg\,L^{-1}}$. Compute the salt concentration of the mixture in ppm.

I am fairly new to chemistry actually this is my first time taking a chemistry course of this level so I'm having quite a hard time. Therefore, if anyone could recommend some websites where I could receive help or even videos, such as khanacademy that would be greatly appreciated!

As for the following question all I know is that $2~\mathrm{g} = 2000~\mathrm{mg}$
and ppm is calculated by:

$$\text{ppm} = \frac{\text{grams of solute}}{\text{grams of solution}}\times  1000000 $$

However, im confused as to how much grams of solute and grams of solution there will be in the mixture after adding the $2~\mathrm{g}$ of salt?

Let's think about it :

You have the formula of the mass concentration $C_m$:

$C_m = \frac{m}{V}$,

where $m$ is the mass in grams of solute and $V$ the volume in liters of the solution.

A first step for you would be to calculate the mass of solute already in the solution : your solution has a volume of $0.5~\mathrm m^3$, which is $V=500~\mathrm L$, with a massic concentration of $C_m=10~\mathrm{mg\cdot L^{-1}}$. Therefore, there is already a mass

$m=C_m~\times~V = 10\cdot10^{-3} \times 500 = 5~\mathrm g$

of salt in your solution.

To this mass, you add $2~\mathrm g$ of salt, so the mass of solute in your solution is now $m=5~\mathrm g + 2~\mathrm g = 7~\mathrm g$.

So, at end you have a mass $m=7~\mathrm g$ of salt.

I let you compute the mass of solution, knowing that $1~\mathrm L$ of water weights $1~\mathrm{kg}$.


Bond Angles of Hypofluorous Acid - Gas and Solid Phases This is the gas-phase structure of hypofluorous acid:

The solid-phase structure however has a bond angle of 101 degrees. Why is there a difference in the solid and gas phase bond angles?

Is this due to packing in the solid-phase? In the solid-phase the molecules should be much closer together than in the gas-phase, so I expect more intermolecular forces and interactions. However, what exactly would be responsible for the angle expansion in the solid phase? From one book I read that bond lengths often differ among the phases.

So what exactly is behind the bond angle expansion in the solid phase?

It is due to hydrogen bonding in the solid but not gas phase, in analogy to what occurs for the water bond angle.

From a VSPER point of view, the oxygen atom of one molecule would form a hydrogen bond with the hydrogen of another molecule, decreasing the repulsion of the lone pairs to be closer to that of the covalently bonded pairs, so the geometry would be closer to tetraherdral in the solid phase than in the gas phase.


How are cathode rays produced? Why cathode rays are produced if pressure of the gas discharge tube is lowered to about $10^{-4}$ atm?
How bombardment of glass of gas discharge tube from cathode ray result in a faint glow?

"Cathode rays" are simply beams of electrons. The name exists because cathode ray devices predate the discovery of the electron. The sharp shadows produced by Crookes tubes meant that something was travelling in a straight line from the cathode down the tube, so they were named cathode rays.

The vacuum doesn't have much to do with the production of electrons, per se, but without high vacuum, the electrons collide with gas particles before they travel very far. If there's enough gas, you get a plasma rather than a beam of electrons. The electron beam is produced by thermionic emission in hot cathodes—electrons are liberated from a hot filament when the thermal energy is enough for electrons to exceed the filament's work function. In cold cathodes, electrons are released by field emission, where a large applied electric field allows electrons to tunnel away from a metal electrode.

The glow of the glass envelope is luminescence—either fluorescence or phosphorescence (or both). Basically, the electrons striking the glass excite components of the glass to an elevated electronic state which releases light upon returning to the ground state. Normal glass isn't particularly luminescent so tubes designed to produce light (like a CRT) are coated with phosphors.


Sulfurous acid tautomerisation In the structure of $\ce{H2SO3}$

there is a lone pair on sulfur.
Why can't sulfur make a bond with one of the hydrogen atoms and a double bond with oxygen to avoid having a lone pair?
Such as given below:

Because here there will be sp3 hybridisation..... no lone pair, 4 sigma and 2 pi bonds.

Most of what I would be saying now is stuff I already highlighted in another answer of mine to a related question.

$\ce{H2SO3}$ exists in two forms, one of which is equivalent to the structure you drew, the other is the structure you proposed. Taking the monoanions (because that was the focus of the other question and I am too lazy to redraw the picture) will give you these two structures:

(All horizontal bars are formal $-$ signs, never lone pairs.)

The crystal structure of $\ce{CsHSO3}$ was solved in 1980 by Johansson, Lindqvist and Vannerberg; they conclude that $\ce{CsHSO3}$ is not cesium hydrogen sulphite but rather cesium sulphonate.[1] See the other answer for details.

In cis-bisbipyridylbishydrogensulphito-κS-ruthenium(II) $\ce{[Ru(bpy)2(HSO3}\unicode{x2D}\unicode[Times]{x3BA}S\ce{)2]}$, a crystal structure solved by Allen, Jeter, Cordes and Durham, the sulphur atom directly coordinates the ruthenium centre which is only possible if it has a lone pair; thus we must be dealing with hydrogensulphite.[2]

Therefore, depending on the conditions both forms are possible and observed.

In case of the free acid, it is likely more stable in one of the two forms. I do not have analytical or thermodynamic data at hand to tell you which tautomer is preferred.

References:

[1]: L.-G. Johansson, O. Lindqvist, N.-G. Vannerberg, Acta Cryst. 1980, B36, 2523. DOI: 10.1107/S0567740880009351.
[2]: L. R. Allen, D. Y. Jeter, A. W. Cordes, B. Durham, Inorg. Chem. 1988, 27, 3880. DOI: 10.1021/ic00295a003.


Comparing explosive properties of mercury(II) cyanate and mercury(II) fulminate I found this question online on Jiskha Homework Help:

Of the compounds mercury(II) cyanate, $\ce{Hg(OCN)2},$ and mercury(II) fulminate, $\ce{Hg(CNO)2},$ one is highly explosive, the other is not. Explain.

My teachers says that most explosives are made of nitro compounds as abnormal expansion in $\ce{N-O}$ bond has been observed in nitro compounds above a certain temperature. So, fulminate will be more explosive as it contains $\ce{N-O}$ bond.
According to Wikipedia, explosiveness of nitro and other nitrogen containing compounds (like $\ce{NI3})$ is because of possibility of forming highly inert and stable $\ce{N2}$ molecules. Thus slight amount of energy causes breaking of relatively weaker $\ce{N-O}$ bonds and highly exothermic formation of $\ce{N2}$ triple bond making it act like an explosive.
Is my teacher correct? The reasoning given on Wikipedia also points to fulminate but some other concept may be at play here. So, which is correct? Also what would be the reaction mechanism if any?

This can be attributed to the formal charges of fulminate ion and cyanate ion. Structural stability of a compound sometimes depend on formal charge.

We observe that formal charges of fulminate ion is higher than cyanate ion for which fulminate ions are said to have a less favorable molecular configurations which leads to facile decomposition of the fulminate ion.
Fulminate vs cyanate stability has been demonstrated by using their corresponding silver compounds. Here is a video demonstration.


What is the order of the catalytic decomposition of hydrogen peroxide? I recently did an experiment to determine the activation energy of the decomposition of hydrogen peroxide using iron (III) chloride as a catalyst. I repeated this experiment at different temperatures. I know how to figure out the rate constant and the rate law, but I am not sure if this is a second-order reaction. Is there any way I can figure out the order of this reaction? (because that would impact the units of the rate constant, k, and the rate law)

Check out this paper. There they studied the decomposition of hydrogen peroxide by iron hydroxide and found that the reaction is a second order reaction for low concentrations of hydrogen peroxide.

In your case, you are working with a soluble iron species which will change the reaction kinetics because the hydrogen peroxide does not need to adsorb to the surface of insoluble iron hydroxide particles. This is assuming you worked under neutral or better slightly acidic conditions, otherwise the iron will precipitate as iron hydroxide.

Therefore, in your case the kinetics should depend on the relative concentration of iron chloride and hydrogen peroxide. As long as the hydrogen peroxide is in excess, I would expect a zero order reaction because the iron will limit the reaction speed. If the iron is in excess, I would expect a second order reaction because you need two molecules of hydrogen peroxide to form molecular oxygen.

How you experimentally determine the reaction order is described on Wikipedia. Briefly, you need to perform the experiment at different concentrations of hydrogen peroxide and measure the initial reaction speed.  Then you plot the logarithm of the initial rates against the logarithm of the concentrations. The slope of the resulting line should be your reaction order.


what's the name of this org. compound? What's the name of this compound? It is a semi product of Reimer-Tiemann reaction.

edit: I think its Sodium salicylaldehyde, but i'm not sure.

That's supposed to be the sodium salt of salicylaldehyde, also known as sodium salicylaldehyde, sodium ortho-formylphenolate, 2-hydroxybenzaldehyde sodium salt or sodium 2-formylphenoxide. CAS number: 3116-83-4.

There's actually an error in the structure. The oxygen should have a negative charge if the sodium is written with a positive charge. The salt as a whole is neutral. What's written here is one lone pair, there should be 3 of those and a negative charge.


Closed shell molecules can have degenerate states? I read a paper which said that $\ce{CO2}$ despite having orbital degeneracy is stable and not subject to the Jahn-Teller effect because it is a linear molecule. But $\ce{CO2}$ is a closed shell molecule, so how can it have orbital degeneracy?

Degeneracy is usually based on symmetry. You can map the binding orbitals for each bond onto the other bond. So you can't distinguish between the molecular orbitals and its energies.

Jahn-Teller doesn't apply for linear molecules because there is no distortion which can reduce the symmetry and reduce the system energy as well.


What is the largest non-polymeric heterostructured molecule possible? I must admit I'm not a chemist but a biologist, so please forgive me if I'm getting some of my terminology wrong. I want to ask a question about the biggest possible complex molecule that can exist in Nature.

I know that certain crystals like quartz and gems like ruby and diamond can grow very large and they could be called molecules. Also, polymers can grow very long with a complex 3D structure.

However my question is about a chemical compound which isn't a polymer but has a complex non-repetitive 3D structure of many covalently bound atoms. If we rely only on synthesis and try to construct the biggest molecule possible how big can we get?

Does the answer vary if we are talking about molecular size, molecular mass, number of atoms, 3D structure (I imagine there can be some very big molecules containing cavities which are lighter and have fewer atoms than other who have more atoms and are heavier but are dense) and so on ?

Or is there a single class or maybe even a single molecular species which combines all these characteristics and can produce the largest molecules no matter what characteristic we use to measure this quantity?

If there is such a class of compounds what it is and how do we synthesize it?

We know of no inherent limitation concerning the size of a molecule. As you pointed out, certain macroscopic entities like diamonds, graphite or charcoal can be considered fully macroscopic molecules to a certain extent (especially diamond). If chemists let their imagination run loose, they can draw whatever size a molecule you wish to create and with the desired complexity.

Whether it can be synthesised is another matter. Even very small molecules, especially natural products, can be very resistant against attempts to synthesise them. The larger a molecule gets, the more mass a rather small sample will have. It will get increasingly tedious but by no means impossible. However, unless you have oil money, Silicon Valley money or a billionaire funding you, you will likely not get the chance to finish what you want.

One of the most well-known large and not inherently repeating (or so Nicolaou says) molecules is maitotoxin whose total synthesis is famously still being attempted by the Nicolaou group.


8a-methyl-1,2,3,4,4a,8a-hexahydronaphthalen-4a-ylium carbocation rearrangement In this reaction after the attack of lone pairs on $\ce{H+}$ ions, a stable $3^{°}$ carbocation is formed. But seeing the six membered ring and the double bonds already present, I can't help but think that there's some way of obtaining a benzene ring through rearrangements. Can someone suggest a mechanism for this?

While there are examples of 1,2-methide shifts, this question has an uncanny resemblance to the dienone-phenol rearrangement whose mechanism was first elucidated by Woodward and Singh in 1950. Dienone 1 under acidic conditions undergoes rearrangement to phenol 6 and not, based on earlier speculation, to phenol 4. The direct 1,2-methide shift (2b --> 3) does not occur but rather the reaction proceeds through the spiro carbocation 5. continued

Applying this approach to the carbocation 7 generated from the alcohol in this question, spiro carbocation 8 forms the predicted tetrahydronaphthalene 10. One way to distinguish between the two mechanisms is via a labeling experiment. All of the carbon label in 7 will retain its location as the red star in 10 if the 1,2-methyl shift mechanism applies. The spiro mechanism will partition the label ~50:50 between the two ring benzylic carbons. For related studies, see reference 2.
1) R. B. Woodward and T. Singh, J. Am. Chem. Soc., 1950, 72, 494.
2) A. J. Waring, J. H. Zaidi and J. W. Pilkington, J. Chem. Soc, Perkin Transactions I, 1981, 1454.


Bond Angles and intermolecular interactions It's fun to consider bond angles of molecules in isolation. For example, what is the number of different bond angles exhibited by the most polar form of $\ce{SIBrClF2}$?

However, can bond angles really be considered in isolation? The method I would employ to determine the number of different bond angles only considers the molecule in isolation; I just consider van der Waals repulsions between the component atoms. Molecules generally do not exist in isolation. We can study molecules in the gas phase as to minimize intermolecular forces, but still, intermolecular forces exist, as it is hard to isolate exactly one molecule.

In the case of $\ce{SIBrClF2}$, I would expect a number of rather strong dipole-dipole interactions among the $\ce{SIBrClF2}$ molecules.

Therefore my question is: are there any cases in which considering solely intramolecular forces leads one astray when determining the bond angles of a molecule? I.e. one might look at a molecule and expect a certain bond angle to be < 180 degrees but find out the angle is in fact > 180 degrees due to intermolecular forces prying the bond angle open?

When people talk about bond angles in molecules, they are usually talking about the equilibrium bond angle in isolation. This can be derived from the energy minimized structure of the molecule in isolation.

As you point out, the angles in a real, non-isolated molecule will fluctuate due to interactions with neighboring molecules. However, the actual angle formed by three successive atoms in a molecule will naturally fluctuate even if the molecule is isolated and has some kinetic energy. Everything's always jiggling around.


Why do different impurities create different colors in diamonds? Traces of nitrogen make diamonds yellow while traces of boron make them blue.  What accounts for the differences in color for a given element?

If you had googled a little bit you would have found this wonderful resource.

  In a yellow diamond, a few carbon atoms per million have been replaced
  by nitrogen atoms, each containing five valence electrons. The
  structure of the diamond crystal does not change significantly, but
  the extra electrons occupy a donor level.

  The nitrogen donor level energy in a diamond is large, peaking at
  about 4 eV. With a concentration of a few nitrogen atoms per million,
  instead of a clean "spike" donor level energy, the nitrogen donor
  level energy broadens into a band because of a number of complex
  factors, including thermal vibrations. This broadened donor energy
  band results in the difference between donor and conduction bands
  being as low as 2.2 eV. The most likely transition allows incident
  light with energy of 4 eV per photon to excite electrons from the
  donor level to the conduction band. However, it is possible for
  electrons to be excited to the conduction band with energy of 2.2 eV
  and upwards. This means that blue and violet light are absorbed from
  the full spectrum of light normally transmitted, and the resulting
  color is yellow.

  Unlike blue boron-doped diamonds, which conduct electricity,
  nitrogen-doped diamonds remain insulators. This is because nitrogen is
  a deep impurity: a relatively high energy (2.2 eV, as compared with
  0.4 eV for boron) is required to excite electrons from the donor energy level to the conduction band, and only a fraction of the
  available electrons are freed to carry a charge.

  An extremely rare green color can result from a higher nitrogen
  content of about 1 atom per 1000 atoms of carbon. At even higher
  nitrogen concentrations, the donor level broadens so that all visible
  light can be absorbed, resulting in a black color.

  Synthetic blue diamonds are created by adding boron as an impurity.
  Boron is trivalent, having three valence electrons; where a boron atom
  replaces a carbon atom in the diamond structure there is one fewer
  electron than usual. This missing electron, or hole, creates an
  acceptor energy level above the valence band. The boron acceptor
  energy is only 0.4 eV, so light of any energy can be absorbed during
  excitation. The boron acceptor band is broadened, and the absorption
  tapers off throughout the visible light energies, resulting in
  stronger absorption at the red end of the spectrum. At a level of one
  or a few boron atoms for every million carbon atoms, an attractive
  blue color results. Natural diamonds of this color, such as the Hope
  Diamond, are rare and highly priced.


On Heisenberg's uncertainty principle Why can't we find the exact position and velocity of a particle? And why is it that if the uncertainty in position is very, very large then the velocity can be determined, and vice versa? Please explain to me.

If you imagine particles to be like billiard balls (the way you're taught in school) then this of course makes no sense. The only way you can understand it is to accept the quantum mechanical description of particles: that they are probability distributions that exhibit both wave-like behaviour and particle-like behaviour depending on how you measure them.

Once you accept that a particle is best imagined as a probability wave smeared out through space then the Heisenberg uncertainty principle is a rather simple mathematical result relating to Fourier transforms. The more concentrated some function $f(x)$ is, the more spread out its Fourier transform $f(k)$ is. And since position and momentum space are Fourier transforms of each other (see here) then this gives rise to the uncertainty principle.

In other words, if you measure the position very precisely (highly peaked $\psi(x)$) then the momentum representation $\psi(k)$ will become very broad (imprecise), and vice versa.


What is the electron configuration for Og with a charge of 2-? I was writing a cool program that determines the electron configuration of elements for me, and I wondered what would happen if you went over 118 — the "limit" of my program.  So, what is the electron configuration after you go over 118 electrons?

According to SCF Relativistic Hartree–Fock Calculations on the Superheavy Elements 118–131 J. Chem. Phys. 53, 2397 (1970), for neutral atoms:

119 $8s$

120 $8s^2$

121 $8p8s^2$

122 $7d8p8s^2$

123 $6f7d8p8s^2$

124 $6f^38p8s^2$

125 $5g6f^38p8s^2$

126 $5g^26f^28p^28s^2$

127 $5g^36f^28p^28s^2$

128 $5g^46f^28p^28s^2$

129 $5g^56f^28p^28s^2$

130 $5g^66f^28p^28s^2$

131 $5g^76f^28p^28s^2$

Later calculations reported in Electronic Configurations of Superheavy Elements J. Phys. Soc. Jpn. 65, pp. 3175-3179 (1996) are different starting with element 122:

122 $8p^28s^2$

123 $6f7d8p8s^2$

124 $6f^28p^28s^2$

125 $6f^48p8s^2$

126 $5g6f^48p8s^2$

127 $5g^26f^38p^28s^2$

128 $5g^36f^38p^28s^2$

129 $5g^46f^38p^28s^2$

130 $5g^56f^38p^28s^2$

131 $5g^66f^38p^28s^2$

However, Electronic structure of eka-thorium (element 122)
compared with thorium J. Phys. B: At. Mol. Opt. Phys. 35 (2002) 1693–1700 finds:

122 $7d8p8s^2$

specifically agreeing with the 1970 paper and disagreeing with the 1996 paper.

See A suggested periodic table up to Z[=]172, based on Dirac–Fock
calculations on atoms and ions Phys. Chem. Chem. Phys., 2011, 13, 161–168 for additional information.


What is the mechanism of ring contraction of 6-bromo-7-methoxy-2,3,4,7-tetrahydrooxepine from seven to five? I have recently come across a paper from here.I have tried to workout the mechanism for the following reaction presented in the paper.

I was able to workout mechanism till acetal formation.It is a ring expansion from six to seven. However,the next step is a ring reduction from seven to five upon heating.

Can someone please explain what is actually happening during ring reduction, and why it happens?

It's in scheme 3 of the paper you cited! The 7-membered acetal undergoes ring-opening to the α,β-unsaturated oxocarbenium ion, then conjugate addition to give the 5-membered tetrahydrofuran. A bit of arrow pushing and addition of solvent gives the other THF product

Not sure the legality of posting screenshots of papers, so I redrew the scheme:


Determine vmax and enzyme concentration (Michaelis–Menten) I am given that the enzyme concentration is $\pu{15 nM}$ and the following data:
$$
\begin{array}{rl}
\hline
S/\pu{mM} & V/\pu{mM s^{-1}} \\
\hline
1 & 0.202 \\
2 & 0.368 \\
5 & 0.729 \\
10 & 1.08 \\
20 & 1.43 \\
50 & 1.77 \\
100 & 1.92 \\
\hline
\end{array}
$$
And calculated that the Michaelis–Menten constant was $K_\mathrm{M} = 9.41$ (unsure about the units).
I then want to calculate $v_\mathrm{max}$ and $[\ce{E}]_0$ for the same reaction given that the reaction rate is $\pu{0.101 mM s-1}$ at a substrate concentration of $\pu{24 mM}.$
I know that once I have $v_\mathrm{max}$ I could just use $v_\mathrm{max} = k_\mathrm{b}[\ce{E}]_0$ to calculate the enzyme concentration, but I am unsure as to how I can calculate $v_\mathrm{max}$ from one set of data. Could someone give me a hint of how to begin this problem?

Michaelis-Menten kinetics is given by
$$V = V_{\max} \frac{S}{K_M + S}$$
Taking the inverse of both sides to linearise the equation,
$$\frac{1}{V} = \frac{1}{V_\max} +\frac{K_M}{V_{\max} S}$$
Taking $y = 1/V$ and $x = 1/S$ in $y = A + Bx$ leads by linear regression to $A = 0.4767$, $B = 4.4754$ (not including units). Therefore,
$$V_\max = \frac{1}{A} = \pu{2.10 mMs^{-1}}$$
$$K_M = B V_\max = \pu{9.39 mM}$$
For the new situation with different $V$ and $S$ and necessarily different $[E_0]$ , $V_{\max}$ would change because $V_{\max} = k_b [E_0]$ but $K_M$ remains constant, thus
$$V_{\max} = V \frac{K_M + S}{S} = \pu{0.101mM s^{-1}} \frac{(\pu{9.39 mM} + \pu{24 mM})}{\pu{24 mM}} = \pu{0.141 mM s^{-1}}$$
You can then calculate $k_b$ by $k_b = V_{\max}/E_0$.


Concentration of dissolved lead(II) bromide Calculate the concentration of $\ce{Pb^2+}$ and $\ce{Br-}$ when $\ce{PbBr2}$ dissolves in water. $K = 7.9 \times 10^{-5}$

$$\ce{PbBr2 <=> Pb^2+ + 2Br-}$$

In most cases we have the concentration of the solute, but here we don’t. How should I be thinking now?

if it’s $1~\mathrm{M}$ for example

$$K = \frac{[x][x]^2}{[1-x]}$$

The given value $K=7.9\times10^{-5}$ actually is the value of the solubility product constant $K_\mathrm{sp}$ of $\ce{PbBr2}$

$$K_\mathrm{sp}=a_{\ce{Pb^2+}}\cdot a_{\ce{Br-}}^2=7.9\times10^{-5}$$

which approximately is (when incorrectly neglecting activity coefficients $\gamma$, the difference of molality $b$ and concentration $c$, and the correct units)

$$K_\mathrm{sp}\approx \left[\ce{Pb^2+}\right] \cdot \left[\ce{Br-}\right]^2$$

When $\ce{PbBr2}$ dissolves in water according to the given chemical equation, the resulting concentrations of $\ce{Pb^2+}$ and $\ce{Br-}$ correspond to

$$\left[\ce{Br-}\right]=2\cdot\left[\ce{Pb^2+}\right]$$

These equations can be used for calculating a value for $\left[\ce{Pb^2+}\right]$.


How is it possible to quantify the strength of a Lewis acid/base? When the $pK_a$ scale is not applicable - when there is no proton/charged species transfer - what should be used to quantify the strength of an acid?

$$\ce{A~  + :\!\!B -> \ \ A\!\leftarrow:\!\!B}$$

$$K_f = \dfrac{\ce{[A\!\leftarrow:\!\!B]}}{\ce{[A][B]}}$$

See sections 4.1 and 4.2 in the following reference for more information:

http://goicoechea.chem.ox.ac.uk/teaching/abse/abse_lecture2.pdf

But $K_f$ is specific to an acid-base pair.

A more universal scale is formed by the Gutmann–Beckett method by observing the phosphorous NMR spectrum of triethylphosphine oxide as it interacts with a sample lewis acid.

An Acceptor Number (AN) scale is formed with AN=0 corresponding to triethylphosphine oxide in hexane (P chemical shift of 41), and AN = 2.21 (d-41), where d is the chemical shift of triethylphosphine oxide interacting with the sample. The greater AN is, the more acidic the compound.


Diluting topical cream (10%) I need to dilute a face cream with 10 % active ingredient to 2.5 % and 5 % but I am confused as to how. I looked into the $$c_1V_1=c_2V_2$$ and $$\text{concentration of a solution}=(\text{amount of solute}/\text{amount of solvent})\times100$$ but not sure which would work.

Would I have to make a standard solution first and place 10 g of the cream into 100 ml of water?

To practically dilute this cream from $10\ \%$ to $5\ \%$, you would need to dilute it to a by a factor of $2$. You can see this by assuming an initial volume for your cream, in this case, $100\ \mathrm{ml}$,
and plugging into the $c_1V_1=c_2V_2$ equation. In this case that means
$$c_1:=10~\%~~~~~\\V_1:=100~\text{mL}\\c_2:=5~\%~~~~~~~$$
Solving for $V_2$ renders
$$~~~~~~~~\frac{c_1V_1}{c_2}=V_2$$
$$\frac{10\ \% \cdot 100~\text{mL}}{5\ \%}=V_2~~~~~~~~~~$$
$$\frac{10\ \% \cdot 100~\text{mL}}{5\ \%}=200~\text{mL}$$
Which means that the volume needed in the final solution needs to be $200~\text{mL}$. It is however, important to note that this does not mean that you need to add $200~\text{mL}$, but rather you need to add $100~\text{mL}$ of solvent to your $100~\text{mL}$ of cream for a final solution volume of $200~\text{mL}$.

Further you may calculate the dilution factor that this correlates to by using the equation

$$DF=\frac{V_\text{final}}{V_\text{initial}}$$

In this case $V_\text{final}=V_2$ and $V_\text{initial}=V_1$, so plugging into this equation renders

$$DF=\frac{200~\text{mL}}{100~\text{mL}}=2$$

You may apply these same equations to making a cream at $2.5~\%$ active ingrediant concentration and should find that if you start with the $100~\text{mL}$ of the $10~\%$ stock, you would need to dilute to a final volume of $400~\text{mL}$, which correlates to a dilution factor of $4$. If you start with $100~\text{mL}$ of the $5~\%$ solution you just made, you would need to dilute to a final volume of $200~\text{mL}$, which correlates to a dilution factor of $2$.

If you want to do this practically, then water may not be the best solvent – to continue to be able to use this as a cream, you may consider using a cream without any active ingrediant in it as your solvent – maybe something like a hand cream?


Mechanism of Water Potential The idea of water potential still confuses me. I understand the general idea: water moves from an area of higher water potential to an area of lower water potential. However, I still do not understand the mechanism by which this occurs.

Take the example of putting salt on a slug. Putting salt on a slug lowers the dilute potential which in turn lowers the water potential on the outside of the slug. This lower water potential on the outside then draws water from the higher water potential of the inside of the slug.

I do not understand how this occurs. All the answers I find is that this just “occurs” because “nature wants to find equilibrium” (such as the second response on this Quora question). That makes sense in theory, but how does the cells of the slug “know” that there is now a lower water potential? What has changed?

If I understand you correctly you are asking why osmosis happens the way it does. And obviously you want a more "real-life-approach" to this than the pure mathematically description via potentials.

Think about a cell filled with water that is divided by a permeable membrane that will only let water molecules pass through it. In this initial state the water molecules on both side will occasionally pass through the membrane switching sides. They do so by pure chance and as the particle-concentration on both sides are equal (there are 100% water particles on both sides) the pass through on both sides with an equal rate leaving the overall particle-distribution unchanged.
Now we go ahead and dilute salt in one half of the box. Leaving aside effects like hydration-hulls forming around the ions (which will magnify the effect described below) the effect of this dilution is that we now have a lower water-particle-concentration on that side. Therefore every now and then when a water particle would have been "thrown" at the membrane to pass through it it's not a water particle but an ion instead that can't pass the membrane. Therefore the rate at which the water-particles switch to the non-dilution-side is lower than the (unchanged) rate in which the water-particles flow into the dilution (by pure statistics). This leads to an increase in the overall particle count on the dilution-side.
This increase in particles also increase the pressure and therefore the rate at which random particles (on the dilution side) are being "thrown" at the membrane until this higher rate ultimately compensates the lower water-particle-concentration. At this point both sides are at an equilibrium again. However this equilibrium is in a state in which there are more water-particles on the dilution-side than there have been in the beginning.

And this is what we call osmosis.


Solubility of CHCl3 in water Why isn't $\ce{CHCl3}$ soluble in water?

Why doesn't $\ce{CHCl3}$ participate in hydrogen bonding in water like $\ce{CH3CN}$ as $\ce{Cl}$ can make the $\ce{C-H}$ bond polar like $\ce{CN}$ does?

Chloroform is soluble in water, though only slightly (~1g/100ml in cold water forming a slightly sweet liquid whose mild anaesthetic affects made it a recreational substance in Victorian times before its toxicity was fully recognised).

Moreover, its hydrogen will form (weak) hydrogen bonds which is probably a better explanation for the solubility in water than the polarity of the molecule.


How to determine electrolyte concentration in solutions? The original question is here

  We've been given the sports drink vs orange juice challenge. To find out which has more electrolytes. I could test for conductivity but that alone wouldn't tell me which has more actual electrolytes in it. So I'm wondering if there's a way I could separate each ion from each other and measure them that way?

I'm posting it here because I'm very curious as to the answer.

Several people have suggested evaporation.

  You can try evaporating the liquids and seeing what's left behind. The electrolytes will be left behind as solids.

However, the issue with evaporation is that it leaves behind all solids - electrolyte or not.

Some people have suggested conductivity measurements.

  Wouldn't conductivity simply be proportional to the concentration of substances that promote conductivity?

However, this objection was raised:

  Well my understanding was that different ions produce different voltages, therefore, there could be a lot of one type of electrolyte producing a large voltage but there could be more electrolytes in the other solution producing a smaller voltage. I'm not sure though as I am in first year and haven't had this completely covered yet. My professor said that conductivity might not give enough information so I may need a second approach as well, I'm not sure what he means exactly.

Is this a valid point? I guess we could empirically test out this assertion that "different ions produce different voltages" by comparing the conductivities of two otherwise identical NaCl and, say, MgBr2 solutions. Wouldn't the MgBr2 have a conductivity measurement that is roughly 1.5 times that of the NaCl solution? Or is it true that the ion's charge affects the conductivity?

Electrolytes turn water into a conductive solution.  Conductivity is affected by how free the electrons are to move.  Conductivity (which is the reciprocal of resistivity) is affected by by concentration, temperature, and how much charge the ions carry.  Conductivity alone will give an equation with both ions and concentrations; however, this equation will have multiple solutions. For some fun understanding this phenomena look at this this link.

Conductivity is different than voltage.  Voltage in this situation is likely referring to a half-cell reaction with a standard hydrogen electrode in a galvanic cell.  Using this approach has the same issue with a multiple solution equation.

Electrolytes are typically ions.  Ion concentration is measured with ion chromatography.  Ion chromatography involves separating the ions from the solution based on various properties to perform a measurement. It is explained here, here, and here.

When electrolytes of drinks are measured it is important to include if the electrolytes are bio-available.  Bio-available electrolytes depend on the activity performed.  An activity such as sports uses a different set of electrolytes than sitting.

Testing bio-available electrolytes requires more work than analytic chemistry. Analytic results both the drink and bodily fluids of a population doing a certain activity are need to draw conclusions about the electrolyte composition of a drink.

Reading a patent for a sports drink ions include $\ce{Na+}$, $\ce{K+}$, $\ce{Mg^2+}$, $\ce{Cl-}$, $\ce{SO4^2-}$, $\ce{PO4^2-}$.  The patent also includes citrate, sucrose, dextrose, ascorbic acid and pyridoxine as these are known to affect the uptake of electrolytes.

If you have access to a university library the values of these compounds for orange juice are available.


Why does sulfonation of alcohols even work? I am now near the end of organic chemistry 2, and I'm trying to understand the patterns of reactions so that I can predict them rather than brute force memorizing all of them. Looking at the sulfonation of alcohols using mesyl chloride, I don't think I would ever predict that. I don't understand why the oxygen should attack the sulfur in mesyl chloride. To me it looks like a nucleophilic substitution reaction, but alcohols aren't even powerful nucleophiles. Not to mention that the end of this reaction creates HCl a very powerful acid. I thought when reactions occur they like to create relatively weaker acids and bases? What is driving the reaction, and what patterns should I be noticing here for future predictions?

There are always two sides of a reaction. In the case of an alcohol attacking a sulphonic acid chloride, we can look closely at the alcohol and say: ‘Oh my, this is a bad nucleophile and we cannot deprotonate it with the conditions we have. Probably, this reaction won’t work’. But we can also look at the sulphonic acid chloride and say: ‘Oh my, this is a very strong electrophile and it has a very good leaving group. This reaction will work like a charm!’

So it is important, that you compare the reactants extensively. In this case, the electrophilicity of the sulphur(VI) is high enough to even allow rather bad nucleophiles to work. On the other hand, oxygen atoms aren’t even that bad a nucleophile. Give them enough time and alcohols will attack via $\mathrm{S_N2}$ — it might just take forever.

We can also take electrostatics into consideration. Oxygen is rather negatively polarised since it has a very high electronegativity. The sulphur atom, on the other hand, is strongly positively polarisied, since it is connected to two formal oxide anions and a chlorine atom — all three are more electronegative than sulphur by a noticeable margin. A rather negative particle such as the alcohol oxygen, will happily attack a rather positive particle such as the sulphonic acid chloride sulphur.

Mechanistically, this attack forms an intermediate pentavalent sulphur which will then displace the best leaving group: The chloride atom. Therefore, even the hydrochloric acid that is produced isn’t as bad as it may look. We loose the chloride first and the proton much later. We aren’t immediately removing a strong acid but rather a good leaving group and the proton just has to deal with it.

However, I will allow the objection that a significant concentration of $\ce{HCl}$ would favour the reverse reaction. Thus, when employed synthetically, a base is usually added that serves no purpose other than to capture the proton and keep it somewhere where everybody can be happy.

Too-long-didn’t-read-version:

Alcohol might be a bad nucleophile, but oxygen is negative.
Sulphonic acid chloride is a very good electrophile and the sulphur atom is positive.
Positive and negative like each other, thus alcohol will attack sulphur.
Chloride is displaced because it is a pretty good leaving group.
The proton just has to do what the others force it to do.
Driving force is the liberation of the chloride and turning a weak $\ce{S-Cl}$ bond into a strong $\ce{S-O}$ one.
Add a base to capture the proton for further enhancement.


Excited helium gas from the colision of alpha particles with beta particles? As alpha particles are made out of 2 protons and 2 neutrons, they are the same as the helium atom with a +2 charge, but with a certain speed. The beta particles are made out of 1 electron at a certain speed. So, theoretically, "mixing" alpha particles with beta particles would create an excited helium atom, which then, would decay to a normal helium atom releasing the kinetic energy through gamma radiation, or simply light. Is this correct? If not, why?

This is very much a non-issue. Alpha particles run at great speed, so do beta particles, and as soon as they cool down, they cease to be called alpha and beta particles. When that happens, they may react all right, but there is more to it.

Everything on Earth (look around you! yes, that too) is made of atoms. Helium has greater ionization energy than any other atom. So an alpha particle, as soon as it cools down enough to react, will snatch electrons from whatever else is nearby and become a neutral helium atom. Beta particles (provided they are cooled down, too) are as good as anything else.

But wait, they both cool down by bumping into matter, and matter contains electrons. You will have your helium atoms, but you'll never be able to tell whether the electrons come from beta particles or from something else. Nor does it really matter, after all.


Copper and water heat capacity concern What mass of copper, originally at $50\ ^{\circ}\text{C}$ must be added to $1\ \mathrm{kg}$ of $10\ ^{\circ}\text{C}$ water to raise the water temperature by $10\ ^{\circ}\text{C}$?

We can work out that the necessary energy gain by the water is $41.8\ \mathrm{kJ}$, but I'm not sure how to work out the mass of copper, since both the mass and the temperature change is unknown in the equation $\Delta E=mc\Delta T$? Or do we assume that the copper cools to $20\ ^{\circ}\text{C}$ also?

There are all elements in your exercise to solve it expect the heat capacities but this is easy to find them.

You are doing an exercise of thermodynamics and this is not mention that the equilibrium is not reached. So if your water heats up only by $10\ \mathrm{^\circ C}$ at the end even in one million years if your system is adiabatic, your solution will still be at $20\ \mathrm{^\circ C}$. And so at the equilibrium the copper must have the same temperature than the water.

So if you use $\Delta H=mc_p\Delta T$, for water $\Delta T=10\ \mathrm{^\circ C}$ and for copper $\Delta T=30\ \mathrm{^\circ C}$.

Are you able to finish now ? :-)


Conductometric titration of Strong Acid And Strong base Consider conductometric titration of Strong Acid with strong base say HCl + NaOH.

My question is why initially the conductance decreases. If H+ of HCl is consumed by OH- of NaOH, forming water and hence conductance decreases.  I agree to these.
But On other hand NaCl is also forming in same amount which can give rise to Conductance. So in my point of view Conductance must remain same or differ slightly. Instead of decreases considerably.

You correctly identify that the number of cations do not change in solution, so from a charge perspective it appears that there should be no significant change in the conductivity.

However, consider two things:

Sodium ions are much heavier and larger than free protons and so should travel through water more slowly. Charge moving more slowly
means lower conductivity.
In addition to their smaller size and lightness, Protons can also "hop" from water molecule to water molecule and so can move through the solution much more quickly and easily. Other water molecules do not act as obstacles

Point 2 is called the Grotthuss Mechanism.


Reaction of octahydroazecine with iodine crystals I really don’t understand why the following reaction would take place first of all.

So this is a sub part of a question wherein we have to compare the basicity of the compounds formed through some reactions. So, I got the rest 3 of them but I have no clue about this one.

My concerns are..

Why would $\ce{I2}$ react in absence of any oxidising agent or catalyst?
Should the presence of $\ce{N}$ influence the reaction.
Is this reaction feasible?

  Ans -

Chlorination and bromination of alkenes are very general reactions, and mechanistic study of these reactions provides additional insight into electrophilic addition reactions of alkenes. Although much less detail is known about fluorination and iodination of alkenes, it is believed that iodination follows the similar mechanistic steps of much studied bromination (Ref.1) to give corresponding 1,2-diiodoalkane. In certain compounds, if neucleophilic center is available within vicinity of cyclic halonium center, the last step of nucleophilic attack of halide ion would be over written by much faster intramolecular cyclization as depicted by following mechanism:

This mechanism is well supported by the examples provided in Ref.1 and Ref.2.

References:

Yoshinao Tamaru, Masato Mizutani, Yutaka Furukawa, Shinichi Kawamura, Zenichi Yoshida, Kazunori Yanagi, Masao Minobe, “1,3-Asymmetric induction: highly stereoselective synthesis of 2,4-trans-disubstituted γ-butyrolactones and γ-butyrothiolactones,” J. Am. Chem. Soc. 1984, 106(4), 1079-1085 (https://doi.org/10.1021/ja00316a044).
Yvan Guindon, François Soucy, Christiane Yoakim, William W. Ogilvie, Louis Plamondon, “Diastereoselective Synthesis of 2,3,5-Trisubstituted Tetrahydrofurans via Cyclofunctionalization Reactions. Evidence of Stereoelectronic Effects,” J. Org. Chem. 2001, 66(26), 8992-8996 (https://doi.org/10.1021/jo010873r).


Why is CO2 more likely to dissolve in a basic solution than an acidic solution? I'm having trouble thinking through this problem. I know that when $\ce{CO2}$ dissolves in water it can release $\ce{HCO3-}$ and $\ce{H+}$ and cause a drop in pH. But can I use this to justify why $\ce{CO2}$ (g) might dissolve more easily in a basic solution than an acidic one? Is the water acting as the base?

A simple way to think about this is as follows (assuming we are talking about an aqueous solution).

The dissolution of $\ce{CO2}$ in water is followed by three chemical reactions:

$$
\begin{align}
\ce{CO2 + H2O &<=> H2CO3}\label{rxn:R1}\tag{R1}\\
\ce{H2CO3 &<=> HCO3- + H+}\label{rxn:R2}\tag{R2}\\
\ce{HCO3- &<=> CO3^2- + H+}\label{rxn:R3}\tag{R3}
\end{align}
$$

Now the more basic the solution becomes, the further reactions \eqref{rxn:R2} and \eqref{rxn:R3} (\eqref{rxn:R3} becomes significant at high pH) lie to the right as hydroxide consumes $\ce{H+}$. As reaction \eqref{rxn:R2} depletes $\ce{H2CO3}$, reaction \eqref{rxn:R1}  is also forced to the right. The net result is more $\ce{CO2}$ residing in solution (dissolved) as the solution becomes more basic.


how are the frequencies at a local maximum of PES like? On the potential energy surface, if you find a local maximum and calculate its frequencies in Gaussian or something like that, will you get all negative frequencies or all positive frequencies?
I know that at a local minimum the frequencies are all positive and that at a saddle point there are both negative and positive frequencies, but what about a maximum point?

The potential energy surface (PES) is a very complex thing and it is hard to be understood (if we can understand it at all). With quantum chemical approaches we can map out this surface. We differentiate between positive, real and imaginary (usually shown as negative) frequencies. There are no negative frequencies. What you may encounter analyzing the PSE, are one of the following extreme points.

Local/Global Minimum
All derivatives have to be positive, the number of imaginary frequencies (NImag) is zero $i=0$. This corresponds to an observable state of the system/ molecule, i.e. it can be trapped and analysed with spectroscopy. Exciting any of these frequencies, the energy of the molecule will increase. But you already knew that.
Transition State
A transition state is a first order saddle point on the PES. The Nimag is exactly one, $i=1$. It connects two local minima with each other. Try to imagine a molecule, like ammonia. This can undergo inversion, i.e. nitrogen inversion. The two pyramidal forms are local minima, which are connected via a planar form (nitrogen is $\ce{sp^2}$ hybridised), which is the transition state. At that point one of the frequencies is imaginary, which means, that it corresponds to lowering the energy while exciting this frequency. All other vibrations would still increase the energy of a molecule.
You cannot observe a state like this via common spectroscopic methods, but it still has a physical meaning. There are various theories in use to describe reactions of molecules. The most popular is the Transition State Theory, which was developed by H. Eyring and M. Polanyi.
Second Order Saddle point
It formally corresponds to a transition between transition states. Its NImag is $i=2$, so there are two vibrations that lower the energy. It usually has no physical meaning, as the molecule would probably decompose before reaching this state.
If one of these vibrations correspond to a rotation of a moiety, e.g. a methyl group, and is therefore close to zero, it corresponds to the so called rotational space and might be within range of the transition state.
$n$-th Order Saddle point
It is like the above a transition between two $(n-1)$-th order saddle points and has even less physical meaning. You sometimes encounter these saddle points as artefacts of the used calculation model. It is strongly advised to not use these states for any explanation.
Maximum
This is one cannot be found. The PES is usually like a well. The borders of this would be total fragmentation of the molecule. Imagine the homolytic dissociation of dihydrogen (a 2D PES). The maximum value of the PES corresponds to superimposed atoms and its energy value is at infinity.

An additional note. A transition state always connects two local minima. And vice versa, two local minima will always be connected via a series of transition states and local minima. It does not make much sense, looking at any other points than these, as they can most likely not interpreted in a physical meaningful way.

Comment by Ath

  What I mean is not global maximum, but local maximum, which doesn't necessarily represent bond breaking...The local maximum I'm interested in is actually the rotation of part of a molecule. I wonder whether the frequencies are all positive or all negative at that point.

There are no maxima on the PES, since the maximum energy corresponds to superimposed atoms. A rotation of moieties is a transition state, or a first order saddle point, and has exactly one imaginary mode.


What effect does hydrogen bonding have on H NMR spectra? My thinking is that an increase in electron density caused by hydrogen bonding should cause the chemical shift of participating atoms to be moved downfield. Am I right or wrong?

Can someone provide an example showing this phenomenon?

Hydrogen bonding is well known to cause large changes in chemical shift values– this is evident if you look up the chemical shift of an alcohol in a data-book (for an aliphatic alcohol this is anywhere between 0.5 and ca. 6.0 ppm depending on solvent/concentration), where a large range of values are given.[1]

In general,  formation of hydrogen bonds causes shifts to move to a higher frequency (higher ppm) due to deshielding.[2] In itself this is counter-intuitive, as in a hydrogen bonding interaction the H is actually receiving electron density from the donor atom (and therefore should be shielded). In the case of NMR however, the actual electron density isn’t whats causing this, but rather the field set up from the presence of the interaction.[#]

Intramolecular vs intermolecular hydrogen bonding

Although hydrogen bonding does cause a change in chemical shift, it is often only evident (/causes an issue) with intermolecular hydrogen bonding (between two molecules). With intramolecular bonding, the system is sufficiently stable that the chemical shift of the proton is consistent (and doesn't change with concentration/solvent, much).

To give an example,[##] consider the 1H-NMR of methanol and hexafluoroacetylacetone:[3]

On the top is the spectra of the neat compounds, where hydrogen bonding is present. On the bottom, the spectra of the compounds with added tetrachloromethane (which breaks down hydrogen bonding). Whilst the intramolecular hydrogen bonding in hexafluoroacetylacetone is unchanged, we can see immediately that the chemical shift of the hydroxyl proton to move.

Applications

Hydrogen bonding in NMR may seem rather inconvenient (and indeed it can be in routine structure elucidation work where the shift keeps on moving depending on how dry the NMR solvent is on any given day), however it has played a vital role in biochemistry and supramolecular chemistry, where the change in proton shifts can be used to determine whether or not multiple components are interacting.

A classic example is the interaction between cytosine and guanidine (C pairs with G in DNA via a hydrogen bonding interaction), which can be observed by NMR when the two compounds are mixed in an equimolar ratio:

[1]: Presch, E.; Buhlmann, P.; Badertscher, M. Structure Determination of Organic Compounds- Tables of Spectral Data; Springer: Berlin, 2009

[2]: Günther, H. NMR Spectroscopy- Basic Principles, Concepts, and Applications in Chemistry; Wiley: Weinheim, 2013

[#]: This is covered in Spin Dynamics by Levitt if you have any interest in the theoretical reasoning behind this

[##]: In this other answer, i’d previously given the example of salicaldehyde, which shows a consistent intramolecular H-bond free from solvent/concentration effects

[3]: Figure reproduced from Breitmaier, E. Structure Elucidation by NMR in Organic Chemistry; Wiley: Chichester, 2002


Haloform reaction in acidic condition I want to know about the reaction mechanism of haloform reaction in acidic condition. I am following a persons doctoral thesis where he used acid medium haloform reaction but unfortunately he has not given any references.

In the book by Clayden et al. you can find a very nice description of the mechanism of the halogenation of enols and the haloform reaction on pp. 533-536. Especially on p. 536 the authors give the mechanistic details of the acid catalysed halogenation of enols and some reasons why multiple halogenation steps on the same carbon atom (as happens in the haloform reaction) occur easily under base catalysis but do usually not occur under acidic conditions. Since I lack the time to write it up myself (and since the authors already did a marvelous job at describing it) I simply post the relevant passage here:


Converting cyclic compounds to linear compounds (possible ?) Can one convert from cyclic convert to linear compound?
Like

$Cyclohexane$ $+$ $?=Hexane$

Even in High Temp and Pressure?

I know it is very tough since we are going from a stable compound to a relatively more unstable compound.But still...???

And can we also convert $Benzene$ to some $Unsaturated Compounds$ ?

And if possible can anyone please tell the reaction mechanism of it?

Thanks

Yes, this can be done.  Unsaturated cycloalkanes can be ring opened using fairly straightforward methods.  For example, heating cyclobutene will produce 1,3-butadiene in an orbital symmetry allowed reaction.  But since your example involved cyclohexane, I'm guessing that you are more interested in the ring opening of saturated cycloalkanes, so that's where I'll focus my answer.

Catalysts are needed to ring-open cycloalkanes.  The catalyst complexes with a cycolalkane bond and then in a second step, hydrogen is added across this bond.  Mechanistically, it is basically the reverse of the hydrogenation reaction (e.g. catalytic hydrogenation of cyclohexene to cyclohexane).  Some of these catalysts are very indiscriminate, so the initially formed linear alkane will rebind to the catalyst and undergo subsequent isomerization and fragmentation reactions.  More recently catalysts have been developed that can open cycloalkanes to the corresponding linear alkane in fairly high yield and relatively free of other hydrogenolysis products. These reactions are run in solution at temperatures slightly above ambient.  Here is a link to a full text article on one such process that uses Lanthanide\Aluminum complexes;  Terbium seems to be the preferred lanthanide.  Yields are typically in the 65-80% range (cyclohexane, 67%; cyclopentane, 77%; cyclodecane, 77%).


What order would these compounds elute in a 3:1 hexanes:ethyl acetate TLC run? I would figure the order would be (in order of smallest Rf to largest) W, Z, Y, X. Is the carboxylic acid more polar than the acetic anhydride?

Making predictions about elution order is always guesswork but based on my experience I would expect acetic acid to streak off the baseline, The benzyl alcohol to be of low Rf but above the acetic acid, the benzyl acetate Y to run high and I really don't know about the acetic anhydride, I suspect it will hydrolyse on the plate to give you more acetic acid so you will not see a separate spot. How are you planning on visualising these?


Different masses for the same element and isotope? I am watching this Khan Academy video on mass defect and binding energy, it uses helium-4 as an example to go through the concept. The person finds the predicted mass (protons + neutrons) to be 4.0318274 amu. He also finds that the experimentally measured mass for He-4 to be 4.00150608 amu.

What's baffling me is, where did he find the experimentally measured mass for He-4? I have been searching for the atomic mass of He-4 all over, and every source I found seems to agree that its mass is, in fact, 4.002602 u ± 0.000002 u, that's a 0.001 difference!

As noted by Ivan in his comment, the difference is likely due to using the helium nucleus $\ce{^4He^2+}$ and not a helium atom. Shame on the Khan Academy video for not making this clear.

However, there is a good reason to only consider the nucleus. The mass defect and binding energy are consequences of the strong nuclear fore, and thus properties of the nucleus.

The choice of helium-4 is convenient: the helium-4 nucleus is the alpha particle - a common product of radioactive decay and thus well-characterized. Thanks to mass spectrometry, we can determine the mass of an alpha particle very precisely to be $4.00150608\ \text{u}$. Here is the source of that mysterious number. Again, shame on Khan Academy for not being explicit.

The electron has a mass of $5.48579909\times 10^{-4}\ \text{u}$, so a helium-4 atom would be expected to have a mass of one alpha particle and two electrons:

$$4.00150608\ \text{u} + 2\left(5.4858\times 10^{-4}\ \text{u}\right) = 4.00260324\ \text{u}$$


Radical Computational Calculations I am trying to calculate the total energies for a particular organic molecule that undergoes radical decomposition. I was able to calculate the neutral molecule with the exact results of a paper I found. My trouble is obtaining the correctly calculated energies for the radical molecules. I am new to computational chemistry and am looking for a bit of guidance on this.

Disclaimer: this is a preliminary answer based on the current state of the discussion with OP; it might be the subject to major changes in the near future.

So.

OP (original poster) is trying to reproduce some relative energies previously computed at ωB97X-D/6-31+G(d,p) at a different M06-2X/6-31G(d,p) level of theory. OP has exact same results for closed-shell molecular species in question, the meaning of which is probably that the resulting M06-2X/6-31G(d,p) relative energies in kcal/mol or kJ/mol are exactly equal to the previously published ωB97X-D/6-31+G(d,p).
OP also mentioned that he has some troubles in reproducing the relative energies for open-shell species, by which he probably means that for open-shell species M06-2X/6-31G(d,p) spits out different numbers in kcal/mol or kJ/mol than ωB97X-D/6-31+G(d,p).

Before we have some numbers let me say few things which hold in generality.

The difference in few kJ/mol in relative energies computed using two different functionals is pretty normal (even at exact same geometry). One should not even expect exactly same relative energies with any two different functionals in the first place.
OP presumably run geometry optimization at M06-2X/6-31G(d,p) but did not mention how different are the equilibrium geometries obtained at this level from those at ωB97X-D/6-31+G(d,p) level published previously. In fact, this would be difficult to do, because I could not find the Cartesian coordinates of the species in question in the Supporting Information for the work published previously. But at least some quick visual comparison has to be done to make sure that equilibrium geometries are at least not totally different.
OP did not mention why M06-2X functional was chosen. Even if the ultimate goal is not just to reproduce the previously published work, it is mentioned there that:

  ωB97X-D, a range-separated double-hybrid functional, has been shown to describe reaction energies accurately.45

But, first of all, ωB97X-D is not a double-hybrid functional. And secondly, the paper quoted there is fairly well-known benchmark of density functional methods done by Lars Goerigkab and Stefan Grimme, and in fact, Goerigkab and Grimme found M062X-D3 and not ωB97X-D to be "statistically the best of all hybrids". Note though that M062X-D3 differs from M062X: the former includes the empirical dispersion correction while the later does not.
With respect to the empirical dispersion correction, today it is usually recommended to include it in DFT calculations, since it is available at almost no (computational) cost but can substantially improve the results. Note also that dispersion correction is already included in case of ωB97X-D functional but has to be "manually" added for the case of M06-2X. In fact, the absence of it might explain some differences in equilibrium geometries and relative energies.


How is the cavity of a cyclodextrin chiral? I am studying the host-guest chemistry of cyclodextrins with achiral aromatic molecules, and I found out that a useful way of studying them is by induced circular dichroism spectroscopy. This technique is used by taking an achiral molecule and placing it in a chiral environment to induce a CD signal. Apparently, the cavity of a cyclodextrin is chiral, but for the life of me I cannot see HOW it is chiral. I know the glucose molecules the ring is made of are D-glucose, but I still see mirror planes in the structure, and I cannot figure out how the cavity is chiral! Help??

It is because the constituent glucose molecules are chiral.

In the image you can see that all the $\ce{CH2OH}$ groups in the center are out of the plane (towards you). These groups are on the upper side of the cavity. If they were below the plane, the cyclodextrin would have other interactions with the host molecule.


Why does covalent bonding satisfy atoms? Having to retake an introductory chemistry course has gotten me thinking. Covalent bonding is where electrons are shared between two atoms that both want more electrons.

Let us take $\ce{O2}$ as an example.

The red and purple oxygen atoms share 2 electrons. This means that they now both have 8. Fantastic!

However, electrons do not sit around statically but rather buzz around randomly. If the electron moves around, then on average having 4 electrons between 2 atoms means 2 electrons each. This means no one is any better off than from when we started.

My question is, seeing as the electron sharing would actually mean that on average each atom still only has 6 electrons, how can they become stable in this covalent bond?

You seem to forget that electrons have spins. And that there are two possibilities for the spin : up or down. No other possibilities exist. Now electrons have the tendency to make pairs with one spin up and on down. Sorry for the image, but it looks as if spins up and down represent two different sexes. And that the electron have the tendency to make "couples" of "up" and "down" together, as in the traditional way of being together. On the average, there exist the same amount of electrons "up" and of electrons "down".

Let's take H atoms. Half to the H atoms have spins up, and half down. If you produce somewhere a lot of free H atoms, they will collide. When they have the same spin, nothing happen after the collision. When they have different spins, they will join and stay together, make a covalent bond, which makes a H2 molecule,  after the collision. After a while all H atoms are groups in H2 molecules

In the next atom, Helium, there are two electrons on the first layer. They are of opposite spins and Helium will not join with any other atom, and not produce any molecule.

For heavier atoms, form Li to Ne, there is enough room for 8 electrons. The first four have the same spin. The next four have the opposite spin. Hopefully you may understand your self that Li makes LiH, Be makes BeH2, etc. But Oxygen has 6 electrons, 4 of a given spin and 2 of the opposite spin. In Oxygen, there are two doublets which cannot form covalent bonds. Only 2 can do it.

I will not develop this theory too long. You may do it your self. It is not too difficult to do it.


Carbon with 5 bonds? I've heard that, even though according to Molecular Orbital Theory there is no chance of having nobel gases bonded to each other, it is not totally impossible. For example, under extreme conditions, Ar2 can be synthesised.

So I am wondering whether a carbon can make 5 bonds if the required conditions are provided.

Carbon cannot have more then 4 double-electron bonds in reasonable conditions. However, in can form a bond with 5 or 6 atoms, like $\ce{Fe6C}$ fragment, where iron atoms form octahedron around the carbon atom. However, the sum of orders of 6 $\ce{C-Fe}$ bonds will be still 4.

The situation is different if we consider exited states. Indeed, it is possible for hight excited state of carbon to be able to form 5 or 6 bonds. The resulting structure, however, will die quickly.

Noble gases are able to form molecules in excited states, or, to be precise, their molecules are stable only in excited state. This kind of molecules is known as excimer (excited dimer).


How to refer to this method of mixing solids? The steps of this process are:

Fill a container with water (or other solvent)
Add mixants (is that the right word?) the container
Stir until all mixants are dissolved/suspended
Boil off the water (or other solvent)

The substances do not end up chemically reacting during this process. A real-world application of this is in small-scale manufacture of Rocket Candy, where this is used to mix KNO3 and Sucrose so the mixture can later be ignited and will burn more smoothly. Another example is making a cake - all the ingredients end up dissolved or suspended in the batter to mix them, and then the water is boiled off during baking. (Yes, there are also a lot of other things that happen during cake-making.)

What is the name of this process, if it has one? If not, what is a succinct way to describe it or state that this is what is used?

I think the closest term would be recrystalisation though this is normally used to dissolve one component (the desired one) to allow the removal of impurities, so I would be hesitant to use it but it is used here to describe the rocket candy example you used: http://www.jamesyawn.net/rcandy/index.htm

With regards to your second example, baking a cake is actually a series of interesting and complex chemical reactions and structural changes. The New Zealand institute of chemistry wrote a nice summary of the chemistry of baking that can be found here: http://nzic.org.nz/ChemProcesses/food/6D.pdf

Hope that helps


Why is the iodide anion a good nucleophile but a poor base? The $\mathrm{pK_{aH}}$ of $\ce I^-$ is very low which indicated that it is not favourable for it to bond with a proton. However, why would it be likely to bond with any other atom (mostly carbon when considering it's nucleophilic properties).

You should clarify whether the iodide anion is a good nucleophile in a polar protic or polar aprotic solvent.

Also let's note that nucleophilicity is a kinetic property, while acidity/basicity are thermodynamic properties. Iodide ion's lack of basicity in water reflects its conjugate acid's ($\ce{HI}$) lack of stability; it's relatively easy to ionize its proton due to the weak hydrogen-iodide bond.

On the other hand, the high nucleophilicity of iodide ion in water reflects the high rate of reaction between iodide ion and some electrophilic atom (other than hydrogen).

The relative lack of basicity of the iodide anion in water is precisely what makes it a good nucleophile. Its lack of affinity toward protons - i.e. the protons of water - physically frees $\ce{I^-}$ to attack electrophilic atoms (not limited to carbon). There is little hydrogen bonding between iodide ion and water.

On the other hand, the relative basicity and thus strong hydrogen-bond forming capabilities of $\ce{F^-}$ makes it a poorer nucleophile in water. Rates of reaction between fluoride anion and some electrophilic atom are slowed because the fluoride anion is trapped (strongly solvated) by water molecules. And there will also be a sizable number of fluoride anions that have completely abstracted a proton from water to make $\ce{HF}$, which isn't nucleophilic at all.

In a polar aprotic solvent, however, where there are no positively polarized hydrogens for a nucleophile to hook onto; $\ce{F^-}$ becomes a much stronger nucleophile. Actually,  $\ce{F^-}$ becomes a better nucleophile than $\ce{I^-}$, likely due to $\ce{F^-}$'s greater charge density.


Conformations of Cyclohexane Is there any such thing as a twist chair conformation?

Also, the boat and the chair conformations are achiral, while the twist boat conformation of cyclohexane is chiral and dissymetric.
What about the half chair conformation? Am I right in thinking that it is achiral because it has a plane of symmetry?

Short answer, no. There is not a "twist-chair."

To convince yourself, it probably helps to make a physical model with a chemistry model kit. If you try to twist the chair, you can't do it without significantly moving at least one atom, in which case, the conformation is basically a "skew" or "twist boat." Try it.

The boat and chair conformations are indeed symmetric and achiral. The twist-boat or skew conformation has point group $D_{2}$ and is indeed chiral.

A normal half-chair will have an axis of symmetry and is chiral.

An envelope conformation will have a plane of symmetry and thus be achiral.

I've found that some of the best depictions of six-member-ring conformations come from pyranose sugars. Because of the hydroxyl groups and oxygen atoms, there are 38 distinct conformations (2 chairs, 6 boats, 6 skew-boats, 12 half-chairs, and 12 envelopes).

From Wikipedia:

Obviously, in less-substituted cyclohexane, there are fewer unique shapes, but it helps to see the different ring-puckering or pseudo-rotation shapes.


Vertical lines in a Pourbaix Diagram I know that because of the absence of the hydrogen ions in a reaction ($n_{H+} = 0 $) , the correspondent E-ph line have to be horizental , but what exactly the types of reactions (if there is) that characterizes the vertical lines of an E-ph plot ?

A Pourbaix diagram is a special kind of phase diagram. As in a phase diagram, the lines of all kinds represent phase boundaries. Each line represents the conditions in which two phases are in dynamic equilibrium. Without information on what the phases enclosed by lines look like, a phase diagram is pretty meaningless. Below is the phase diagram of water. Imagine how hard it would be to interpret without the labels. We still need to go look up what is meant by Solid IX - is it ice IX or ice 9?

Here is the iron-carbon (steel) phase diagram. Again, the phases are labeled.

In a Pourbaix diagram, the axes are pH and applied electric potential $(E_0)$ in volts instead of temperature and pressure (pure substance) or temperature, pressure (sometimes omitted), and composition for mixtures. The lines represent boundaries, but the "phases" are now the predominant iron-containing ion in solution. These diagrams seem to be primarily for aqueous systems, and so there is never an absence of $\ce{H+}$ ions. They may not be needed in the redox equation, but they are in solution. Otherwise, how would we have different pH along the horizontal axis?

Below is the Pourbaix diagram for aqueous solutions of iron. Notice that not all lines are horizontal or vertical. Each line represents a set of conditions ($E_0$ and $\text{pH}$) where two ionic species are in dynamic equilibrium.

For example, at $E_0=1.0\ \text{V}$ and $\text{pH}=2$, the predominating species is $\ce{Fe^{3+}(aq)}$. If we increase the $\text{pH}$ to $5$, the predominating iron species becomes $\ce{Fe2O3}\cdot\ce{nH2O}(s)$, rust. The reaction might be (we increase the pH by added $\ce{OH-}$):

$$\ce{2Fe^{3+}(aq) +6OH- (aq) -> Fe2O3(s) + 3H2O(l)}$$

Note that this reaction is redox neutral - all elements remain in the same oxidation state. Vertical lines represent redox neutral transitions/reactions.

If we start at the same place $E_0=1.0\ \text{V}$ and $\text{pH}=2$, and change $E_0$ from $1.0 \ \text{V}$ to $0.4 \ \text{V}$, the predominating species becomes $\ce{Fe^{2+}(aq)}$. When we lower $E_0$, it is like adding electrons to the system (it is really encouraging electrons to flow through the system in a specific way), we get this reaction, which is pH-neutral.

$$\ce{Fe^{3+}(aq) + e- -> Fe^{2+}(aq)}$$

As you noted, there is no $\ce{H+(aq)}$ in this reaction. Horizontal lines are pH-neutral transitions/reactions.

A diagonal or curved line is neither pH or redox neutral. For example, let's go from $E_0=0\ \text{V}$ and $\text{pH}=6 \ (\ce{Fe^{2+}(aq)}$ to $E_0=0\ \text{V}$ and $\text{pH}=10 \ (\ce{Fe2O3\cdot nH2O(s)}$ crossing a diagonal line. The reaction now involves a change in oxidation state and in pH. Since we are traveling horizontally, the electrons lost by $\ce{Fe^{2+}}$ must go somewhere.

$$\ce{2Fe^{2+}(aq) + 4HO- (aq) -> Fe2O3(s) + 2H2(g) + H2O(l)}$$


What do (2S,3S) and similar notations mean? When my textbook is discussing stereoisomers with more than one chiral centers, it's using notation like (2S,3S) being an enantiomer with (2R,3R) and so forth. I understand R and S configuration, and I'm guessing there are two items in each brackets to represent the two chiral centers (so 2S, 3S would at least mean that both chiral centers have an S configuration), but I don't understand what the numbers preceding the configuration mean. What does the 2 in 2S mean and what does the 3 in 3R mean? I was unable to find an explanation in the text and it seems that their meaning is assumed. Can I get some help?

When naming a organic compound, S/R-notation comes if there is a chiral atom present. The S/R-notation always accompanies a numerical such as (2S,3R)- where the numerical refers to the chiral carbon position in the molecule according to the IUPAC numbering while R and S representate the orientation of the chiral atom according to the Cahn–Ingold–Prelog priority rules. For example, let's look at glyceraldehyde molecules where all this stereochemistry started from:

The IUPAC name of glyceraldehyde is 2,3-dihydroxypropanal. However, it has a one chiral carbon at $\ce{C}$2 and hence has two stereoisomers. The priority order of groups attached to chiral carbon is $\ce{-OH \gt -CHO \gt -CH2OH \gt -H}$. Hence you can look at the $1 \rightarrow 2 \rightarrow 3$ priority group rotation through the bond of chiral carbon and least priority group, which is $\ce{H}$ here (though $\ce{C-H}$ bond). Accordingly, the rotation for the first compound is clockwise as shown in the image, thus it carries (R)-notation. Since chiral carbon is 2 according to the IUPAC rules, the complete notation is (2R)-. Similarly, the mirror image is caring (2S)-notation since it has counterclockwise rotation as depicted in the image. Therefore complete names of these two isomewrs are (2R)-2,3-dihydroxypropanal and (2S)-2,3-dihydroxypropanal (Note that these isomers are called enantiomers).
In addition, if a $\ce{H}$ in $\ce{C}$3 is replaced by another different group, say it is $\ce{CH3}$ group, it will also become chiral and can have either (S) or (R) rotation. Suppose you replace $\ce{H}$ in $\ce{C}$3 of (2R)-2,3-dihydroxypropanal by a $\ce{CH3}$ group making it has (S) rotation. By addition of $\ce{CH3}$ group make it 2,3-dihydroxybutanal. With correct stereochemistry, its correct name is (2R,3S)-2,3-dihydroxybutanal.


Why does equilibrium exist? I've read an article that equilibrium exists when $G$ (Gibbs free energy) becomes minimum. If formation of products decreases Gibbs free energy, why won't the reaction proceed to completion so that Gibbs energy becomes minimum.

This is because the Gibbs free energy also depends on the concentration of the reactants and products. Hence, for a reaction, if the energy change is negative it moves forwards, forming more product, which also changes the concentrations of the species involved. The Gibbs energy change at this new concentration will be less negative than that in the initial case, but the reaction still proceeds since the reaction is accompanied by a negative change of free energy.

A point comes, where the concentrations of the reactions and the products is such, that the Gibbs energy change now becomes zero for the particular reaction. This is the equilibrium condition (minimum free energy, maximum entropy). If the energy change is positive, the reverse reaction becomes favorable, again inching towards the zero energy change condition, i.e. equilibrium. So, the reaction is already at the minimum Gibbs energy when it is at equilibrium. Further proceeding of the reaction will be accompanied by a positive change, and hence a rise in energy.

For how it depends on the concentrations, see these questions

Why does the reaction quotient use the products (multiplications) of reactants and products, rather than their respective sums?
Derivation of relationship between equilibrium constant and Gibbs free energy change


Chirality of symmetrically substituted biphenyl I am facing trouble with the following problem: is 2,6-dimethyl-2',6'-dinitro-1,1'-biphenyl chiral?

My idea is that there is a axis of symmetry along the line joining the two benzene molecules. Hence it should be achiral.
However my book states it otherwise.

No, the biphenyl above is not chiral.
2,2′,6,6′-tetrasubstituted biphenyls are nonplanar at room temperature in order to avoid steric clashes between the substituents. This can in general give rise to chirality depending on the substituents, such as in this example.

However, this chirality is only true under certain conditions, namely:

The substituent groups must be sufficiently bulky in order to prevent rotation about the C–C single bond that connects the two rings. If this rotation occurs freely, then the two enantiomers will be able to interconvert, and chirality will not be observed (under typical conditions)

Both pairs of substituents on the phenyl rings must be non-identical; otherwise, the molecule will possess a plane of symmetry, rendering it achiral even if rotation about the C–C bond is restricted.

Here, since the two pairs of substituents on both rings are the same (i.e. two methyl groups on one ring, and two nitro groups on the other), the second criterion is not fulfilled. As a result, the molecule is not chiral.


Synthesise the CBZ derivative of glycine from glycine and benzyl chloroformate Question
  The benzyloxycarbonyl group (CBZ) is widely used to protect amines, especially in aminoacid chemistry. A procedure to synthesise the CBZ derivative of glycine from glycine and benzyl chloroformate is given in the description below.

  Glycine ($0.1~\mathrm{mol}$) was dissolved in aqueous sodium hydroxide ($50~\mathrm{mL}$ of a $2~\mathrm{M}$ solution)
  and the solution was cooled in an ice bath. Benzyl chloroformate ($1.2~
\mathrm{eq}$) and aqueous
  sodium hydroxide ($25~\mathrm{mL}$ of a $4~\mathrm{M}$ solution) were added dropwise simultaneously to the
  glycine solution at $0°\mathrm{C}$ dropwise over thirty minutes. The mixture was stirred for an extra
  ten minutes, then warmed to room temperature. The solution was extracted twice with
  ether. The combined organic layers were acidified to $\mathrm{pH}=1$. The precipitate was filtered,
  washed with a little cold water and dried to give carbobenzoxyglycine ($18.5~\mathrm{g}$), m.p. $119-120°\mathrm{C}$.

  (a) Relative to glycine, how many equivalents of sodium hydroxide are used in this reaction? Explain why this number was used.

  (b) Can sodium carbonate ($\ce{Na2CO3}$) be used in place
  of sodium hydroxide? Explain why it can or cannot be used.

  (c) What inorganic by-product will be formed (write its chemical name)?

  (d) The procedure given includes a serious mistake, which would result in a yield of $0\%$.
  What is the mistake, and how might it be corrected?

My answers

(a) I mentioned $1~\mathrm{eq}$ of sodium hydroxide as sodium hydroxide reacts with glycine in a 1:1 ratio.

(b) I mentioned that sodium carbonate cannot be used in place of sodium hydroxide as sodium Carbonate is not basic enough to catalyst the reaction.

(c) Sodium chloride

(d) I mentioned that during the process when the combined organic layers were acidified to $\mathrm{pH}=1$, it could destroy the product by hydrolysing it hence giving a yield of $0\%$.

I'm unsure whether my answers are correct though, could anyone explain? Also, are there any resources on the web that contain practice questions of a similar nature to this?

I agree with Ron's answer except for d. Using an extra equivalent of NaOH is the right procedure since you need to "mop up" the HCl that is produced when $\ce{CbzCl}$ reacts with the amino group. What is not correct is this:
When you wash the reaction mixture with ether you get rid of any impurities and your product should stay in the aqueous phase. So you have to acidify the aqueous layer and not the ether one. pH 1 should be fine for the Cbz unless you leave it there and go away for the weekend.


Lewis Structure on (CH3)2NO? I've been taking an chemistry class online (for fun only) and one of the questions in the homework threw me a bit. Here's the wording:

"Draw the Lewis structure for a nitrogen atom attached to. two methyl groups ($\ce{CH3}$) and one oxygen atom."

OK, so I know that $\ce{NO}$ is a free radical due to the odd combined valence, i.e., the octet rule isn't satisfied. Will that carry over when the $\ce{NO}$ bonds with the two methyl groups?

Here's the sketch of what I did:

Lastly, what's the name of this compound? This is not part of the HW question; I'm just curious.

Name of the compund: Dimethyl nitroxide
Lewis structure:

The properly way to determine the Lewis structure, based on this example, is:

Total valence electrons: $4\cdot2 + 5 + 6 + 1\cdot6 = 25$
Total electrons needed for octets/doublets: $8\cdot4+2\cdot6 = 44$
Total shared/bonding electrons: $44 - 25 = 19$ (In other words, there is only nine single bonds and one electron left.)
Total electrons in lone pairs: $\text{Step 1} - \text{Step 3} = 25 - 19 = 6$ (In other words, the are only 3 pairs of lone electrons.)

Furhter information on how to draw Lewis structures, please follow the link
Further information of related Lewis structures and how this knowledge is applied on real top science, I recommend seeing this reference: Moreira, G.; Charles, L.; Major, M.; Vacandio, F.; Guillaneuf, Y.; Lefay, C. and Gigmes, D. Beilstein J. Org. Chem., 2013, 9, 1589 - 1600.


Why does bond breaking in ATP release energy? We always come across that formation of bonds releases energy, and bond breaking requires energy, but in the case where ATP converts into ADP or AMP (bond breaking), energy is released. Why?

Further, how bond formation releases energy?

In most chemical reactions where bonds are broken, other bonds are formed. Take your example of the hydrolysis of ATP. A bond between two phosphate groups breaks, but one of the phosphate groups forms a new bond with the oxygen of water. Whether energy is released or taken up (whether the reaction is exothermic or endothermic) depends on the sum of the energies associated with breaking and making bonds in that reaction.

  How does bond formation release energy?

In the simplest case in the form of kinetic energy, heating up the reaction mixture. In the case of ATP hydrolysis in biological processes, the energy is sometimes converted into mechanical energy (muscle contraction), used to run pumps (transmission of signals in the nervous system), or other processes that would not go forward on their own.