Why is the groundstate electronic degeneracy of monoatomic O(g) 5? A problem in my Pchem textbook states that the electronic degeneracy of an oxygen atom is 5. My reasoning leads me to believe that the degeneracy is actually 6 (S = 1, L = 1, so (S+1)(L+1) = 6. This would indicate there are two unpaired electrons and a pair in the P orbitals...

The only way I can reason a 5 degeneracy is if the ground state has a pair in the d-orbitals, which I do not believe is the case.

S = 1, L = 1, so (S+1)(L+1) = 6

You are using here not just alternative physics, but also alternative math. For, even if the formula were right, you should get $2 \cdot 2 = 4$ and not $6$.

Anyway, the right formula for the degeneracy is $(2S + 1)(2L + 1)$ which being used for the ground $^1D$ state of oxygen atom (i.e. for $S = 0, L=2$) indeed gives $1 \cdot 5 = 5$.


Why are wavefunctions with same principal quantum number and different azimuthal quantum number orthogonal The value  turns out to be zero . This is understandable since they are completely independent of each other and also perpendicular in real space. But also turns out to be zero . How do I justify the orthogonality of the s and the p levels?

The mathematical answer to this specific question lies in the fact that s and p orbitals are simultaneous, non-degenerate eigenfunctions of the quantum mechanical $\hat{L}^2$ operator, which is hermitian:

$$\begin{align}
\hat{L}^2\left|\mathrm{s}\right> = [l(l+1)\hbar]^2 \left|\mathrm{s}\right> &= 0 \\
\hat{L}^2\left|\mathrm{p}\right> = [l(l+1)\hbar]^2 \left|\mathrm{p}\right> &= 2\hbar^2 \left|\mathrm{p}\right>
\end{align}$$

since $l = 0$ for s orbitals and $l = 1$ for p orbitals.

It is a simple proof that the eigenstates of hermitian operators that correspond to different eigenvalues are necessarily orthogonal. Suppose, for a hermitian operator $A$, we have $A\left|m\right> = a_m\left|m\right>$ and $A\left|n\right> = a_n\left|n\right>$ with $a_m \neq a_n$. Then:

$$\begin{align}
\left\langle m \middle| A \middle| n\right\rangle &= \left\langle n \middle| A \middle| m\right\rangle^* & &\text{(definition of hermiticity)} \\
a_n\left\langle m \middle| n\right\rangle &= (a_m\left\langle n \middle| m\right\rangle)^* & &\text{(extracting the eigenvalues)} \\
&= a_m\left\langle n \middle| m\right\rangle^* & &\text{(eigenvalues of hermitian operator are real)} \\
&= a_m\left\langle m \middle| n\right\rangle & &\text{(definition of inner product)} \\
(a_n - a_m)\left\langle m \middle| n\right\rangle &= 0 \\
\left\langle m \middle| n\right\rangle &= 0 & &(a_m \neq a_n)
\end{align}$$

You can likewise show that any pair of atomic orbitals in the hydrogen atom are orthogonal, using whichever of the three operators $\hat{H}$, $\hat{L}^2$, $\hat{L}_z$ that you need. All atomic orbitals are simultaneous eigenfunctions of these three operators, and have different eigenvalues which correspond to the quantum numbers $n$, $l$, and $m_l$:

$$\begin{align}
\hat{H}\left|n,l,m_l\right> &= E_n\left|n,l,m_l\right> = -\frac{m_e e^4}{2(4\pi\varepsilon_0)^2 \hbar^2 n^2}\left|n,l,m_l\right> \\
\hat{L}^2\left|n,l,m_l\right> &= L^2\left|n,l,m_l\right> = [l(l+1)\hbar]^2\left|n,l,m_l\right> \\
\hat{L}_z\left|n,l,m_l\right> &= L_z\left|n,l,m_l\right> = m_l\hbar\left|n,l,m_l\right>
\end{align}$$

Since no two orbitals have the same three quantum numbers, no two orbitals can have the same eigenvalues for all three operators. Consequently, there must be at least one operator which can be used to show that those two orbitals are orthogonal.


Calculating solubility of iron(III) hydroxide in water The following example in Skoog's Analytical Chemistry uses mass-balance and charge-balance equations to calculate the solubility of iron(III) hydroxide in aqueous solution.

  $$
\begin{align}
\ce{Fe(OH)3 (s) &<=> Fe^3+ (aq) + 3OH- (aq)} \\
\ce{2H2O (l) &<=> H3O+ (aq) + OH- (aq)}
\end{align}
$$

  $$
\begin{align}
K_\mathrm{sp} &= [\ce{Fe^3+}][\ce{OH-}]^3 = 2\times 10^{-39}\\
K_\mathrm{w} &= [\ce{H3O+}][\ce{OH-}] = 1\times 10^{-14}
\end{align}
$$

  $$[\ce{OH-}] = 3[\ce{Fe^3+}] + [\ce{H3O+}]$$

After writing down the equations for the relevant equilibria and the corresponding equilibrium and mass balance equations (as seen above), the problem made an approximation that puzzled me:

  Step 7a. Make approximations As in Example 11-5, assume that $[\ce{H3O+}]$ is very small so that $[\ce{H3O+}] \ll 3[\ce{Fe^3+}]$ and

  $$3[\ce{Fe^3+}] \approx [\ce{OH-}]$$

How can we make an approximation that $[\ce{H3O+}] \ll 3[\ce{Fe^3+}]$ when $K_\mathrm{sp} \ll K_\mathrm{w}$? Shouldn't it be the other way around, actually?

Please explain the reasoning for this step in the process.

In my version of the book the original assumption (step 7a) is shown to lead to a contradiction when later checked (logically, this form of proof is called reductio ad absurdum or proof by contradiction). The point is that you should 1) learn not to be afraid to apply assumptions, but that you should then 2) verify that the original assumption was indeed acceptable. This is a strategy repeatedly encouraged in the book.

To show how this leads to a contradiction,

make the assumption, so that $ \ce{3[Fe^{3+}]} = \ce{[OH-]}$
solve for $S = \ce{[Fe^{3+}]}$  based on the $K_{sp}$ and the assumption
Compute $\ce{[OH-] = \ce{3[Fe^{3+}]}  }$, following the original assumption
Compute $\ce{[H+]}$ from $K_w$ and $\ce{[OH-]}$
Check the accuracy of the assumption

It is by the way not difficult to solve the problem exactly, by computing the $\ce{[OH-]}$ as follows:

$$ \ce{[OH-]}=\mathrm{\sqrt\frac{K_w+\sqrt{K_w^2 +12K_{sp}}}{2}}$$

and then applying the expression for the solubility constant to compute the solubility.
However the point of the problem is to illustrate how to follow this line of reasoning when solving problems (the use and verification of assumptions).


Atmopsheric Gas IR-Spectra I am looking for normal resolution IR-spectra of: $\ce{O2}$ and $\ce{N2}$ at high concentrations and; common minor gas in atmosphere: $\ce{H2O}$, $\ce{CO}$, and $\ce{CO2}$ at environmental concentrations. I have found some reference on internet (mostly from Green House Effect articles), but I cannot found it on a free, accurate and reliable source even on the free part of NIST databases.

Where can I find good references for such spectra?
Thank you.

Completely symmetrical diatomic molecules like $\ce{O2}$ and $\ce{N2}$ are not IR-active because they have a zero dipole moment, even when oscillating within their one vibrational mode. Hence, there is no change in dipole moment possible, so these molecules will not be detectable by IR. The other molecules you mention, however, are IR-active, and I was able to find spectra of all three in the NIST database.

There are other spectroscopic methods that can detect IR-inactive molecules, however. Raman spectroscopy may be applicable, and I'm dimly aware of sophisticated methods based on absorptions in electronically excited states, but I don't know nearly enough about the subject to offer any concrete information on those techniques.


How did hypophosphoric acid get its name? Why does hypophosphoric acid has a strange name? My knowledge of inorganic nomenclature tells me that the prefix "per-" should be used with "-ic" acids and the prefix "hypo-" goes with "-ous" acids, which isn't the case here. Why is this so?

Just as part of an additional observation, two other things which I find strange about $\ce{H4P2O6}$ are:

I think this is the only oxy-acid of phosphorus having a $\ce{P-P}$ bond.
It has phosphorus in $+4$ oxidation state.

You are most of the way to the reason yourself:

$$\begin{array}{|c|c|c|}\hline
\text{name} & \text{formula} & \text{oxidation number of P}\\ \hline
\text{phosphoric} & \ce{H3PO4} & +5\\
\text{phosphorous} & \ce{H3PO3} & +3\\
\text{hypophosphorous} & \ce{H3PO2} & +1\\ \hline
\end{array}$$

The "hypo-" prefix means "beneath" or "less than". Hypophosphoric acid, $\ce{H4P2O6}$, has oxidation number of phosphorous as +4, which is in between phosphoric acid and phosphorous acid. Thus, "hypophosphoric" refers to "beneath" phosphoric (but above phosphorous) acid.


What do chemical constituent measurement units 'g%g' and 'µg g⁻¹' mean? I'm trying to understand the following data being presented in this paper measuring the mineral content of the Cannabis sativa plant (leaves and inflorescences):

I need to interpret these data readings in terms of simple percentages.

What would the first two examples - sodium and potassium - translate to in simple percentage / ppm etc.?

$\pu{g\%g}$ literally means grams per cent grams or grams of mineral per $\pu{100g}$ of sample. So, in population 1 there is $\pu{2.33g}$ of $\ce{K}$ per $\pu{100g}$ of leaves. Not bad when you consider cooked spinach only has about$\pu{0.05\%}$ $\ce{K}$! Eat your greens, kids. About $\pu{2\%}$ $\ce{K}$ in plants is average across a broad spectrum of plants.

$\pu{ug g-1}$ is micrograms of mineral per gram of sample. Population 1 has 23 micrograms of $\ce{Na}$ per gram of leaves.

You should be able to convert these to percents or ppm, as needed, and check in the text as to whether this refers to dry weight of plant or not. I'm guessing with this multivariate array of mineral analyses, the origin of particular cannabis crops could be determined.


Non-Commercial Thermal Insulation I am trying to create a small 10 to 20cm cube that can conserve the heat in 100 to 200ml of 60-90c water in a small Pyrex glass for about an hour. The problem is that I can not use commercial insulators like foam, bubble wrap, and the like. I have searched Google for ideas, but all the results are commercial. I also can not heat or cool the device beforehand, and I have to have a 2cm hole in the top.

I have tried cotton balls and a packed cotton balls with felt mixture, and cotton balls worked the best.

Often the key with stopping the conduction of heat is to provide as many phase transitions as possible. Most thermal insulators achieve this by having voids of Air which is achieved in a variety of ways, bubble wrap, fibre glass etc.

For a "non comercial" I assume they mean stuff you have lying around the lab?

To attempt this I'd layer blue roll / cotton wool / glass wool with tin foil. The more layers the better.


Calculate the number of moles produced per hour per liter of air? A reaction of importance in the formation of smog is that between ozone and nitrogen monoxide described by

$$
\ce{O3(g) + NO(g)->O2(g) + NO2(g)}
$$

the rate law for this reaction is:

$$
\text{rate of reaction} = k\left[\ce{O3}\right]\left[\ce{NO}\right]
$$

Given that $k = 2.98 \times 10^6\ \mathrm{M^{–1}\cdot s^{–1}}$ at a certain temperature, calculate the initial reaction rate when $\left[\ce{O3}\right]$ and $\left[\ce{NO}\right]$ remain essentially constant at the values $\left[\ce{O3}\right]_0 = 5.34 \times 10^{–6}\ \mathrm{M}$ and $\left[\ce{NO}\right]_0 = 7.99 \times 10^{–5}\ \mathrm{M}$, owing to continuous production from separate sources.

So, I have found that by using the rate of reaction equation

$$
\text{the rate of reaction} = $0.00127\ \mathrm{M/s}
$$

Now I’m wondering how to calculate the amount of substance of $\ce{NO2(g)}$ produced per hour per liter of air.

Right now, my idea is that you can take

$$
\left(1.27\times 10^{-3}\ \mathrm{M}\over\mathrm{s}\right)\left(60\ \mathrm{s}\over{1\ \mathrm{min}}\right)\left(60\ \mathrm{min}\over{1\ \mathrm{h}}\right) = \ce{NO2} \,\text{per hour per liter of air}
$$

Is this the correct way to go about it? I was thinking that the seconds and minutes would cancel leaving just $\mathrm{M/h}$ and then if you multiply by 1 liter it would be $\mathrm{M/(h\times L)}$.

That is the correct way to go about it. In your calculations the rate of the reaction should include L because M is defined as moles/L. (1.27×10^-3 moles/Ls)(3600s/Hour) = 4.572 moles/Lh NO₂ produced.


What is the state of stationery phase in paper chromatography? Paper chromatography is a kind of partiotion chromatography where stationery phase is water, and I think it should be , because in partiotion chromatography solid is not stationery phase. but as per Wikipedia and this site:http://www.chemguide.co.uk/analysis/chromatography/paper.html

Stationery phase in paper chromatography is paper(a solid) . So which one is correct? Is paper chromatography really a kind of partition chromatography or adsorption chromatography?

The picture is from my textbook: NCERT class 11 ,pg-353

It's a bit of both, actually, and also ion-exchange chromatography.
The cellulose and other components of paper do exhibit adsorption and ion exchange properties with some analytes, but as paper chromatography is usually performed, partitioning between the flowing solvent and trapped water is the major mode of separation.
Paper chromatography typically uses water-saturated solvents and enclosed chambers where water vapour can keep water trapped in the paper (the paper is equilibrated with this vapour before development).
Since cellulose is very hydrophilic, immiscible or poorly miscible solvents are usually used, which keeps the water stuck to the cellulose as it has much higher affinity for water than the solvent.

This is an early paper showing the apparatus used for paper chromatography of amino acids and this book details the theory involved starting on page 4.


Getting 3,4-dimethylhex-3-ene from but-2-ene How can one get 3,4-dimethylhex-3-ene from but-2-ene ?

$\hskip2in$

Using $\ce{Br_2}$ then two equivalents of $\ce{CH_3CH_2MgBr}$, I can get the corresponding alkane (3,4-dimethylhexane) but I don't see how to do it while preserving the double bond...

Your approach, which uses $\ce{Br2}$ to convert 2-butene into 2,3-dibromobutane and react that with ethylmagnesium bromide $\ce{CH3CH2MgBr}$ has two flaws:

This approach removes the alkene and there is no obvious way to get it back (there is a way)
More seriously, Grignard reagents are pretty terrible for $\text{S}_N 2$ reactions. They are very strong bases, so elimination can compete, but more importantly, they can undergo metathesis:

$$\ce{R-Br + R'-MgBr <=> R-MgBr + R'Br}$$

Assuming your approach worked to generate 3,4-dimethylhexane, you could carefully radically brominate it to produce 3,-bromo-3,4-dimethylhexane, and I am sure you can get back to the alkene from there.

A better approach uses two molecules of 2-butene. Do you see how you can break your target molecule up into two 4-carbon pieces? Convert one into an electrophile and the other into a nucleophile. It might take two reactions in each case. If you choose your electrophile correctly (think about what makes a great combo with a Grignard reagents), you will form the $\ce{C-C}$ bond between the two pieces and you will still have a functional group left that you can use as a handle to direct the elimination reaction to regenerate the alkene.


Breaking of carbon chains due to annihilation of electrons using positron beams I and my friends were taking part in a competition(physics and chem expertise needed) and we were having a few doubts about something. I was hoping if anyone of you can help in the clarification of the doubt.
We use a positron beam(from the DESY II particle accelerator) directed towards a polymer(eg polythene), which will annihilate a few valence electrons of the atoms in the chain. We’ve not found any source to support our theory that bonds will break although they should as electrons are being annihilated which support bonding. We searched and found nothing because no one else has had a similar experiment before.
We also wanted to know if we annihilate an inner core electron then will it lead the de-excitation of the valence electrons further leading to destabilization of bonds.
We can’t confirm how the electron donors will react with carbocations/radicals formed, also since there are many possibilities and we are unable to consider all of them.
The momentum of the positrons can be set to any value between 0.5 GeV/c and 6 GeV/c.
We needed some confirmation to the feasibility of the experiment.
All we need is to break the carbon chain, any other ways to do that would be appreciated.

This a problem of radiation chemistry. It has been thoroughly studied in the years 1950 - 1960 - 1970 with gamma rays. When intense source of gamma rays strikes water, electrons are extracted from the water molecule. And some of these electrons are coming from the valence bond O-H. The bond is then weakened, and may even be broken. Usually the electron soon comes back to its original atom. This produces individual H atoms and OH radicals. Usually H and OH soon recombine and produce again H2O a few nanoseconds later. But if the irradiation is intense, they may recombine with similar neighbors. And this produce dissolved H2 and H2O2 molecules. So the water sample contains more and more dissolved H2 and H2O2. Titration of H2O2 in solution is possible, and may be used to determine the doses and the intensity of the radiation absorbed by the sample.

The same experiments have been done with hydrocarbons. After irradiation, hexane, for example, is partly converted into dodecane, with a lot of different isomers. The distribution of these isomers allows to determine where the initial breaking in hexane happened, in other words which C-H or C-C bond has been initially broken, and in which proportion. Why don't you read the publications by Ausloos or Gäumann or others in this old literature. Today nearly nobody is interested any more in this chemistry, which is considered to be "finished".


Why isn't water acidic? The definition of an acid is a compound with a hydrogen cation and a non-metal anion. Water is a hydroxide bonded to a hydrogen atom. This hydroxide has a net negative charge, too, since the negative oxygen's charge is stronger than the positive hydrogen's. Water has a pH of 7, though, meaning that it's neutral. Why isn't water acidic

If you look at what happens in the equilibrium dissociation of water, both a hydronium ion (Arrhenius acid) and a hydroxide ion (Arrhenius base) are formed. Since these two ions are  formed in equal quantities ($\ce{10^{-7} mol/L}$ each in pure water), the water will be neither acidic nor basic.

$$\ce{H2O_{(l)} + H2O_{(l)} <=> H3O+_{(aq)} + OH^{-}_{(aq)}}$$  (from wikipedia)

Wikipedia can explain more about acids and their definitions if you are interested: http://en.wikipedia.org/wiki/Acid.


Friedel craft alkylation halo-carbon bond cleavage In this reaction, why is the $\ce{C-F}$ bond broken instead of the $\ce{C-Cl}$
bond, even though $\ce{C-F}$ bond is stronger than the $\ce{C-Cl}$ bond?

In Friedel-Crafts alkylation, the halogen-carbon bond breaking depends on the strength of bond between halogen atom and the Lewis acid used. The more stronger is the bond, the easier it is to generate a carbocation for the reaction.( In all cases, the carbocation exoplicitly needn't have to generate, a strong $\delta$+ charge on the electrophilic carbon is sufficient enough to have a $\ce{S_E2}$ or $\ce{S_E1}$ type reaction. Here, the Lewis acid used is  $\ce{BCl_3}$ where there is a vacant  $\ce{2p_z}$ orbital on the  $\ce{B}$ atom,(Assuming, the molecule is oriented in a triangle in the  $\ce{xy}$-plane.)and the Lewis base is the given reagent 1-chloro-3-fluoropropane., which has two Basic centre. But donation through  $\ce{F}$ will correspond to a stronger  $\ce{2p$\pi$ -2p$\pi$}$ overlap with Lewis acid,rather than weaker  $\ce{3p$\pi$ -2p$\pi$}$ overlap in the case of donation through  $\ce{Cl }$ atom. Thus bonding with  $\ce{F}$ will correspond to a very huge amount of  $\ce{-I}$ effect of the adjacent carbon atom which will weaken the bond between them by pulling the electron pair towards  $\ce{F}$ and easily favour the attack by the $\pi$-electron cloud of Benzene ring.Also, the repulsion between $\pi$-electron cloud and the  large sized $\ce{Cl}$ atom is very less during the formation of  the above compound. But, if it would have bonded through  $\ce{Cl}$, there won't be a great electrophilic nature of the adjacent carbon due to less effective bonding in the case of  $\ce{Cl}$(corresponding to even lesser  $\ce{-I }$ effect),so attacking will correspond to a highly energetic transition state, and also, there will be a slight electrostatic repulsion with the  $\ce{F}$ atom of the  $\ce{$\pi$ }$-electron cloud.
Due to such above mentioned phenomena and energetically favoured conditions, bonding through cleavage of  $\ce{C-F}$ bond is more preffered over cleavage of  $\ce{C-Cl}$ bond.


Is there 'no actual equilibrium'? If you look at the concentration verses time graph of typical reactions, you can see that the concentration of reactants and products, when time goes towards infinity, becomes the equilibrium concentration. But! It never seems to reach that concentration (it converges towards that , but never reaches it). Which means that there is no 'real equilibrium??

I think this is true in terms of theory, where the number of molecules is considered to be infinite, while in reality, the number of molecule is finite. Then, in reality, it reaches equilibrium??

Yes there is actual equilibrium. The Achilles and the turtle kind of argument fails for three reasons, one of which you mentioned yourself:

Like you said, the number of molecules is finite, so the matter is not infinitely divisible; there must be a limit.
More importantly, nobody really cares about one molecule. Anyway, it's not like the system is going to hit just the right number with one-molecule precision. No, there are going to be mighty fluctuations involving millions of molecules, and they never stop (that is, unless you freeze the system down to absolute zero, in which case reaching the equilibrium may indeed become problematic). Like anything with molecules, equilibrium is a statistical thing, and once we're within certain error margin from the theoretical figure, we might just as well consider the goal achieved.
We may circumvent the argument altogether by playing with the conditions. Say, we want to get a saturated solution of some salt in water, and it just seems to take forever. But wait, what if we just heat the thing up a bit (which would make the solubility go up), dissolve the exact amount we need, and then cool it down again? Here is your saturated solution in a finite time. You may even make it supersaturated if you'd like.


How does monoethanolamine (MEA) increase pH? I know pH of a solution increases with increase concentration of $\ce{OH-}$.

I'm wondering: when adding MEA, is the $\ce{OH-}$ group from the MEA breaking off, or is it the whole MEA molecule taking a hydrogen from $\ce{H2O}$ leaving the $\ce{OH-}$?

It is the latter case where "the whole MEA molecule is taking a hydrogen from $\ce{H2O}$ leaving the $\ce{OH-}$". Alcohols very seldom acty as arrhenius bases adding $\ce{OH-}$ to solution. One exception is triphenylmethanol but I cannot think of a second that isn't a variation of the given example. Instead the unbonded electron pair of the amine will abstract a hydrogen from water to form an ammonium ion leaving a hydroxide counter ion and thus raising the pH.


How were old style stereographic structures produced? Old papers on chemical structures often contained "3D" renderings of the molecule by using 2 2D images and requiring the reader to go cross-eyed to get the 3D image. A good example I recently found was figure 2 from Carter and Kraut, 1974, which I copied below:

However, this paper is from 1974, which feels like ancient history for 3D graphics. Since I have unrelated interests in molecular biology and 3D graphics, I was wondering how they produced these images so long ago, and if the code to produce them is floating around somewhere. I know there are programs to do it for me because I have seen modern images of proteins that do this, I want the code because I might try to build it into my crappy 3D rendering project that I'm wasting too much time on.

Created in FORTRAN by Carroll K. Johnson, of the Oak Ridge National Laboratory (ORNL), and first released in 1965, ORTEP (Oak Ridge Thermal-Ellipsoid Plot Program) rapidly became a favorite of crystallographers and protein crystallographers to produce illustrations of structures for conference presentations and publications. A key strength of ORTEP was its capacity to generate stereoscopic images automatically.

For source of above and more information see: History of Visualization of Biological Macromolecules

For the ORTEP program itself, check ONRL Report #3794: https://web.archive.org/web/20170210061629/http://web.ornl.gov/info/reports/1965/3445600599779.pdf (big file warning)


Silver electroplating In most of silver electroplating (or electrorefining) methods, it is mentioned that silver anode is oxidized in aqueous solution of silver nitrate, and that silver is deposited at the cathode (regardless of cathode material, let's assume it's also silver for the sake of simplicity). What I don't get is why would silver anode be oxidized before $\ce{OH-}$? The reduction potentials are ~0.8 V for silver and ~0.4 V for $\ce{OH-}$, at standard conditions.

Hydroxide is bad for this process. You will have additional problems if hydroxide anion is present.

Silver cation reacts with hydroxide anion to form silver hydroxide, which spontaneously decomposes into silver oxide:

$$\begin{aligned}\ce{Ag+ + OH-}&\ce{ -> AgOH}\\
\ce{2AgOH}&\ce{ -> H2O +Ag2O}\end{aligned}$$

In addition to shutting down the oxidation of silver, hydroxide precipitates any silver cation that does dissolve.

For this reason, the electroplating process is probably run at a slightly acidic pH, and the competing oxidation reaction is not of hydroxide, but of water (reduction potential 1.229 V):

$$\ce{2H2O -> O2 + 4H+ + 4e-}\ E^\circ = +1.229\mathrm{\ V}$$

So long as you don't apply too much of an overpotential to the electroplating operating, silver will oxidize before water.


Are elemental compounds composed of different isotopes polarised? Are elemental compounds composed of different isotopes polarised? I'd expect any such effect would only really be noticeable for very light elements such as hydrogen.

I'm not sure which way the charge would be distributed and whether a deuterium protium pair would be more like deuterium protide or protium deuteride.

When taken isotopes into account, properties of compounds change. The polarisation of the compound included. Your hunch is correct, this probably only matters for very light compounds, or more precise compounds where the mass difference between the isotopes is large. The most extreme case is probably $\ce{HT}$, which has roughly double the mass of $\ce{H2}$.

I found an article by M. Trefler and H. P. Gush on the electric dipole moment of $\ce{HD}$ which explains the presence of the dipole moment as follows:[1]

  The center of positive charge oscillates along the axis of the molecule during a molecular vibration because the excursions of the proton are greater than those of the deuteron; since the electrons do not follow exactly the motion of the positive charge, a weak, vibrating dipole moment results.

There are more articles of the dipole moment of $\ce{HD}$, I simply did not have the time to go through all of them. I also think the above already makes the point you were interested in.[2]

A very thorough theory is presented by W. R. Thorson et. al.,[3] which essentially explains the dipole moment of HD as follows:

  [...] ; it arises directly from asymmetry of the electronic Hamiltonian, provided that we cast. that Hamiltonian in an appropriate set of canonical coordinates. The model of this Hamiltonian as having a pair of nuclei with slightly different effective charges is particularly pleasing, since we can link that directly to the orbital-size and binding-energy effects which we know must be associated with the difference in electronic reduced masses for D and for H.

M. Trefler and H. P. Gush found the dipole moment to be very small, i.e. $\mathbf{p}=(5.85\pm0.17)\times 10^{-4}~\mathrm{D}$. Hence to speak of deuterium protide or protium deuteride is a stretch.

Since the effect is quite small in $\ce{HD}$ already, one can probably assume it will not have a significant effect on the polarisation in other than symmetric diatomic molecules.

However, with isotopes a few notable applications can be achieved. One of the prime examples is “chiral acetate”, (R)-deuterotritioacetic acid $\ce{HDTC-COOH}$, which helped elucidate biosynthetic transformations. For a quite comprehensive, educational article see [4].

References

M. Trefler and H. P. Gush; Phys. Rev. Lett. 1968, 20, 703.
W. R. Thorson, J. H. Choi, and S. K. Knudson; Phys. Rev. A 1985, 31, 22.
(a) S. M. Blinder; J. Chem. Phys. 1960, 32, 105. (b) S. M. Blinder: The dipole moment of HD. http://hdl.handle.net/1811/7809 (c) J. Bradley Nelson and G. C. Tabisz;
Phys. Rev. Lett. 1982, 48, 1393. (d) G. Herzberg; Nature 1950, 166, 563. (e) etc. see google scholar
Addison Ault; J. Chem. Educ. 2003, 80 (3), 333.


Why can you sanitize paper money notes with a Clothes Iron? I have some Canadian and British bank notes. The first issue is that I don't know if they're paper or plastic.
Second, I don't understand this answer. Why's it safe  to iron paper bills? Won't the heat or pressure ruin the paper bill?

You can iron paper bills, like the American ones, on rather high heat. Just the same iron you use for your clothing, at the hottest setting, but no steam. It will kill all viruses.
Plastic bills can be hand washed, with the same soap you would use on your hand, you can dry them the same way you dry your hands, with a clean cloth or even the hot air dryer you use for your hands. Or you can let them air dry.
These bills can withstand some heat but I have heard not enough to iron them clean.

In short: This is possible because «paper money» actually is not referring to the paper like paper from the office supply for (your) desktop printer or xerox, but to cotton paper.
Because of the cotton, if folded and forgotten in a pocket of your jeans close to $\pu{40 ^\circ{}C}$, they may physically survive a pass in the washing machine with a mild non-bleaching detergent (including the spinning).  They obviously suffer from the dye if e.g., the jeans blead, or mechanically if you try to unfold them after if passing a dryer.  Of course different to the ones made of polymers as (but not limited to) in Australia.


Safest way to conduct a reaction with bromine under reflux conditions I am conducting an electrophilic aromatic bromination reaction according to literature. The substrate and intended substitution pattern do not matter, but the reported procedure requires the dropwise addition of bromine at room temperature followed by refluxing overnight. (The solvent’s boiling point is very close to bromine’s so it may be assumed that both compounds reflux.) The reaction is not conducted under inert atmosphere and as with every organic bromination I expect HBr to be a (gaseous) by-product.
Obviously, the reaction is being conducted in a fume hood whose front opening is closed as far as possible. Nonetheless, I want to limit the amount of bromine vapour that gets released into the surrounding atmosphere. I came up with the following setup:

Figure 1: A two-necked flask with a dripping funnel and reflux condenser. The glassware on top of the reflux condenser should be a drying tube (no accurate clip-art was found) containing damp $\ce{Na2S2O3}$ between cotton wool.
The dripping funnel is obviously used to add bromine dropwise to the reaction mixture and is sealed on top. The reflux condenser is open to the atmosphere through a drying tube; I filled the drying tube with sodium thiosulphate to quench any bromine vapour that makes it up there.
I did not attach the coolant flow to the reflux condenser until after the dropwise addition. This may have been a mistake, as bromine vapours had already diffused all the way up and penetrated the cotton wool and thiosulphate. It seems that these were far less effective in holding back any vapour than I had anticipated. Luckily, our coolant system is closed-circuit and the temperature adjustable. Thus, I set the cooling temperature to $\pu{0 ^\circ C}$ which seemed to reduce bromine’s vapour pressure sufficiently to clear the air in the drying tube. However, I believe there must be a better solution.
How can I run this reaction in reflux conditions while preventing bromine vapour escaping as much as possible? I had thought of sealing the entire reaction apparatus, but the reaction gives a gaseous side product (HBr) which could cause an undesired increase in pressure. I had also considered using a balloon to close off the reaction atmosphere, but I fear that both bromine and HBr will react with the rubber and destroy the balloon rapidly. What other methods exist that I may have missed?
Side note: the literature I am following was published in 1987. As expected of this era, they do not note specific precautions against bromine release.

Check if you may skip using elemental bromine as starting material altogether.  An alternative approach may be running the bromination in a heterogenic, biphasic setup with the couple $\ce{NaBrO3/NaHSO3}$ in water (e.g., 1998JOrgChem6023, 2000OrgProcRes Dev.30), or $\ce{NaBr}$ / sodium perborate (e.g., 1998SynthCommun925), or pyridinium tribomide (e.g.,2005JChemEduc306), or bromodimethylsulfonium bromide prepared by mixing aqueous $\ce{HBr}$ with DMSO (1997JOC4321, often at rt).  They share a slower release of bromine, and because of this, lower amount and concentration of reactive bromine and attenuate the hassle with corrosive gases.
If staying with liquid bromine:

Trapping unconsumed bromine with thiosulfate per se is a good idea, but the reaction of bromine gas with solid $\ce{Na2S2O3}$ in the drying tube is both incomplete, bears the potential that the drying tube gets congested, and tends to be messy to clean.  You could end up with a closed vessel, which isn't safe; and even dangerous as you generate HBr in stoichiometric amounts.  Thus passing the exhaust gas across scrubbers filled with aqueous thiosulfate solution has a much higher chance of win.  Don't forget the safety flask (different reaction, but same principle).

The the normal balloons from the toy shop does break under bromine, as well as HBr gas too quickly.

Beside @Mathew Mahindaratne's comment to dissolve bromine in chloroform, using acetic acid as solvent provides you a wider margin of temperature.  (If your substrate can withstand the acid.)


When is it better to use coupled-cluster (CC) over configuration interaction (CI) calculations? My experience with different theoretical methods in computational chemistry primarily comes from use and reading papers. My experience is that people almost always use CCSD instead of CISD if they wish to do calculations with single and double excitations.

My understanding is that truncated CI expansions suffer from not being size-extensive and also do not converge very quickly. I do not think coupled-cluster theories suffer in the same way.

Why, however, do people seem to prefer CCSD over CISD? What are the big advantages it has? Is any of it just a habit because CCSD(T) is the preferred method when possible, and so it's easier to make comparisons if these calculations come around in the future and can't be done in some current study?

Also, are there any systems for which it would definitely make more sense to choose a truncated configuration interaction calculation over a coupled-cluster method? Probably systems with some kind of low-lying excited state I would guess...

I'm not looking for a huge amount of math here, although that's totally fine. I would just be fine with the facts and any useful references.

CCSD is preferred in nearly all cases because size-extensivity/size-consistency is a critical property to have for practical chemistry, and on top of that, more of the correlation energy (the part of the energy missing from Hartree-Fock) is captured by CCSD. CISD only accounts correlation energy for up to double excitations and not at all for higher excitations. CCSD accounts for correlation up to double excitations and then approximates all higher excitations as combinations of the single and double excitations it has already calculated. This gives much more accurate numbers for pretty much all systems and provides for size-extensivity/size-consistency. The only time you would really be better off doing CISD than CCSD is if you wanted to do full CI on a 2-electron system, in which case CCSD is not as efficient.

So in short, CCSD is preferred because it almost always gives better, more usable numbers than CISD for similar or better costs (they both scale as ~$n^6$).


Supersaturated gas solution Can you make a supersaturated gas solution?
As we know solubility decreases when we increase the temperature, So If I do that, will the gas molecules escape the solution or will they stay in the solution and make a supersaturated solution? Or is it just a plain fact that you can't make supersaturated solutions except with solids solutes?

I am kind of leaning towards escaping, because it is hard to keep gas molecules intact with water molecules at high temperatures.

Every can of soda, coke etc. is supersaturated with carbon dioxide after opening, i.e. relieving the pressure in the can.

The surface tension of a gas bubble adds to the outside pressure, and more pressure always increases solubility of gases, as you know from Le Chatelier.

Same with gas supersaturated with a condensable gas or solvent with too much solute, you have to overcome the surface tension of small particles or droplets before precipitation can take place.


Isothermal vs. adiabatic compression of gas in terms of required energy I am working on an instruction manual of sorts to be used with an introductory course in thermodynamics. As an example of problem solving, I attempt to answer the following question:

  If you have little energy available, would you rather use an
  isothermal or an adiabatic process to compress a gas?

My analysis is as follows:

The work required to compress a gas from volume $V_0$ to volume $V$ is $$W=-\int_{V_0}^{V}PdV$$
For the isothermal compression, the ideal gas law, $P=\frac{nRT}{V}$, is used and inserted into the equation above:
$$W=-\int_{V_0}^{V}\frac{nRT}{V}dV=-nRT\int_{V_0}^{V}\frac{dV}{V}=nRT\ln\left(\frac{V_0}{V}\right)$$
For the adiabatic compression, $PV^\gamma=P_0V_0^\gamma \iff P=P_0\left(\frac{V_0}{V}\right)^\gamma$ is valid. Thus, the required work is $$W=-\int_{V_0}^{V}P_0\left(\frac{V_0}{V}\right)^\gamma dV=-\frac{P_0V_0^\gamma}{-\gamma+1}\left(V^{-\gamma+1}-V_0^{-\gamma+1}\right)=\frac{PV-P_0V_0}{\gamma-1}$$
       $$W=-\frac{P_0V_0-PV}{\gamma-1}$$

This result indicates that the magnitude of the required work depends on the properties of the gas in question. Thus, it may vary which of the two compression processes requires the lowest amount of energy.

Now, finally, to my question:

Intuitively I would assume that the isothermal work is usually lower than the adiabatic work, as compression lowers the volume of the system and therefore usually increases the temperature if heat exchange with the surroundings is not allowed. Isothermal compression requires heat transfer to the surroundings to maintain constant temperature, lowering the pressure of the system and thus lowering the resistance to compression compared to the adiabatic compression (where heat exchange is not allowed).

Is the isothermal work actually smaller than the adiabatic one in most cases or is my argument flawed?

To solve this, try to use what I call the "graphical apparatus". For an isothermal process:
$$
\begin{align}
PV&=\text{constant}\\
P\mathrm{d}V&=-V\mathrm{d}P\\
\frac{\mathrm{d}P}{\mathrm{d}V}&=-\frac{P}{V}\\
\end{align}$$
for adiabatic process:
$$
\begin{align}
PV^\gamma&=\text{constant}\\
\frac{\mathrm{d}P}{\mathrm{d}V}&=-\gamma\frac PV
\end{align}$$
Therefore, starting at the same point on a P-V graph, the curves for an adiabatic and isothermal processes will diverge and the adiabatic curve will have a steeper slope. For the same reduction in volume (the graph in the picture is for expansion, not for contraction. In case of contraction, the curves will be reversed, i.e. adiabatic curve will be above the isothermal curve, and will enclose greater area under it for the same reduction in pressure), more area will be enclosed by adiabatic, and since the area $\int P\mathrm{d}V$ gives the work required, isothermal work is smaller than adiabatic for the same reduction in volume.

Your argument is correct. To provide more mathematical support to it, you can observe the fact that it is both increase in temperature and reduction in volume which increases the pressure in adiabatic process and only reduction in volume increases pressure in isothermal process. The exponent of volume in adiabatic equation ($PV^\gamma=K$) is $\gamma>1$, as compared to 1 in the isothermal equation. Hence, pressure change is more sensitive to volume change and is larger in magnitude.


How does Rosuvastatin 5 S-lactone form in human body? I've searched lots of free resources online but cannot find the answer.

the top compound is Rosuvastatin.
Some websites say it is metabolized by the enzyme CYP2C9,
but I can't figure out how a CYP450 mediated reaction will produce Rosuvastatin 5 S-lactone. And I tried thinking about an intramolecular esterification, but it does not look like the answer. Perhaps I missed something?

Rosuvastatin is metabolized to a limited extent by CYP2C9 to form an N-desmethyl metabolite and the other metabolite is the rosuvastatin lactone.

The acid moeity (highlighted part) is likely to undergo glucuronidation first and subsequently lactonisation to form the rosuvastatin lactone

Unlike other statins, rosuvastatin undergoes relatively little metabolism by hepatic CYPs with no significant involvement of CYP3A4.

  The metabolism of statins is complex, involving the inter-conversion
  of acid/lactone by various pathways. The acid forms are converted to
  the corresponding lactones via a glucuronide intermediate.

Here is the proposed, generalised metabolic pathway of statins. The acid forms of statins are converted to the corresponding lactones via a glucuronide intermediate via an elimination reaction.

Sidenote for some statins: (After metabolism by CYP3A4 some of the resulting lactone metabolites are converted to the corresponding acid forms:

References

Focus on Statin research. B. A. Wong: Drug Metabolism and Transporter profiles of Statins.
Foye’s Principles of Medicinal Chemistry 7th ed


Does NH4 anion exist? $\ce{BF3}$ can react with $\ce{F-}$ to give $\ce{BF4-}$ where we have for $\ce{B}$ in this molecule the 2s orbital and the three 2p orbitals  filled totally (hybrid orbitals).

In case of $\ce{NH3}$ we have the 2s orbitals and the three 2p orbitals of $\ce{N}$ filled;

So i am asking this: can the hydride ion $\ce{H-}$ make a dative bond with the 3s orbital of $\ce{N}$ (which is vacant) to form $\ce{NH4-}$? If not, why?

The 3s orbitals are far too high in energy to participate in chemical bonding interactions. Lower in energy are the empty σ* antibonding orbitals of the N–H bonds, which can still be energetically relevant in further reaction chemistry. As a thought experiment, it might be possible for a hydride anion to donate electron density into an empty N–H σ* orbital in a sort of Lewis acid-Lewis base interaction. In fact, this is a hallmark of so-called `hypervalent' bonding of heavier main group elements. Rather than using d-orbitals (as is often incorrectly taught to intro chem students), these acid-base interactions are how the phosphorus(V) center of PF$_5$ bonds to the fifth fluoride ion, forming what is known as a three-center four-electron bond.

Why isn't this type of bonding seen with NH$_3$ and H$^–$ to form NH${_4}^-$? These `hypervalent'-type interactions involving 3c-4e bonds depend on the outer atoms having a high-enough electronegativity to stabilize a positive charge at the central atom. In the case of the hypothetical NH${_4}^-$ anion, this just isn't the case. Hydrogen is in fact far less electronegative than nitrogen.


How does activated carbon trap impurities? I've tried and I've tried to find an answer to this all over the web, but I can't seem to find an in depth explanation.

To the best of my knowledge, activated carbon has a very microporous structure and uses a process called adsorbption to "trap" particles. This isn't very in depth though and I'd like to better understand the process of what's going from a scientific standpoint. How does the carbon/charcoal keep the adsorbed material attached or trapped within the carbon?

Never forget to interpret in terms of physical/chemical forces, by order of energy: van der Waals, hydrophobic (pseudo forces), hydrogen bonds, disulfide bridges, covalent, electrostatic (for the most frequent).

Indeed, in charcoal, there is a large number of $\pi$-electrons that can make bonds, in particular. And yes, the porosity is important.

But perhaps could you have a look at Boehm H. P. Some aspects of the surface of carbon blacks and other carbons, Carbon, 1994, 32(5), 759-769. DOI 10.1016/0008-6223(94)90031-0.


What is (approximately) the minimum exchange current for a potentiometric measurement? Is there a rule of thumb for what should be the minimum exchange current to detect a certain redox couple in an aqueous solution?

I would like to make a rough calculations to see whether any of my hypothesis is feasible but I have no idea about the order of magnitude an exchange current should have in water. I could find a few examples for "normal" exchange currents for pH measurements or $\ce{Fe^{3+}/Fe^{2+}}$ but no minimum.

Background: I am trying to figure out, why one can measure a negative ORP in fermentations. There has been a lot of work done on the role of the microorganisms and their exofacial sulfhydryl groups. However, the question of WHAT is actually measured/happening at the electrode has not been answered.
I want to tackle this as part of my thesis and I can think of two mechanisms:
A) Direct interaction of sulfhydryl groups with the electrode surface
B) Mediation, e.g. via Fe or Mn ions.
My concerns about A are that cells are pretty big compared to ions so I am not sure whether enough cells could ever interact with the electrode to make this scenario realistic. In the case of B I am not sure if the concentrations are high enough.

Apologies if my language is too sloppy.

Pfeiffer at al. defines sensor effective (SERC) and sensor ineffective redox couples (SIRC) in relation to the biggest net exchange current found in the system $\text{i}_{net}$.

$\text{SERC} \geq 0.01~\text{i}_{net}$

$\text{SIRC} < 0.01~\text{i}_{net}$

DOI: 10.1016/0169-7722(92)90041-C


Effective chemical for protecting hide glue from microbial attack One problem with hide glue is that it can be attacked and degraded by bacteria and fungi.

To protect hide glue from this kind of deterioration I am interested in identifying an effective anti-microbial. Toxicity against beetle larvae would be a plus. Note that hide glue is typically prepared at 140F and so the chemical must be stable at that temperature. Also, any kind of volatility is out of the question because it needs to last for a long time (100 years).

The main chemical I am considering right now is triclosan, but would be interested in any other ideas.

Silver salts are used to kill bacteria, algae and fungi.  Different types of clothing are 'laced' with silver to kill the bacteria that cause body odor.  Silver spoons were suspected historically to have self-sanitizing properties.  I ran a small analytical chemistry lab that was exempt from most emission oversight, but we could not dispose of silver waste into the sewer as it would destroy the 'good' bacteria at the water-treatment plant.  Different silver compounds, often phosphates, are used to strike a balance between required toxicity and durability/longevity.  If I were starting your project, I would at least start with a good literature search along these lines and some small scale proof-of-concept experiments if it seemed at all feasible.

I do want to be clear however, that I know of no reason to expect silver to be toxic to beetle larvae, at least not at the concentrations that would provide anti-microbial protection. Maybe you would need a second additive for that purpose


Why is ammonia more basic than acetonitrile Why is $\ce{NH_3}$ more basic than $\ce{CH3CN}$?

I tried to reason it out with the hybridization of nitrogen in both the molecules. In $\ce{NH_3}$ nitrogen is $\mathrm{sp^3}$ hybridized while in $\ce{CH3CN}$ nitrogen is $\mathrm{sp}$ hybridized. Hence it follows that nitrogen in $\ce{CH3CN}$ is more electronegative as compared to that in $\ce{NH_3}$, which makes $\ce{NH_3}$ more basic. Is this the main reason for $\ce{NH_3}$ to be more basic or is there any other reason?

p orbitals are higher in energy than s orbitals.
sp3 orbitals with 75% p-character are therefore higher in energy than sp orbitals with 50% p-character.
This means that the lone pair in ammonia is higher in energy than that in acetonitrile.
Higher energy lone pair corresponds to greater basicity.

No need to bring electronegativity into the picture.

For a similar question see Relative acidities of alkanes, alkenes, and alkynes


Identification of a compound C10H12O with 1H and 13C NMR data I was given this problem to do for Organic Chem II.  I know that it's a benzene (unsat # 5), with 2 para-substituents due to the doublets in the 7 region of the HNMR.  I know that the oxygen is not an OH due to the absence of a D2O shake.  There's probably an alkene in one substituent to finish out the unsaturation number.  However, I can't figure out what the 2 substituents are - but I'm guessing one's either an ether or ketone!  Can anyone help?

Like you said, the 2 doublets at around $\pu{7 ppm}$ belong to a para-substituted benzene. The three single-proton multiplets around $\pu{5-6 ppm}$ belong probably to a single substituted alkene. A singlet and a doublet between $3$ and $\pu{4 ppm}$ belong to $\ce{CH3-{}}$ and $\ce{-CH2-{}}$ groups, respectively (Note that the signal that belongs to the $\ce{-CH2-{}}$ group is a doublet, indicating that it is adjacent to a carbon with only a single proton, so it is probably connected to the alkene).

The absorption maximum at $\pu{275 nm}$ gives us a hint that the disubstituted phenyl ring of our compound is not conjugated to the $\ce{C=C}$ bond.

Concerning the oxygen atom: it is definitely neither a carbonyl group (indicated by the absence of an intense peak at $\gt \pu{180 ppm}$), nor a hydroxy group. So it probably is an ether.

So far, we have two possible candidates:

Allyl para-tolyl ether
1-allyl-4-methoxybenzene

The chemical shift of the $\ce{CH3}$-group is shifted downfield compared to the $\ce{CH3}$-shift in toluene (where it is around $\pu{2.3 ppm}$, while this compound has $\gt \pu{3.7ppm}$), and thus more likely belongs to $\ce{CH3-O-Ar}$ protons. So, our compound is 1-allyl-4-methoxybenzene.

However, the multiplets of the vinylic protons in the $\ce{^1H}$-NMR spectrum don't fit this compound well. The allyl group should give two dd signals for the terminal protons and a ddt for another vinylic proton. But the $J_2$ constant for terminal protons is very small ($\lt \pu{2 Hz}$) and can't be seen on some old NMRs, what would lead to the simple two doublets.

Why do we see dddd instead of ddt for $\ce{R-CH=CR2}$ proton? I'm not sure, but maybe this is because the $\ce{-CH2-{}}$ protons are diastereotopic. Then it makes sense.


Plastic membrane that lets water in from the outside only? I'm looking for a type of plastic membrane that would let water through it from the outside, but wouldn't allow water to escape once it's inside the membrane.

Ideally available in a roll.

Does such a plastic membrane exist?

The comments above have gently hinted that it would be a violation of thermodynamics for a passive membrane to force a substance in a single direction. Molecules have an equal probability of moving through the membrane in either direction, unless you apply a force to make them move in a specific direction, such as electro-osmosis or active water transport in plants.

Consider that if such a membrane as you posit existed, you could use it to pump water from a lower level to a higher, and then power a water-wheel: perpetual motion. Sadly, Maxwell's demon is getting a bit arthritic to shuttle water molecules, and the dampness only aggravates its condition.


How is entropy change defined for a chemical reaction? I know that:

ds = dq/T for reversible processes.

Can I write change in entropy of a system in which a chemical reaction is happening using this equation. Basically, my question is about whether a chemical reaction happening in an open vessel(isobarically and isothermally) be treated as a reversible process?

The simplest way is to calculate  $\Delta$H and $\Delta$G of the reaction, then calculate the difference $\Delta$H - $\Delta$G. And you divide this difference by the temperature T. The result is $\Delta$S of the reaction.


How can I rationalise the different basicities of cyclohexylamine, piperidine and morpholine? Compare the basicity of the following compounds:

Comparing compounds I and II
Since there is no conjugated system, there is no resonance at play and we only have to discuss hyperconjucation and inductive effect.
The conjugate acid of compound I has one α-hydrogen while the conjugate acid of  compound II has four α-hydrogens. So, according to hyperconjugation, the conjugate acid of compound II is more stable than the conjugate acid of compound I and hence the basicity of compound II is more than that of compound I.
Further, the conjugate acid of compound I has one α-carbon, two β-carbons, two γ-carbons and one δ-carbon. While the conjugate acid of compound II has two α-carbons, two β-carbons and one γ-carbon.
Alkyl groups are electron releasing groups and, since the conjugate acid of compound II has its carbon atoms closer to the amine than in the case of compound I, it is further stabilised, and hence compound II is more basic.
Comparing compounds II and III
The conjugate acid of compound III has an ether group attached to the amine while that of compound II has an alkyl group to the amine. Since ether is an electron withdrawing group and alkyl is an electron releasing group, the former will destabilise the conjugate acid compound III while the latter will stabilise the compound II, and hence compound II is more basic than compound III.
So, I know that compound II is the most basic, but I still don't know how to compare compounds I and III.
Can anyone help?

You are right that (I) < (II) and (II) > (III), but my explanation for (I) < (II) is different from yours: nitrogen atom in (II) has more alkyl group than (I), therefore the its conjugated acid is more stable.

As inductive effect decrease fast along the carbon chain, even though cyclohexyl group have more carbons, its effect is still weaker than two alkyl group (you can treat piperidine as ethylpropylamine, they're similar). Inductive effect is usually stronger than hyperconjugation.

In (III), both two branches have -C effect, as oxygen is an electronegative element, which destabilises the conjugated acid. In (I), the only alkyl group has +C effect, which stabilise the conjugated acid. Therefore, (I) is a stronger base than (III)

Hence, we have the sequence of basicity: (II) > (I) > (III).

We can look up pKb of each substance to check the result:

Cyclohexylamine (I): 3.36
Piperidine (II): 2.78
Morpholine (III): 5.64

So, my comparison is rational.


Validity of ionic resonance structures What's wrong with ionic resonance structures? I asked one professor about them once and all he commented was "I've seen them too, and I think they're wrong."

Another professor rejected an ionic resonance structure showing the "H+" ion and asked me what phase the experiment was run in, in an attempt to figure out why the authors reported a resonance contributor with a H+ proton. He specifically asked if the experiment was run in the gas phase.

I think this guy forgot what resonance structures are because I know that he emphasizes that bare protons do not exist in solution, and he was probably thinking about how there could be a bare proton shown in a resonance structure if the molecule is in solution. The resonance structure I showed him was this (which I could see him picturing in solution, given that it is nitrous acid.

However, resonance structures are not discrete forms of the molecule in question - they're just contributors and show that there is partial ionic character; that there is a highly electronegative atom that is withdrawing electron density.

So I ask:

1) Are ionic resonance structures valid?

2) If so, why might they be valid, or what counterarguments might you present to someone who argues they are invalid?

I can't imagine anyone arguing that ionic resonance structures are invalid.  The often contribute significantly to the structure of a molecule and help explain the various physical aspects of the molecule.  The following ionic resonance structure contributes significantly to the description of the carbonyl compound.  $\ce{NaCl}$ is not totally ionic and is described by a blend of both the ionic (major) and covalent (minor) resonance structures shown below.

Resonance structures with a positive hydrogen (proton) are important in explaining hyperconjugative effects.


How to determine oxidising and reducing agents in the reaction between ammonia and hypochlorous acid? How do I figure out what is the oxidizing agent and the reducing agent in the following reaction:

$$\ce{\overset{-3}{N}\overset{+1}{H}_3 + \overset{+1}{H}\overset{-2}{O}\overset{+1}{Cl} <=> \overset{-1}{N}\overset{+1}{H}_2\overset{-1}{Cl} + H2O}~?$$

I assigned known oxidation numbers for the elements, but I'm not sure how to proceed with the rest.

Is $\ce{NH3}$ the reducing agent? Or nitrogen? Going from $-3$ to $-1$
and $\ce{HOCl}$ the oxidizing agent, chlorine going from $+1$ to $-1$?

You're on the right track with assigning oxidation numbers to each element. And you've assigned them successfully.

To successfully complete the task you need to define a few terms:

What's oxidation?

What's reduction?

What's the oxidizing agent? Try thinking of this in terms of what oxidation is, and then try figuring out what an oxidizing agent would do.

What's the reducing agent?


How to make silicic acid, potassium silicate or calcium silicate? I am a student of tropical agriculture. I am taking the class of agronomic chemistry and plant nutrition. I want to make use of silicate source I have at home for e.g. zeolite, wood ashes, diatomaceous earth, silica sand. I have hydroxide of calcium, potassium and sodium. Can I use any of these to make silicate for use on plant nutrition ?

You can make the silicates by fusing the hydroxides with silica. Potassium silicate fertilizer grade are successfully produced by direct fusion of silica ($\ce{SiO2}$) and potasium compounds($\ce{KOH}$ and $\ce{K2CO3}$) in furnaces at temperatures up to melting point of mixture. The fusion temperature reaches around 1350 °C.
$$\ce{SiO2 + 2 KOH ->[\Delta] K2SiO3 + H2O}$$
$$\ce{SiO2 + K2CO3 ->[\Delta] K2SiO3 + CO2}$$
Calcium silicate is prepared by reacting calcium hydroxide and silica. This paper1 discussed the reaction at 30-85 °C.
References

https://www.matec-conferences.org/articles/matecconf/pdf/2016/21/matecconf_bisstech2016_01021.pdf
REACTION BETWEEN SILICA AND CALCIUM HYDROXIDE SOLUTIONS. I. KINETICS IN THE TEMPERATURE RANGE 30 TO 85°, S. A. Greenberg, The Journal of Physical Chemistry 1961 65 (1), 12-16, DOI: 10.1021/j100819a005


What is the relation between surface tension and initial molecular harmonic? Initial harmonic is measured as Hartree/Bohr^2 ( like IHarmonic=n in Gaussian Software ). As I am from physics background I am used to look at features from their dimension point of view:

Actually surface tension is the only physical feature that I know is compatible with Force / Distance. I was wondering if there is any relation between the surface tension and initial harmonic.
Update: Another physical feature is spring constant which is not irrelevant also !

Surface tension arises from the fewer neighbors that molecules at the surface have, relative to the number of neighbors for molecules in the bulk.

For example, consider a simple model of rigid identical spheres in a hexagonal closest packing.  In the bulk, there are 12 spheres in contact with a given sphere, which could be thought of as 6 spheres at vertices of a regular hexgon surrounding the given sphere, 3 spheres in a triangle below and 3 spheres in a tringle above.

At a surface, at least 3 of the neighboring spheres would be absent.

Considering an interaction energy between neighboring spheres, the loss of interaction energy (due to fewer than 12 neighbors per sphere) will be proportional to the surface area of the system of spheres.  The minimum energy of the system will be when surface area of the arrangement of spheres is minimized.

For the above reason, the units of surface tension are Energy/surface area, which the OP correctly points out is the same as force/distance.

From the above explanation, particularly that even for rigid spheres always in direct contact the Energy/surface area (force/distance) relationship arises, it can be understood that surface tension has no particular relationship with a force constant between molecules (or between atoms of a molecule).  Instead, surface tension is related to the energy difference between a molecule with maximum neighbors and a molecule with fewer neighbors.

Considering a simple harmonic oscillator, energy vs distance is a parabola.  There is no horizonal asymptote.  Regardless of what the force constant is, there is always infinite energy required to remove a neighbor. In reality, intermolecular force will approach zero at infinite distance between molecules. It is the difference between energy at equilibrium seperation and infinite separation that is relavent to surface tension, and force constant does not provide this information.

The terms "initial molecular harmonic" and "initial harmonic" are not standard chemistry terms and only have meaning in the context of a particular computer program.  "Initialharmonic" or "Iharmonic" refers to a force constant that can be for linear stretching, bending about a vertex that is an atom, or dihedral bending, etc.


Justify assuming first-order kinetics for a complex For the hydrolysis of $trans-\ce{[Co(en)2Cl2]Cl}$ (and cis) we measured reaction rates via spectrophotometry. Plots of change in absorbance $(\ln(A-A_\infty))$ were used to determine the rate law and the rate for the reaction. Using my data, I created the first-order plot seen below:

The question is for the report was:

Justify assuming first-order kinetics

My explanation was that first-order reactions can be expressed as $$[\ce{A}]=[\ce{A}]_0\operatorname{e}^{-kT}$$
and because the concentration decreases exponentially over time, we can assume its first order. Other rate laws I looked up didn’t show an exponential function, so this seemed like sound reasoning, but I was told I was wrong.
What would be the justification for thinking this is first-order?

What would be the justification for thinking this is first-order?

The KISS principle :)

Before you start to model complex kinetics, begin with a simple model. If it fits, consider your job done.

Good enough is good enough!

(Imagine that you perform therapeutic drug monitoring. It's not your job to elucidate the exact metabolism of a drug here, you just want to be sure that the patients receive the right dose.)


What is the best way to calculate the mass percentage of Calcium Carbonate from water? Where I live it's believed that local tap water is really bad for drinking because of the amount of limestone. As a science experiment for school I wanted to approximately calculate the percentage of calcium carbonate in tap water so I can make the conclusion: Is that amount of drinking $\ce{CaCO3}$ actually the problem or something else?

This question might be similar to DK4's question but I'm trying just to calculate not extract, only by using items you can find in a typical house (or in a local shopping center).

If calcium carbonate isn't the issue what is?

Edit: As pointed out in comments, percentage of $\ce{Mg}$ is also applied.

Water contains negligible amount of $\ce{CaCO3}$ or $\ce{MgCO3}$, as these are almost insoluble in water.
What you are interested in are bicarbonates $\ce{Ca(HCO3)2 + Mg(HCO3)2}$. Their content is known as the temporary or carbonate water hardness that can be expressed as $\ce{CaCO3}$ equivalent.
The complement to the carbonate hardness is the permanent hardness, formed mostly by $\ce{CaSO4, MgSO4, CaCl2, MgCl2}$. The sum of the temporary and permanent hardness is called the total hardness.
Bicarbonates ($\ce{HCO3-}$) are not threat to health themselves, as they convert to $\ce{CO2}$ in the stomach and are present in blood as the component of the blood $\ce{pH}$ buffer.
What can be a subject of health concern is the total content of $\ce{Ca^2+}$ (and in lesser extent $\ce{Mg^2+}$). It is again expressible as $\ce{CaCO3}$ equivalent.
Calcium/magnesium content is not measurable by resources usually found in households or local shop center.
The best way is to ask the water provider for the data.
The 2nd best way, close to home-like determination, is to obtain the set for aquarists from a pet shop for determination of carbonate and total water hardness. I am a former aquarist, and this was available even in 1980s in the former communist Czechoslovakia.
Determination is done by counting  the drops of the colored solutions, mixed with the tested volume of water, until the color changes.
Carbonate (in fact bicarbonate) hardness determination is based of reaction of an acid with bicarbonates: $$\ce{HCO3- + H+ -> H2O + CO2 ^}$$
Drop counting ends when yellow color (methyl orange dye indicator) changes to red. The dye turns red by the first excess of the added acid, as has red color in acidic solutions.
Total hardness is based on reaction of $\ce{Ca^2+ + Mg^2+}$ with ethylendiamintetraacetic acid (EDTA) in alkaline solution, enriched by the addition of the dye indicator Eriochrome black T,
$$\ce{Ca^2+ + H2EDTA^2- -> 2 H+ + CaEDTA^2-}$$ The EDTA first reacts with free $\ce{Ca^2+ + Mg^2+}$. The drop count ends when red turns to blue, as EDTA captures $\ce{Ca^2+ + Mg^2+}$  bonded to the Eriochrome black T.
As result, you obtain the carbonate (temporary) and total (temporary + permanent) hardness in 2 possible units: $\pu{mmol/l}$ or $\pu{dGH}$ (the degree of General Hardness).
$\pu{1 mmol/l = 5.6 dGH}$.
$\pu{1 mmol/l}=\pu{100 mg/l}= 0.01\%$, as molar mass of $\ce{CaCO3}$ is cca $\pu{100 g/mol}$.
The typical total hardness of drinking water is $\pu{1\!-\! 3 mmol/l}$,  i.e. the equivalent of $\pu{100\!-\! 300 mg/l}$, resp. $0.01\!-\! 0.03\%$ of $\ce{CaCO3}$.


Can we repair a dead body? The following passage has been extracted from the book "Organic Chemistry-10th edition by Solomon and Frhyle":

  "The growth of living things from microbes to elephants rests on organic reactions, and organic reactions provide the energy that drives our muscles and our thought processes"

Is the above statement true that, organic reactions provide energy that drives our thought processes? Is there any other process by which we are given energy to drive our thought processes?

If you all agree with the statement, then when a person is called "dead" (or after a while being called dead), I hope the reactions will cease (because he can't perform "thought processes"), isn't it?

If these all points are agreed, is their a possibility of repairing a dead body by restarting the reactions immediately?

I agree that, it is not as easy as I said, there is a complexity, but I believe we all can do some research "together" and try to do a "little" contribution to save ourselves.

So, can we repair a dead body?

No, you cannot. There is no way to reset the transient pulses in the neurons that carry the essence of thought processes. The information is already gone.

You will find the neural synapses all in the discharged state. All thought process, all memory, all of what makes a person is gone.


Alkali metal carbonates, their thermal stability and solubility From Li to Cs, thermal stability of carbonates increases. From Li to Cs, due to larger ion size, hydration enthalpy decreases. So, solubility should decrease from Li to Cs. But, experimentally, order is reverse. I cannot wrap my head around this.

Lattice energy is also decreasing down the group. hence thermal stability is decreasing. But since lattice energy is decreasing squarely and hydration energy is decreasing linearly, the effect of decrease in lattice energy is higher than the effect of decrease in hydration energy. So if energy required to break lattice is very less, then the lattice of the bottom elements in alkali group will break easily compared to the lighter ones. Thus if not lattice form, then they prefer the aqueous state and hence they love to stay dissolved and not form lattice. So solubility increases.


What are the products in a reaction between a metal and an acid? When magnesium metal and hydrochloric acid react, hydrogen gas is released (and not chlorine gas). However, when metals such as copper is placed in nitric acid, NO(g) is released and not hydrogen. So, how do we know that chlorine gas is not released in the first reaction? Does this have to do with reduction potential?

Nitric acid ($\ce{HNO3}$) is a strong oxidizing agent. The hydrogen gas ($\ce{H2}$) produced in the reaction with metal is oxidized to $\ce{H2O}$, in turn evolving nitric oxides like $\ce{NO2}$ and $\ce{NO}$. Even magnesium can be observed evolving nitric oxides when reacted with $\ce{HNO3}$.

Additionally, you know that chlorine gas isn't released by the reaction of $\ce{HCl}$ with magnesium because you would see it evolved from the solution as a pale, yellow-green gas. Instead, only clear hydrogen gas is evolved.


Are alkynes more electrophilic than alkenes? I think that since alkynes have a greater electron density, they would be less electrophilic than alkenes.

Which functional group is more prone to nucleophilic attacks and why? An example of a known reaction would be greatly appreciated.

Here are what my notes from Bruice's Organic Chemistry textbook says:

"The clouds of electrons surrounding the sigma bond makes an alkyne an electron-rich molecule. They are therefore nucleophiles that react with electrophiles.

Thus alkynes, like alkenes, undergo electrophilic addition reactions because of their weak pi bonds. The same electrophilic reagents that add to alkenes also add to alkynes.

Despite being less stable than an alkene, an alkyne is less reactive than an alkene, since reactivity depends on ΔG ‡ , which depends not only on the stability of the reactant but also on the stability of the transition state.

When a proton adds to an alkene, an alkyl cation is formed. When a proton adds to an alkyne, a vinylic cation is formed. A vinylic cation has a positive charge on a vinylic carbon. It is less stable than a similarly substituted alkyl cation because the positive charge is on an sp carbon, which is more electronegative than an sp 2 carbon of an alkyl cation. This makes it less able to bear a positive charge."


Proof that BF3 exists? What I know is, elements with octet are stable, But In BF3, Boran doesn't get octet, So, How come BF3 molecule exist? If it exists, How do we know whether it is BF3 or not? Is there any proof that shows BF3 exists?

Boron trifluoride, as described in Wikipedia is proved to exist the way everything in chemistry is ultimately proved, by experiment.  The octet rule is really only a model that describes the electronic structure of some but not all molecules.  Boron trifluoride is one of those (many, actually) molecules for which the octet rule is not really correct.  The actual electronic structure is not just a single Lewis structure but a combination of multiple contributing structures, some of which do satisfy the octet rule and some of which don't.


What's the meaning of the $-\equiv-$-like bond symbol? In the following image, $\mathrm{CH}_2$ is connected to benzene rings with a $-\equiv-$ like symbol. What's the meaning of this symbol?

This is a triple bond between two $\ce{C}$ atoms, i.e. $\ce{-C#C -}$. It is drawn that way because the carbon atoms in a triple bond are $\mathrm{sp}$-hybridised and thus exhibit a linear arrangement.


Ester as a side product in the synthesis of bromoalkane from alcohol. Is it correct? About 5 months ago, I answered a very interesting question albeit a very simple one.

  Why do we need sulfuric acid in creating alkyl bromide and not in creating alkyl iodide from alcohol and hydrogen halide?

The question was why sulfuric acid is used to convert any alcohol to bromoalkane but not iodoalkane. Jan's answer quite pointed out the explanation:

  hydrogen iodide is sufficiently acidic to protonate the corresponding    alcohol and drive the reaction. More acids are not
  needed.
  hydrogen bromide is not acidic enough to drive the reaction. Adding    sulphuric acid protonates the alcohol partially and allows the
  reaction to proceed.
  in the case of weakly soluble alcohols and hydrogen iodide, phosphoric acid may be used instead since it does not oxidize iodide (while sulfuric acid does).

Though I am satisfied with Jan's excellent answer, I further pointed out that sulfuric acid may react with hydrobromic acid to form bromine(aq.) and sulfur dioxide. This bromine may oxidize alcohol to form aldehyde and carboxylic acid and they two will form ester. This was my proposed explanation but I couldn't find any experimental result to back my explanation and heavily relied on theory.

Now recently Nurdrage, a renowned chemist famous for his synthesis of pyrimethamine has uploaded a video on Youtube about the synthesis of bromoethane and bromohexane from ethanol and hexanol using sulfuric acid. He also explained the concern of forming bromine as side product from the reaction of hydrobromic acid and sulfuric acid. He showed a purification technique by adding sodium bicarbonate to react with bromine to form sodium hypobromide.

Now my question is:

Is my explanation of forming ester instead of bromoalkane correct enough? Is there any possibility of forming minute amounts of ester? The reaction may be slow but it can happen before any purification is done (reacting bromine with sodium bicarbonate). If so, what can be done to remove any ester or prevent the reaction of ester formation?

Also related: Why is alcohol boiled with HBr for preparing alkyl bromide?

Technically it could form, but it would be hydrolyzed. If it did form, it would be in negligible quantities. Lots of other stuff could form, and they do, but in miniscule amounts ... bromine oxidized the alcohol to the aldehyde. The aldehyde undergoes aldol condensation. Another alcohol undergoes elimination. The compounds  react in a hetero diels-alder. So dihydropyrans can also form ... Yes, but not in quantities that are observable.


Uranium enrichment via effusion through a porous membrane The following problem is from Principles of General Chemistry, Silberberg, 1st edition:

  One way to utilize naturally occurring uranium ($0.72\%$ $\ce{^235U}$ and $99.27\%$ $\ce{^238U}$) as a nuclear fuel is to enrich it (increase its $\ce{^235U}$ content) by allowing gaseous $\ce{UF6}$ to effuse through a porous membrane. From the relative rates of effusion of $\ce{^235UF6}$ and $\ce{^238UF6}$, find the number of steps needed to produce uranium that is $\ce{3.0 mol}\ \%$ $\ce{^235U}$, the enriched fuel used in many nuclear reactors.

Elsewhere in the text the following ratio is given:

  $$\frac{\text{Rate}(\ce{^235UF6})}{\text{Rate}(\ce{^238UF6})} = \sqrt\frac{M(\ce{^238UF6})}{M(\ce{^235UF6})} = 1.0043$$

I've found two different ways to solve this that give different answers.

Solution 1 based on similar problem in LibreTexts Chemistry:

$$\begin{align}
1.0043^n &= \frac{0.030}{0.0072} \\[5pt]
n &= 333
\end{align}$$

Solution 2 based on similar problem in Principles of Modern Chemistry by Oxtoby, Gillis, Butler:

$$\begin{align}
1.0043^n &= \left(\frac{3.0}{97.0}\right) \left(\frac{99.27}{0.72}\right) \\[5pt]
n &= 338
\end{align}$$

Which of these approaches is correct? 2 makes more sense to me but 1 is closer to the answer given in the book (332).

The second approach looks more correct to me. The first approach is using raw percentages without converting to a ratio between isotopes, and the ratio rates is going to be directly related to the ratio of isotopes, not raw percentages.

From the comments, ed:
Well, $1.0043$ is the ratio between rates, so each stage will cause the ratio between isotopes to change by this factor.
I simple set up:
$$\frac{0.72}{99.27} \times 1.0043^{n} = \frac{3}{97}$$ which is the same equation as method 2.

$\frac{0.72}{99.27}$ is the initial ratio (before any enrichment). You need to increase it successively by $1.0043$ until you get to $\frac{3}{97}$, the desired mol%.


Stacking two square (2D) layers to make a 3D close packed structure My textbook states these ways of stacking 2D layers to make 3D close packed structures:

Square close packing layer over Square close packing layer (though not written explicitly, the illustration imply stacking done in a non staggering manner), generating Simple primitive cubic unit cell lattice.
Hexagonal (2D) close packing layer over Hexagonal (2D) close packing layer (here, again not stated explicitly, but the illustrations imply it is done in staggering manner so that one layer just fits right in the depressions of other layer) giving two different close packed structures: hcp and ccp

However, they did not mention the two cases arising naturally (out of curiosity  maybe):

a) Square close packing layer over Square close packing layer done so that the second layer fits just right in the depressions of the first layer, i.e. in a staggering manner so the layer pattern is of 'ABAB...' type.

b) Hexagonal close packing layer over Hexagonal close packing layer in a non-staggered manner, which might generate a lattice with the primitive hexagonal unit cell, layer put in the pattern 'AA...' type.

The case (b) was easier to work out its unit cell's type, however I am unable to comprehend the 3D structure of what type of unit cell case (a) can generate.

My question is, are these cases of lattice arrangements possible? If yes, then what are their respective unit cells?

In the picture, the cubic face-centered packing is shown, with layers of identical atoms colored in red and blue for better contrast. The unit cell shown with thin lines is the conventional face-centered setting. It has red spheres on the corner, and on the center of the bottom face (the faces not normal to the plain of the paper have blue spheres in their centers).

The unit cell shown with thick dashed lines is what you were asking about. It has square close packing of the red spheres, with square close packing of the blue spheres shifted below. (The third unit cell vector is not shown, and is the same for both, from red to red layer).

So to answer your question, just tilt your head by 45 degrees, and you have your "close square" packing rather than face-centered cubic. As you can see, the three unit cell vectors will not be of equal length anymore, so the unit cell is no longer cubic, and it is body-centered instead of face-centered. The volume is smaller by a factor 2, and the number of atoms per unit cell is smaller by a factor of 2 (2 atom per cell instead of 4 atoms per cell).

In exploring this question, I made use of this site: https://scilearn.sydney.edu.au/fychemistry/iChem/solids.cfm?page=ccp&type=cubic


Alkylation of benzene I was solving a question where I had to determine a way to convert benzene into n-propyl benzene.
I figured out a way by first performing Friedel-Craft acylation with propanoyl chloride and then performing Clemmensen reduction to form the desired compound.
On checking the solution,  there were 2 alternative ways portrayed:
One was the same method I had used.
The other involved reaction of benzene with n-chloropropane in the presence of $\ce{FeBr3}$.
However wouldn't there be carbocationic rearrangements taking place that convert the carbocation to isopropyl carbocation?
Please tell me what prevents the rearrangements from happening in the second case.

Initially one would assume that cumene (isopropyl benzene) would be the major product. However, kinetic control does have a say in this reaction, according to Gilman and Means.1

At - 2°, n-propyl bromide with benzene and aluminum chloride gave n-propylbenzene, identified as its sulfonamide. Genvresse (6) obtained both n-propylbenzene and isopropylbenzene by conducting the reaction at reflux temperature. Konowalow (7) found that below $0^\circ$, n-propyl chloride gave n-propylbenzene, while from $0^\circ$ to reflux temperature it gave mixtures of n-propylbenzene and isopropylbenzene. More recently it has been shown that at $-6^\circ$, 60% of the monopropylbenzene was n-propylbenzene, and 40% was isopropylbenzene. At $35^\circ$, this ratio was reversed.

a low temperature would facilitate the production of larger amounts of n-propylbenzene due to kinetic control. However, this is still not a very efficient method of preparing n-propylbenzene, and acylation followed by Clemmensen/Wolff-Kishner reduction would be more suitable here.
References:

Gilman, Henry, R. N. Meals, “Rearrangements in the Friedel-Crafts Alkylation of Benzene” Journal of Organic Chemistry 1943, 08(2), 126–146. doi:10.1021/jo01190a003.


How to simplify these system of equilibrium equations What is the pH when $0.025$ mol $\ce{NiCl2}$ is added to $\pu{1.00 L}$ of a $\pu{0.100 M}$ solution of $\ce{HCN}$, if the formation constant for $\ce{Ni(CN)4^2-}$ is $K_f=\pu{1.0e22}$, and the $K_a$ for $\ce{HCN}$ is $\pu{4.9e-10}$?

My approach is as follows:

1) $\ce{\frac{[H+][CN-]}{[HCN]}=4.9*10^{-10}}$

2) $\ce{\frac{[Ni(CN)4^{2-}]}{[Ni^2+][CN-]^4}}=1.0*10^{22}$

3) $\ce{[H+][OH-]}=10^{-14}$

4) $\ce{0.025=[Ni^{2+}] + [Ni(CN)4^{2-}]}$       .............  (mass balance for $\ce{Ni^{2+}}$)

5) $\ce{[CN-] + 0.05 + [OH-] + 2*[Ni(CN)4^{2-}] = 2*[Ni^{2+}] + [H+]}$ ........... (charge balance)

6) $\ce{0.1 = [CN-] + [HCN] + 4*[Ni(CN)4^{2-}]}$ ............. (mass balance for $\ce{CN^{-}}$)

Since the equilibrium constant for the formation of the complex ion is very large, I assume that $\ce{[Ni(CN)4^{2-}] >> [Ni^{2+}]}$, so $\ce{[Ni(CN)4^{2-}] = 0.025 M}$, from the mass balance for $\ce{Ni^{2+}}$.

However, I am not sure what is the next step and what other assumption to make.

Since the equilibrium constant for the formation of the complex ion is very large, I assume that $\ce{[Ni(CN)4^{2-}] >> [Ni^{2+}]}$

From the comments:

  [comments:] That assumption is incorrect. The equilibrium constant seems large, but the exponents are high, so it is misleading. In fact, more of the cyanide is in the form of HCN than in complex with nickel, and there is very little free cyanide.

OP replied:

  How may I make a better assumption to solve this problem?

In the absence of nickel, the pH is around 5 and HCN is the major species, with the concentration of cyanide in the micromolar range (standard weak acid problem). When you add nickel, the concentration of cyanide gets lowered further. Because the free cyanide concentration contributes little to the total, we can combine the two reactions to see whether there is more complex or more HCN and to determine the pH:

$$\ce{4HCN + Ni^2+ <=> Ni(CN)4^2- + 4H+}$$

The equilibrium constant is

$$K_\mathrm{combined} = \pu{5.8e−16}$$

So for this reaction, reactants are favored. HCN and nickel concentrations hardly decrease compared to the initial concentrations. If we call the concentration of the complex $x$, we get:

$$K_\mathrm{combined} = \frac{x (4x)^4}{[\ce{HCN}]^4 \cdot [\ce{Ni^2+}]}$$

Solving for $x$, we get $\pu{2.24e−5}$, confirming that reactants are favored. From this we can calculate the pH as

$$\mathrm{pH} = -\log 4x = 4.05 $$

Checking the solution

Upon adding nickel, the pH turned a bit more acidic. That makes sense as a micromolar concentration of nickel complex formed, releasing some more protons from HCN. We can calculate the cyanide concentration from either one of the given equilibrium constants, and verify that it is miniscule. Then, we can verify that all reactions are at equilibrium (at pH 4, hydroxide is a minor species as well, and has little impact on the exact solution).


How does a Brunauer–Emmett–Teller device function? The Brunauer–Emmett–Teller (BET) theory is comprehensible.

My question is more geared towards what is physically and chemically happening within a BET device to get a reading?  How does a BET device actually go about measuring the physisorption or chemisorption of a material?  Why does this process tend to take multiple hours?

BET analysis on zeolithes represents a typical example.  Dependent on the size of the particles you have, you select a glass cell of about a centimetre; for a powder a smaller cell, for more coarse pellets you don't want to grind a bigger one.  You need to outgas the samples first, i.e. by applying vacuum and elevated temperatures already adsorbed gases and moisture is removed from the surface of the material to be analyzed as well as the inner walls of the sample cell. This already may take time to be accomplish -- because you do not want to destroy or alter the material in question by too harsh conditions.

The actual measurement requires the sample cell being kept at a constant temperature.  An initial determination takes place while the sample cell is immersed in a Dewar of liquid nitrogen.  This low temperature provides better conditions for the adsorption experiment.

A first blank is measured with Helium gas, entered via a graduated piston. Helium typically does not adsorb on surfaces (at this temperature).  This allows to determine the dead volume of the measurement cell.  In the second step, Nitrogen gas enters the measurement cell, again in a controlled manner (graduated piston, pressure control).  Again, this takes time as you want to establish the equilibrium between the adsorbent and the amount of gas adsorbed in function of gas pressure and sample temperature.
A controlled entry of Helium gas is necessary as a second determination of the dead volume to complete the calibration eventually leading to the read-out for our substance at given temperature.

Of course, dependent on what kind of isotherme (BET/Langmuir) is to be determined, Nitrogen gas as a load may be replaced by other gases, like methane.  Beside determination of the surface area of a solid, the adsorption properties may be determined in a series of more elevated temperatures, too.


What is the correct discharge reaction in nickel-cadmium cell? In school textbook

$$\ce{Cd +2Ni(OH)3->CdO + 2Ni(OH)2 + H2O}$$

Wikipedia

$$\ce{2NiO(OH) + Cd + 2H2O->2Ni(OH)2 + Cd(OH)2}$$

Chemwiki

$$\ce{Cd +NiO2 + 2H2O->Cd(OH)2 + 2Ni(OH)2}$$

Which reaction should I follow?

The ChemWiki page got updated on 22nd Jan 2017, an year after this question was posted, and the reaction in question has since been updated. The reaction on the ChemWiki page is now identical with that on the Wikipedia page. This question now has no significance whatsoever but for the sake of the time and effort of the answerer, I am keeping it undeleted

Note that the three equations are different only in whether the nickel and cadmium oxides are hydrated (and thus where water appears in the equation). Also, the Chemwiki equation is not balanced because nickel oxide is $\ce{Ni2O3}$ and not $\ce{NiO2}$.

The correct equation for the ChemWiki version is:

$$\ce{Ni2O3 + Cd + 3H2O -> 2Ni(OH)2 + Cd(OH)2}$$

Many metal oxides react with water to form hydroxides. This kind of behavior is probably most frequently taught with the alkali metals and alkaline earth metals:

$$\begin{align}
\ce{2NaO + H2O} &\ce{-> 2NaOH}\\
\ce{CaO + H2O} &\ce{-> Ca(OH)2}\end{align}
$$

However, other metals undergo these kinds of reactions. For example, the chemistry of rust involves the equilibrium between iron oxides and iron oxide hydroxides:
$$\begin{align}
\ce{FeO + H2O} &\ce{<=> Fe(OH)2}\\
\ce{Fe2O3 + 3H2O} &\ce{<=> 2Fe(OH)3}\end{align}
$$

  Nickel

The three ionic nickel species in the reactants are different only on the degree of hydration:

$$\begin{align}
\ce{Ni2O3 + H2O} &\ce{<=> 2NiO(OH)}\\
\ce{NiO(OH) + H2O} &\ce{<=> Ni(OH)3}\end{align}
$$

  Cadium

The two ionic cadmium species in the products differ only by degree of hydration:

$$
\ce{CdO2 + H2O<=> Cd(OH)2}
$$


Does the van der Waals equation remain valid when repulsive intermolecular forces dominate? The van der Waals equation for a real gas is:

$$RT =\left(p+\frac{a}{V_\mathrm{m}^2}\right)(V_\mathrm{m}-b)$$

We have understood this formula by saying that $a$ is the term which is for force of attraction between the gas molecules and thus the pressure of a real gas will be less than that of an ideal gas.

Therefore some factor must be added to the pressure of a real gas to make it equal to ideal gas pressure so that it can be used in $p_\mathrm{ideal}V_{\mathrm{ideal}}= nRT$.

Also, if we see the compressibility factor graph , suppose for $\ce{N2}$:

(Source: TutorVista)

If $Z < 1$, then attractive forces are dominating. We see that after a particular temperature, the value of $Z$ becomes greater than $1$, implying that the repulsion forces are dominating in $\ce{N2}$.

Now, since repulsive force dominates, the pressure of the real gas will increase and be greater than that in ideal case.

To find the $p_{\mathrm{ideal}}$ now, we will need to deduct some value from the $p$ of real gas, which means $a$ should be negative. However, this is not so as $a$ can never be negative.

No value can ever be deducted from $p$ in the van der Waals equation.

Then how does the van der Waals equation still work and give the correct data?

You need to consider the $b$ term, as well as $a$.

Rearranging the van der Waals equation to solve for $p$ gives

$$p = \frac{RT}{V_\mathrm{m} - b} - \frac{a}{V_\mathrm{m}^2}$$

You are right that a positive $a$ will always decrease the pressure, compared to the ideal gas result.

But a positive $b$ will act to increase the pressure compared to the ideal gas result.

At high pressures (small molar volumes), the effect from finite molecular volume ($b$) outweighs the effect of intermolecular attraction ($a$).


Boiling methanol in a microwave Some protocols (metabolomics) use boiling methanol for metabolites extraction and cleaning glassware. Is it safe to boil methanol in a microwave?

It is possible and is not unheard to perform an extraction assisted by microwaves (even "dry ones" e.g.).  Either you seal the reaction vessel with a special cap to clamp, and monitor temperature and pressure evolution in function of time and power applied by the dedicated microwave oven.  This has the potential advantage of both rapid heating, and "pressure cooking".  (Depending on the material, using the reaction tube sealed a twice may be discouraged, though.)

Or you opt-in for a system where the the reaction flask is in the microwave oven, and the reflux condenser passes across the walls of the oven.  While in principle ovens designed for use in a kitchen may be used, e.g.,

(source)

it is both safer and more reproducible to use ovens dedicated to this (especially: focussing on the spot where your reactor is situated) and providing proper shielding and monitoring of the process parameters.

(source)


Understanding chemical kinetics For a chemical reaction
$$\ce{2 A + 2 B -> C + D}$$
the order of reaction is one with respect to $\ce{A}$ and one with respect to $\ce{B}$. The initial rate of the reaction is $\pu{4e-2 mol L-1 s-1}.$ When $50\%$ of the reactants are converted into products, the rate of the reaction would become
(A) $\pu{2e-2 mol L-1 s-1}$
(B) $\pu{1e-2 mol L-1 s-1}$
(C) $\pu{4e-2 mol L-1 s-1}$
(D) $\pu{2e-1 mol L-1 s-1}$

This is the first time I am attempting kinetics problems. Is the question asking the concentration when the system has reached its half-life?
Could I use the half-life formula here?
My attempt at a solution:
I used the rate law which I equated to the definition of rate.
I have serious conceptual issues which I hope to clear up. So is this this right?
$$\frac{1}{c(t)} - \frac{1}{c_0}  = kt$$
Would this formula solve the above problem?

Is this the right formula to use to solve the above problem

Yes, this is it, if the initial concentrations of the reagents are the same (which is not mentioned explicitly in the problem statement, but probably is implied, because otherwise the problem would be unsolvable.)

Indeed, as A and B are spent in equal amounts, their concentrations are bound to be the same not only initially, but at all times: $[\ce{A}]=[\ce{B}]=c(t)$. Therefore $\dot c=-k\cdot[\ce{A}]\cdot[\ce{B}]=-k\cdot c^2$, which you seem to already know how to solve.

Update. Come to think of it, you don't actually need to solve for $c(t)$. You know that the rate is proportional to the first power of $[\ce{A}]$, and also to the first power of $[\ce{B}]$, and that both $[\ce{A}]$ and $[\ce{B}]$ have decreased by 50%; what would happen to the rate?

The concept of half life does not apply here, since the decay is not exponential.

Also, welcome to Chem.SE, and pay attention to the formatting.


How to perform solvent extraction of a solid Suppose I have a solid compound from which I want to extract a target chemical I know is soluble in acetone.  So I crush and mix the compound in acetone, dissolving the target and suspending the detritus.

What are methods for filtering the detritus from the solution? I have frequently seen coffee filters used for this purpose.  Are commercial coffee filters considered "lab-grade?"  Or, suppose that the target chemical has some affinity for cellulose, what alternatives exist for suspension filtration?
Now, having the target chemical in solution I assume it is easiest to extract it by evaporating the acetone.  Is there a process for this that doesn't leave any of the target stuck to the glass or whatever was containing the solution?

What are methods for filtering the detritus from the solution? I have frequently seen coffee filters used for this purpose. Are commercial coffee filters considered "lab-grade?" Or, suppose that the target chemical has some affinity for cellulose, what alternatives exist for suspension filtration?

The method that comes to mind for this task is a Soxhlet extraction. Commercial coffee filters are not "lab-grade." Soxhlet extraction thimbles are available in cellulose.

  Now, having the target chemical in solution I assume it is easiest to extract it by evaporating the acetone. Is there a process for this that doesn't leave any of the target stuck to the glass or whatever was containing the solution?

Yes - rotary evaporation, in conjunction with Soxhlet extraction.


Models for diastereoselectivity in nitronate additions In the scheme below (taken from the referenced Chem. Eur. J. paper on the synthesis of Tamiflu), there is a step in which a nitronate anion (the anion derived from 5) is added into an electrophile 6, undergoing what is essentially a Michael addition.

This Michael addition sets the nitro stereocentre, as the nitronate can evidently react from one of two faces. This isn't entirely unusual, the same thing happens with enolates during aldol reactions, and the facial selectivity is explained using various models depending on whether the reaction goes via an open or closed TS (i.e. Zimmerman Traxler vs open chain models).

Clearly the nitronate anion is able to exist in several conformations, however if one draws the Newman projection, then the conformation leading to the product is the one in which H is perpendicular to $\ce{C=N}$ as if were the large group (not the most sterically favored since the ester and the alkyl chain are clearly larger).

I'm hoping someone might know the name of a model that explains the selectivity (in the same way that a Zimmerman Traxler TS rationalises a chelated aldol and the Felkin Ahn model may be used to rationalise the stereochemistry from addition into chiral aldehydes). The closest thing I can think of is the Houk model, however, this itself doesn't give the correct product and I can find no literature from Houk's group that mentions this class of reactions.

  Ref: H. Ishikawa, T. Suzuki, H. Orita, T. Uchimaru, Y. Hayashi, Chem. Eur. J. 2010, 16, 12616. DOI: 10.1002/chem.201001108.

  High-Yielding Synthesis of the Anti-Influenza Neuraminidase Inhibitor (−)-Oseltamivir by Two “One-Pot” Sequences

Dave Evans taught me that you should generally analyze these types of problems by drawing out allyl conformations.

I've drawn the nitronate to minimize 1,3-allylic strain. Now, the top face of the nitronate is largely blocked by the t-butyl carboxylate. I strongly suspect that having a bulking alkyl group there is important to getting good selectivity.

This means that the electrophile will react on the bottom face, and that will lead to the desired stereochemistry depicted in 7 with the nitro group down on the ring.


Why does the internal energy increase when ice melts to give water? Ice melts at 0 °C to give water at 0 °C. The internal energy apparently increases. Thermal energy is transferred to ice causing this to occur. Since the temperature doesn’t change, all the thermal energy is used to increase the potential energy(is this correct?).
Now I know that water is MORE dense than ice. Considering that, shouldn’t the potential energy DECREASE when ice melts? Consequently, shouldn’t the internal energy decrease?

Potential energy refers in this case primarily to the energy associated with intermolecular hydrogen bonds between water molecules. The total strength of H-bonds for each water molecule is on average weaker in the liquid. The density is a function of the geometry of the solid or liquid required to accommodate the bonds. Water is odd in that it acquires a spacious bonding arrangement in ice, but this does not mean that bonding is any weaker for that matter. On the contrary, you might say that the geometry is special in order to accommodate those strong intermolecular bonds.
To complement a point made in another answer: the change in internal energy  during a thermodynamic tranformation is the sum of the heat transferred to and work done on the system. Note that work done by a system (here the water) decreases its internal energy. In the case of ice melting under atmospheric pressure, the volume contraction means that the internal energy increases due to work done on the system by the surroundings. This increase in internal energy is on top of the gain due to a transfer of heat from the surroundings.
Note however that the free energy is unchanged during the process (if carried out reversibly), or, equivalently, the joint entropy of system and surroundings is constant. So the transfer of heat to the system (which reduces the entropy of the surroundings) is compensated by an increase in the entropy of water with melting.