Addition of hydrogen bromide to 1,3-butadiene: thermodynamic and kinetic control The reaction of one equivalent of hydrogen bromide with 1,3-butadiene gives different products at under different conditions:1

The addition of hydrogen chloride also gives rise to similar products.2 I can see that the low-temperature product is formed via an electrophilic addition, with Markovnikov regioselectivity. But how is the other product formed, and why does the major product differ with temperature?

References

1 Kharasch, M. S.; Margolis, E. T.; Mayo, F. R. The peroxide effect in the addition of reagents to unsaturated compounds. XIII. The addition of hydrogen bromide to butadiene. J. Org. Chem. 1936, 01 (4), 393–404. DOI: 10.1021/jo01233a008.

2 Kharasch, M. S.; Kritchevsky, J.; Mayo, F. R. The addition of hydrogen chloride to butadiene. J. Org. Chem. 1937, 02 (5), 489–496. DOI: 10.1021/jo01228a010.

Thermodynamic and kinetic control

This is a classic example of the concept of thermodynamic versus kinetic control of a reaction. Take a look at this energy profile diagram.1 The horizontal axis is a reaction coordinate, and the vertical axis represents Gibbs free energy.

In this scenario, the starting material $\ce{A}$ can react to form either $\ce{B}$ or $\ce{C}$. The formation of the product $\ce{C}$ has a lower activation energy, which means that it will form faster:

$$E_{\mathrm{a},\ce{AB}} > E_{\mathrm{a},\ce{AC}} \qquad \Longrightarrow \qquad k_\ce{AB} < k_\ce{AC}$$

If we keep the temperature sufficiently low, the molecules of $\ce{C}$ - which are inevitably formed faster - will probably not have enough energy to overcome the reverse activation barrier. The forward reactions $\ce{A->B}$ and $\ce{A->C}$ are, under such conditions, effectively irreversible. Since the formation of $\ce{C}$ is faster, it will predominate, and the major product formed will be $\ce{C}$. This is known as kinetic control; $\ce{C}$ is the kinetic product.

Now, what happens if we heat it up? At higher temperatures, $\ce{C}$ is still going to be the product that is formed faster. However, it also means that all the reactions will be reversible. This means that molecules of $\ce{C}$ can revert back to $\ce{A}$ - and since the system is no longer limited by temperature, it will try its best to minimise its Gibbs free energy, which is the thermodynamic criterion for chemical equilibrium. This means that, as the most thermodynamically stable molecule, $\ce{B}$ will be predominantly formed.2 The reaction is said to be under thermodynamic control, and $\ce{B}$ is the thermodynamic product.

A simple definition is that the kinetic product is the product that is formed faster, and the thermodynamic product is the product that is more stable. This is precisely what is happening here. The kinetic product is 3-bromobut-1-ene, and the thermodynamic product is 1-bromobut-2-ene (specifically, the trans isomer).

A disclaimer

Note that not every reaction has an energy profile diagram like this, and not every reaction has different thermodynamic and kinetic products! If the transition state leading to the formation of $\ce{C}$ were to be higher in energy than that leading to $\ce{B}$, then $\ce{B}$ would simultaneously be both the thermodynamic and kinetic product. There are plenty of reactions in which the more stable product (thermodynamic) is also formed faster (kinetic), so do not assume that every reaction fits this mould!

The reaction mechanism

The first step is the protonation of one of the $\ce{C=C}$ double bonds. In butadiene (1), both double bonds are the same, so it does not matter which one you protonate. The protonation occurs regioselectively to give the more stable carbocation:

The more stable cation is not only secondary, but also allylic, and therefore enjoys stabilisation via resonance (or conjugation). This is depicted in the resonance forms 2a and 2b above.

This allylic carbocation, more properly denoted as the resonance hybrid 2, has two carbons which have significant positive charge, and the bromide ion (here denoted as $\ce{X-}$) can attack either carbon. Attacking the central carbon, adjacent to the site of protonation, leads to the kinetic product 4; attacking the terminal carbon, distant from the site of protonation, leads to the thermodynamic product 4.

There are some people who write that 3 results from attack of $\ce{X-}$ on resonance form 2a, and 4 from attack of $\ce{X-}$ on resonance form 2b. This is not correct! Resonance forms do not separately exist, and they are not distinct species that rapidly interconvert. As such, one cannot speak of one single resonance form undergoing a reaction.

With that out of the way, let's examine why 4 is the thermodynamic product, and why 3 is the kinetic product.

The thermodynamic product: trans-1-bromobut-2-ene

It is perhaps simple enough to see why 4 is more stable than 3. It has an internal, disubstituted double bond, and we know that as a general rule of thumb, the thermodynamic stability of an alkene increases with increasing substitution. So, compared to the terminal, monosubstituted alkene 3, 4 is more stable.

Both the trans isomer 4 as well as the cis isomer 5 can be formed via attack of the nucleophile at the terminal carbon, and both are disubstituted alkenes. However, the trans isomer 4 is more stable than the cis isomer 5, because there is less steric repulsion between the two substituents on the double bond. As such, 4 is the thermodynamic product.

The kinetic product: 3-bromobut-1-ene

There are quite a few explanations on the Internet,3–6 and not all of them are correct.

The worst possible argument, which I have thankfully not seen yet, goes something along the lines of: resonance form 2a, being an allylic secondary carbocation, is more stable than resonance form 2b, which is an allylic primary carbocation. Therefore, resonance form 2a exists in greater proportion, and the nucleophile preferentially reacts with it, leading to the formation of 3.

However, this is patently incorrect, as discussed earlier! Individual resonance forms do not exist, and on top of that, such an argument suggests that we are looking for the more stable intermediate. In fact, we should be looking for the more stable transition state. The carbocation is an intermediate, and not a transition state.

The most common argument that I have seen is actually: since resonance form 2a is more stable than 2b, it contributes more towards the resonance hybrid 2. As such, the positive charge on the internal carbon is greater than the positive charge on the terminal carbon. The nucleophile, being negatively charged, is more strongly attracted to the more positively charged or more electrophilic carbon, and therefore attack there occurs faster (the transition state being stabilised by greater electrostatic interactions).

That's actually a very sensible explanation; with only the data that has been presented so far, we would not be able to disprove it, and it was indeed the accepted answer for quite a while.

The catch

In 1979, Nordlander et al. carried out a similar investigation on the addition of $\ce{DCl}$ to a different substrate, 1,3-pentadiene.7

This experiment was very ingenious, because it was designed to proceed via an almost symmetrical intermediate:

Resonance forms 7a and 7b are both allylic and secondary. There is a very minor difference in their weightages, arising from the different hyperconjugative ability of $\ce{C-D}$ vs $\ce{C-H}$ bonds,8 but in any case, it is not very large. Therefore, if we adopt the explanation in the previous section, one would expect there not to be any major kinetic pathway, and both 1,2- and 1,4-addition products (8 and 9) would theoretically be formed roughly equally.

Instead, it was found that the 1,2-addition product was favoured over the 1,4-addition product. For example, at $-78\ ^\circ\mathrm{C}$ in the absence of solvent, there was a roughly $75:25$ ratio of 1,2- to 1,4-addition products. Clearly, there is a factor that favours 1,2-addition that does not depend on the electrophilicity of the carbon being attacked!

The authors attributed this effect to an ion pair mechanism. This means that, after the double bond is protonated (deuterated in this case), the chloride counterion remains in close proximity to the carbocation generated. Immediately following dissociation of $\ce{DCl}$, the chloride ion is going to be much closer to $\ce{C-2}$ than it is to $\ce{C-4}$, and therefore attack at $\ce{C-2}$ is much faster.

In fact, normal electrophilic addition of $\ce{HX}$ to conjugated alkenes in polar solvents can also proceed via similar ion pair mechanisms.9,10 This is reflected by the greater proportion of syn addition products to such substrates.11

Notes and references

1 Smith, M. B. March's Advanced Organic Chemistry, 7th ed., p 272

2 This does not mean that all of $\ce{A}$ will be converted to $\ce{B}$; the reaction is still an equilibrium, and equilibria always go forward and backward. In general, the minimum system Gibbs free energy ($G_\mathrm{syst}$) will occur at a certain proportion of $\ce{A}$, $\ce{B}$, and $\ce{C}$. However, since $\ce{B}$ has the lowest Gibbs free energy, it will be formed in a greater proportion than $\ce{C}$. See, for example, the colourful graph in this answer of mine.

3 Thermodynamic and Kinetic Control. http://www.masterorganicchemistry.com/2012/02/09/can-opener-economics/ (accessed August 8, 2017).

4 Addition of Hydrogen Halides to Dienes. http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch10/ch10-4-1.html (accessed August 8, 2017).

5 Electrophilic Attack on Conjugated Dienes-Kinetic and Thermodynamic Control. https://chem.libretexts.org/Core/Organic_Chemistry/Conjugation/Electrophilic_Attack_on_Conjugated_Dienes/Electrophilic_Attack_on_Conjugated_Dienes-Kinetic_and_Thermodynamic_Control (accessed August 8, 2017).

6 1,4-Addition. http://www.ochempal.org/index.php/alphabetical/a-b/14-addition/ (accessed August 8, 2017).

7 Nordlander, J. E.; Owour, P. O.; Haky, J. E. Regiochemistry of the addition of hydrochloric acid-d to trans-1,3-pentadiene. J. Am. Chem. Soc. 1979, 101 (5), 1288–1289. DOI: 10.1021/ja00499a045.

8 Because of the larger reduced mass and lower zero-point energy, a $\ce{C-D}$ bond is stronger and therefore less willing to donate electron density into an adjacent empty $\mathrm{p}$ orbital. This is the origin of some secondary kinetic isotope effects; in our case, it means that 7a is marginally less stable than 7b.

9 Fahey, R. C.; McPherson, C. A. Mechanism of the hydrochlorination of tert-butylethylene and styrene in acetic acid. J. Am. Chem. Soc. 1969, 91 (14), 3865–3869. DOI: 10.1021/ja01042a030.

10 Addition of $\ce{HX}$ to butadiene in the gas phase gives approximately a $1:1$ ratio of 1,2- to 1,4-addition product, suggesting that an ion pair mechanism (which would favour the 1,2-addition product) does not operate. See: Mascavage, L. M.; Chi, H.; La, S.; Dalton, D. R. Surface-catalyzed hydrochlorination of alkenes. The reaction of the gases hydrogen chloride and 1,3-butadiene. J. Org. Chem., 1991, 56 (2), 595–601. DOI: 10.1021/jo00002a021.

11 For more details and references to primary literature, see Carey, F. A., Sundberg, R. J. Advanced Organic Chemistry - Part A: Structure & Mechanisms, 5th ed., pp 478–482.


ELI5: What is the effect of ultrafine Teflon (PTFE) particles in rats and mice I am attempting to research the results of case studies regarding polytetrafluoroethylene [PTFE] (commercially known as Teflon). More specifically I am attempting to make sense of this Abstract on Acute pulmonary effects of ultrafine particles in rats and mice. Not being a chemist I'm having a hard time understanding the results and hoping someone can explain in laymens terms. More specifically, the abstract mentions;

We used ultrafine Teflon (polytetrafluoroethylene [PTFE]*) fumes (count median diameter [CMD] approximately 18 nm) generated by heating Teflon in a tube furnace to 486 degrees C to evaluate principles of ultrafine particle toxicity that might be helpful in understanding potential effects of ambient ultrafine particles. Teflon fumes at ultrafine particle concentrations of approximately 50 micrograms/m3 are extremely toxic to rats when inhaled for only 15 minutes.

That seems straight forward enough. They heated up Teflon in a tube, and made rats inhale it, who subsequently died. Then it goes on to say;

We found that neither the ultrafine Teflon particles alone when generated in argon nor the Teflon fume gas-phase constituents when generated in air were toxic after 25 minutes of exposure. Only the combination of both phases when generated in air caused high toxicity, suggesting the existence of either radicals on the particle surface or a carrier mechanism of the ultrafine particles for adsorbed gas-phase compounds.

This is the part thats unclear to me. The Teflon particles alone, I think means the compound before they heat it, but what is the 'fume gas-phase when generated in air'? How is that different then the fumes they made the rats inhale (from which they died after 15 minutes)? What does the 'combination of both phases when generated in air' mean?

Aging of the fresh Teflon fumes for 3.5 minutes led to a predicted coagulation resulting in particles greater than 100 nm that no longer caused toxicity in exposed animals.

How would coagulation affect the toxicity of the fumes? Does it change the chemical structure of what the animals inhale somehow making it less toxic? Naturally speaking, I would assume 'more' particles would result in higher levels of toxicity, not less.

I think the authors' choice of words might have confused you a bit.

We found that neither the ultrafine Teflon particles alone when generated in argon nor the Teflon fume gas-phase constituents when generated in air were toxic after 25 minutes of exposure.

What the authors mean is that when they used only ultrafine Teflon particles on the rats it did not kill them, nor did the Teflon fume gas-phase constituents when the rats were breathing that air for 25 minutes.

Only the combination of both phases when generated in air caused high toxicity, suggesting the existence of either radicals on the particle surface or a carrier mechanism of the ultrafine particles for adsorbed gas-phase compounds.

The individual substances, that is, ultrafine Teflon particles and Teflon fume gas-phase constituents were non-toxic for the rats. When the two were combined together in the air and the rats were made to breathe in that air was the point that they died, meaning that when these substances were combined in the air, some radicals$^{[1]}$
were being produced or some form of carrier mechanism was taking place.

The Teflon particles alone, I think means the compound before they heat it.

No, they meant that the ultrafine Teflon particles after heating. Alone meant these Teflon particles, by themselves, are not toxic.

what is the 'fume gas-phase when generated in air'

Gas-phase is simply a way of referring to the gaseous physical state. See here$^{[2]}$ for the definition. The phrase fume gas-phase refers to the fumes that the author observed on heating the Teflon to the temperature at which it turned into gaseous phase (phase is the word for referring to the physical state of a substance) while producing fumes.

How would coagulation affect the toxicity of the fumes? Does it change the chemical structure of what the animals inhale somehow making it less toxic? Naturally speaking, I would assume 'more' particles would result in higher levels of toxicity, not less.

Its not about more or less here. After some time, coagulation takes place causing the Teflon to take a solid form in the whichever utensil the researchers were using. What they mean by the coagulated Teflon being non-toxic is that in whichever way they made the animals (they have not specified which animal though) ingest it, it did not kill them. The quantity of the Teflon, the form in which it was ingested and by which animal was it ingested has not been specified, at least I couldn't find any clear reference in the abstract and I don't know how to access the full paper.
References:-

IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). Online version (2019-) created by S. J. Chalk. ISBN 0-9678550-9-8. https://doi.org/10.1351/goldbook.
IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). Online version (2019-) created by S. J. Chalk. ISBN 0-9678550-9-8. https://doi.org/10.1351/goldbook.


Acidity of oxonium ion How was the pKa value (-1.7) for the oxonium ion determined? Why does it correspond to the minimum pH value accessible in water? What reaction does it represent?

There is a nice little article in the Journal of Chemical Eduction Negative pH Does Exist pointing out examples of negative pHs in water, for example pH = -3.6 in mine water in California.  So -1.7 is not the lowest possible pH in water. Keep in mind that pH is defined in terms of hydrogen ion activity rather than concentration.  This is emphasized in Paradoxes: Demonstrating That It Is Not True That pH ≡ -log[H+].  At high acid concentation, hydrogen ion activity greatly deviates from hydrogen ion concentration.  For example, in 16 molar HCl, hydrogen ion activity is 678!

And see Negative pH and Extremely Acidic Mine Waters from Iron Mountain, California (no paywall), which describes pH values even below -4.

In Hydronium ion activity in strongly acidic media. Remarkable agreement between independent estimates Am. Chem. Soc. vol. 95, pp 3055–3057, 70% sulfuric acid is found to have a pH of -9.8 (negative 9.8)!

However, ignoring activity and the true definition of pH, considering that pure water has a concentration of 55M, if all the water molecules were instead hydronium ions, -log(55) = -1.7.  Then in articles like Who Knows the Ka Values of Water and the Hydronium Ion? this value is attributed to the pKa of hydronium, with the supposed justification that:
$K_a = \frac{[\ce{H3O+}][\ce{H2O}]}{[\ce{H3O+}]} = [\ce{H2O}]$

The article New point of view on the meaning and on the values of
Ka(H3O+, H2O) and Kb(H2O, OH-) pairs in water is a serious consideration of the true meaning of the pKa of hydronium, and points out that the -1.7 value for the pKa of hydronium has no justification.

In conclusion, pH can be below -1.7 and the pK of hydronium isn't really -1.7.


How should I think about potential energy surfaces for reactions of complex systems? I've recently started doing some computational chemistry, studying the reaction dynamics of some moderately complex organic and inorganic reactions. I'm doing things like finding maxima, minima, TS's, saddle points, geom. optimizations, and finding conformational isomers of all of the above. I'm really enjoying myself.

One thing I would like to better understand, though, is how to think of the potential energy surface (PES) of these sorts of systems. The way I see it, a PES is just a potential energy profile (PEP) with one or more dimensions beyond energy and reaction coordinate. I understand that you cannot really compute a complete PES for a complex system. Most of the PES's I've seen are for simple systems like this solvent study of an SN2 reaction:

(R. Otto et al. 2011, Nature Chemistry 4, 534-538)

This is simple enough to understand. They have added the dihedral angle as another dimension of their picture of the reaction dynamics. But what about a large system, especially a non-rigid one? Since there are so many more possible ways the system can move, does that mean its 'true' PES is polydimensional? I like to think of these mechanistic studies of big reactions as us 'exploring the PES', even though we cannot see the whole thing. How should I think about this surface? Am I understanding this concept properly?

In order to understand what the PES really is, it is useful to see where this concept comes from.

A system of $M$ nuclei and $N$ electrons should be described by a multidimensional wavefunction $\Psi(\vec{r}_1,\dots,\vec{r}_N,\vec{R}_1,\dots,\vec{R}_M)$, where $\vec{r}$ denotes electronic coordinates and $\vec{R}$ denotes nuclear coordinates (we neglect spin for simplicity). Within the Born-Oppenheimer approximation, the total wavefunction is split in an electronic part $\phi$ and a nuclear part $\chi$:
$$
\Psi(\vec{r}_1,\dots,\vec{r}_N,\vec{R}_1,\dots,\vec{R}_M) = \phi(\vec{r}_1,\dots,\vec{r}_N;\vec{R}_1,\dots,\vec{R}_M)\chi(\vec{R}_1,\dots,\vec{R}_M)
$$
where the electronic wavefunction depends parametrically on nuclear positions (as it will be clear in a moment).

If you inject the Born-Oppenheimer ansatz in Schrödinger equation for a system of interacting nuclei and electrons and you integrate out electronic degrees of freedom you get an equation for the nuclear wavefunction only
$$
\left[ -\sum_I \frac{1}{2M_I} \nabla^2 + \sum_{I<J}  \frac{Z_IZ_J}{|\vec{R}_I-\vec{R}_J|} + E(\vec{R}_1,\dots,\vec{R}_M) \right]\chi = E_\text{tot} \chi
$$
where $E(\vec{R}_1,\dots,\vec{R}_M)$ is the solution of the electronic problem
$$
\left[ -\sum_i \frac{1}{2}\nabla_{i}^2 +   \sum_{i<j} \frac{1}{|\vec{r}_i-\vec{r}_j|} -\sum_{i,I} \frac{Z_I}{|\vec{r}_i-\vec{R}_I|} \right]\phi = E(\vec{R}_1,\dots,\vec{R}_M)\phi
$$
As you can easily see the solution of the electronic problem (both the energy and the wavefunctions) depend parametrically on nuclear coordinates. The potential energy surface is the potential energy appearing in the equation of the nuclear wavefunction, i.e.
$$
U(\vec{R}_1,\dots,\vec{R}_M)=E(\vec{R}_1,\dots,\vec{R}_M) + \sum_{I<J}  \frac{Z_IZ_J}{|\vec{R}_I-\vec{R}_J|}.
$$
It is clear that finding the PES is a formidable task, since you have to solve the electronic problem for each possible set of nuclear coordinates. So, yes the PES is a multi-dimensional function of nuclear coordinates. If you have $M$ nuclei, your PES will be a function of at least $3M-6$ coordinates (once you remove translation and rotation of the whole system).

Since a multi-dimensional function is impossible to visualize, usually we study the PES by considering few interesting degrees of freedom (as the O-C distance and O-C-I angle in your example). Therefore, as you say, you can think as if you are exploring the PES locally and in a reduced space.

ADDITIONAL INFORMATIONS

Since the nuclear wavefunction is sharply picked, (heavy) nucley can usually be considered as classical particles. Thus instead of the nuclear quantum Hamiltonian
$$
H_n = -\sum_I \frac{1}{2M_I} \nabla^2 + \sum_{I<J}  \frac{Z_IZ_J}{|\vec{R}_I-\vec{R}_J|} + E(\vec{R}_1,\dots,\vec{R}_M)
$$
you can consider its classical counterpart
$$
H_n^\text{cl} = \sum_I \frac{\vec{P}_I^2}{2M_I} + \sum_{I<J}  \frac{Z_IZ_J}{|\vec{R}_I-\vec{R}_J|} + E(\vec{R}_1,\dots,\vec{R}_M) = \sum_I \frac{\vec{P}_I^2}{2M_I} + U(\vec{R}_1,\dots,\vec{R}_M)
$$
where $\vec{P}_I$ is the momentum of the nucleus $I$. From the laws of classical mechanics you can easily verify that the force acting on nucleus I is simply given by
$$
\vec{F}_I = -\nabla_{\vec{R}_I} U(\vec{R}_1,\dots,\vec{R}_M).
$$
So even if you consider the nuclei as classical particles, the PES has the same meaning, i.e. is the potential felt by the nuclei. Here you can easily apply a classical picture. A transition state is simply a state along your reaction coordinate where you have a local maxima of the PES for the reduced coordinates you are considering(as the C-O bond in you example).

I never looked to this into details but I think computational programs climb the PES uphill from the initial and final state of your system (i.e. they follows -$\vec{F}_I$) in order to find the TS.


Diffusion rate of gases What is the effect of temperature on diffusion rate of gases? Do the diffusion rates of gases decrease with an increase in temperature? Why?

I just read it somewhere that the diffusion rate of gases decreases when temperature increases. So i just wanted to know if its true.

Temperature is directly proportional to diffusion rate.

Diffusion, being the dispersion of molecules throughout a space, is affected by the speed of molecules and the density of the space through which the molecules are being dispersed.

An increase in temperature increases the speed at which molecules move at. This is because the average kinetic energy of the gas molecules is proportional to the absolute temperature of the gas. Increased speed will yield faster and more even diffusion.

Average Kinetic Energy of gas molecules = $\frac{3}{2}RT$.

Therefore, decrease in temperature will cause molecules to slow down (due to reduction in average kinetic energy). Hence rate of diffusion decreases.

I hope this helps!


Why is measurement in molarity preferred over molality? NCERT Chemistry Part 1 states that:

"The osmotic pressure method has the advantage over other methods as pressure measurement is around the room temperature and the molarity of the solution is used over molality."

I do not understand how the second statement is really an advantage. Infact, if I had to guess, I would choose molality to be the easier of the two to measure as molarity might involve change in volume upon mixing.

Molality is used for those people trying to understand how the melting point or the boiling point of water or any other substance, changes when they add a given amount of salt or any other impurity to one kilogram of this pure substance. They don't care about the volume of the obtained solution. They will not keep the solution once obtained. As an example, they just want to check if it is correct to say that adding $1$ mole of a solute to $1$ kg water lowers the freezing point by $1.86°$C, whatever the nature of the solute.

Molarity is used for another group of people. They don't mind about the process of dissolution of a substance in a solvant. They don't care about the volume of water to be used to make $1$ liter of the solution. They just want to know the composition of the solution they have to use, i.e. how many moles (or grams) of a solute are present in a given volume of this solution. If for a given use, they must have a solution containing exactly $9$ g NaCl per liter, they don't care about the way of dissolving the salt. All they need to know is whether this solution can be used for making an injection, or any other use.


How to synthesise 2-aminoethyl benzene Starting with benzene, describe the synthesis of 2-phenylethan-1-amine providing all the reagents and conditions.

I am not really great at organic chemistry and my knowledge is pretty limited.

I am currently thinking of forming toluene by Grignard reagent and then reacting with $\ce{Cl2}$ to form benzyl chloride. Then react with sodium cyanide to form benzyl cyanide and then react it with $\ce{HCl}$ and $\ce{H2O}$ to form an amide. Then reduce it with $\ce{LiAlH4}$ to form the compound.

I am not sure if that is correct as it seems a bit complicated for the synthesis of a seemingly simple compound. Also I am not sure on the exact conditions for the chlorination step.

I was also thinking of acylation with ehtanoyl chloride followed by Wolff-Kishner reduction to form ethyl benzene. However I am not sure on how to from the $\ce{NH2}$ group.

Several thoughts:

I can't see how you form toluene with a Grignard - it's certainly possible though
Radical chlorination of toluene with $\ce{Cl2}$ isn't the cleanest reaction.
To reduce the nitrile to the amine, you don't need to go through the amide.

Also, from ethylbenzene, it's very hard to introduce that amine group cleanly. You effectively just have a compound with C-H groups and you have to somehow introduce some kind of functional group into the non-benzylic position. Try to think backwards instead of forward!

The amine you want can be obtained by direct reduction of a nitrile either by simple $\ce{LiAlH4}$ reduction or hydrogenation:

That arrow essentially means "can be made from".

Of course, that's easy to make using a nucleophilic substitution reaction with $\ce{CN-}$ and benzyl chloride (for example):

To get from benzene to benzyl chloride, you could use a Blanc chloromethylation. So your forward synthesis would be:

and you might want to use catalytic $\ce{KI}$ in the nucleophilic substitution step. Of course, for such a simple compound, there are tons of ways to synthesise it. The amine could be made from reduction of an azide, or by a reductive amination, and so on. I strongly doubt my proposed route is the most efficient. Still, I think it should work.

Theoretically, you could use a Friedel-Crafts alkylation and a radical bromination to get to benzyl bromide instead:

I don't like the alkylation step though.


How do I determine crystal field splitting (for example, for [Ni(CO)4])? How to know how many electrons are there in $T_{2g}$ and $E_g$ orbitals?

I think rather than a broad answer, an example would help. So, consider the compound $\ce{Ni(CO)4}$ and how to calculate CFSE for it?

Firstly, what geometry does the complex have?

Once you know the geometry, find the spin-state of the $d$-electrons, as well as the number of $d$-electrons applicable.

Find the crystal field diagram relevant to your geometry. Fill in the $d$-electrons according to the Aufbau principle and to Pauli's exclusion principle.

For your example, tetracarbonylnickel has a tetrahedral geometry and is a $d$10 compound. Carbon monoxide is a strong field ligand causing a high spin complex (note that since $d$ = 10, all of the $t_{2}$ and $e$ orbitals are full and hence spin-state is irrelevant).

As per the Aufbau principle, you start building up from the bottom, placing two electrons in the bottom $t_2$ orbitals and three in the top three $e$ orbitals, giving:

$e$: ↑ ↑ ↑

$t_2$: ↑ ↑

The remaining five electrons go in each half-filled orbital to give 2 filled $t_2$ orbitals and 3 filled $e$ orbitals:

$e$: ↑↓ ↑↓ ↑↓

$t_2$: ↑↓ ↑↓

To find the CSFE, the $t_2$ orbitals each stabilise by $\frac{3}{5}$ and the $e$ destabilize by $\frac{2}{5}$. Since $2 * \frac{3}{5} - 3 * \frac{2}{5}$ = 0, the CFSE = 0.

For octahedral complexes, the $e_g$ and $t_{2g}$ orbital populations can be used analogously to find the stabilisation energy.


Metal phosphate buffers A calibration solution contains $\pu{0.02M}$ $\ce{Cr^3+}$ and $\pu{0.02M}$ $\ce{Cu^2+}$. In order to control the solution $\mathrm{pH} = 12$, a phosphate buffer is used with the weak acid being $\ce{HPO4^2-}$ and the weak base being $\ce{PO4^3-}$. The maximum permissible change in $\mathrm{pH}$ is 0.2 units, how many moles of strong acid can be buffered per mole of weak base present?
  $\ce{H3PO4}$: $K_\mathrm{a3}$ = $4.2 \times 10^{-13}$

I have tried to formulate an equation for the reaction but I'm not sure which metal ion is/if both are taking part and therefore finding it hard to produce a $K_\mathrm{a}$ expression.

I think phosphate buffer is the solution here. Reason:

$K_\mathrm{sp}$ of $\ce{CrPO4}$ at $\pu{25 ^\circ C}$ is $2.4 \times 10^{-23}$ while that of $\ce{Cu3(PO4)2}$ at $\pu{25 ^\circ C}$ is $1.4 \times 10^{-37}$, thus both ions will precipitate since concentration of each ion is $\pu{0.02 M}$.


Rotational degrees of freedom (3N-5 and 3N-6) In spectroscopy we described the electric energy with the approximative separability of internal motions as:

\begin{equation}
E=E_e+E_v+E_r+E_{ns}
\end{equation}

(energies: electronic, vibratory, rotatory; nuclear spin (neclected))

With the Born Oppenheimer approximation of nuclei and electrons you get a formula which describes the degrees of freedoms for vibration motions:

3N-5 for linear molecules
3N-6 for non-linear molecules

This formula can be understood by:

the coordinates of 3N atoms.
3 degrees of motions for the three translation of the centre-of-mass motion
2 (for linear) respective 3 (for non-linear) degrees of freedoms reduced for the rotation of the whole molecule

I guess trans buta-1,3-diene is here described as a linear molecule. [added: in the rigid rotor approximation]

First I was thinking why we are so much interested on the number of freedoms for vibrations. We consider only the rotation of the whole molecule but not around bonds.

Why do we not consider rotation about a bond axis, e.g. for pentane a rotation of an ethyle around the propyle group? (rotation about the C(2)-C(3) axis)

May be because it's much another energy range as rotational energy of about 1-100 cm-1 (typical period of 0.1-10 ps).

Rotations around bonds are typically termed "internal rotations", and represent one of the most common problematic cases for the rigid-rotor-harmonic-oscillator (RRHO) model of internal molecular motion.  This is because RRHO assumes that any vibrational amplitudes are "small," and internal rotations are most definitely not small!  Such rotations still involve the internal degrees of freedom of the molecule, though, and thus (as Jan notes) are considered as part of the 'vibration' of the molecule, not its 'rotation.'

Internal rotations typically have some energy cost involved (as in your pentane example), and so cannot be treated as "free" rotations. They are usually termed "hindered rotations," and there is a great deal of literature studying them.  Some citations I know offhand:

Classic Pitzer (J Chem Phys 5 469, 1937), and Pitzer & Gwinn (J Chem Phys 10 428, 1942)
Independent-rotations approximation for treating hindered internal rotation: Pfaendtner (Theor Chem Acc 118 881, 2007)
Other / more-advanced treatments of internal rotation:

Hecht (J Chem Phys 26 31, 1957)
Strekalov (Chem Phys 355 62, 2009)
Vansteenkiste (J Chem Phys 124 044314, 2006)
McClurg (J Chem Phys 106 6675, 1997)

One species of particular interest to me on this topic has been nitromethane, which has an extraordinarily small barrier to rotation around the $\ce{C}-\ce{N}$ bond (see Strekalov, above), such that at ambient temperature it can be considered essentially a free rotation.


Why is the boiling point of rectified spirit lower than alcohol absolute? The boiling point of rectified spirit is ($78.1~\mathrm{^\circ C}$) while that of absolute alcohol is ($78.3~\mathrm{^\circ C}$). I know its too small, but shouldn't the addition of impurities increase the boiling point?

What qualifies as an impurity? Is it that solids like salt only can be called an impurity? — which increase the boiling point of water?

Ethanol and water have weaker intermolecular interactions than water with water and ethanol with ethanol. For this reason ethanol and water molecules can more easily escape from the liquid mixture, called azeotrope, to the gaseous phase than it is the case for the pure alcohol. Consequently the azeotrope has a lower boiling point.

What is considered as an impurity depends on your use case. E.g. 96% pure ethanol may qualify as extra fine spirit.


Hydrogen Peroxide as an Oxidizer Why is HOOH such a good oxidizer?

I understand that the oxygens in HOOH are in a negative 1 oxidation state, and that oxygen generally adopts an oxidation state of negative 2.

I also know that oxidizers themselves are reduced, and that reduction entails a drop in oxidation state. This suggests to me that oxygen is more stable when found with a negative two oxidation state rather than a negative one oxidation state. This makes sense; oxygen is the second-most electronegative element on the periodic table, so it likely would be okay with a more negative oxidation state.

And from organic chemistry I understand that the O-O linkage in peroxides is relatively weak; light can rather easily homolytically cleave the O-O linkage. And HOOH is generally stored in dark bottles as to block light. This says something about thermodynamic stabilities to me - that HOOH has relatively weak bonds and that it would be enthalpically favorable for the oxygens to form other, stronger, bonds. Is this how HOOH is used in bombs? The O-O bond is weak; bond formation is exothermic, thus providing a way for the O-O bond to break and rearrange into stronger bonds releases vasts amount of energy?

I'm trying to tie together redox with thermodynamic stabilities.

From your argument I assume the real question you want to ask is "Why HOOH is such a good oxidizer compared to molecular oxygen?".

From thermodynamical point of view, molecular oxygen is a very potent oxidizer. However when you are talking about reactivity, you should always take a look at kinetics, not only at thermodynamics.

One important part of the answer to your question is that not HOOH a good oxidizer, but O2 is a very bad oxidizer from kinetic point of view. Molecular oxygen is in the triplet state, therefore its reactions are often spin-forbidden. Especially the oxygen->superoxide->peroxide steps require often level crossings between different spin surfaces. One very important reason why transition metal complexes catalyse oxidation in biology is this: as they often have electrons to flip and large spin-orbit coupling, the level crossings became easy. There is literally tons of literature on this in bioinorganic chemistry.

Note that in flames and in high temperature burning that reaction is essentially a huge variety of radical reactions, so the mechanism is completely different from the oxidation reactions in aqueous solutions.

Molecular oxygen has also 4 electrons to take when turned to oxides. This sometimes require complicated mechanism where every single step is thermodynamically and kinetically is feasible, and there is always a oxidizable component around.

Also note, that HOOH itself need catalyst to be an effective catalyst, like in Fenton reactions and similar.


Would it be possible to produce a non-toxic atmosphere using a coal fire and frozen oxygen? A Pail of Air is a classic science fiction short story by Fritz Leiber from 1951 describing how a family survives on an Earth that has left its orbit around the sun, using desparate low-tech methods: a shelter isolated by many layers of blankets and a coal fire to thaw frozen oxygen gathered in the eponymous pails.

I am curious how realistic this scenario is from a basic chemistry and biology perspective. I think not for a number of reasons, but don't know enough to quantify these factors:

Would a coal fire produce enough energy to thaw the oxygen the fire itself uses?
Would the resulting atmosphere (at sufficient pressure for the oxygen to support human life) contain toxic levels of carbon dioxide?
...or carbon monoxide?
How does oxygen at lower pressures affect flammability? Wouldn't it all go up in a blaze at the drop of a hat?

Yes, it is possible to produce a non-toxic atmosphere using a coal fire and frozen oxygen.

I'll answer your questions last to first.

No, low pressure oxygen will not make flammability worse, fires require something to bond with oxygen (often carbon, causing carbon dioxide). Low pressure will make things less flammable as there is less oxygen to bond with.

Yes, using an open coal fire in a semi-enclosed space will gain toxic levels of carbon dioxide, carbon monoxide and sulfur dioxide (among a few others) quickly. However with a minor modification from a open fire to a fire in a old fashioned chimney pipe stove (venting outside the hut) would eliminate this.

Lastly, Definitely Yes, a coal fire will produce more than enough energy to thaw the oxygen it uses. To quote figures 1 pound of coal will produce roughly 12,500 btu's (British Thermal Units) Assuming the temperature of the frozen oxygen we are thawing is -455°F (the "temperature of deep space") and about 3 degrees below the freezing temperature of helium. To thaw 1 gallon of the frozen oxygen will require roughly 900 btu's (if your stove has a 80% efficiency) and produce 860 gallons of pure oxygen at 80°F and atmospheric pressure. Coal will consume 140% of it's weight in pure oxygen, so for 1 pound of coal 135 gallons of your oxygen will be consumed. Leaving you with 11,500 btu's extra heat and 725 gallons of oxygen. The average Human consumes 25 gallons of oxygen per day making this plenty to survive on.


Why is sand/glass polar? How does lattice structure change polarity? The molecular geometry of sand is linear. Dipole moment is thus cancelled.

But we all know from the meniscus water makes with glass that the dipole-dipole/H-O attraction  between sand and water is stronger than the hydrogen bond between water molecule. maybe I am wrong

My guess is either that electron lone pair makes it polar or the lattice structure does.

I'm not sure where you get the idea that glass is linear. What makes a glass is the lack of long range ordering, though in fused silica, the approximate tetrahedral structure of $\ce{SiO4}$ is maintained (this is a 2D section, the fourth oxygens are not shown):

 Source

In normal glasses, there is no significant net dipole moment across the entire structure, but the reason we observe strong adhesive forces between water and glass is a surface phenomenon—in the bulk of the glass, every silicon atom is connected to four oxygen atoms and each oxygen atom to two silicon atoms, but what happens at a boundary/surface? There are oxygen atoms that only have one bond. When these are exposed to an aqueous solution, some will bear a negative charge, others will pick up a proton from water. The theory behind it is pretty complicated, but you can think of it as a simple acid-base reaction—at low pH more will be protonated:

$$\ce{Si-OH <=> Si-O- + H+}$$

In any case, protonated or not, silanol groups are very happy to associate with water, accounting for the concave meniscus.


Why must the rate law be experimentally determined? I am familiar with the idea that rate law must be determined experimentally. I am also familiar with the idea that there is no formulated way for coming up with the rate law besides just experimentation.

My question is: why isn't the rate law of:

$$ aA + bB \rightarrow \mbox{something}$$

just:

$$r = k[A]^a[B]^b$$

?

I mean, the equilibrium constant expression involves something to that effect, and the equilibrium constant expression should be precisely talking about when the rates for each direction are equal. So if the rate law is not the above, then why does the equilibrium have that expression?

EDIT: A bit off topic, but how is it possible for something to be zeroth order?

Since rate of a reaction depends actually upon the way reactants react, there is not surety that the quoted stoichiometric coefficients may be the same for rate of a reaction, the rate depends on the way or mechanism of the reaction, and since all reactions have a bit different mechanism no strict formula could be quoted. It is only experimentally derived and can be theoretically approached if we know that it is an elementary (simple step) reaction or it falls under a already classified or studied reaction.

Zero Order, what if a reactant is in very large excess, now the rate is not limited or dependent on its concentration, also it's concentration is almost constant, so it is also merged into the constant and hence we have zero order, some adsorption reactions are maybe zero order too, it is because it actually depends on the surface area of the substance and it's constant.


What would be the physical interpretation of the equation dG = Vdp − SdT? I was trying to understand the derivation of the formula $G=G^\circ+RT\ln (p/p^\circ)$ starting with the following relationships:
$$
\begin{align}
\mathrm dU&=T\,\mathrm dS-p\,\mathrm dV\tag{1}\\
\mathrm dH&=\mathrm dU+p\,\mathrm dV+V\,\mathrm dp=T\,\mathrm dS+V\,\mathrm dp\tag{2}\\
\mathrm dG&=\mathrm dH-T\,\mathrm dS-S\,\mathrm dT=V\,\mathrm dp-S\,\mathrm dT\tag{3}
\end{align}
$$
Under isothermal condition $\mathrm dG=V\,\mathrm dp$ $(\mathrm dT=0)$. Therefore, the change in $G$ from the standard state is given by:
$$
\begin{align}
\int\mathrm dG&=\int V\,\mathrm dp=\int\frac{nRT}p\,\mathrm dp\tag{4}\\
G-G^\circ&=\ln p-\ln p^\circ\tag{5}\\
G&=G^\circ+\ln\frac p{p^\circ}\tag{6}
\end{align}
$$
But I'm confused about the equation $\mathrm dG=V\,\mathrm dp-S\,\mathrm dT$. What does this physically mean?
I'm assuming $V\,\mathrm dp$ is some form of work done by the system, but that suggests $G$ of the system increases while it loses energy doing the work. I also have the same problem for $\mathrm dH=T\,\mathrm dS+V\,\mathrm dp.$ Can someone please explain what they really mean?

For a continuous flow system operating at steady state in contact with a constant temperature reservoir at the same temperature as that of the entering fluid, and with negligible change in kinetic energy and gravitational potential energy, the reversible shaft work per mole that the system does on the surrounding is equal to the integral of -vdP, where v is the fluid molar volume.  So $\Delta G$ per mole is equal to minus the shaft work per mole.


Structure of xenon hexafluoride The central atom has a hybridization of $\mathrm{sp^3d^3}$. Thus, its structure should be pentagonal bipyramidal.

Why is it not that but a distorted octahedron?

This is one of the many reasons why hybridisation including d-orbitals fails for main-group elements.

Xenon in $\ce{XeF6}$ is not hybridised at all. Instead of invoking populated core d-orbitals or energetically removed d-orbitals (remember the aufbau principle: the next shell’s s-orbital has a lower energy than the d-orbitals you are proposing to include in hybridisation!) xenon just offers its three p-orbitals $\mathrm{p}_x, \mathrm{p}_y$ and $\mathrm{p}_z$ for four-electron-three-centre bonds. These 4e3c bonds can be understood using the following two mesomeric structures:

$$\ce{F^-\bond{...}Xe^+-F <-> F-Xe^+\bond{...}F-}$$

Each $\ce{Xe-F}$ bond has a bond order of ½, and for each fluorine there is another with a bond angle $\angle(\ce{F-Xe-F}) \approx 180^\circ$ as part of the same 4e3c bond.

Also note that this means that xenon’s lone pair is comfortably located in the $\mathrm{5s}$ orbital.


Can the electron configuration of Te be written that way? Normally, the electron configuration of Te is known as:

$$\begin{aligned}  {[Kr]} 5s^2 \ce{4d^10} 5p^4 \end{aligned}$$

Then, one day I was asked in a exam if this can be written also as:

$$\begin{aligned}  {[Kr]} 5s^2 \ce{4d^10} 5p^3 6s^1\end{aligned}$$

I answered that it couldn't. But my answer turned out to be wrong.

If this is correct, why is it allowed to be written like this? And what general case can I learn when being asked such tricky questions?

The second configuration allows Te to take advantage of the relative stability of half-full shells. Having the 5p and 6s orbitals both half-populated is more stable than having 4 electrons in the 5p orbital.

Similar behavior is what allows carbon to form 4 bonds. The $\ce{s^1p^3}$ configuration is more stable than the $\ce{s^2p^2}$.

Tellurium shows the behavior while oxygen does not as a result of the overlap between the 5p and 6s orbitals in the larger atoms. The same degree of overlap does not occur with the 2p and 3s orbitals.


Is there a characterisation for conjugated dienes similar to cis-trans? As shown above,we know there are two configurations possible for conjugated dienes.

Are they given any special names like we have cis and trans?
This came up because while studying the Diels Alder reaction, it occured to me that maybe only the 'cis' like dienes will go through the reaction.

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) supports various stereodescriptors that are used to denote specific conformers, in particular concerning conjugated dienes:

  P-94.3.1.3 Stereodescriptors s-cis and s-trans

  The spatial arrangement of two conjugated double bonds about the intervening single bond is described as ‘s-cis’ if ‘synperiplanar’ and ‘s-trans’ if ‘antiperiplanar’. This descriptor should not be applied to other systems such as N-alkyl amides (use ‘E/Z’ or ‘sp/ap’).


Significance & applications of polarizability for nucleophiles One of the reasons given for the greater nucleophilicity of iodine ion compared to fluorine ion is the greater polarizability of the large electron cloud of iodine. The other reason given is that it is less crowded by polar protic solvent molecules.

However, when the solvent is changed to a polar aprotic solvent the nucleophilicity order is reversed.

If polarizability had any significant role to play, the order would've not been completely reversed. It seems to me that the only reason iodine ion is more nucleophilic than the fluoride ion is because it is less crowded by the solvent.

Now, this leads me to believe that polarizability is not a very important factor and should only be used when comparing nucleophiles of the same basicity, but of different size (and hence polarizability) in a polar aprotic spent.

But why then, for uncharged nucleophiles does polarizability dominate over basicity in determining the order of nucleophilicity (regardless of the solvent used)?

The examples the site you mentioned gave of uncharged nucleophiles were: the hydrides of Group V and VI. They say that the bigger atoms can have a more effective overlap of atomic orbitals and thus, the bigger the uncharged nucleophile, the stronger it is.

I would like to dispute this reasoning and answer your question regarding this trend.

How we should see a nucleophile should always primarily be in terms of electron donation because a nucleophile donates electron density to electrophiles to form bonds. This is a very important consideration.

For bigger atoms, the valence electron pair, which is being donated to the electrophile, is further away from the nucleus and is more shielded from the nuclear charge. Thus, it is easier for donation to take place. This should be the reason why uncharged nucleophiles with bigger atoms are more nucleophilic; that is, they are more capable of donating the electron pair of theirs. My explanation of the trend has nothing to do with polarisability.

The reasoning stated is incorrect because although bond formation is easier for bigger atoms, it does not mean that the overlap of atomic orbitals is more effective. In fact, it is likely less effective because the atomic orbitals used are more diffused (Recall that as the principle quantum number n increases, the orbitals become more diffuse in terms of electron density). Thus, their reasoning is nonsensical.

Hope I have clarified everything you were not sure of.


Thermodynamic versus kinetic reactions In a lecture, teacher said

"Nucleophilicity is about 'kinetic', and basicity is about 'thermodynamic'"

but I saw the picture below from another source.

Here, both pathways start with an acid/base reaction (it's about basicity) and then nucleophilic addition happens (it's about nucleophilicity). as a result, 'kinetic' and 'thermodynamic' product come up.

So here's my question.

Are the terms 'thermodynamic' and 'kinetic' that are used above in two cases different?

The terms are used in a different context to mean slightly different things.

The kinetic product of a reaction is the one which is formed the fastest. In the case shown above this is the 1,2 product which results from formation of the most stable cation, which has a lower activation energy for formation.

The thermodynamic product is the most stable product. In this case it is the 1,4 product because it has a more substituted double bond. However, it forms from a less stable carbocation and so the activation energy is higher and it is formed slower than the kinetic product. Raising the temperature favours the thermodynamic product because the activation energy becomes less important. Another way to get the thermodynamic product is to run the reaction under equilibrating conditions, allowing the product to convert back to reactants. This will eventually form more of the more stable product.

Nucleophilicity is often referred to as a kinetic property because it has to do with how well a species can attack an electrophile and is affected by factors such as steric hindrance.

Basicity is often referred to as a thermodynamic property because it is to do with the position of the acid-base equilibrium: $$\ce{B + H+ <=> BH+}$$


Method to synthesize KAu(CN)2 (or KCN)? Every reasonably effective method of electroplating $\ce{Au}$ onto another metal's surface (e.g. $\ce{Cu}$,$\ce{Ni}$,$\ce{Ag}$, etc.) involves use of $\ce{KAu(CN)2}$ or in some cases $\ce{KAu(CN)4}$ in acidic, neutral, or (in almost all cases less effective) basic solution. Ideally it should be $60$ to $80~\mathrm{^\circ C}$ with a certain current density applied (c. $0.5~\mathrm{A/dm^2}$). In the past I have tried a direct route of dissolving the gold in warm $\ce{HCl / HNO3}$ as the deposition solution to little effect (a brownish/blackish tint is left on the receiving surface with very little tint of gold). The short story is that I have found the use of cyanide unavoidable.

$\ce{KAu(CN)2}$ runs $157/g from Sigma-Aldrich so I would like to synthesize it, if possible.

I have:

1g solid $\ce{Au}$
Excess $\ce{KNO3}$, $\ce{KI}$
Various $\ce{C-N}$ organic compounds, e.g. urea, uric acid, etc.

I think my best route is indirect — to first synthesize $\ce{KCN}$ and then form the product:

$$\ce{4 Au + 8 KCN + O2 + 2 H2O → 4 K[Au(CN)2] + 4 KOH}$$

...and follow by vacuum filtration on the solid precipitate. Which begs the question of the best method for safe synthesis of $\ce{KCN}$.

Much of my information comes from this difficult-to-find and expensive book, "Properties of Electrodeposited Metals and Alloys: A Handbook" by Safranek. It's got quite a lot of proprietary data. Also Kohl's chapter on gold electrodeposition.

I read that you are a computational chemist at university, so I assume you went through chemical training. This is a good thing. But please remember that cyanides are potent poisons and that protonated cyanide ($\mathrm{p}K_\mathrm{a}(\ce{HCN}) \approx 8$) aka hydrogen cyanide is a poisonous gas.

If you are not a trained chemist (or if you are one but do not feel $100~\%$ safe about using cyanides), stick away from them!

With that in mind this leaves two methods of acquiring $\ce{K[Au(CN)2]}$ — either buying it directly from YourSuppliers™ or by synthesising it. My preferred route of choice would definitely be to buy it from YourSuppliers™, as that is the safest way possible — and always always ensure that it remains in an alkaline solution.

I have no idea about the experimental procedure to synthesise $\ce{K[Au(CN)2]}$ according to the reaction you propose, but it didn’t escape my notice that you are generating $\ce{4KOH}$. This means, I’ll have to assume that the reaction runs under acidic conditions. If that is the case, do not use that method unless you have access to a proper and tested fume hood! If there are other procedures out there that employ alkaline reaction conditions you may (as a trained chemist!) decide to use them at your own risk.

Assuming you did find a method that employs alkaline conditions, the question remains where to get the cyanide from. Well, SigmaAldrich gives me a price of 38 € for $100~\mathrm{g}$ which means that the chemical is very cheap. Note that acquisition of potassium cyanide may well be regulated in your jurisdiction — and rightly so!

Whatever you do, do not attempt to synthesise $\ce{KCN}$ (or any other cyanide for that matter) at home!

Remember that a certain percentage of the population can sense $\ce{HCN}$ which they perceive as an almond-like smell. Don’t attempt to test which part you belong to. If the concentration of $\ce{HCN}$ is such that you can smell it, you are in severe danger!

This bears repeating for future visitors:
Do not do this unless you are a trained chemist and have access to a proper fume hood! Never, ever try this at home! Never work with cyanide in the open air or in any other place where unsuspecting passers-by may be endangered!


Why does cyclobutadieneiron tricarbonyl behave aromatically? It is said that $\ce{(\eta^4-C4H4)Fe(CO)3}$ can undergo electrophilic substitution reactions. Therefore, it displays aromaticity.
For the iron atom, it has $8$ electrons in its outer shell initially and it receives $6$ electrons from three carbonyls. It achieves the stable state of $18e$ after bonding with cyclobutadiene. Therefore, the oxidation state of iron should be zero, and the $\pi$ electrons of cyclobutadiene in the complex is still $4$.
But this result contradicts with Huckel's rule, which states that the $\pi$ electrons should be $4n+2$. Where did I go wrong?

In the paper "(Cyclobutadiene)iron Tricarbonyls - A Case of Theory before Experiment", Organometallics 2003, 22, 2-20 https://pubs.acs.org/doi/pdf/10.1021/om020946c, at the page 12 nice explanation of bonding in this compound is provided:

  The electronic structure of (cyclobutadiene)iron tricarbonyl has been the subject of many papers. He I and He II low-energy photoelectron spectra provided useful information.77,78 The eight observed bands in the lowenergy PE spectrum of  (cyclobutadiene)iron tricarbonyl in the range 7.65-20.31 eV all were assigned.78 The
  calculations (ab initio SCF MO) showed that there is a net negative charge on the cyclobutadiene ligand that results from π back-bonding from the iron atom into an
  antibonding MO of the ligand (δ bond). Better agreement between the experimental assignments and theoretical calculations was obtained by Chinn and Hall using generalized MO calculations with configuration interaction.79 A simple textbook approach to the bonding is shown in Figure 8.80 It is assumed that it is triplet state cyclobutadiene that is involved with unpaired electrons in the two degenerate ψ2 and ψ3 molecular
  orbitals. These interact with two singly occupied iron orbitals, generating two covalent bonds.


Why is the melting point of hydrogen fluoride lower than that of ammonia? The melting point of hydrogen fluoride is -83.6°C, as compared to that of ammonia, which is -77.73°C. How does this make any sense?

Both hydrogen fluoride and ammonia show hydrogen bonding, which "elevates" the boiling points of these two compounds when compared to the rest of the pnictogen and halogen hydrides. However, I was under the impression that hydrogen bonding increased in strength the more elctronegative atom the hydrogen atom is bound to, explaining why the boiling point of ammonia is significantly lesser than that of hydrogen fluoride (-33.34°C and 19.5°C respectively).

However, this makes no sense when looking at melting points of the two elements.

The additional energy involved here is the stabilization energy, a form of potential energy that is stored in the more stable arrangement of a solid compared to the liquid form. This energy is determined by many factors:

Size and shape of the individual particles.
How they can 'pack' into a solid arrangement.
etc...

SO, there is clearly something different about how Ammonia is able to arrange in solid form that makes the solid stable for a greater range of energy added.
When we look at a phase change diagram the slope and horizontal component of the line determine total energy needed for that phase change. This will be different for each substance considered, and this is where your discrepancy comes from.


Which ultra-high temperature ceramics make a good electrical insulator? I am looking for materials with melting point above 3,000 °C and yet is a good electrical insulator. Generally metal carbides have high melting point but a lot of them are electrical conductors.

Thorium dioxide gets you 3390°C.  Have fun getting it, $$$ and also the thorium is radioactive.  You also risk some nonstoichiometry, due to thorium having variable oxidation states, and that could make the compound somewhat conductive.  Magnesium oxide weighs in at 2800°C; but is more stable, cheaper, nonradioactive and having no variation in stoichiometry.


How to re-suspend silver from the dried out silver paste? Kindly suggest any method or solvent to re suspend silver from old dried out silver paint/paste. It's a 25 years old "Dotite©" D-550 bottle. Strangely there is no expiry date mentioned on the bottle.

According to SPI Supplies:

  If the Dotite Silver Paint dries out, unlike some other silver paints, it can be re-suspended with the addition of SPI #05004 Silver Paint Thinner followed by an ultrasonic bath treatment.

Generally, ketones are used as diluents, also for conductive paints (acetone, methyl ethyl ketone (MEK), methyl isobutyl ketone (MIBK)).


Why Heating of certain salts like Lead salts produce crackling sound? I recently discovered that heating of certain salts produces crackling sound and that is called Decrepitation. But I couldn’t figure out the reason why that happens moreover I couldn’t find anything satisfactory on the internet.

Frankly speaking, you're reading too much into this phenomenon. Have you ever heard the satisfying crackling noise when you burn logs of wood in a bonfire? I'm sure you must've got the answer by now:
Decrepitation is more of a physical process rather than a chemical process.
When we write chemical equations, we treat compounds as single molecules, but in reality this is not so.
You specifically mentioned lead nitrate. You must be aware that it is a  white/colorless crystalline salt. To expound further,  solid lead(II) nitrate exists in a face centered cubic structure. Coming back to your question, when we heat Pb(NO3)2, we don't heat a single molecule, we heat a sample of the salt.
When we do so, the crystal structure breaks as the entropy increases. The deformation of the crystalline structure causes sound energy to be evolved.
TL;DR: Breaking down of large crystals into smaller crystals and/or intermediates causes the crackling sound known as decrepitation.


Difficulties understanding how salt bridge works A picture for reference:

Difficulty 1: I don't get why a salt bridge is needed. People say it's to maintain the overall neutrality. But in both cases where  salt bridge is a tube of solution and a membrane, two half cells can be seen as one. So as long as no thing leaves either of the two half cells, the whole is always neutral.

Difficulty 2: as in the picture, cations go to the right cell, and anions go to the left. I can understand that anions go the left because the number of zinc ions increase as the reaction proceed. But in the right half cell, the electrons are being absorbed, its charge cancelled by that of the copper ion; I don't see any need for sodium ions to move to the right.

Regarding Difficulty 1)  When the zinc bar (unconnected) is first immersed into the solution, some zinc ions immediately dissolve away.  Now the zinc bar has extra electrons and the solution has extra positive zinc cations.  This is entropically acceptable, however, because cations in the solution (both zinc cations and hydronium cations) can align close to the surface of the zinc metal (creating what's called the 'Helmholtz layer').

One might imagine that as soon as the zinc bar was connected, the extra electrons in the zinc would drain to the other side of the cell.  But they don't because the electrostatic pull of the zinc and hydronium cations in the left-beaker solution is holding them back.

There's an analogous situation in the right cell:  the copper metal can't accept any electrons because of the electrostatic repulsion of the sulfate and hydroxide ions in right-beaker solution.

Regarding Difficulty 2) When the cell is running, there is an electrostatic gradient across the membrane where the salt bridge meets the right-side solution (among other places, for more info see note 1).  Cations (such as sodium) are driven out of the salt bridge;  anions (such as sulfate) are drawn into the salt bridge.  Sodium cations are smaller and diffusion more quickly than the larger, slower sulfate anions. Overall, many more sodium ions will leave the salt bridge compared to the number of sulfate anions that will enter the salt bridge.  There is an analogous situation at the membrane where the salt bridge meets the left-beaker solution.

Note 1, Electrostatic Gradients) When the cell is running, the zinc bar has a slightly negative charge and the copper has a slightly positive charge.  If we let ourselves ignore the metal wire for the moment, this creates an electrostatic field throughout  the cell very much like the classic "electrostatic field around adjacent positive and negative charges in a uniform media".  Notice that in the diagram of that situation, there are electrostatic field lines pointing from the positive charge straight-across to the negative charge.  Another way to describe this situation would be to say that there is an electrostatic gradient between the two charges.  If you drop a much smaller, and otherwise free-floating positive charge any where in the neighborhood around those two charges, it will move down electrostatic gradient to the negative charge:  the same way rain-water flows downhill, (in that case due to a gravitational gradient).  Now imagine there was an imaginary plane exactly halfway between the two charges:  you could now also say that there is an electrostatic gradient across that plane, from the one side to the other. If you dropped a positive charge right beside the plane, but on the same side as the bigger positive charge, the electrostatic gradient would drive the positive charge across the plane.  Similarly, if you dropped a negative charge next to the plane,  on the same side as the bigger negative charge, then again the electrostatic gradient would drive the negative charge across the plane.

In the electrochemical cell problem, the electrostatic field lines are going to be more complicated because there are two materials (air and water / salt-bridge) with different permittivity.  Long story short: the electrostatic fields will actually be much stronger in the water than in the air.  The strongest part of the electrostatic field will go from the copper electrode, across the right-side solution, through the salt bridge, and across the left-side solution to the zinc.   So, among other places, when the cell is running, there will be an electrostatic gradient across the membrane where the salt bridge meets the right-side solution.


The rate of change in pH v.s Temperature I have somewhat established that there is a linear relationship between change in pH of carbonic fizzy drinks over time. I also realise that the pH increases linearly faster for the same carbonated fizzy drink when the temperature of the drink is changed (increased).

Now the problem is, when I graph the relationship between the rate of pH increase (y-axis) and the temperature (x- axis), I get an exponentially increasing relationship. I do not know if this relationship is true, but even if it was true I do not know how to explain why this is true.

Are there any equations I can manipulate to show a mathematical relationship between rate of pH increase and temperature of the same carbonated fizzy drink?

You are dealing with temperature variation of carbonic fizzy drinks. So, the situation is somewhat tricky.

Lets approach the problem step by step.

If you raise temperature of neutral water, the rate of decomposition of water will increase. So, the water will contain more protons at higher temperature than at lower temperature, although the neutrality is maintained. Therefore, the pH of neutral water will decrease with temperature. pH of water at 100°C is ~6.126.
Carbonic fizzy drinks contain sodium & potassium bicarbonate salts, which ionize completely in aqueous solution to yield alkali cations & bicarbonate anions. The initial pH is between 3 and 4. (1)
Bicarbonate anion can capture a proton from the system to yield carbonic acid. Carbonic acid is unstable and thus decomposes even at room temperature to produce carbon-dioxide, & thus the bubble! So, even at room temperature gradually the solution turns less acidic, then neutral & finally alkaline. So, here you are not increasing temperature & just letting time to flow, & the pH rises gradually. Therefore, I can expect 'linear relationship between change in pH of carbonic fizzy drinks over time'.
When you raise temperature of carbonic fizzy drinks, the pH increase gets faster because the decomposition of carbonic acid is very much temperature sensitive. It is worthwhile to mention that the rapid pH rise is somewhat countered by the pH lowering due to water's decomposition, but that effect is much less significant as water's decomposition is much less sensitive to temperature. In this case, I can certainly expect deviation from linear behavior (nonlinear or exponential increase) aided by temperature influenced decomposition of carbonic acid.

You may be able to arrive at a mathematical relationship by combining the temperature dependency of carbonic acid & water decomposition.


Is propanol 1-2 diol 178 the same as propylene glycol? I clean vinyl records in an ultrasonic vinyl record cleaning machine the manual states it uses propanol 1-2 diol 178 as a surfactant.  I think this is the same chemical as propylene glycol. Is it?

Yes, “propanol 1-2 diol 178” is the same compound as propylene glycol and appears to be a fabricated name used in a manual for a proprietary cleaning system by Kirmuss Audio.
According to the article from Analog Corner magazine [1, p. 22] (emphasis on the footnote):

The Kirmuss Audio KA-RC-1 comes with accessories: 60 mL of an antibacterial/antistatic/antifungal spray; a goat hair brush applicator; an optical-quality microfiber cloth; a larger microfiber cloth (decorated with rabbits) to cover the work space; a circular felt mat to support the record during hand cleaning; a combination carbon fiber/parastatic felt brush; and a bottle of stylus cleaner.
All of this costs $870—far less than at least one competing system and only
slightly more than the iSonic machine and its inferior record-spinning mechanism.
Granted, the accessories are not costly, but no matter how you figure it, Kirmuss doesn’t seem to be in this for the money: He’s not looking to lose money, but clearly he’s not trying to make a killing.
From what I’ve witnessed, he’s more concerned with waging a worldwide war against fungus.
The antibacterial/antistatic spray is identified in the manual as a “98-99% distilled water, 1-2% propanol 1-2 diol 178 mix.” That last one is propylene glycol,4 an organic solvent used in, among other things, pharmaceutical preparations and as a food additive.

4 That chemical name seems made-up, and Kirmuss at first disputed that the ingredient was nothing more than propylene glycol, but a subsequent correspondence made it clear.

Reference

Michael Fremer. The Seven “M”s of the Charles Kirmuss Vinyl Restoration System. Analog Corner, July 2019. (PDF, accessed 2020-09-03)


Thermodynamic quantities for the freezing of water I am trying to do the following problem:

Freezing of water at $273~\mathrm K$ and $1~\mathrm{atm}$
Which of the following is true for the above thermodynamics process?
p) $q=0$
q) $w=0$
r) $\Delta S_\mathrm{sys}<0$
s) $\Delta U=0$
t) $\Delta G=0$

My answer was r, s, and t:

Since the reaction happens at constant temperature, internal energy is constant (yes for s)
Since the reaction is open, it is isobaric in nature; and since the volume changes due to the phase change the work done is not zero (no for q)
Heat must be exchanged (first law of thermodynamics) (no for p)
Randomness decreases so $\Delta S_\mathrm{sys} < 0$ (yes for r)

But, the answer given is q, r, and t. I don't understand why q is correct but s is not.

s is not correct because U is not just a function of temperature but also a function of chemical composition.  The internal energy of liquid water is higher than the internal energy of ice.  If you want to think of it in terms of temperature, then just view the heat capacity as being infinite at the phase transition (Dirac delta function of T), such that the integral over the phase change is equal to the heat of fusion.  As far as q is concerned, I guess they are assuming that the change in volume from solid to liquid is negligible.


Dipole moment of ferrocene conformations Which conformation of ferrocene has a zero dipole moment?

I thought that the answer should be the staggered conformation because it would have the cyclopentadienyl groups oriented away from each other. However, the answer given is the eclipsed form. I can’t understand why that happens.

To find out whether a compound can have a nonzero dipole moment, consider its symmetry. A dipole moment is a vector which must be transformed onto itself by every symmetry operation present in the molecule’s point group.

It should be easy to see that te eclipsed conformation has five planes of symmetry perpendicular to the cyclopentadienyl rings ($5\sigma_\mathrm v$). Another sixth plane of symmetry is sandwiched between the two rings and passes through the iron atom ($\sigma_\mathrm h$). The only vector that will be transformed onto itself by all these planes of symmetry is the null vector $\vec{0}$. Therefore, this conformation has zero dipole moment.

The staggered conformation also has five planes of symmetry perpendicular to the ring planes as the eclipsed conformation ($5\sigma_\mathrm v$). It also features a centre of symmetry in the iron atom ($i$) as well as improper rotation ($S_{10}$). The inversion centre alone is enough to determine that only the null vector $\vec 0$ will be transformed upon itself. Therefore, this form also has no dipole moment.

Both you and the book are correct.


Reaction of sulphur and nitric acid I would like to ask a question about Sulfur in the reaction with Nitric Acid, $\ce{HNO3}$ to produce Sulfuric Acid, $\ce{H2SO4}$

I read that as Nitrogen and Sulfur are reduced and oxidised respectively, the reaction equation can be constructed as follows:

$\ce{S + 6HNO3 -> H2SO4 + 6NO2 + 2H2O}$

Now, when I read an article on wikipedia regarding sulfur, the introductory paragraph states the following:

  Under normal conditions, sulfur atoms form cyclic octatomic molecules with a chemical formula $\ce{S8}$.

So why is sulfur represented in the equation as $\ce{S}$ and not $\ce{S8}$?

The most stable allotropic form of sulfur exists in the formula of S8. In other words, sulfur occurs naturally in S8 formula. Multiply all the stoicheomtric coefficient by 8 (other than S) to obtain the balanced chemical equation


How do I calculate the concentration of sulphuric acid by a titration experiment with sodium hydroxide? In my latest chem lab the objective was to create a primary standard of $\ce{NaOH}$ and use it to determine the concentration of sulfuric acid.

The first part of the lab was determine the molarity of the $\ce{NaOH}$ solution through a series of titrations.

A sample of KHP (abbreviated form of $\ce{KHC8H4O4}$) was placed into a flask with approximately $\pu{25 ml}$ of water.
Phenolphthalein was added to the flask as the indicator. $\ce{NaOH}$ was then titrated into the flask with a burette. From multiple titrations of this sort I was able to calculate the molarity of $\ce{NaOH}$.

Below I have included part of my table and calculations (Note: $\pu{1 mol}$ of KHP is equal to $\pu{1 mol}$ of $\ce{NaOH}$ in this experiment. If I have made any mistakes please tell me).

Trial 1:
Mass of KHP in flask $= \pu{0.5108 g}$
Volume of $\ce{NaOH}$ added to flask $ = \pu{21.73 ml}$
Calculation of molarity of $\ce{NaOH}$ for trial 1:
Molar mass of KHP $ =  \pu{204.23 g/mol}$
$\pu{0.5108 g}/\pu{204.23 g/mol} = \pu{0.002501 mol}$ of KHP, which is equal to $\pu{0.002501 mol}$ of $\ce{NaOH}$.
Molarity of $\ce{NaOH} = \pu{0.002501 mol}/\pu{0.02173 L} = \pu{0.1151 M}$

I did 3 other trials like this (in total 4) and calculated the average molarity of $\ce{NaOH}$ to be $\pu{0.1159 M}$.

The second half of the lab is the part I had trouble with.
We were given a sample of $\ce{H2SO4}$ with an unknown concentration. I took $\pu{10 ml}$ of this $\ce{H2SO4}$ and mixed it with $\pu{100 ml}$ of distilled water. This new diluted solution of $\ce{H2SO4}$ (I will refer to it as solution 2 now) was the solution used in the trials to determine the molarity. So $\pu{25 ml}$ of solution 2 was added to a flask with a few drops of phenolphthalein. A titration using $\ce{NaOH}$ (the same $\ce{NaOH}$ as used in the previous section) was performed.

My task is to now figure out the concentration of the original $\ce{H2SO4}$ solution. I have tried 2 different methods. The first method I attempted seems so flawed I didn't bother to put it on (it didn't even make sense to me). Each method seems incorrect and have yielded drastically different results. Below I have provided a sample of my table and one of my attempts to solve for the molarity of $\ce{H2SO4}$.

The net ionic equation of this procedure is:
$$\ce{H2SO4 +2NaOH <=> Na2SO4 + 2H2O}$$

Trial 1:
Volume of diluted acid (solution 2) in flask: $\pu{25.00 ml}$
Volume of $\ce{NaOH}$ added to flask: $\pu{23.81 ml}$

Attempt 1 at finding molarity:
Moles of $\ce{NaOH}$ added to flask: $\pu{0.02381 L} \cdot \pu{0.1159 M} = \pu{0.0027596 mol}$ $\ce{NaOH}$

Amount of of $\ce{H2SO4}$: $0.0027596/2 = \pu{0.0013798 mol}$ $\ce{H2SO4}$ (The 2 came from the net ionic equation above)
Molarity of diluted $\ce{H2SO4}$ (solution 2): $\pu{0.0013798 mol}/ \pu{0.025 L}= \pu{0.054172 M}$
(I may be using the wrong volume, is it possible that I have to add the $\pu{25 ml}$ to the $\pu{23.81 ml}$ and divide by $\pu{0.04881 L}$?)

\begin{align}
C_1V_1 &= C_2V_2\\
C_1&=?\\
V_1&= \pu{0.01 L}\\
C_2&= \pu{0.054172 M}\\
V_2&=\pu{0.1 L}\\
\text{Therefore:}\\
C_1&=(\pu{0.054172 M})\cdot(\pu{0.1 L})/(\pu{0.01 L})\\
C_1&=\pu{0.54172 M}\\
\end{align}
Molarity of original/stock $\ce{H2SO4}$.

Molarity of diluted $\ce {H2SO4}$ (solution 2): $\pu{0.0013798 mol}/\pu{0.025 L} = \pu{0.054172 M}$
  (I may be using the wrong volume, is it possible that I have to add the $\pu{25 ml}$ to the $\pu{23.81 ml}$ and divide by $\pu{0.04881 L}$?)

No, you used the correct volume since you want to know the concentration of the $\pu{25 mL}$ that you added to the flask.

  $\pu{0.54172 M}$ - Molarity of original solution of $\ce {H2SO4}$.

Correct.


What do the subscripts 1/2 and 3/2 for the p-orbitals refer to? What does $p_{1/2}$ and $p_{3/2}$ mean when referring to electron configuration?

I've seen sub shells and I'm assuming it has something to do with spin. I am asking this in regard to the  $8p_{1/2}$ and  $8p_{3/2}$ shells and the way they get filled at different energy levels?

Essentially, this derives from Dirac’s relativistic quantum mechanics.

In nonrelativistic quantum mechanics—the Schrödinger equation that you are probably highly familiar with—p orbitals are triply degenerate while d orbitals are 5-way degenerate. However, nonrelativistic quantum mechanics does not include the electronic property known as spin unless it is introduced in a separate step.

Dirac’s relativistic quantum mechanics solves this problem by introducing $c$ as a finite constant. Almost magically, his equations directly result in a quantum number for the spin, $\vec s$. However, the result is that the azimuthal quantum number no longer behaves like a true quantum number (i.e. it no longer describes stationary states) but thankfully the sum $\vec j = \vec l + \vec s$ does. Therefore, instead of considering just one $\vec l$ as a descriptor to decide between s, p and d orbitals, we are stuck with $\left |\vec l - \vec s\right |$ and $\left | \vec l + \vec s\right |$. For $\vec l = 1$ (a p orbital) this gives us the values of $\vec j = \frac12$ and $\vec j = \frac32$. There are two p orbitals of the $\mathrm p_\frac32$ type and one of the $\mathrm p_\frac12$ type. These are no longer degenerate, i.e. $\mathrm p_\frac12$ has a lower energy than the $\mathrm p_\frac32$ ones.

The orbital of the $\mathrm p_\frac12$ type happens to be spherical in nature. For a more thorough background, I recommend Nicolau’s extensive answer which I mostly paraphrased in the above paragraphs and which contains links to further materials.


Geometrical Isomerization Can you explain to me:

What are Cis-Trans, E-Z and Syn-Anti Isomerization?
What is the difference between them?
How many functional groups is cis-trans isomerism applicable to?

I  know mostly about cis trans and E-Z but have no idea about syn- anti isomers. Correct me if needed: I think cis and trans isomers respectively occur when identical atoms or groups are on the same or opposite side of the double bond respectively. E and Z isomers are for trans and cis in complex cases where more functional groups are involved.

Also, I couldn't follow the Wikipedia Link.

cis and trans are older notations that express roughly same concept of E and Z.

Explicitly, E and Z have absolute meaning and are assigned based on precedent, where the Z is assigned to the case where the highest priority substituents are on the same side of the alkene/double bond and E being opposite.  These notations explicitly refer to a pair wise relationship.  It is not normally said '... the group E to the ...'

cis and trans refer to substituents of context, with cis being substituents on the same side of the alkene/double bond, and trans being opposite.  Frequently these refer to the most important nomenclaturally, but not always.  Additionally cis and trans are used in discussions such as '...the group trans to the ...' describing a relation of one object relative to the other.  cis and trans are no longer considered IUPAC-correct nomenclature, but still have some usage for descriptive discussion.


Does chemistry change under high voltage? Do chemical reactions change when you charge the entire reaction vessel plus or minus $\pu{1 MV}$ or more?  Is there a name for such chemistry?

I was looking at "electrochemistry" expecting to see a host of information on charged reactions (either statically or with a constant charge applied) but it all just appears to be focused on batteries, not so much on synthesizing novel materials.

Long story short, no, they don't.

Charging a vessel to 1 MV is not a big deal, if you look at it from the inside. To put things into perspective, imagine a vessel of about 10 cm across, which is kinda OK for a flask. Imagine it spherical to make the calculations easier.

Now, the capacitance of a sphere is $C=4\pi\varepsilon_0R$, which evaluates to 5.56 pF (yes, "p" stands for "pico").
So charging your vessel to 1 MV implies putting $5.56\;\mu\rm C$ of electric charge on it.
That's $3.48\cdot10^{13}$ extra electrons (or some negative or positive ions, that doesn't matter) ...
... which may seem big until you compare it to the amount of water that would fit into the vessel: $1.75\cdot10^{25}$ molecules.

So it is like 1 electron per quite a few billions of molecules. That would not make much of an influence. The molecules will just sit there as usual, feeling nothing; most of them have never seen an electron in their entire lives, and have no idea that their vessel is being charged.

Things will change a great deal if you touch the vessel from the outside, but that's just ordinary electrochemistry which you seem to put out of scope of your question.


What are the oxidation states of galium and arsenic in GaAs semiconductor? According to the book I'm referring to ("S.M. Sze Semiconductor Devices"), gallium has an oxidation state of +1 and arsenic has that of -1. This should not be possible as each of these elements forms a covalent bond with four atoms of the other element (structure of GaAs semiconductor). For that to be possible, shouldn't both of them have one electron in $\ce{3s}$ subshell and three electrons in $\ce{3p}$ subshell?

In general, gallium can have +1 and +3 oxidation states, though +1 is a very rare condition. In its turn arsenic possible oxidation states are -3, +3 and +5. As it is stated in Wikipedia, gallium oxidation state in gallium arsenide is +3. Once again, arsenic has -3 oxidation state in arsenides or intermetallic compounds. Therefore, gallium arsenide oxidation states are +3 for Ga and -3 for As. This is logical because the electronegativity of As is higher then Ga. As for Simon Sze, he is an electrical engineer and physicist and not a chemist. From the mathematical point of view (-3)(+3) is the same thing as (-1)(+1).


Confused on the peak intensities in this mass spectrometry problem A problem from Introduction to Spectroscopy by Pavia et al. [1, p. 137]:

*4. An unknown hydrocarbon has a molecular ion peak at $m/z = 84,$ with a relative intensity of $31.3.$ The $\ce{M + 1}$ peak has a relative intensity of $2.06,$ and the $\ce{M + 2}$ peak has a relative intensity of $0.08.$ What is the molecular formula for this substance?

I normalized $31.3$ to $100\%,$ so the relative intensity $I$ was calculated as follows:
$$
\begin{array}{lr}
\hline
\text{Peak} & I/\% \\
\hline
\ce{M^+} & 100.00 \\
\ce{(M + 1)+} & 6.58 \\
\ce{(M + 2)+} & 0.25 \\
\hline
\end{array}
$$
The textbook is saying the molecular formula is $\ce{C6H12}.$ To me the answer doesn't make sense because if I calculate the $\ce{(M + 1)+}$ intensity using this formula from my textbook given for $\ce{C_xH_y}$:
$$[\ce{M + 1}] = 1.1x + 0.015y$$
Thus for $\ce{C6H12}$:
$$[\ce{M + 1}] = 1.1 × 6 + 0.015 × 12 = 6.78\%$$
However, $6.78\%$ is off from $6.58\%$ by about $0.2\%$. Am I missing something or is there some sort of acceptable tolerance for peak intensity differences?
Reference

Pavia, D. L.; Lampman, G. M.; Kriz, G. S.; Vyvyan, J. R. Introduction to Spectroscopy, 5th ed.; Cengage Learning: Stamford, CT, 2015. ISBN 978-1-285-46012-3.

Such problems in chemistry don't work out to infinite precision like pure math problems. You have to apply some chemical knowledge to the problem too.
The book formula is obviously an approximation to the exact solution. To find the exact values expected you'd need to use combinatorics.
Considering the chemistry:

7 carbons have a mass of 84, so a hydrocarbon molecule can't have that many carbons since there wouldn't be any hydrogen atoms.

5 carbons have a mass of 60, but that would mean at most 12 hydrogen atoms for an aliphatic molecule, so the mass could be at most 72.

Thus there must be 6 carbons in the molecule.


Estimating standard Gibbs free energy of a molecule In order to estimate the standard Gibbs free energy of a molecule at room temperature and normal pressure, we use Maxwell Boltzmann average over  rotational conformers.

Question

Should we consider cis/trans isomers besides rotational conformers? Is there any criterion for it?

You will need to calculate the partition functions for each type of motion and then calculate the thermodynamic quantities from these.

As to internal rotation, if the energy for this is far greater than $k_BT$, such as with a C=C bond, then treat it as a torsion and use this frequency. If the barrier to rotation is small compared to $k_BT$ then treat it as a free rotor but reduce the partition function by n if there are n barriers. In the intermediate case you will have to find the barrier (say by spectroscopy) and fit the data to a model of the potential (usually of a cosine form)and then numerically solving the Schrodinger eqn. to find energy levels.

If your temperature is so high that cis-trans reaction occurs then you should probably have to treat this as a mixture.


Computing energy of Guassian basis set Actually, I'd like to compute decay rate using :

$$ \Gamma = 2\pi \ | \langle ab|V|cd \rangle|^2 \delta\ (\epsilon_f-\epsilon_i) $$

where a, b, c, d are the MOs and V is coulomb interaction, and $\epsilon_f$ and $\epsilon_i$ are final and initial energy of the system.

I am interested in computing this value regarding to a special atom (for example atom A) which means MOs should be expandes in terms of basis set:

$$ \Gamma = 2\pi \ \sum_{\mu, \nu,\lambda, \sigma \ \in A  }C_{\mu,a}\ C_{\nu,b}C_{\lambda,c}C_{\sigma,d}|\langle \mu\nu|V|\lambda \sigma \rangle|^2 \delta \ (\epsilon_{\mu}-\epsilon_{\nu}-\epsilon_{\lambda}+\epsilon_{\sigma}) $$

where $\mu, \nu, \lambda, \sigma$ are states in Gaussian basis sets, and $\epsilon$ are corresponding energies.

It seems a straightforward calculation, but I am completely confused because I don't know how I can calculate the energy of basis sets! I think the energy of the basis set is not meaningful.

Somebody, suggests using a numerical atomic orbital basis set to compute this kind of decay rate. (for numerical atomic orbital, the energy of states is meaningful) but I also don't know how to compute this kind of basis set!

Is there any suggestion to resolve this problem?

First, these Coulombic interaction integrals can be simplified to $ \langle \mu\nu|\frac{1}{r_{12}}|\lambda \sigma \rangle $, which is the common two electron integral in the AO basis (your atom-centered gaussian basis functions).
You are correct that energies are only well defined for eigenfunctions of the Hamiltonian, and thus basis set functions (the choice of which is arbitrary) do not have a well-defined energy. Therefore, the most straightforward workaround for this is to calculate the two electron integrals in the AO basis, forming the rank 4 tensor of two electron integrals, $g(\mu,\nu,\rho,\sigma) = \langle \mu\nu|\frac{1}{r_{12}}|\lambda \sigma \rangle $. Then, you must (approximately) solve the Schrödinger equation with some level of electronic structure theory (e.g. CCSD, MP2, etc.) to solve for the molecular orbital coefficient matrix, C. The a$^{th}$ column of the matrix C is the weightings of the AO basis (atom-centered gaussian basis functions) to give the a$^{th}$ molecular orbital with well-defined energy $\epsilon_a$.
Finally, you should rotate $g(\mu,\nu,\rho,\sigma)$ with the coefficient matrix C to yield $g(a,b,c,d)$:
$$g(a,b,c,d)=C_{\mu a}C_{\nu b}g(\mu,\nu,\rho,\sigma)C_{\rho c}C_{\sigma d}$$
Note that, in this form, this change of basis scales as $N^8$, where N is the number of basis functions used, but can be implemented efficiently (by doing one rotation at a time) to scale as $N^5$. See the Crawdad programming tutorial on MP2 (site seems to be down right now) for more discussion on tensor rotations. Although it was done for MP2, it works identically for you.


Conductivity of a semiconductor at high temperatures I understand that increasing the temperature of a semiconductor excites electrons in the valence band into the conduction band. Hence, the resistivity of a semiconductor decreases with temperature.

Is there a high enough temperature where the conduction band would be saturated from electron with the valence band and the resistance of the semiconductor would start to increase like a regular metal?

Assuming the seminconductor couldn't melt, as temperature got aribtrarily high would the seminconductor's resistance get arbitrarily low?

A number of different things are going on in this scenario.

First, the carrier concentration does increase steeply, although for Si at its melt temperature it is still only about 10$^{20}$/$cm^{3}$, much lower than a metal.

In addition, the band gap is decreasing with temperature (not from carriers, but from changes to the lattice spacing). From a graph in Sze's Physics of Semiconductor Physics, Si goes from 1.12eV at room temperature to a little under 0.9eV at 900K. Extrapolation (probably a bad idea) means that somewhere above 2700K there will be little to no band gap. However, this assumes the physics extrapolates linearly, which it doesn't. (Well, liquid silicon is metallic. However, there is still a large increase in conductivity at the melt point).

Finally, the phonon population increases with temperature, which will increase scattering, decreasing the conductivity (just like a metal). This can be seen from a graph of carrier mobility (also in Sze) with temperature. The mobility drops roughly as T$^{-2.2}$ or so.

Getting the balance of all of these just right under extreme extrapolation, well, that is hard. It is actually easier to point to papers on the electrical characteristics of undercooled liquid silicon (since that has been measured), than to superheated solid silicon (which has not been observed). - for ev


Kinetic Energy Evaluation Integral Evaluation Program I'm reading Ostlund's Modern Quantum Chemistry. In Appendix A, the kinetic energy integral is evaluated using the Gaussian Basis functions to be

  $$
\left(A\left| -\frac{1}{2}\nabla^2 \right| B\right) =
\alpha\beta/(\alpha + \beta)[3 - 2\alpha\beta/(\alpha + \beta) |\mathbf{R}_A - \mathbf{R}_B|^2][\pi/(\alpha + \beta)]^{3/2} \\
\times \exp [-\alpha\beta/(\alpha + \beta)|\mathbf{R}_A - \mathbf{R}_B|^2]
\tag{A.11}\label{kin-en.int}
$$

So, in the integral evaluation the Gaussian functions themselves are not used, but in the computer program  he is evaluating the integral using

T11=T11+T(A1(I),A1(J),0.0D0)*D1(I)*D1(J)

The function T() is calculating the equation above \eqref{kin-en.int}.

I can't understand why he is multiplying by D1 and D2 which are the Gaussian functions themselves.
$$g_\mathrm{1s}(\alpha) = (2\alpha/\pi)^{3/4}\mathrm{e}^{-\alpha\mathbf{r}^2}$$

The equation for the Kinetic Energy matrix element that you quote is for two unnormalised 1s Gaussians. The d factors contain the normalisation factors and the contraction coefficients - look more carefully at the code, you have got it slightly wrong what the d's represent.

Talking of the code please, Please don't use that as a model for your own Fortran - that style is getting on for 50 years out of date!


How do I setup the electrodes for cyclic voltammetry on a capacitor made on dry copper sheets? I'm trying to measure capacitance of a cell I'm creating. It's composed of two copper-sheets with nano-material coated on one side, sandwiched with a membrane and wetted with $\ce{KOH}$. All of the three electrode setups I've found on the internet are for solution based voltamettry, but this cell is not in a solution. The setup that my labmates had yesterday had the reference electrode, $\ce{Ag|AgCl}$ setting in a solution, and not connected to the cell, which doesn't make any sense to me.

What reference electrode should be used to do this kind of measurement, and what should the schematic look like for our setup?

The reason reference electrodes are needed in 3 electrode solution cells is that you're trying to control the interfacial potential of the working electrode (basically from the surface to a point very near the electrode in solution). The wiki's kinda lousy, but the diagram explains essentially what the potentiostat is doing: It's controlling the counter electrode (the op-amp output), to set the potential from the working electrode (ground) to a point just into the solution and the reference electrode is used to read the potential and provide feedback to the amplifier to correct the output. This way, any extra solution resistance between the counter and reference electrodes doesn't reduce the interfacial potential (known as iR drop in electrochemical circles) because you're measuring the potential near the surface. In other words, we're interested in applying a potential across $Z_2$ precisely, but we don't care about what's happening with $Z_1$.

Now, since you have a capacitor, what you're trying to do is control the potential of one plate versus the other plate, so you want the point where you're getting feedback to be one plate, and the other plate is ground—one plate is the "reference electrode" and the other is the "working electrode". Your capacitor is $Z2$ and there's no solution, so there's no $Z1$. (In some cases, it still makes sense to use a separate cable for the reference input, but it's probably fine to just short them together at the inputs) Many potentiostats will work with the counter electrode and reference electrode inputs shorted together (which is how one typically connects a 2 electrode cell), but a resistor between them can help prevent it from oscillating by increasing the time constant of the system.

You're not really using a reference electrode in this case because you don't care about measuring the applied potential versus some known potential, just in controlling the potential across the capacitor. In a cell in which we're interested in studying redox reactions and such, it's important to know how the potentials relate to one another and to compensate for the various potentials that arise when electrodes are dipped in electrolyte solutions.

If you're really interested in the details of potentiostats, Bank has a good primer with lots of details.


Differences in Gibbs activation energies for rotation around a bond Question
Why is there a difference in energy in the rotational barrier between these two structures?

My reasoning is such:

Due to conjugation between the O electrons and the pi- double bonds, the bond has actually single bond characteristics and is therefore weaker than a conventional double bond alkene. But that doesnt explain the difference between the molecules? Can someone please explain?

In both molecules, there is π delocalization across the $\ce{C=C}$ bond which extends across the two carbonyl groups on the alkene. In order to maximize the overlap of the p orbitals involved in this conjugated system, it means that this portion of each molecule has to be flat. It is not a problem for the conjugated system to be flat in molecule B, but in molecule C you have three bulky methyl groups near the conjugated system. Steric clashes in molecule C between the t-butyl group and the carbonyl group on the same side of the $\ce{C=C}$ bond mean the carbonyl will rotate out of the plane of the conjugated system to avoid "bumping into" the t-butyl group, lowering the overlap between the p orbitals in the conjugated π system and thus destabilizing the molecule.

Because molecule C does not experience as much favorable pi electron delocalization as  molecule B, molecule C sits at a relatively higher energy than molecule B. The transition states of the two molecules, however, are not so different in energy. The Gibbs energy of activation is defined as the difference in Gibbs energy between the transition state of a reaction and the reactants (where our "reaction" is a rotation about the $\ce{C=C}$ bond). As molecule C sits at a higher energy relative to molecule B, but the transition states are relatively similar, it makes sense that there is a lower barrier to rotation about the $\ce{C=C}$ bond in molecule C.


Change in pH when boiling off acidic water Assume 1 L of deionized water at pH 2 at ambient pressure open to atmosphere. I boil off 500 mL. I believe the pH will not change because to the same extend as water boils off it will also boil off H+ ions. Correct?

$\rm H^+$ ions (as well as any other ions, for that matter) do not boil off. Acid does. Depending on the nature of the acid, it might boil off faster or slower than water, and hence your pH might shift up or down.

$\ce{HCl}$, in particular, forms a high-boiling azeotrope with water at 22%, so the solution of pH 2 (which is pretty dilute) would form a vapor enriched with water, and the liquid phase would get enriched with acid. So despite the seemingly volatile nature of $\ce{HCl}$, it will mostly stay in the solution and make it progressively more acidic as it boils.