How to specify atomic carbon terms in the coupled and uncoupled representation? So, we know, that the atomic carbon in the electronic configuration $1s^22s^22p^2$ has the following terms

$${}^1S, {}^1D, {}^3P$$

My question is - how can I correctly specify these terms in the terms of coupled and uncoupled representations?

My attempt

So, in the case of terms, we're only considering the orbital angular momentum, not the spin. Because of that, we can describe the single terms in the coupled representation $\left|L, M_L\right>$ which correspond with the linear combination of microstates, i.e. the uncoupled representations $\left|m_{l1}, m_{l2}\right>$ using Clebsch-Gordan coefficients.

For ${}^1S$ term it's pretty easy, as $L=0$ and $M_L=0$ (as described in this answer):

$$\begin{align}{}^1S: |L = 0, M_L = 0\rangle &= \frac{1}{\sqrt 3} |m_{l1}= 1, m_{l2} = -1\rangle + \frac{1}{\sqrt 3} |-1, 1\rangle - \frac{1}{\sqrt 3} |0, 0\rangle\\ &= \frac{1}{\sqrt{3}} \left| 8 \right> + \frac{1}{\sqrt{3}} \left| 11 \right> - \frac{1}{\sqrt{3}}\left| 14 \right >\end{align}$$

In the last expression there are wavefunctions specified with indices from the microstate table below.

But further it gets somewhat more tricky - both ${}^3P$ and ${}^1D$ will contain multiple states.
$P$ corresponds with $L=1$ and so $M_L \in \left\{ -1, 0, 1 \right\}$. I suppose, that its coupled representations are $\left| L=1, M_L=-1\right>, \left| L=1, M_L=0\right>, \left| L=1, M_L=1\right>$.

$$\begin{align}
{}^3P: \left| L=1, M_L=-1\right> &= \frac{1}{\sqrt{2}}\left| -1, 0 \right> + \frac{1}{\sqrt{2}}\left| 0, -1 \right>\\
&= \frac{1}{\sqrt{2}}\left| 2 \right> + \frac{1}{\sqrt{2}}\left|5 \right>\\
\left| L=1, M_L=0\right> &= \frac{1}{\sqrt{2}}\left| 1, -1 \right> - \frac{1}{\sqrt{2}}\left| -1, 1 \right>\\
&= \frac{1}{\sqrt{2}}\left| 3 \right> - \frac{1}{\sqrt{2}}\left|6 \right>\\
\left| L=1, M_L=1\right> &= \frac{1}{\sqrt{2}}\left| 1, 0 \right> - \frac{1}{\sqrt{2}}\left| 0, 1 \right>\\
&= \frac{1}{\sqrt{2}}\left| 1 \right> - \frac{1}{\sqrt{2}}\left|4 \right>
\end{align}
$$

${}^1D$ corresponds with $L=2$ and $M_L \in \left\{ -2,-1,0,1,2 \right\}$.

$$\begin{align}
{}^1D:\left| L = 2, M_L = -2 \right> &= \left| -1, -1 \right> = \left| 15\right>\\
\left| L = 2, M_L = -1 \right> &= \frac{1}{\sqrt{2}}\left| 0, -1 \right> + \frac{1}{\sqrt{2}}\left|-1, 0 \right> \\
&= \frac{1}{\sqrt{2}}\left| 10 \right> + \frac{1}{\sqrt{2}}\left| 12 \right>\\
\left| L = 2, M_L = 0 \right> &= \frac{1}{\sqrt{6}}\left|1, -1 \right> + \sqrt{\frac{2}{3}}\left| 0, 0 \right> + \frac{1}{\sqrt{6}}\left| -1, 1 \right> \\
&= \frac{1}{\sqrt{6}}\left| 8 \right> + \sqrt{\frac{2}{3}}\left|14 \right> + \frac{1}{\sqrt{6}}\left| 11 \right> \\
\left| L = 2, M_L = 1 \right> &= \frac{1}{\sqrt{2}}\left| 1, 0 \right> + \frac{1}{\sqrt{2}}\left| 0, 1 \right>\\
&= \frac{1}{\sqrt{2}}\left|7 \right> + \frac{1}{\sqrt{2}}\left|9 \right> \\
\left| L = 2, M_L = 2 \right> &= \left| 1, 1 \right> = \left| 13 \right>
\end{align}
$$

Is this the right approach or do I understand it incorrectly?

Table of microstates

I'll try to illustrate the procedure for two of the nine states that form the term symbol $^3P$. This term symbol has $S = 1$ and $L = 1$, which makes for three possible values of $M_S$ and three possible values of $M_L$, for a total of nine states. These states are "coupled" in both the spin and the spatial dimensions:

$$|S,M_S,L,M_L\rangle = |1,+1,1,+1\rangle, |1,0,1,+1\rangle, \cdots$$

The microstates in the table are states which are "uncoupled" in both dimensions:

$$|m_{s1},m_{s2},m_{l1},m_{l2}\rangle = |{+1/2}, {+1/2}, {+1}, {+1}\rangle, \cdots$$

To go from the term symbols ("doubly coupled" - don't use this terminology, I made it up) to the microstates ("doubly uncoupled"), we have to convert both spin and spatial components from coupled to uncoupled representations.

Thankfully, the spin and spatial components are separable, so we can write

$$|S,M_S,L,M_L\rangle = |S,M_S\rangle \cdot |L,M_L\rangle$$

Uncoupling the spin part is simple enough:

$$\begin{align}
|S=1,M_S=+1\rangle &= |\alpha(1)\alpha(2)\rangle \\
|S=1,M_S=0\rangle &= \frac{1}{\sqrt 2}\left\{|\alpha(1)\beta(2)\rangle + |\beta(1)\alpha(2)\rangle\right\} \\
|S=1,M_S=-1\rangle &= |\beta(1)\beta(2)\rangle \\
\end{align}$$

where $|\alpha\rangle$ denotes $m_s = +1/2$ and $|\beta\rangle$ denotes $m_s = -1/2$, as usual.

For the spatial parts, you will need to use the Clebsch–Gordan coefficients to expand $|L,M_L\rangle$ in terms of the "uncoupled representations" $|m_{l1},m_{l2}\rangle$. You already seem to understand this. For illustrative purposes we'll use

$$|L = 1, M_L = +1\rangle = \frac{1}{\sqrt 2}(|m_{l1} = +1, m_{l2} = 0\rangle - |m_{l1} = 0, m_{l2} = +1\rangle)$$

and we'll simplify the notation by denoting $m_l = +1, 0, -1$ as $p^+, p^0, p^-$ respectively, so:

$$|L = 1, M_L = +1\rangle = \frac{1}{\sqrt 2}\left\{|p^+(1)p^0(2)\rangle - |p^0(1)p^+(2)\rangle\right\}$$

Now, say we wish to find the microstate corresponding to $|S,M_S,L,M_L\rangle = |1,+1,1,+1\rangle$. What we need to do is to expand both spin and spatial components in terms of their respective uncoupled representations, then multiply them together again.

$$\begin{align}
&|S=1,M_S=+1\rangle \cdot |L = 1, M_L = +1\rangle \\
&\qquad = |\alpha(1)\alpha(2)\rangle \cdot \frac{1}{\sqrt 2}\left\{|p^+(1)p^0(2)\rangle - |p^0(1)p^+(2)\rangle\right\} \\
&\qquad = \frac{1}{\sqrt 2} \left\{|p^+(1)\alpha(1)p^0(2)\alpha(2)\rangle - |p^0(1)\alpha(1)\,p^+(2)\alpha(2)\rangle\right\}
\end{align}$$

Notice how this is a (suitably antisymmetrised) state where one electron is in $p^+$ with spin up, and the other electron is in $p^0$ with spin up. This corresponds exactly to microstate 1.

Although the microstates are uncoupled states, it doesn't mean they aren't antisymmetrised. By virtue of quantum indistinguishability, they have to be antisymmetrised: that's why most of the microstates are linear combinations of "composite states" such as $|p^+(1)\alpha(1)p^0(2)\alpha(2)\rangle$ (such a state on its own is not physically permissible, whereas the microstates are physically permissible states).

For an uglier but more instructive case, let's look at $|S,M_S,L,M_L\rangle = |1,0,1,+1\rangle$.

$$\begin{align}
&|S=1,M_S=0\rangle \cdot |L = 1, M_L = +1\rangle \\
&\qquad = \frac{1}{\sqrt 2}\left\{|\alpha(1)\beta(2)\rangle + |\beta(1)\alpha(2)\rangle\right\} \cdot \frac{1}{\sqrt 2}\left\{|p^+(1)p^0(2)\rangle - |p^0(1)p^+(2)\rangle\right\} \\
&\qquad = \frac{1}{2} \left\{ |p^+(1)\alpha(1)p^0(2)\beta(2)\rangle - |p^0(1)\beta(1)p^+(2)\alpha(2)\rangle + |p^+(1)\beta(1)p^0(2)\alpha(2)\rangle -  |p^0(1)\alpha(1)p^+(2)\beta(2)\rangle \right\}
\end{align}$$

Notice the first two terms in this expansion describe an (antisymmetrised) state where one spin-up electron is in $p^+$ and one spin-down electron is in $p^0$. This corresponds to microstate 7. Likewise, the second two terms correspond to microstate 9. All in all, we can write:

$$|S = 1, M_S = 0, L = 1, M_L = +1\rangle = \frac{1}{\sqrt 2}(|7\rangle + |9\rangle)$$

Using this procedure you can work your way through the rest of the states in $^3P$ (and indeed, $^1S$ and $^1D$ too). As a final example, let's look at one of the five $^1D$ states:

$$\begin{align}
&|S = 0, M_S = 0, L = 2, M_L = 0\rangle \\
&\qquad = |S = 0, M_S = 0\rangle \cdot |L = 2, M_L = 0\rangle \\
&\qquad = \frac{1}{\sqrt 2}\left\{|\alpha(1)\beta(2)\rangle - |\beta(1)\alpha(2)\rangle\right\} \cdot \\
&\qquad\qquad\qquad \left[\frac{1}{\sqrt 6}\left\{|p^+(1)p^-(2)\rangle + |p^-(1)p^+(2)\rangle \right\} + \sqrt{\frac{2}{3}}|p^0(1)p^0(2)\rangle \right] \\
&\qquad = \frac{1}{\sqrt{12}}\left\{ |p^+(1)\alpha(1)p^-(2)\beta(2)\rangle - |p^-(1)\beta(1)p^+(2)\alpha(2)\rangle \right\} \\
&\qquad \qquad + \frac{1}{\sqrt{12}}\left\{ |p^-(1)\alpha(1)p^+(2)\beta(2)\rangle - |p^+(1)\beta(1)p^-(2)\alpha(2)\rangle \right\} \\
&\qquad \qquad + \frac{1}{\sqrt{3}}\left\{ |p^0(1)\alpha(1)p^0(2)\beta(2)\rangle - |p^0(1)\beta(1)p^0(2)\alpha(2)\rangle \right\} \\
\end{align}$$

The first line of this expansion corresponds to $(\sqrt{1/6}|8\rangle)$; the second to $(\sqrt{1/6}|11\rangle)$; and the third to $\sqrt{2/3}|14\rangle$. So, your expansion was correct; however, be careful, because the uncoupled representations of the spatial wavefunctions

$$|m_{l_1} = +1, m_{l2} = -1\rangle$$

do not themselves correspond to the microstates. Instead, it is only by multiplying this with the spin wavefunction can you obtain a wavefunction that corresponds to the microstates.


Is there a pressure dependent function for freezing points of water and carbon dioxide? I'm trying to determine whether or not certain compounds form solid ice at certain atmospheric pressures. These pressures vary significantly, from 0.001 atm to 800 atm. I understand that there is no equation relating the freezing points of all elements, but is there one or two that relate to carbon dioxide and water?

Ideally I'd like a Python function (or library) or equation where the input is atmospheric pressure or kPa.

Or, failing that, if someone could point me in the direction of a phase diagram for both compounds where I can input specific atmospheric pressure values and read a specific temperature for freezing point, that would be helpful as well.

For ice I melting pressure as a function of temperature:

$$\pi = 1- 0.626000 \times 10^6 (1- \theta^{-3}) + 0.197135 \times 10^6 (1- \theta^{21.2})$$

where

$\pi =$ (pressure in MPa)/0.000611657 MPa

$\theta =$ (temperature in kelvins)/273.16K

Source is International Equation of the Pressure along the Melting and along the Sublimation Curve of Ordinary Water substance (no paywall).

For carbon dioxide:

$$\frac{p_m}{p_n} = 1 + a_1 (\frac{T}{T_t}) + a_2 (\frac{T}{T_t})^2$$

Where:

$T$ is temperature in kelvins

$T_t =$ 216.592K

$p_m$ is the melting pressure in MPa

$p_n =$ 0.51795 MPa

$a_1 =$ 1955.5390

$a_2 =$ 2055.4593

Source is A New Equation of State for Carbon Dioxide... (no paywall)

Check original sources before using equations in case I made typo(s).


Side Products of Dow's process In Dow's process of manufacture of phenol, $\ce{PhCl}$ is fused with $\ce{NaOH}$ at high temperature and pressure. What are the possible side products? And what is the intermediate formed?

I think the intermediate will be benzyne. After the formation of benzyne, $\ce{OH-}$ nucleophilic attack takes place. the reaction mixture also contains some phenoxide ions. So Diphenyl ether is a possible side product (Formed by nucleophilic attack of $\ce{PhO-}$). But the answer given species two side products: Diphenyl ether and p-phenyl phenol. How can I explain the second product?

Before we can consider the reaction, we must consider the conversion of the reactants into the reacting species:

Now, we may consider the reaction occurring between these two species:

The product of this reaction is now easily converted into para-phenyl phenol by an acid workup.


Permanent suspension of cocoa powder into water I have failed in permanently suspending cocoa powder into water.  I have tried heat (212'f) and pressure (40 psi) into a sugar water solution (50/50).  Also, powdered milk and evaporated milk have failed to yield a permanent suspension.  There was another method of Irish moss to coagulate the whole thing together, but that failed as well.  I need the bonding/ emulsifying agent to be edible, delicious and shelf stable.  I am a low level cook and do not have the chemistry knowledge to do this myself.
Thank you for reading.

Sounds like you need a food-grade surfactant. Maybe you could try lecithin?

The problem is that cocoa powder is mostly made of stuff that is not soluble in water (hydrophobic). So, to create the emulsion, you will need to "package" the powder molecules in something they are soluble in. You could either use a surfactant directly, or build up in layers using fat molecules and proteins, with a surfactant around the whole package.

Lecithin would be a good starting choice since it is FDA approved and used in lots of products already.

I would try this:

Add lecithin in gradually larger amounts until the suspension worked (you will have to try different heat/pressure and emulsification treatments - consider high-pressure homogenization as well.)
Add some whole milk (we are looking for the fat molecules and proteins) and see if you can find an acceptable milk/lecithin ratio that tasted good and that was still stable.

If that doesn't work, your other option is to modify the cocoa powder granule surface so that it is hydrophilic. That means you will need to add polar functional groups. One way to do that is with an acid (citric, for example), but you will add a sour taste which might be hard to get rid of. There is a patent for an approach using enzymes, but I don't know which one they used (it just refers to "starch degrading," "protein degrading," and "cell-wall degrading" enzymes), or how well it works.


Mechanism of pseudo buffer pH maintenance Some solutions, like concentrated HCl, act as buffers even though they lack the buffering component. What is the cause for such a behaviour?

I do wonder what your question is, but I have a hunch. Assuming that what you’re asking is essentially:

  In a concentrated $\ce{HCl}$ solution, the $\mathrm{pH}$ will more or less be the same value even if we add external $\ce{H+}$ or $\ce{OH-}$. Why?

Then there are multiple effects at work:

Aquaeous $\ce{HCl}$ contains a buffering component — the $\ce{H3O+}$ (or closest relative) ion. It is the most acidic ion that can be present is solution, so it ‘buffers down’ the acidity of said solution to what the corresponding $\ce{H3O+}$ acidity would be.
$\ce{HCl}$ in itself is an acid, but at the same time $\ce{Cl-}$ is also a (very weak) base. For high total $\ce{HCl}$ concentrations, the ratio of $\ce{HCl}$ to $\ce{Cl-}$ is at equilibrium. If we add additional $\ce{H+}$ to the mixture, we are effectively shifting that equilibrium back to the $\ce{HCl}$ side, reducing the proton count and thus buffering the $\mathrm{pH}$. Therefore also, if we add a proton scavenger (colloquially also known as ‘base’), then that will shift the equilibrium to the $\ce{Cl-}$ side, and as long as there is still enough $\ce{HCl}$ around, the total proton concentration won’t change notably.

In essence, everything that can act as a Brønsted acid or base will buffer at a specific $\mathrm{pH}$ value, even if that $\mathrm{pH}$ value is far from what one would typically call ‘buffered’.


How can bromine water be used to distinguish between benzoic acid and salicylic acid? I have come to learn that bromine water can distinguish benzoic acid from salicylic acid. I know that salicylic acid on reaction with bromine water forms 2,4,6-tribromophenol. I think that benzoic acid will form m-bromobenzoic acid. So with this respect how can bromine water distinguish the two compounds?

Salicylic acid is partially activated towards electrophilic aromatic substitution by the hydroxyl group, and slightly deactivated by the carboxylic acid. Overall, it is activated towards the substitution. Benzoic acid is deactivated by the carboxylic acid, so overall, it is deactivated towards substitution. Hence, salicylic acid will react without the need of a catalyst (and upon decarboxylation, the reaction will actually become faster), whereas benzoic acid needs a catalyst, such as $\ce{FeBr3}$, to react.

Hence, salicylic acid will react with bromine water, but benzoic acid won't. Physically, bromine water is orange due to $\ce{Br2}$, but if the bromine reacts, then the orange color will no longer be present. Thus, the solution will turn colorless upon contact with salicylic acid, and will stay orange in contact with benzoic acid. Furthermore, depending on the scale and concentrations, you may also observe the evolution of $\ce{CO2}$.


How to thicken Paint I'm non technical and have a question that you can hopefully help me out with as I've been searching online and so far haven't found any answered
I'm looking for a product that will thicken the following paint
polymarineshop - hypalon paint for avon inflatables
Any help at all would be greatly received
Best wishes
Ant

Diisocyanates are used in some coatings and polymerize in contact with moisture. Flexithane contains a non-reactive polyurethane (already reacted diisocyanate), and so is simply a solution of rubber in a mixture of solvents (SDS: http://www.hmgcoatings.co.uk/msdspdfs/Flexithane.pdf).
The literature claims dry to touch in 5 minutes, ready for a second coat in 20 minutes at 20C. One way to thicken the paint without adding anything that would destroy its performance would be to evaporate some of the solvent. In a can about half-filled with the paint, stir well, with plenty of ventilation (as advised by the SDS) for a few minutes. At £49.66 to £75 per 500 mL, it could be be an expensive experiment if you don't stir well enough and you get lumps or skin in the paint when you try to apply it.


Resonance of diazomethane I understand the diazomethane resonance forms that have two double bonds between $\ce{C=N}$ and $\ce{N=N}$ in the structure and also the form that has one single between $\ce{C-N}$ and triple between the two nitrogens ($\ce{N#N}$) in the structure.

Why is not possible for diazomethane to have a double bond between $\ce{C}$ and $\ce{N}$ as well as a triple bond between $\ce{N}$ and $\ce{N}$ as a resonance form?

As OP suggested there are three resonance structures of diazomethane contributing to the structure:

If you want to neutralize the formal charges of last resonance structure, you can put a double bond on negatively charged carbon and adjacent positively charged nitrogen to get the fourt resonance structure. However, traditionally we don't do that because it make the middle nitrogen hypervalent (or pentavalent nitrogen). On other word, it gave that nitrogen 10 valance electrons in its outer cell, which can't have in a 2nd row element.

It's just like resonance structures of nitrate ion, in which central nitrogen contains a full positive charge in all three resonance structures and two of three oxygen atoms contain full negative charges. So far, every one accept those resonance structures without questioning hypervalency of the central nitrogen atom until recently.

In 2015, Durrant has published his work on quantum mechanical calculations, suggesting possibility of hyper valency in diazomethane and some other possible molecules (Ref.1). According to its abstract:

  From the inception of Lewis' theory of chemical bonding, hypervalency has remained a point of difficulty that has not been fully resolved by the currently accepted qualitative definition of this term. Therefore, in this work, a quantitative measure of hypervalency has been developed. The only required input is the atomic charge map, which can be obtained from either quantum calculations or from experiment. Using this definition, it is found that well-known species such as $\ce{O3, \color{blue}{CH2N2}}$ and $\ce{ClO4−}$, are indeed hypervalent, whilst others such as $\ce{XeF4, PCl5}$ and $\ce{SO4^2−}$, are not. Quantitative analysis of known species of general formulae $\ce{XF_n^m−, XCl_n^m−}$, and $\ce{XO_n^m−}$ shows that there are no fundamental differences in chemical bonding for hypervalent and non-hypervalent species. Nevertheless, hypervalency is associated with chemical instability, as well as a high degree of covalent rather than ionic bonding. The implications for accepted Lewis structure conventions are discussed.

The experimentally determined atomic charges from electron densities obtained by X-ray crystallography have been compared with theoretical values obtained by QTAIM analysis of the output from quantum calculations. In order to calculate the overall electron count at the $\ce{X}$ atom, the author has defined a parameter called the valence electron equivalent, $\gamma$, as ‘the formal shared electron count at a given atom, obtained by any combination of valid ionic and covalent resonance forms that reproduces the observed charge distribution’. Mathematically:

$$Q(\ce{X}) = \displaystyle \sum_{i=1}^n{a_iq_i(\ce{X})}; \ \text{and then, } \ \gamma(\ce{X}) = \displaystyle \sum_{i=1}^n{a_i\gamma_i(\ce{X})}$$

where $Q(\ce{X})$ and $\gamma(\ce{X})$ are the overall charge and valence electron equivalent of atom $\ce{X}$, $q_i(\ce{X})$ and $\gamma_i(\ce{X})$ are the formal charge and electron count of each contributing resonance form, and $a_i$ are the coefficients required to reproduce $Q(\ce{X})$.

According to the paper:

  It follows that for any given atom $\ce{X}$, if $\gamma_i(\ce{X}) = 8$, the atom obeys the original Lewis octet rule. If $\gamma_i(\ce{X}) \lt 8$, the atom obeys the ‘modified octet rule’, but if $\gamma_i(\ce{X}) \gt 8$, neither form of the octet rule is obeyed and the atom is hypervalent. In such a case, no combination of formally charged Lewis octet-compliant structures can reproduce the observed charge distribution and it is necessary to include a contribution from an ‘expanded octet’ structure.

In the case of central nitrogen atom of diazomethane, $\gamma_i(\ce{N}) = 10$ ($ \gt 8$), thus it is considered to be hypervalent (Note: for nitrate ion: $\gamma_i(\ce{N}) = 8.30$ and hypervalent as well). The author has compared the results of $\gamma$ calculations on a variety of species including diazomethane, using quantum method 5 plus the QTAIM, Natural Bond Orbital (NBO), Hirshfeld, and Mulliken charge models. All of these methods agree with the findings, giving value of $\gamma_i(\ce{N})$ of diazomethane $10, 9.91, 9.78,$ and $9.56$, respectively.

References:

M. C. Durrant, "A quantitative definition of hypervalency," Chemical Science 2015, 6, 6614-6623 (https://doi.org/10.1039/C5SC02076J).


Anhydrous magnesium sulfate to replace epsom salt? Can anhydrous magnesium sulfate be used instead of epsom salt for feeding plants? Do hydrates and anhydrous salts become the same when dissolved in water?

Yes, they become the same, with some rare exceptions. There is usually difference in the dissolving kinetics, but the final result is the same.

Sometimes there are different thermal effects, Hydrates have sometimes tendency to cool the solution, while anhydrates to warm it. It is caused by released hydration energy of the anhydrate.The typical example is calcium chloride.

The obvious thing is to recalculate the needed amount for the contained water.


Arrows used in chemical reactions A point was raised about arrows used in a chemical reaction. I think we can all agree that $\ce{->}$ arrow in the reaction

$$\ce{A + B -> C} \tag{1}$$

means that the reaction goes in one direction. But what of

$$\ce{A + B <-> C} \tag{2}$$

$$\ce{A + B <=> C} \tag{3}$$

$$\ce{A + B <--> C} \tag{4}$$

$$\ce{A + B <=>> C} \tag{5}$$

$$\ce{A + B <<=> C} \tag{6}$$

Is there any IUPAC or American Chemical Society guideline about how double arrows are supposed to be used?

Examples of usage:

  (1) Used for example:
  "Removal of Triphenylphosphine Oxide by Precipitation with Zinc Chloride in Polar Solvents",
  Donald C. Batesky, Matthew J. Goldfogel, and Daniel J. Weix,
  J. Org. Chem., 2017, 82 (19), pp 9931–9936

  "Cation Radical Accelerated Nucleophilic Aromatic Substitution via Organic Photoredox Catalysis",
  Nicholas E. S. Tay and David A. Nicewicz*
  Department of Chemistry, University of North Carolina at Chapel Hill, Chapel Hill, North Carolina 27599-3290, United States
  J. Am. Chem. Soc., 2017, 139 (45), pp 16100–16104

  (4) Used for example:
  "Cation Radical Accelerated Nucleophilic Aromatic Substitution via Organic Photoredox Catalysis",
  Nicholas E. S. Tay and David A. Nicewicz*
  Department of Chemistry, University of North Carolina at Chapel Hill, Chapel Hill, North Carolina 27599-3290, United States
  J. Am. Chem. Soc., 2017, 139 (45), pp 16100–16104

Previous questions in Chemistry:

In his answer user Chemgoat noted that this question had been asked before:

What are the correct equilibrium arrows?

A comment on that question points to yet another

What is the difference between "reaction in both directions" and "equilibrium"?

The second one, $(2)$, $\ce{<->}$, indicates resonance structures.

The difference between the second two is explained in another post -

What are the correct equilibrium arrows?

Loong basically says that $\ce{<-->}$, $(4)$, indicates forward and reverse elementary steps happening in equilibrium, that is that those are the only two things actually reacting with each other. On the other hand, $\ce{<=>}$, $(3)$, indicates a net reaction, that is that there may be multiple steps in between, leading to this net equation. He does a more in depth explanation of this in his post.

Case $(5)$ indicates a net equilibrium where products are favored, and $(6)$ indicates a net equilibrium where reactants are favored.


Nitrogen dioxide dimerization $\ce{NO}$ can't dimerize and that is explained because the free electron is in an antibonding orbital. The same is true for $\ce{NO2}$, but it can dimerize, how?

Having an electron in an antibonding orbital means that the electron exists in a high energy orbital.  If anything, this means the electron will be "motivated" to react and form a bond, particularly if bond formation lowers the energy of the electron.  An electron in an antibonding orbital can form a bonding (as opposed to antibonding) bond with positive overlap.


why are halogens considered deactivating groups while methoxyl groups are activating? Something's wrong with how I think of this:

methoxyl has the partial negative charge on oxygen which means it's activating. However, halogens are more electronegative yet they're considered deactivating. Why?

On my lecture notes, I have one section that says
"if Q is an electron withdrawing group then attack on the ring is slowed because this leads to additional positive charge on the ring".

While another section says "all meta directing groups have either a partial or full positive charge on the atom directly attached to the aromatic ring."

Because fluorine has such a high electronegativity, its inductive withdrawing effect is much larger than any resonance electron-donating effect.

The other halogens have poor orbital overlap with the pi system of an aromatic ring, which precludes any appreciable resonance electron-donation (this is why you almost never see a C-X double bond). Therefore, their inductive capabilities lead to a net deactivation of the aromatic ring.

This is why an alkoxy group is more activating  than a halogen. Oxygen does have a higher electronegativity than does chlorine, bromine or iodine. However, oxygen's orbital overlap with carbon allows it to more readily donate into the ring's pi system. The larger halogens aren't able to do this (to an appreciable extent).


Can R-3-methyl-4-heptanone be enantioselectively synthesised from 4-heptanone? I wonder if it is possible to do the following reaction with its enantioselectivity:

I am trying to do it via enolate, but I don't know how to make only the R product or if it's even possible to distinguish between R or S simply using enolates.

There's another way?

Thanks!

Enders SAMP/RAMP hydrazone chemistry will do this(1).

The Wikipedia article on this technique is a good introduction:

(1): Corey, E. J.; Enders, D. (1976). "Applications of N,N-dimethylhydrazones to synthesis. Use in efficient, positionally and stereochemically selective C-C bond formation; oxidative hydrolysis to carbonyl compounds". Tetrahedron Letters. 17 (1): 3–6.


Do hybrid orbitals exist in unbonded molecules? What would they look like? For example, the ground state of a neutral carbon atom could be notated as:

$$ [\ce{He}] \underset{\ce{2s}}{[\uparrow \downarrow]} \underset{\ce{2p}}{[\uparrow \vert \uparrow \vert \; \;]} $$

I know $\ce{s}$ orbitals are shaped like spheres so the two electrons in the $\ce{s}$ orbital should probably form a spherical shape.

But how should the electrons in the $\ce{p}$ orbitals be distributed?

I think in the end result a neutral carbon atom is either shaped in a trigonal-planar arrangement or a linear arrangement.

So do hybrid orbitals exist in unbonded molecules? What would they look like?

I think this would have interesting results for collisions between atoms and how probable certain reactions are.

Hybrid orbitals do not exist. Individual atoms have electronic configurations which can be explained by considering atomic orbitals. Molecules have electronic configurations which can be explained by considering molecular orbitals. Hybrid orbitals are just one mathematical way of arriving at molecular orbitals from combining atomic orbitals. They are a pure mathematical fiction and have no physical meaning (this does not mean they aren't useful).

Some schools or textbooks teach that in order for a reaction to occur energy is required to 'promote' electrons into hybrid orbitals to create some sort of excited atom prior to bonding. This does not happen in real life and is not a useful way of looking at bonding. What does happen is that as atoms approach each other, their atomic orbitals start to overlap significantly, giving rise to a new set of molecular orbitals (in quantum mechanics you are changing the potential in which the electrons reside and so the allowed energy states change). If the electrons in the new molecular orbitals are lower in energy than in the atomic orbitals then the reaction will proceed.

As for the electron distribution of atoms - it is spherically symmetric. The two $\ce{2p}$ electrons will reside in different orbitals because this does not require pairing of their spins (which is a higher energy state). However, it is meaningless to ask which $\ce{p}$ orbital they are in because all three orbitals are degenerate and so are indistinguishable.


For a copper/copper sulfate half reaction, do sulfate anions move to the anode and lose electrons? In my (high school) textbook, there's an example on finding the charge on 1 mole of electrons, which involves performing the electrolysis of an aqueous copper (II) sulfate using a copper anode and a copper cathode. I've been given to understand that in electrolysis, the negative ions move to the anode, and lose electrons to the anode.

However, in this example, the copper anode loses mass as the copper goes into solution as copper ions. What I don't understand is, since the sulfate ions in this case would go to the anode, wouldn't it combine with the copper ions to form copper sulfate? Or wouldn't it donate two electrons to the copper ion so it becomes copper metal again? Since the sulfate ion moves to the anode, shouldn't this happen?

I've been given to understand that in electrolysis, the negative ions move to the anode, and lose electrons to the anode.

This is one way of achieving charge balance, but there are many other ways.

Charge balance of half-reactions

Half-reactions have electrons either as reactants (reduction half-reaction at the cathode) or as products (oxidation half-reaction at the anode). For the cell in question, the oxidation half-reaction at the anode is:

$$\ce{Cu(s) -> Cu^2+(aq) + 2e-}$$

There is no physical state associated with the electrons, but they are removed through the wire. So if electrons appear in the half-reaction, you need some ions to balance it out. In a oxidation, we produce electrons, so either we need to produce cations at the same time or consume anions (or have ions on both sides of the equation, but of unequal charge). The ions (in solution) are also responsible for the charge transport between anode and cathode.

Oxidation state imbalance of half-reactions

At least one atom (as element, simple ion, or part of a compound or complex ion) has to undergo a change in oxidation state. The species that balance the charges, however, might or might not undergo a change in oxidation state. In the half-reaction in question, copper changes oxidation states, and the copper ions balance out the charge of the electrons so that both sides of the half-reaction have equal charge (zero, in this case).

To give a different example, here is a half-reaction involving lead:

$$\ce{Pb(s) + HSO4^-(aq)-> PbSO4(s) + H+(aq) + 2e-}$$

Here, the lead gets oxidized (from element to +II), but remains at the electrode. The charge balance is achieved by hydrogen sulfate ions traveling to the electrode and remaining at the electrode as part of the solid lead(II)sulfate, and the released hydrogen ion.

Direction of ion flow

In the reaction in question, copper cations travel from anode to cathode in the solution to balance the charge transport due to electrons traveling from anode to cathode via the wire. We can write the equation showing this explicitly by combining the half-reactions and keeping anode and cathode species labeled.

$$\ce{Cu(ano) + Cu^2+(cat) + 2e-(cat) -> Cu^2+(ano) + 2e-(ano) + Cu(cat)}$$

Because we can't have electrons or copper ions accumulate in or at the electrodes, we need a transport mechanism for both species (through the wire and through the solution, respectively).

Here are two more examples with different ion transport.

Water electrolysis

$$\ce{4H+(aq) + 4e- -> 2H2(g)}$$
$$\ce{2H2O -> O2 + 4H+(aq) + 4e-}$$

At low pH, the cathode consumes hydrogen ions while the anode produces hydrogen ions. At neutral pH, you would say the cathode consumes hydrogen ions and the anode consumes hydroxide ions produced from auto-dissociation of water (i.e. anions travel to the anode and cations travel to the cathode - the classic case giving anions and cations their name).

Lead-acid battery

$$\ce{Pb(s) + HSO4^-(aq)-> PbSO4(s) + 2e- + H+}$$
$$\ce{PbO2(s) + 3H+(aq) + HSO4^-(aq) + 2e- -> PbSO4(s) + 2H2O}$$

The element that changes oxidation state is lead in various solid compounds (elemental, lead(IV) in lead oxide and lead(II) in lead sulfate). The ion transport is due to hydrogen sulfate going from solution to both electrodes, forming lead sulfate, and hydrogen ions going to the cathode to combine with oxygen to form water (and hydrogen ions being released from hydrogen sulfate at the anode). So here, ion transport is accomplished with anions going to both electrodes, but an excess of cations (hydrogen ions) going to the cathode. To emphasize, hydrogen sulfate anions are traveling to the cathode. This is an counter example to "anions travel to the anode".

Cells with salt bridge

None of the cells in my three examples had salt bridges. If half-reactions have separate solutions connected by a salt bridge, ions don't travel the entire way between electrodes; instead, redox-inert ions such as potassium cations and chloride anions negotiate the charge transport between solutions of the half cells.


Is ciprofloxacin acidic or basic? I would expect an aqueous solution of ciprofloxacin to be basic for the following reasons:

The carboxyl group has pKa = 6.09 and the secondary amino group has pKa = 8.74. Because 8.74 is further from neutral pH than 6.09, I would expect the pH of an aqueous solution to be above 7.
The tertiary amino groups are connected to an aromatic ring, so they should be very weakly basic and not affecting pH much.

However, the PubChem Compound Summary says the following about ciprofloxacin: "pH range for the 1.0% aqueous concentrates in vials is 3.3 to 3.9; the pH range for the 0.2% ready-for-use infusion solutions is 3.5 to 4.6" (see https://pubchem.ncbi.nlm.nih.gov/compound/ciprofloxacin#section=pH)

Is an aqueous solution of ciprofloxacin solution really acidic and if it is, what is the reason?

I also found an article citing several estimates of the pKa:s and the consensus seems to be that the amino group is stronger than the carboxyl group. (Babić et al. Determination of pKa values of active pharmaceutical ingredients. Trends in Analytical Chemistry, Vol. 26, No. 11, 2007.)

Here is a picture of ciprofloxacin from Wikipedia:

Your predictions are right; the molecule has more basic character.

The drug itself is practically insoluble in water, because the big organic molecule is near it's isoelectric point at the pure water's pH. However by protonating or deprotonating it and thus making it charged it's solubility is greatly enchanced.

This is also the case with those medical ready-for-use infusions:

  ...CIPRO IV solutions are available as sterile 0.2% ready-for-
  use infusion solutions in 5% Dextrose Injection. CIPRO IV contains lactic acid as a solubilizing agent and hydrochloric acid for pH adjustment. The pH range for the 0.2% ready-for-use infusion solutions is 3.5 to 4.6.

Reference


Type of reaction between (+)-bis salt of (+)-phenylsuccinic acid and HCl Im confused about what type of reaction occurs between the (+)-bis salt of (+)-phenylsuccinic acid and HCl. Would this be considered an acid base reaction or a neautralization reaction?

MY THOUGHTS: I know that an acid-base reaction occurs between (+)-phenysuccinic acid and (-)-proline to make the (bis)+ salt, im confused at to whether such salt is an acid or a base, or neither... If its an acid and an acid, it would be a neutralization reaction mechanism. If it were a base, it would follow an acid-base reaction mechanism. Any thoughts on if either of these make sense based on such info?

The reaction between the bis salt of proline and phenylsuccinic acid will give free phenyl succinic acid and proline hydrochloride as HCl is a much stronger acid than phenylsuccinic acid. All you are really doing is exchanging one acid for another so it is hardly an acid/base reaction.


How to activate polystyrene with carboxyl group? Is there a simple (that is one that requires few and not very exotic reagents) chemical reaction to "activate" or "functionalize" the polystyrene of a simple microtitre plate?
In other words, I would like to add carboxyl groups to the surface of a common ELISA plate to then use ethyl-carbodiimide hydrochloride and N-hydroxysuccinimide to immobilize a target on the bottom of the well.

It is certainly possible to nitrate polystyrene powder with nitric & sulfuric acid mixture and reduce the nitro groups to aniline with $\ce{SnCl2/HCl}$ as described in here1. How effective this would be for a microtitre plate I cannot say, but chemistry is an experimental subject.
Reference

Zhao, H. & Guo, J. & Bian, L. & Wei, Z. & Deng, K.. (2009). Synthesis of nitro and amino linear polystyrene. Hecheng Shuzhi Ji Suliao/China Synthetic Resin and Plastics. 26. 22-26. (link)


What are the best cleaning agents for carbon residues on glassware? Introduction

Recenly I wanted to isolate some ascorbic acid (Vitamin C or chemically - $\ce{C6H8O6}$). I've put the filtered solution (I had to filter it because of the binder) in a flask and heated it. It didn't go as planned because then I didn't know that ascorbic acid decomposes at high temperatures, but I still decided to keep the product of the reaction. I left it on a hotplate at a high temperature and it all went back to carbon because it was there for too long (It was dehydrated).

The cleaning agent that I know of - $\ce{NaOH}$

I used a saturated solution of $\ce{NaOH}$ to clean it but there were still some flakes stuck on the wall that I couldn't reach.

It worked well for me - cleaned ~90% of the residue.

The question

Are there any other cleaning agent that'll work better than aqeous Sodium Hydroxide?

Are there any other cleaning agent that'll work better than aqeous sodium hydroxide?

Yes, but I would not recommend them to anybody running experiments at home!

Back in the days in the lab we used arm-deep tanks filled with a saturated solution of potassium hydroxide in isopropanol to soak dirty flasks. You need long (!) chemical-resistant gloves and once in a while, the whole mess has to be discarded and replaced. Down the sink is not an option ;-)

And there is piranha solution, a mixture of 30% hydrogen peroxide and concentrated sulfuric acid. While is is effective, it is infamous for going off with a bang, see this report.

To sum it up: Stay with the sodium hydroxide solution. Don't risk your health and don't screw up your own environment to rescue a spoiled flask.


Representation of chemical formulas as strings? Is there a standard manner of representing chemical formulas as linear strings?

For example, $H_2O$ can be represented with the string H2O. Isotopes are usually specified with a number before the element. But this must be preceded by a space, or it can be get mixed with the number right before: O4 13CH, which is equivalent to $O_4{^{13}C}H$.

I'm just wondering if there is any set of accepted conventions when representing chemical formulas as linear strings.

If I understand your question correctly, the simplified molecular-input line-entry system (SMILES) and the IUPAC International Chemical Identifier (InChI) are two examples of "human-readable" string notations for chemical formulas your are looking for.


How exactly are molecules detected in space? Recently the sugar glycoaldehyde was detected in a star system 400 light years from Earth. How exactly are molecules detected in space? I am aware that spectroscopy is used to detect them, but I don't know the details.

The research article corresponding to the observation to which the question is referring is Detection of the simplest sugar, glycolaldehyde, in a solar-type protostar with ALMA. Note that the molecule is glycolaldehyde rather than glycoaldehyde.

ALMA is the Atacama Large Millimeter/submillimeter Array, a radio telescope array in the Atacama desert.  The extremely low humidity of the Atacama desert minimizes water vapor interference.

This observation of glycolaldehyde in space is not the first.  It was earlier observed by Greenbank radio telescope, in West Virginia, USA, for example.  The Greenbank observation is reported in INTERSTELLAR GLYCOLALDEHYDE: THE FIRST SUGAR.  Glycolaldehyde was detected through radio signals in the 71-103 GHz range that are photons emitted when the molecule transitions from an excited rotation state to a lower energy rotational state.

In the ALMA observation, the molecule was detected based upon radio signals near 220 GHz and 690 GHz.  These correspond to photons emitted by the molecule when transitioning from an excited rotational state to a lower energy rotational state, but are a different set of transitions from those observed at Greenbank, including some for excited vibrational states.

I'll try to add more information later, there are a wide variety of spectroscopic techniques: absorbance vs. emission; rotation, vibration and electronics transitions, and many different frequency ranges.


Inflection point, slope, and equivalence point in a titration My textbook, Harris 7th Edition Quantitative Chemical Analysis, says:

  In titrations involving 1:1 stoichiometry of reactants, the equivalence point is the steepest point of the titration curve. This is true of acid-base, complexometric, and redox titrations as well.

My question is, does this apply to all acid-base titrations (strong-strong and weak-strong)? My intuition seems to say that it only applies to strong-strong, but I may be incorrect.

Your intuition is not sufficient in this matter. The equivalence point is the steepest point of any titration curve, however weak or strong is the acid and the base.


How to determine the structure of organic molecules without spectroscopy I wonder how chemists worked out the structures of organic molecules before spectroscopy (especially NMR) was invented?

Surely, we could do functional group tests, but the correct connection of atoms and function groups would be hard to tell when the number of carbon atoms get larger. E.g. How could they tell one compound was 2-hexanol rather than 3-hexanol? How could Emil Fischer and others worked out the correct structures of carbohydrates with so many chiral centers? And god forbids, how could the chemists worked out the correct structures of natural products with fused rings (eg the cholesterol family)? It's not a trivial task even with 2D NMR!

One thing that was well known, even in the early days, was elemental analysis. It is Justus von Liebig’s contribution to have worked out the sum formulae of hundreds of natural compounds. This generally worked by combusting a given sample of a compound, capturing the vapours, absorbing the water with desiccants such as calcium chloride, absorbing carbon dioxide by potassium hydroxide (forming a carbonate), weighing the different parts and thereby calculating the mass of carbon and hydrogen in a given sample. Tests for nitrogen and sulfur also existed, but I can’t remember them off the top of my head. Oxygen content was indirectly calculated as being the remainder.

Chemists since the advent of their science had done everything to generate pure compounds. It was widely accepted even then that a pure compound was one with a defined melting or boiling point, and liquids could also be analysed by their refractive index and density. This provided a rather good library of methods for realising ‘is my compound pure?’, and ‘is it the same as $\ce{X}$?’

Another very old concept was that of valencies. By analysing samples of known and simple elemental composition, it was realised that certain elements bonded to a certain number of other elements, e.g. that a hydrogen would always require a bonding partner of value 1. Somebody (I can’t remember who but I could look it up in my lecture notes) proposed a valency of 4 for carbon. Together with a valency of 2 for oxygen, you had most pieces of the puzzle in hand.

It was recognised by Liebig and Wöhler, that isomers existed, i.e. compounds of the same elemental composition but distinct. (This is also a consequence of the ammonium cyanate experiment Wöhler performed to create urea, breaking the wall between organic and inorganic chemistry and proving vis vitalis to be inexistent — both are isomers of each other.) So there must have been something important in the structure of molecules that needed to be discovered.

A lot of further clues then came from reactions that transformed known compounds into other known compounds. For example, oxidation would transform some known compound $\ce{C3H8O}$ into acetone $\ce{C3H6O}$. This also lead to Liebig’s theory of radicals — where radicals is something a modern chemist would refer to as functional group. With these clues and radicals, the task of putting the puzzle together to pinpoint certain structures became solveable.

From then on, it all depended on whether the initial assignments were correct (they mostly were). You would go ahead either breaking down unknown compounds by known reactions after having established their functional groups or resynthesising them from known fragments in other known reactions to arrive at a final product.

Working out the difference between hexan-2-ol and hexan-3-ol would have been a none-trivial task; that is true. One variant seems to be starting from hex-2-ene and adding water to the double bond. You would get two different products and with the knowledge of structure in hand, you would realise that they had to be isomers of each other and had to be the 2- and 3-isomer respectively. How did you get hex-2-ene? Oh, maybe you managed to dehydrate hexan-2-ol into two different products, hex-1-ene and hex-2-ene. How did you get hexan-2-ol in the first place, then? Addition to hex-1-ene sounds reasonable. How to get that? Dehydration of hexan-1-ol. That final one was easily established as a special hexanol, since it could be oxidised to hexanoic acid, not only to a ketone. It is all possible but requires a lot of work. Think how lucky we are that we can record an NMR spectra within minutes (or a complete set over night) and solve the structure the next day!

It is entirely possible that for the cholesterol family, the exact ring connections were initially not known, only that there is some type of a fused ring system that can be modified in certain positions (not all!). It would have taken extra time to break down the ring system accordingly. And Emil Fischer’s work on carbohydrates probably warrants one or two questions of its own.

And then, crystallography happened, and all of a sudden was it possible to see structures. It must have seemed like the dawn of a new era.


Effect of pKa value of acid component in Mitsunobu Reaction? Mitsunobu Reaction is used to established C-S, C-C, C-N, C-O bond formation by condensation of acid component (Nuclephile) with primary and secondary alcohol in the presence of triphenyl phosphine (Ph3P) and diisopropyl azadicarboxylate (DIAD).

Literature suggest that Mitsunobu reaction is only possible if The nucleophile (or pronucleophile) is normally a relatively acidic compound containing an O-H,
S-H, or an N-H group with pKa<15, preferably below 11.

My question is that, why Mitsunobu reaction can't occur with acidic compound having Pka value more than 15 (pKa>15)?

I figured out the answe

The desired Mitsunobu reaction proceeds through path (a). On the contrary, in case of the reaction with less acidic HA (pKa>15) , the hydrazo anion (4) attacks the alkoxyphosphonium (5) directly to afford alkylated the hydrazine derivative (6) as a by-product path(b).

Since the anion (4) is not efficient in deprotonating the weakly acidic HA. In order to overcome these drawbacks, the acidic component should have pKa<15 to get desire product.


How to calculate the frequency of cyclohexane conformational interconversion? My book mentioned that the energy barrier for cyclohexane to reach the half-chair conformation is $50.6\ \mathrm{kJ\ mol^{-1}}$. It says that from this value, it was calculated that cyclohexane undergoes $10^5$ flips per second.

How did they calculate this? We can assume that all the conformational states possible will be of the same energy as the chair conformation (so they all encounter the same energy barrier).

I'm more interested in the principles behind the calculation – why the equation works – than just seeing the calculation done.

Look up application of the rate of reaction equation given barrier height and energy available at room temp, $kT$ (Boltzmann's constant).

$k = A \mathrm e^{ - \frac{E_\mathrm a}{RT} }$

http://research.cm.utexas.edu/nbauld/teach/cyhexconf.jpeg

Twist-boat leads to mathematically perfect geometric chirality,
Symmetry: Culture and Science 19(4) 307-316 (2008).


Lewis structure of Cl2SO Why is it that the Lewis structure of $\ce{Cl_2SO}$ has a total of $24$ valence electrons? I thought that the number of valence electrons should be $2\times7 + 1\times6 + 1\times6 = 26$.

Sulfur is sometimes described as a hypervalent atom, or one that has more than eight valence electrons. This is possible to some extent; sulfur can indeed utilize its d-orbitals for bonding. However, recent quantum mechanical calculations suggest that the extent of this utilization is negligible. Therefore, your book is most likely representing the $\ce{Cl_2SO}$ molecule as a "charge-separated" molecule.

EDIT: Upon further thought this still makes no sense. Charge is conserved and thionyl chloride should always have 26 electrons (no matter how they are distributed). So perhaps, due to the high degree of ionic character in the $\ce{S-Cl}$ and $\ce{S-O}$ bonds, the missing two electrons are "dispersed" among these ligands. (This is speculation). Perhaps your book is genuinely in error. Perhaps they did not count a lone pair on the central sulfur atom or otherwise miscounted the number of electrons?

EDIT 2: Here's a picture of two possible Lewis structures of thionyl chloride; as we can see, the book agrees that the thionyl chloride molecule is better represented as a "charge-separated" molecule. Nonetheless, all the electrons are present.

EDIT 3: Here's a picture of thionyl chloride which makes it seem as if the molecule only has 24 valence electrons. NB: the lone pair is still there and its presence is implied through the lack of any indication of a non-zero formal charge on the sulfur atom.


Macgyvering a Spectroscope I am currently playing a dungeons and dragons campaign in which there are some new elements. My party, being the nerds that we are, is looking to experiment with them and attempt to discover new ones using a spectroscope. The problem is that we need a solvent that will dissolve our sample. I was wondering if there were any suggestions as for what to use, or how to obtain it, as the game takes place in a medieval time period. Acetone would be best, but I'm not sure that it's possible to extract that from nature, although it does occur naturally. Thanks!

The medieval "nerd" in a D&D campaign might use acids to get minerals into solution.

Sulfuric acid ($\ce{H2SO4}$, vitriol oil) was initially prepared by heating iron(II)-sulfate ($\ce{FeSO4 · 7 H2O}$, iron vitriol) in a retort. The emerging sulfur trioxide was collected an led into water.

Nitric acid ($\ce{HNO3}$) was prepared by heating mixtures of potassium nitrate ($\ce{KNO3}$, saltpetre) with iron vitriol.

Hydrogen chloride ($\ce{HCl}$) was available by the reaction of vitriol oil with halite ($\ce{NaCl}$).

Mixing hydrochloric acid with nitric acid would then give aqua regia.

Another method to get the cations of minerals with low solubility in water into solution is the so called Aufschluß.

Here, one would either melt the mineral with

a mixture of soda and potash, or
a mixture of sulfur and potash

in a crucible and eventually leach out the now soluble material with water.

Admittedly, our D&D alchemist still doesn't have a spectroscope :\

In order to characterize the metals in the samples, he could use his Lötrohr, a kind of blow pipe and blow the flame of his oil lamp onto the sample place on a piece of charcoal or a gypsum plate. From the colour of the flame and/or the colour of the residue formed, he could draw some conclusions on the metal.

If our medieval science hero is a bit ahead of his epoch, he would also carry some borax in a bag and create borax pearls for a bead test.


What do these mineral classification symbols mean? For my birthday I got an awesome mineral specimen case, as pictured below. What piqued my interest was the notations in the lid where the minerals are described with 5 specifications: C (obviously for Colour) then S, H/SG (I'm guessing specific gravity), L and CL.

Can anyone shed any light on what these abbreviations mean? Additionally, do the coloured squares mean anything or are they just for identification (they match colours painted on some of the samples).

In addition to being a non-practicing chemist, I'm also a non-practicing geologist, so look no further for an answer you can trust.

Your minerals are classified as follows: C is for color (sample, gross), S is for streak (color of powdered mineral, essentially, on an unfinished streak plate), H is hardness on the Mohs scale, SG is specific gravity, L is luster, and CL is cleavage (how the sample breaks along planes dictated by the crystal habit of the mineral). See this reference for a printable chart.

The color-coding is certainly used to mark the samples in case they get jostled or mixed up and need to be matched to their appropriate slots: those codes are not related to the mineral classification.


Can I change viscosity of an HDPE polymer? I want to make composite with HDPE and wood, but when I melt it I have problem to mix them, because this polymer have high viscosity.

If you use an extruder to prepare your composite material, try to use HDPE with a melt flow index: MFI of 0.1 g/10 min.


Can in any case the faster step of the reaction be rate determining? I found a sentence in book which states the reverse step of this reaction ( I forgot what was it!)  has the faster step as rate determining .

Even Rate determining step-Wikipedia  states:

  In chemical kinetics, the rate (or velocity) of a reaction mechanism with several steps is "OFTEN"  determined by the slowest step .

which I think  'OFTEN' means not 'everytime' so,

if faster step of the reaction is rate determining. Then why it is?

I can provide you an example.
The oxidation of formate ion by peroxydisulfate in water solution:
$\ce{HCOO^{-} +  S_2O_8^{2-} -> CO_2 + 2SO_4^{2-} + H^+}$
has the following mechanism:

As you can see, the first step is the slowest, but by using the rate-determining step approximation you wouldn't arrive at the correct rate law which is: $r=k[\ce{HCOO^-}]^{1/2}[\ce{S_2O_8^{2-}}]$. The  first  reaction  is  very  slow,  so  most  of  peroxydisulfate  is consumed  in the third reaction. The correct rate law can be obtained by applying the steady-state approximation to the two radical species.
(This isn't the true mechanism but it is good for showing the point. A  more complex  mechanism  includes  the  formation  of  OH  radicals  and  several  chain termination reactions. That’s why the given rate law is valid only for a limited range of reactant concentrations.)


Are there compounds that melt under their own weight? Is there a compound that can melt under its own weight? That is, a block of such material remains solid when put horizontally but starts to melt when put vertically or remains solid in microgravity, but melts in normal gravity?

The question is motivated by a documentary which I saw on TV some years ago where they claimed that a metal block (indium or its alloy) melts under its own weight. It is suspicious though because it was quite a crank documentary which featured alleged "red mercury" for instance.

Sure. Even discounting non-equilibrium conditions, any substance which expands upon freezing will by symmetry melt under pressure, even if the pressure is caused by supporting its own weight. There's a simple example: water! A sufficiently tall self-supporting block of ice standing on a flat surface will create enough pressure at its bottom that it could melt in temperatures between approximately $-21.9\ ^\circ \mathrm{C}$ and $0\ ^\circ \mathrm{C}$, according to the phase diagram for water. This has very important implications for the study of glaciers, for example. However, this process requires a very large amount of pressure, and thus a very tall block of ice. Using an approximate value for the freezing point depression under pressure of $0.0072\ ^\circ \mathrm{C\ atm^{-1}}$, it would take an ice block about $7.5\ \mathrm{km}$ tall for the base to melt at a temperature of $-5\ ^\circ \mathrm{C}$ or higher.

I thought indium was among the substances which expanded upon freezing, but it actually turns out to contract by 2.5% when frozen, so this mechanism would not work (though there could be something else to consider).  Some other materials which could theoretically melt under their own weight in some regime of temperatures and pressures are gallium, bismuth and silicon. There might be some very specific conditions under which this effect could be achieved in a column of around $20\ \mathrm{cm}$, though pressure usually only slightly alters the freezing point, so it's likely that the material would have to be kept only slightly under its melting point at standard pressure for this to happen.


Figuring out formula of a hydrocarbon from amount of products obtained and reactants used One mole of a hydrocarbon is combusted. The products obtained are cooled down to STP and occupy a volume of $89.6\ \mathrm l$. Oxygen required for combustion was $145.6\ \mathrm l$ at STP. Find the molecular formula of the hydrocarbon.

Let the hydrocarbon be $\ce{C}_x\ce{H}_y$. The combustion reaction can be written as:
$$
\ce{C_xH_y + O2 -> H2O + CO2}
$$
Keep in mind that this is not yet balanced.
We can now find out the amount of reactants and products. Since one mole of any gas occupies $22.4\ \mathrm l$ of space at STP, we can say that we have:
$$
\frac{89.6\ \mathrm l}{22.4\ \mathrm l/\mathrm{mol}} = 4\ \mathrm{mol}
$$
of products. As for the oxygen reacted:
$$
\frac{145.6\ \mathrm l}{22.4\ \mathrm l/\mathrm{mol}} = 6.5\ \mathrm{mol}
$$.
We can now deduce that the hydrocarbon will react with $13\ce{O}_2$. Therefore, the equation becomes.
$$
\ce{C_xH_y + 13O2 -> H2O + CO2}
$$
But, There are $x$ $\ce{C}$s one the left hand side so the right hand side also has $x$ $\ce{C}$s:
$$
\ce{C_xH_y + 13O2 -> H2O + xCO2}
$$
There are $y$ $\ce{H}$'s on the left hand side. Same for the right hand-side:
$$
\ce{C_xH_y + 13O2 -> y/2H2O + xCO2}
$$
Since the amount of $\ce{O}$ has to balance as well, we have $26$ oxygens on the LHS and therefore $26$ on the RHS. Which means:
$$
26 = \frac{y}{2} + 2x \implies \frac{y}{2} = 26 - 2x
$$
We have already figured out that $4\:\mathrm{mol}$ of products are obtained. i.e.:
$$
4 = \frac{y}{2} + x = 26 - 2x + x = 26 - x \implies x = 26 - 4 = 22
$$
But this would mean that $y$ is negative. Where did I go wrong?

From your interpretation I have to say that giving the stochiometric coefficent of $13$ is wrong because this number is not related to the molecules of the species (therefore the suspect that you multiplied the moles obtained by two is where lies the main error).

Since we are talking about a combustion, we can safely assume that every reagent is completely oxidised, so we can write the complete reaction with the generic coefficients:
$$
\ce{C}_{\color{blue}{x}}\ce{H}_{\color{blue}{y}(g)}+\left[\color{blue}{\frac{y}{4}}+\color{blue}{x}\right]\ce{O2_{($g$)}}\to\color{blue}{\frac{y}{2}}\ce{H2O}_{(g)}+\color{blue}{x}\ce{CO2_{($g$)}}
$$
Note that the rule of given saturated hydrocarbons implies that $y=2x+2$, but this is left intentionally generalized because the presence of specific double/triple bonds is unknown.

The purpose of writing this equation is to specify the relation between the reactants, so now we can assume the molar relation between the oxygen needed to the first hydrocarbon to obtain a simple equation in terms of $x$ and $y$:
$$
n_{\ce{O2}}=\left[\frac{y}{4}+x\right]n_{\ce{C}_{x}\ce{H}_{y}}
$$
Since, as you calculated, there are needed $6.5\:\mathrm{mol}$ of oxygen to oxidise $1\:\mathrm{mol}$ of the unknown hydrocarbon; we can substitute those values inside the above equation, giving:
$$
6.5=\frac{y}{4}+x\tag{1}
$$
After the complete combustion, the total amount of moles is indicated as you derived correctly, and there can be expressed in relation to the unknown reagent, and then added together to obtain the second equation as follows:
$$
n_{\ce{H2O}}=\frac{y}{2}n_{\ce{C}_{x}\ce{H}_{y}}\quad n_{\ce{CO2}}=x\:n_{\ce{C}_{x}\ce{H}_{y}}
$$
Since the total volume is additive, then (because $\ce{C}_{x}\ce{H}_{y}=1:\mathrm{mol}$):
$$
4\:\mathrm{mol}=n_{\ce{H2O}}+n_{\ce{CO2}}=\frac{y}{2}n_{\ce{C}_{x}\ce{H}_{y}}+x\:n_{\ce{C}_{x}\ce{H}_{y}}=\frac{y}{2}+x
$$
It now produces the second needed equation:
$$
4=\frac{y}{2}+x\tag{2}
$$
Solving the equations $(1)$ and $(2)$ gives the values of $x$ and $y$ needed to derive the brute formula for the unknown hydrocarbon:
$$
y=-10,\quad x=9
$$
Since the result is negative for one term, we can assume that the water produced is not in the vapor phase, hence we eliminate water term inside the equation $(2)$, $y/2\:n_{\ce{C}_{x}\ce{H}_{y}}$, giving the immediate value of $x$ because there is no effort given by the water:
$$
x=4,\quad y=10\to \ce{C}_{4}\ce{H}_{10}\quad\mathrm{butane}
$$
This result is describing a saturated hydrocarbon, because it follows the rule $\ce{C}_{n}\ce{H}_{2n+2}$.


How can I choose the right symmetry in CAS calculation using MOLCAS? With $C_{2\mathrm{v}}$ symmetry and a closed shell configuration, how can I choose the occupation for each irreducible representation in the input file?

To perform a CASSCF calculation in MOLCAS, you have to supply a "standard" SCF calculation, specifically the orbitals, in a manner that the CASSCF module of MOLCAS can read them and use them as the one-determinantal basis for the following CASSCF procedure. As your system is closed-shell, a standard Hartree-Fock calculation performed beforehand using MOLCAS will do the trick. So, before you may run the RASSCF module of MOLCAS to perform a CASSCF calculation, you need to run the SCF module of MOLCAS to perform a Hartree-Fock calculation.

If this Hartree-Fock calculation is performed utilizing the $C_{\mathrm{2v}}$ symmetry of the molecule it will yield a set of orbitals, some for each symmetry group, i.e. $\mathrm{a_1, b_1, a_2, b_2}$. As the calculation is closed-shell the occupation will either be zero or 2. If your issue is the detection of symmetry in MOLCAS, please consider the following:

MOLCAS handles symmetry for all calculations via the GATEWAY module, you can specify the generators for the symmetry you deem the molecule to have there. This specification using generators might seem a little peculiar if you are used to programs like Gaussian or Turbomole, however, there are many resources to be found explaining it. See for example here:

Lecture on using symmetry in MOLCAS
Manual page on "Computing high symmetry molecules"

For C2V, if I'm not mistaken, the following keywords for GATEWAY might suit you:

&GATEWAY
coord=yourStructre.xyz
basis=cc-pVDZ
group=xy y

Now, assuming the symmetry was correctly recognized during the run of the GATEWAY submodule, and you have successfully performed a HF calculation utilizing the molecular symmetry, you can the proceed to chose your active space. Picking your active space is actually the tricky part of a CASSCF calculation where experience and chemical intuition come into play. I cannot give a full tutorial on choosing an active space, as this is a huge subject of its own. You can find some information here:

MOLCAS Manual - Tools for selection of the active space
MOLCAS FAQ: How do I choose my active space in a CASSCF/CASPT2
calculation?
Chemistry StackExchange:
How to choose proper active space in method such as CASPT2 and CASSCF?

Once you have chosen the orbitals you want to include in your active space, you need to identify their "number" in the Hartree-Fock calculation in order to be able to specify them. The "number" is of course given due to the energetical ordering of MOs ascending in energy. If your Hartree-Fock calculation yielded for example

10 occupied orbitals of $\mathrm{A_1}$ Symmetry
8 occupied orbitals of $\mathrm{A_2}$ Symmetry
6 occupied orbitals of $\mathrm{B_1}$ Symmetry
4 occupied orbitals of $\mathrm{B_2}$ Symmetry

and you have determined (via chemical intuition, experience and black magic) that the inclusion of the energetically highest two occupied orbitals of each symmetry and the energetically lowest two unoccupied orbitals of every symmetry will yield a fine active space, your input to the RASSCF module of MOLCAS for a ground state calculation could look like this:

&RASSCF
 Symmetry
 3
 Charge
 0
 Spin
 1
 Inactive
 8 6 4 2
 Nactel
 8 0 0
 Ras2
 4 4 4 4
 LumOrb

Please notice the Symmetry keyword for the RASSCF module which is set to "3" in my example above. A $C_{\mathrm{2v}}$ symmetry yields 4 symmetry groups. Different symmetries will yield a different number if symmetry groups, for example $C_2$ symmetry will give you two ($\mathrm{A}$ and $\mathrm{B}$). The number of values used as input for the keywords Inactive and Ras2 correspond to the number of symmetry groups present. You have to specify the desired symmetry of the wavefunction via the Symmetry keyword of the RASSCF module, or else the default, which is $\mathrm{A_1}$, will be calculated.

The total inputfile for a CASSCF calculation in symmetry may look like this:

&GATEWAY
coord=yourStructre.xyz
basis=cc-pVDZ
group=xy y
&SEWARD
&SCF
&RASSCF
 Symmetry
 3
 Charge
 0
 Spin
 1
 Inactive
 8 6 4 2
 Nactel
 8 0 0
 Ras2
 4 4 4 4
 LumOrb
&GRID_IT
 All; Ascii

I hope this helped you.


What is the unit of pH? I'm making some graphs and I have to label the axes. I want to be extra careful and put the units in even though the meaning of $\text{pH}$ is well known. But I have a problem (though a simple one): $\text{pH}$ is a minus logarithm (base 10) of concentration of hydrogen ions (or rather their activity). What is the unit then, is it $[-\log(\text{mol}/\text{L})]$? What should I write, could you help me?

The real definition of the $\text{pH}$ is not in terms of concentration but in terms of the activity of a proton,

\begin{equation}
  \text{pH} = - \log a_{\ce{H+}} \ ,
\end{equation}

and the activity is a dimensionless quantity. You can think of the activity as a generalization of the mole fraction that takes into account deviations from the ideal behaviour in real solutions.
By introducing the (dimensionless) activity coefficient $\gamma_{\ce{H+}}$, which represents the effect of the deviations from the ideal behaviour on the concentration, you can link the activity to the concentration via

\begin{equation}
  a_{\ce{H+}} = \frac{\gamma_{\ce{H+}} c_{\ce{H+}}}{c^0} \ ,
\end{equation}

where $c^0$ is the standard concentration of $1 \, \text{mol}/\text{L}$.
If you ignore the non-ideal contributions you can approximately express the $\text{pH}$ in terms of the normalized proton concentration

\begin{equation}
  \text{pH} \approx - \log \frac{c_{\ce{H+}}}{c^0} \ .
\end{equation}

In general, there can be no logarithm of a quantity bearing a unit. If however you encounter such a case it is usually due to sloppy notation: either the argument of the logarithm is implicitly understood to be normalized and thus becomes unitless or the units in the logarithm's argument originate from using the mathematical properties of logarithms to divide the logarithm of a product which is by itself unitless into a sum of logarithms: $\log(a \cdot b) = \log(a) + \log(b)$.


How to round correctly when calculating the average and standard deviation? Is my Chemistry book wrong on the significant figures in this exercise?
Why did they round to 0.16 instead of 0.1580 ?

Also why did they round down the means?

I figured it out.

The means are rounded to only two sig figs because its subtraction.
And significant figures follow different rules under subtraction/addition vs multiplication/division.

And its 0.16 for the same reason that the means are only two sig figs.
Because its addition, not multiplication/division.

derp, two hours down the drain


Is phosphorus a stereogenic centre in this phosphate ester? My textbook says that in the above diagram that there is only 2 stereoisomers. I am guessing this is due to the stereogenic carbon atom and not due to the phosphorus atom which apparently isn't a stereoisomer.

However, it states that the above molecule has 4 stereoisomers. This must indicate that in this case the phosphorus atom, it is a stereogenic and gives rises to 2 stereoisomers.

Why is the phosphorus atom in A not a stereocenter while the phosphorus atom in B is a stereocenter?

In compound A, the negative and double bonded oxygens bound to the phosphorus are equivalent:

$\hspace{5.1cm}$

In compound B, they are not equivalent:

$\hspace{7.5cm}$


How can I make my yeast experiment more chemistry-oriented? For a school chemistry design lab I tried to find the optimal sugar concentration for fastest yeast fermentation. I put the same amount of yeast and sugar water solutions with different concentrations of sugar together and measured the carbon dioxide produced using a water displacement apparatus. My teacher approved the topic, which is why I went along with it, but our teachers are often too lenient with approving topics, and this will be moderated so I can't just squeak through.

My main problem is that I feel it is too biology-oriented. There is very little chemistry in it. What can I do (preferably with yeast because I still have quite a bit and it wouldn't seem like a complete change of topic to my teacher)

It sounds like a perfect chemistry experiment to me. It sounds good enough to even be one of my AP chemistry labs. As an objective for the lab, you could try to determine the amount of CO$_2$ gas produced per amount of sugar you use, (with all other ingredients in excess). I'm pretty sure the equation for fermentation is C$_6$H$_{12}$O$_6$ → 2 C$_2$H$_5$OH + 2 CO$_2$ (although note that common granulated sugar has the equation C$_{12}$H$_{22}$O$_{11}$ and decays into two glucose molecules when combined with water).Why not test a certain amount of sugar (like .1 moles) and see if you yield the theoretical .2 moles of CO2. You could use the ideal gas laws (PV=nRT) to determine the moles of CO2 product you yielded. Compare what you yielded to the theoretical amount you should've yielded. If you yield less, then you could include in your report/conclusion including why you think your yield was short.

The lab would involve gas laws, chemical equations, and percent yield, all of which are chemistry topics/principles. What is your current level of chemistry experience? If you haven't learned of anything I mentioned, I can help explain it better and point you in the right direction if you'd like.


How does the Lennard-Jones potential relate to energy? Considering the Lennard-Jones potential,

$$V(r)=\frac{A}{r^{12}} − \frac{B}{r^{6}}$$

How does this relate to force or energy? I suspect that there is some physics equation I'm forgetting. I know that $F=−\nabla(V)$, but that's a partial derivative with respect to time, and I don't see how time comes into effect here. Also, the derivative of energy (work) with respect to displacement is force, but I'm not sure how to connect that to potential.

An example application that stimulated this question: given the interaction potential above, how would you find the interaction energy between two systems?

We have some nomenclature issues here.

In physics, the electric potential and gravitational potential refer to energy per unit something. In the gravitational potential, it's usually energy per unit mass, e.g. kJ/kg. In the electric potential, it's usually J/C (more commonly known as a Volt).

Here, though, we're not talking about a potential in that sense. Rather, I suspect (though I may be wrong) that potential was once a shorthand for potential energy surface, which is a way to think about the potential energy of a system with respect to various parameters. In the case of an L-J potential, there is only one parameter, the distance between the two particles of the system.

As mentioned in my comment though, I don't understand what you mean by two systems. The Lennard-Jones potential describes the energy of one system. If you put two of these systems together, the potential becomes ill-defined.

Added later: I found today that the Wiki page on thermodynamic potential says the same thing, namely that the "thermodynamic potential" is actually an energy instead of a potential. It quotes an ISO/IEC document that I don't have access to. I'm still not sure of the origin of this nomenclature.


Can pthalic acid or 1,2-Benzenedimethanol be possible products of Cannizzaro reaction of phthalaldehyde? I know the main product that is ortho-hydroxymethyl benzoate ion but I think that phthalic acid and catechol must be formed also in very small amounts but I am unable to get any confirmation from the internet. Please help me with the products.

The Cannizzaro reaction is the disproportionation of an aldehyde, which do not have an $\alpha$-hydrogen, to an equimolar mixture of corresponding primary alcohol and carboxylic acid salt. For example, original Cannizzaro's experiment:
$$\ce{2C6H5-CHO ->[aq NaOH] C6H5-CH2OH + C6H5-COO^-Na+}$$
Since there is no $\alpha$-hydrogen, the mixture does not undergo an aldol condensation, which is the characteristic of this reaction. The Cannizzaro reaction is an example of a typical redox reaction, which is considered to be one of the most important synthetic reactions of organic chemistry prior to the discovery of $\ce{LiAlH4}$ in 1946 (Ref.1).
The proposed reaction mechanism is as follows (Ref.1):

According to Ref.1, the hydride transfer must be from one molecule of benzaldehyde to the other molecule (when the reaction is performed in $\ce{D2O}$, no deuterated $\ce{CD2-O}$ is found in benzyl alcohol). The formation of benzyl benzoate was suggested and supported (e.g., Ref.2), evidence in Ref.1 doubts this proposal.
The Cannizzaro reaction of phthalaldehyde is studied in aqueous base ($\ce{NaOH}$) to obtain sodium o-hydroxymethylbenzoate as the product. The studies include the kinetics of the reaction, which have been measured over a range of $\ce{NaOH}$ concentration at $\pu{40 ^\circ C}$. This kinetic study proposed the hydration of phthalaldehyde to hemi-ketal monooxoanion is one of the first steps in the mechanism (hemi-ketal formation is depicted in the insert box of above image). The dioxoanion formation is depend on the concentration of $\ce{NaOH}$ solution (Ref.3). The hydride transfer from mono- or di-oxoanion follows, which is the rate determining step (determined by the isotope effect). The observed rate is based on following equation:
$$k_\mathrm{obs} = \frac{a[\ce{OH-}]}{1 + b[\ce{OH-}]}$$
where $a= \pu{(1.38 \pm 0.02)\times 10^{-3} M-1s-1}$ and $b= \pu{0.99 \pm 0.02 M-1}$ (the reaction follows this equation for $\pu{0.05-2.0 M}$ $\ce{NaOH}$). There are no other product has been observed.
I understand your concern of intra- versus inter-molecular hydride transfer. If intermolecular transfer occurs, at least 1,2-bis(hydroxymethyl)benzene and 1,2-benzenedicarboxylic acid should form as minor components. Since hydride transfer is the rate determining, intramolecular hydride transfer is much faster than that of intermolecular version. Therefore no chance of minor product formation is possible since all other steps are in equilibrium.
Note: I assume your suggestion of forming phthalic acid and catechol as byproduct is mistaken for 1,2-bis(hydroxymethyl)benzene and 1,2-benzenedicarboxylic acid (phthalic acid). Keep in mind that there is noway of breaking two aromatic-CHO bonds and making two aromatic-O bonds to form catechol molecule.
References:

C. Gardner Swain, Arnet L. Powell, William A. Sheppard, Charles R. Morgan, "Mechanism of the Cannizzaro reaction," J. Am. Chem. Soc. 1979, 101(13), 3576–3583 (https://doi.org/10.1021/ja00507a023).
Arthur Lachman, "The Benzyl Rearrangement v. Cannizzaro's reaction," J. Am. Chem. Soc. 1923, 45(10), 2356–2363 (https://doi.org/10.1021/ja01663a017).
Robert S. McDonald, Christine E. Sibley, "The intramolecular Cannizzaro reaction of phthalaldehyde," Canadian Journal of Chemistry 1981, 59(7), 1061-1067 (https://doi.org/10.1139/v81-157).


What is the mathematical reason behind the gradual increase in osmotic pressure? Question:

  Arrange the following in increasing order of osmotic pressure, $0.1~\mathrm{M}$ cane sugar, $0.1~\mathrm{M}$ $\ce{NaCl}$ solution, $0.1~\mathrm{M}$ $\ce{H2SO4}$.

Solution: As the number of particles/effective concentration of  the solutions increase in the following order:

Cane sugar $(0.1~\mathrm{M})$ > $\ce{NaCl}$ $(0.2~\mathrm{M})$ > $\ce{H2SO4}$ $(0.3~\mathrm{M})$

So the osmotic pressure which is related to concentration by the relation $P=cRT$ varies as:

Cane sugar > $\ce{NaCl}$ > $\ce{H2SO4}$

Is the following reasoning correct?

That formula for the osmotic pressure is known as the van 't Hoff law. It is described in some detail at the Wikipedia page for osmotic pressure, which also includes a brief (albeit incomplete and rather sloppy) derivation. To reach the van't Hoff law

$$\Pi = cRT$$

you have to approximate $\ln x_v$ by a Taylor series $(\ln x_v = \ln (1 - x_\mathrm{solute}) \approx -x_\mathrm{solute}$ which holds for small values of $x_\mathrm{solute})$ to get

$$\Pi = \frac{x_\mathrm{solute}RT}{V_\mathrm{m}}$$

The Wikipedia page uses $V$ to refer to the molar volume of the solvent (which should really be represented by $V_\mathrm{m}$).[1] In any case, $x_\mathrm{solute} = n_\mathrm{solute}/n_\mathrm{tot}$ and $V_\mathrm{m} \approx  V/n_\mathrm{tot}$ such that

$$\frac{x_\mathrm{solute}}{V_\mathrm{m}} \approx \frac{n_\mathrm{solute}}{V} = c_\mathrm{solute}$$

and if the solute is a 1:1 electrolyte, such as $\ce{NaCl}$, then you have to multiply by two to account for the total number of solute particles.

I am not sure about the term "effective concentration" (I personally have never seen it, and if I had to, I would probably write something like the "total concentration of solute particles") but the logic used is correct as long as you are in the regime where the van 't Hoff law is applicable i.e. very dilute solvent.

Lastly you should probably use $<$ instead of $>$ in your ordering to be clear.

[1] In the actual derivation this should strictly be the partial molar volume of the solvent

$$\overline{V}_{\!\!\mathrm{m,solvent}} = \left(\frac{\partial V}{\partial n_\mathrm{solvent}}\right)_{\!n_\mathrm{solute}}$$

but this can to a good extent be approximated by the molar volume of the pure solvent.


Calorimetry: Temperature Increase due to stirring of reactants I am not yet in university, but I hear a common first-year chemistry lab/practical is to do calorimetry taking into consideration the small (and usually negligible) rise in temperature due to the reaction container being stirred.

What is the standard method for estimating/calculating this increase in temperature, given the rate at which the stirring occurs and all the other necessary information? Is this possible to obtain an answer using molecular kinetics or are these calculations done solely on guesswork?

This is addressed experimentally, not through theory or computation. You run one experiment starting with the reactants, measuring the change in temperature over time. Then, you run a control with a similar solution that does not react (e.g. the product mixture), setting the mixer to the same speed. This will allow you to correct for any effect not due to the reaction. The stirring will be a minor contribution (depending on the tools used, not measurable at all). Other effects are large and always have to be corrected for (for example, if the reaction needs to be started with a spark and an ignition charge).

  What is the standard method for estimating/calculating this increase in temperature, given the rate at which the stirring occurs and all the other necessary information?

The work done to stir is based on the viscosity of the fluid. The faster you spin, the more work is necessary. In the classic experiment by Joule mentioned in the comments, a special stirring mechanism was used to have a large effect on the temperature. The experiment served to figure out the specific heat capacity of water (what energy is required to raise the temperature of a certain mass of water by a certain amount). You can read more about the experiment here.


Friedel-Crafts reaction of anisole? We know that phenol and aniline, despite being strongly activating, are unable to to undergo Friedel-Crafts reactions because they form complexes with $\ce{AlCl3}$, thanks to the lone pair on nitrogen and oxygen atoms.

However, according to my notes (and Google) anisole does undergo Friedel-Crafts reactions. How can it be possible, why is there no complex formation this time?

Also, I wanted to know how can we say that a certain group like $\ce{OH}$ will surely form a complex? Is containing a lone pair a sufficient condition?

Friedel-Crafts acylation of anisole with acetyl chloride and aluminium chloride as catalyst is a standard laboratory procedure to synthesize acetanisole.

The fact that this reaction is possible does not imply that there is no complex formation with $\ce{AlCl3}$. On the contrary, it is quite probable that such complexes do form due to the interaction of the methoxy oxygen with $\ce{AlCl3}$.

Therefore, and to compensate the formation of complexes with the resulting ketone, $\ce{AlCl3}$ is added in more than two-fold excess in the above synthesis.


How to quantify the difference between two xyz files for the same molecule? Sometimes I encounter xyz files with slightly different coordinates, for example, before and after optimization.

What is the easiest way to compare how different are they?

In the first approximation, I would define the difference as a minimum of the sum of squares of differences between individual coordinates, where the minimum is taken over translations/rotations of one of the representations.

What you describe is to calculate the RMSD after superposition. It is not necessary to do a minimization, there is a classic algorithm to get the best superposition directly: Kabsch algorithm. Sometimes, only a subset of coordinates will be superimposed, either because coordinates are missing in one of the structure, or because some parts of the structure show larger differences and interfere with superimposing the similar parts of the structures.

In some cases, where the structures are very different from one another but there are domains of conserved structures, you could also compute differences of pairwise distances and plot them as a distance difference matrix. Distance differences within the conserved domains will be small, while distance differences from domain to domain will be large.

When you do a superposition and RMSD calculation for the first time, it is nice to be able to check the superposition. Many structure viewers have the capability to visualize the superposition while also calculating the RMSD, for example the "compare" command in jmol.


How can you tell how much actually reacted in an acid base reaction If you facilitate a weak acid base reaction. i.e. $\ce{NaOH + H2SO4}$. All of it may not react and the solutions will not neutralize. How can you determine how much actually reacted?

The procedure is called titration. It consists in adding a pH indicator (that changes color depending on pH) to the solution and adding measured quantities of base or acid until the indicator changes color and determining how much reacted from the amount used.

An alternative is to use a pH meter instead of an indicator specially in automated analysis.


What is the major product in the reaction of 2,3-dichlorobutane with sodium amide in liquid ammonia? I am not able to decide which among 1,3-butadiene and 2-butyne is the major product of this reaction. Will there be even a major product out of the two or will there be a mixture?

According to me, the product formed from the 1st mechanism should be the major product as it is more stable (two double bonds in a conjugate alkene are more stable than a triple bond). But I also think that product formed from mechanism II can be formed faster as the hydrogen being picked up in the first step of mechanism II is more acidic (due to the electron withdrawing effect of electronegative chlorine atom) as compared to the hydrogen being picked up in mechanism I.

Another factor, which seems to favor the formation of product by mechanism I, is the picking of hydrogen atom in the 2nd step which is easier from an sp3 carbon than from an sp2 carbon (in mechanism II).

These are both reasonable mechanisms, and the question outlines well the factors favoring each. In favor of mechanism I:

Low temperature suggests kinetic deprotonation
Statistically more terminal hydrogens than internal hydrogens

In favor of mechanism II:

Small base suggests thermodynamic deprotonation

In these cases where there are conflicting factors, I agree with @Lighthart that the best answer is to actually look at the experimental results. That will give an indication of how to weigh the various factors.

In this case, I do not find an exact hit in SciFinder, but the reaction of  2,3-dibromobutane does come up in two older reviews by Bergstrom and Franke. I do not have access to Franke's review, although SciFinder has a note that "Bromine represents any halogen."

The SciFinder result for the Bergstrom review is deceptive. There is a comment "generalized reaction, halogen and R-groups can vary," and I don't find the exact result (or reasonable analogs) in the review.

More broadly, there are seven results for an internal, vicinal (1,2) dibromide reacting with sodium amide to give an alkyne. One reliable example is the synthesis of stearolic acid from Organic Syntheses.

Similar SciFinder searches for the 1,3-diene product come up empty.

Based on this research, I believe the alkyne would be the predominant product.


Original concentration calculation homework A student performs the following gravimetric analysis of iron ions in a water system
$$\ce{Fe^2+  (aq) + CO3^2-  (aq) <=> FeCO3 (s)}$$
If this student used excess carbonate and obtained the following data, what was the original concentration in moles/L) of $\ce{Fe^2+}$ in the water sample?

Total Volume of Solution: 100.00 mL

Mass of $\ce{FeCO3}$ collected: 23.758 grams.

Change 100 ml to 0.1 liters $x + 115.583 = 23.758 \times 0.1 = 0.02~\mathrm{l}$
- would this be how you solve this problem?

The mass of $\ce{FeCO3}$ is given as $m_{\ce{FeCO3}}=23.758\ \mathrm g$.

You already found that the molar mass of $\ce{FeCO3}$ is $M_{\ce{FeCO3}}=115.85\ \mathrm{g\ mol^{-1}}$.

For a pure sample, molar mass $M$ is defined as
$$M=\frac mn$$
where $m$ is mass and $n$ is amount of substance.
Thus, the amount of $\ce{FeCO3}$ can be calculated as follows:
$$\begin{align}n_{\ce{FeCO3}}&=\frac {m_{\ce{FeCO3}}}{M_{\ce{FeCO3}}}\\[6pt]
&=\frac {23.758\ \mathrm g}{115.85\ \mathrm{g\ mol^{-1}}}
\end{align}$$

The chemical equation
$$\ce{Fe^2+ + CO3^2- <=> FeCO3}$$
indicates that $1\ \mathrm{mol}$ of $\ce{Fe^2+}$ yields $1\ \mathrm{mol}$ of $\ce{FeCO3}$. Therefore,
$$n_{\ce{Fe^2+}}=n_{\ce{FeCO3}}$$

The volume of the solution is given as $V=100.00\ \mathrm{ml}=0.10000\ \mathrm l$.

The concentration $c$ is defined as
$$c=\frac nV$$
where $n$ is amount of substance and $V$ is the volume of the solution.
Thus, the concentration of $\ce{Fe^2+}$ can be calculated as follows:
$$\begin{align}c_{\ce{Fe^2+}}&=\frac {n_{\ce{Fe^2+}}}V\\[6pt]
&=\frac {n_{\ce{Fe^2+}}}{0.10000\ \mathrm l}\end{align}$$


Are there any semiempirical methods that work with heavy alkaline earth metals? I did some research. Found some semiempirical methods that work even with heavy lanthanide compounds (RM1). But could not find any references to semiempirical methods compatible with the heavier members of the alkali earth group: Sr, Ba and Ra. If there is such method, does any open source / free software implements it? I'm learning to use Gamess-US, but the only semiempirical methods implemented in this package are MNDO, AM1, PM3 and RM1 (refs file).

The implementations of MNDO, AM1, and PM3 along with PM6 and PM7 in MOPAC include Sr and Ba, but not Ra. MOPAC is free for academic use, but modern versions of it are not open source. Most of these models have an open-source implementation in SPARROW (GitHub). An entirely different class of semiempirical models (GFNx) is available in the open-source XTB code (GitHub), which also includes Sr and Ba, but not Ra. The lower portions of the periodic table (lanthanides and beyond) are of less general interest to scientists and have less reference data for fitting purposes, so there is lower demand and higher difficulty in covering this portion of the periodic table. The only parameterization that I know of that covers Ra is the QUASINANO2013 parameter set (J. Chem. Theory Comput. 2013, 9 (9), 4006–4017) in DFTB, which is proprietary and only available in the commercial implementation of DFTB by SCM.

If you only need Ra to behave like a +2 cation with a slightly larger ionic radius than Ba, then you probably could modify the parameter set of one of the aforementioned models and tune the repulsion parameters of Ba to increase its ionic radius in a few reference ionic systems. That would of course take a bit of effort, in both learning what the parameters of the model mean, where they are stored in a particular simulation code, identifying some suitable reference data, and tuning one or more parameters to fit the data.

I'm sorry that there isn't a simple, affirmative answer to your question, but I hope that this helps anyways.


Is this a counterexample to ΔU = 0 J for isothermal processes? In class, I have learned that for isothermal processes the following is true at all times.

$$\Delta U=q+w=0\text{ J}$$

But suppose I lift an object against the force of gravity. Then I (the surroundings) have done work on the object (the system) without exchanging heat (neglecting friction).

Now, I claim that because some nonzero work has been done on the system, and no heat has been exchanged, the change in internal energy is nonzero:

$$\Delta U=w\neq0\text{ J}$$

Are my assumptions about internal energy incorrect or is there something totally wrong about my approach?

The situation you cite calls for the use of the more general form of the first law of thermodynamics (which I assume your textbook provides), given by $$\Delta U+\Delta (KE)+\Delta (PE)=q+w$$where KE is the organized kinetic energy of the system and PE is the potential energy.  Applying this to the problem at hand gives $$\Delta (PE)=w$$and the change in internal energy is zero.  So the thing that was wrong with your approach was not using the proper form of the first law.


How do I purify the resulting compound after a nitro- to amine-group reduction? I have an aromatic compound that I reduced with either $\ce{Fe + HCl}$ or $\ce{Sn + HCl}$.

How would you choose to go about extracting the compound?

I'm thinking to dry off all the $\ce{HCl}$, redissolve in water, and react out the iron salts to Iron Hydroxide, which can be filtered out.

Then dry out again, dissolve in something like acetone then filter out the sodium chloride?

It is very easy to purify the amino compounds from the reaction mixture of reduction of nitro- to amine- group. Firstly, we will do  workup after completion of the reaction by extraction method with organic solvents  either with DCM or with ehtylacetate, then organic layer wlii be separated, and then dried with anhydrous sodium sulphate and the solvent will be evaporated under reduced pressure by applying rotary evaporator. Then after making slurry by adding little amount of silica gel we will pack slurry in Column-chromatography silica gel (100-200 mesh) by using eluent as Ethylacetate:Hexane (20% to 60%),  the amino- group compond will be purify. This is so because nitro- group is less polar compared to anino- group. So, nitro- compound will come first from column chromatograhy and amino- compound will come letter in pure form.


How can I find out how much caffeine is in my decaf tea and coffee? I know that the decaffeination process is not perfect and that it leaves some caffeine. I've looked online and there was a company in 2009 that offered caffeine test strips but they seem to be no longer offered.
Another approach would be to use a GCMS, but used GCMS devices are expensive and it takes alot of effort to keep them in calibration (at least it did when I was in college).
Are there services where I can send a sample off and they will tell me how much caffeine they contain? Failing that, how would I reasonably test how much caffeine is in my tea and coffee?

You may try to extract the caffeine with DCM and determine its concentration by spectrophotometry.


Jahn-Teller distortion of [Fe(H2O)6]2+ I am wondering whether Fe(II) hydrated to an octahedral complex would undergo a Jahn-Teller distortion.

As far as I am aware, this distortion comes about because of asymmetry in filling orbitals, in this case the t2g and/or eg orbitals.

I am also aware that the distortion is most obvious for asymmetry in the eg  orbitals because these interact most with the ligands.

However, I deduced the configuration t2g4, eg2 for [Fe(H2O)6]2+, which has no asymmetry in eg but does in t2g.

My question is can it distort because of this asymmetry? Would it be possible that promotion to an excited state takes place for the distortion to then be possible, with an overall stabilising effect?

(1) It can and will distort. In general, any non-linear molecule with an orbitally degenerate ground state is susceptible to a first-order JT distortion. $\ce{[Fe(H2O)6]^2+}$ falls under this category.

However, the Jahn-Teller theorem has its roots in symmetry and group theory. These can only tell you whether an integral is zero or whether it is nonzero - i.e. whether there is no distortion, or whether there is a distortion. They cannot make any prediction as to the extent of the distortion, or the exact form of the distortion.

The ground-state distortion in $\ce{[Fe(H2O)6]^2+}$ is small, which may be rationalised as being due to an unsymmetrical occupancy of the $\mathrm{t_{2g}}$ orbitals, which point between ligands.

(2) No, the energy required to promote an electron is far larger than any stabilisation that can be derived from a JT distortion.


How do you calculate the correct charge on F after F2 dissociates? When you ionize $\ce{F2}$ to $\ce{F2^5+}$, it quickly dissociates. Experiments show that the products of dissociation are $\ce{F^2+}$ and  $\ce{F^3+}$.

I have tried to model the dissociation of $\ce{F2^5+}$ using Gaussian$09$ (DFT), specifying a 10 Å separation between products.

No matter which functional I use ($\omega$B97X, PBE, M06,etc.), I get a $+2.6$ charge on one $\ce{F}$ ion and $+2.4$ on the other. Why don't these functionals give $\ce{F^2+}$ and $\ce{F^3+}$ ions the corresponding formal charges as shown by experiment? And how do you get the correct charges?

If you have two fragments that are very far apart, certain quantum chemistry methods can cause spurious effects from including both fragments in the same calculation.  A well known example of this is dissociation curves for $\ce{H2}$ with Restricted Hartree Fock.

By using a density based method, we are less likely to localise the electrons on any one atoms in integer quantities.  A wavefunction method such as Hartree Fock, which places integer electrons in well defined molecular orbitals is more likely to create this localisation.  If you need higher accuracy, MP2 or coupled cluster may be of more use.

In fact I have achieved Mulliken charges of +2 and +3 on the two different atoms at 10 angstrom separation using Hartree Fock, with a cc-PVTZ basis set in Gaussian 09.

If you insist on using DFT you should use some chemical/physical intuition to partition this system.  From experiment, i.e. time of flight spectrometry, we know that the atoms have integer charge when they strike a detector, and dispersion interactions over that distance are going to very small compared to the coulomb attract of the other atom as a point charge.  So instead you can calculate the energies of the energies of the individual $\ce{F^n+}$ for $n=0,1,2,3,4,5$.  The combination of atoms with total charge $+5$ and the lowest energy, considering the classical coulomb attraction between them, is probably the product of the dissociation.

Gaussian input used:

%chk=FF_HF.chk
%nproc=1
%mem=4GB
# UHF/CC-PVTZ NoSymm

F2 5 plus Ion

5 2
 F      0.00000000    0.00000000   -5.00000000
 F      0.00000000    0.00000000    5.00000000