How to calculate the amount of chlorine emitted yearly by a leak in the air conditioning system? A leak in the air conditioning system of an office building releases $12~\mathrm{kg}$ of $\ce{CHF2Cl}$ per month. If the leak continues, how many kilograms of $\ce{Cl}$ will be emitted into the atmosphere each year?

My working out is as follows:

\begin{align}
12~\mathrm{kg} &= \mathrm{CHF}_2\mathrm{Cl} \\
\text{Cl mass in compound} &= \frac{35.45}{86.448}\times100\% = 41\% \\
\end{align}

$41\%$ of $12~\mathrm{kg}$ is $4.92~\mathrm{kg}$, so I multiplied it by $12$ to get the annual result.

$$
4.92~\mathrm{kg} \times 12(\text{months}) = 59.04~\mathrm{kg\ year}^{-1}
$$

I don't have access to the answer so I'm not sure whether or not my answer is correct. Also, this process seems a little to simple to be correct. I say this because the question prior to it was very difficult. Can someone please tell me if I am right, and if I'm not, where did I go wrong?

For the sake of correct formalism, here the answer as it should be given in the textbook:

The mass flow rate of the leaking substance, under the assumption of linear scaling, is
$$\dot{m}_{\ce{CHF2Cl}} = 12 ~\mathrm{kg\, month^{-1}} = 144~\mathrm{kg\, a^{-1}}\; .$$

The mass flow rate of chlorine, as contained in the leaking substance, is then calculated as follows:
$$\dot{m}_{\ce{Cl}} = \dot{m}_{\ce{CHF2Cl}} \cdot \frac{M_{\ce{Cl}}}{M_{\ce{CHF2Cl}}} = 144~\mathrm{kg\, a^{-1}} \cdot \frac{35.453~\mathrm{g\,mol^{-1}}}{86.47~\mathrm{g\,mol^{-1}}} = 59.04~\mathrm{kg\, a^{-1}}$$

For the end result to the question "how many kilograms of chlorine will be emitted into the atmosphere each year?", please note that the mass flow rate still has a dependency on time
$$ \dot{m} = \frac{\mathrm{d} m}{\mathrm{d} t}\; , $$
which means that we have to do an integration so we can calculate the mass:

$$ m(\tau) = \int_0^\tau \dot{m}\,\mathrm{d}t = \dot{m}\,\tau$$

For one year, we get
$$ m_{\ce{Cl}}(1~\mathrm{a}) = \dot{m}_{\ce{Cl}}\cdot1~\mathrm{a} = 59.04~\mathrm{kg}\; , $$
which means that your attempt at answering your own question were very good, on spot and totally valid.


Does PCl3 react with ester as it does with carboxylic acid to produce an acid halide? I did not find it anywhere in a book but I do wonder if it might react.

Yes it does. All carboxylic acid derivatives undergo nucleophilic substitution reactions. So esters, acid anyhyrides, amides all on reacting with $\ce{PCl3}$ will give acid chlorides.


Empirical formula of Glauber's salt I'm not too sure on how to get the empirical formula for Glauber's salt because water ($\ce{H2O}$) is included.

The question is:

  Glauber's salt contains $14.3\%$ sodium, $10\%$ sulphur, $10.9\%$ oxygen (not in water) and $55.8\%$ water. Find this substance's empirical formula.

This is what I did. I got $\ce{Na4SO.H2O}$ as my answer.

Your approach seems correct, however you seemed to have made a few computation errors. First of Sodium's molar mass is $ 23 \ce{g/mol}$ and not 11.

So assuming a total mass of 100g, we can re-interpret the given percentages as follows:

mass of sodium = 14.3 g

mass of sulphur = 10 g

mass of oxygen = 19.9 g

(note: the percentage you provided seems incorrect, the percentages add up to 91, not a 100. I've changed this data point to get the correct result.)

mass of water = 55.8 g

Dividing these by their respective molar masses to get no. of moles:

no. of moles of sodium  = 0.62

no. of moles of sulphur = 0.31

no. of moles of oxygen  = 1.2

no. of moles of water   = 3.1

The ratio we get is 2:1:4:10. Thus, the empirical formula is $\ce{Na_2SO_4.10H_2O}$


Can quantum entanglement affect the chemistry of molecules? I recently heard a talk from a physicist about experiments on simple two-atom molecules in vacuum that showed that certain behaviour of the molecules was explained by quantum entanglement. Now this experiment was about creating holes in the inner electron shells of the molecule, so pretty much in the realm of physics and not chemistry. It was also performed in vacuum, and the lecturer himself said that in solution such an effect would likely be destroyed by frequent collisions in the solvent.

I was wondering if there are any examples of chemical behaviour of molecules that can only be explained by quantum entanglement. Does it ever play a role on the scale of chemistry, or is this effect just not relevant on anything beyond the small moleculs in vacuum stage physicists tend to use?

I've not really read anything on it, but a short while ago research on the role of entanglement in excitonic transfer in the photosynthetic light harvesting apparatus caused quite a stir.

Here is an ArXiV preprint of a paper on entanglement in an antenna protein from C. tepidum. It was published in Nature Physics. The authors note the extraordinary result of entanglement being observed in an inherently quite 'noisy' system.


Why can't Pd/C and H2 reduce both the alkene and carbonyl portions of α,β-unsaturated carbonyls? Why is it that the major product of the reduction of chalcones the ketone and not the monoalcohol? In other words, Why isn't the major product a benzyl alcohol?

From what I understand, catalytic hydrogenation can be used to reduce carbonyls as well as alkenes.

My TA told me that nucleophilic hydrides are preferred for reducing carbonyls. Why wouldn't hydrogenation work as well? Does it have to do with resonance involving the carbonyl since in a chalcone, the carbonyl is adjacent to an aromatic ring and is also conjugated with the alkene? Could the reason that catalytic hydrogenation can't effectively touch the α,β-unsaturated carbonyl be the same reason catalytic hydrogenation can't effectively reduce carboxylic acids, esters, and amides — all of which are also resonance stabilized?

My first thought had to do with heats of hydrogenation and how resonance-stabilization found in carboxylic acids and its derivatives reduce the heat of hydrogenation. However, later I found a resource online implying that all pi bonds - even the delocalized, resonance-stabilized ones found in benzene — could be reduced through catalytic hydrogenation with enough time.

In addition, it was noted elsewhere that chalcones could be completely reduced to the benzylic alcohol, although not with ease.

This leads me to believe that there is an activation energy barrier that's impeding the hydrogenation of certain substrates within the confines of a 3 hour undergraduate lab period … am I on the right path? Sterics, perhaps? It was noted that the more highly substituted an alkene is, the slower it is reduced because of the difficulty in getting a highly substituted alkene to approach the catalyst's surface with the $\ce{M-H}$ bonds in an appropriate manner.

You hit it right on the nose. The real key piece of information is that given enough time, all the unsaturated bonds will be reduced. This tells you that though the reduction is thermodynamically favorable, it is the difference in the energy barriers ($\ce{\Delta \mathrm{G^{‡}}}$) that prevents the carbonyl reduction from occurring at the same rate as the alkene reduction. This means to produce the alcohol faster, we must manipulate the kinetics of the reaction.

To understand why the carbonyl-reduction transition state is higher in energy, we should consider the differences between the carbonyl and alkene bonds. The bonds are much more polarized, and as such the carbonyl $\pi$ bond is considerably stronger than alkene ($93\ \mathrm{kcal\ mol^{-1}}$ vs. $63\ \mathrm{kcal\ mol^{-1}}$)$^{\mathrm{[1]}}$. This means that at low temperatures, only a small fraction of the molecules have enough energy for the carbonyls to associate onto the $\ce{Pd/C}$ surface. Catalytic hydrogenation of carbonyl compounds (aldehydes, ketones, and especially esters) requires high temperatures and pressures to increase the presence of both the substrate and hydrogen on the catalyst.

$^{\mathrm{[1]}}$ Fox, M. A.; Whitesell, J. K. Organic chemistry; Jones and Bartlett: Sudbury, MA, 1997.


Spontaneous/Non-Spontaneous Reactions and Reversible Reactions My chemistry textbook states "Reversible reactions constitute a limiting case between spontaneous and non-spontaneous processes."

Does this mean that some of the reversible reactions are spontaneous and others are not? Or that the reversible reactions are sort of intermediary between said two types of reactions? Or something else?

"Reversible reactions constitute a limiting case between spontaneous and non-spontaneous processes." So what does this mean?

To me this sentence is meaningless gobbledygook and should be ignored.  That is, this sentence is unhelpful and trying to understand it will just make you more confused and less able to master chemistry.  Here is a short summary version of the points I make in the "Advanced discussion" below:

All reactions can be viewed as reversible from a mathematical standpoint, if the reacting system is "big enough".  Reactions that are highly spontaneous in the forward direction mean that at equilibrium, the number of "reactant" molecules will be very very small (but not zero!).
Many times, "big enough" would mean astronomically big, such as bigger than the Earth.  In these cases, reactions can be regarded as "irreversible", although there is no strict, universally agreed upon boundary between "reversible" and "irreversible".
Whether a reaction is spontaneous isn't that much related to whether it is "reversible".  Don't worry about understanding the connection between these concepts until you have understood each concept independently.

Advanced discussion.

The meaning of spontaneous

Spontaneous processes have negative Gibbs free energy changes ($dG < 0$) and also result in positive entropy changes in the universe ($dS_{univ} > 0$).  However, $dG$ is a function of state.
For chemical reactions, this means that $dG$ depends on the concentration of the reactants and products, i.e.

$$dG = dG^\circ + RT \ln Q$$

where $Q$ is the reaction quotient.

Thus, if we start off with all "product" and no reactant, then the "reverse" direction of the reaction will be spontaneous.  But if we start off with all reactants and no product, then the forward direction is spontaneous.

The meaning of reversible

The short answer is that for chemical reactions, "reversible" does not have a precise thermodynamic definition.

$$dG = dG^\circ + RT \ln Q = -RT \ln K + RT \ln Q = RT \ln \frac{Q}{K}$$

where $K$ is the equilibrium constant.  People will usually call reactions with a "large" equilibrium constant irreversible and those with a "small" equilibrium constant reversible.  But the exact definition of "large" and "small" isn't well-defined.  Many people might say $\frac{Q}{K} \leq .01$ represents an "irreversible" reaction.  For chemical reactions with very large equilibrium constants, the size of the system required to ensure that an equilibrium can be reached can get very large, as explained here.


Spectroscopic transition frequency I have difficulty understanding the first sentence of the Introduction in the paper linked below. Can someone explain to me what is spectroscopic transition frequency and thermal bath?

  The question addressed here is how the modulation of the spectroscopic transition frequency Ω(t) by fluctuations of the thermal bath is related to the observed spectral line shape.

http://pubs.acs.org/doi/abs/10.1021/jp5081059

In an isolated molecule /atom the intrinsic width of the transition is limited by the lifetime of the upper level and this can be huge ( seconds) and so the line-width can be minuscule by time-energy 'uncertainty'. If the decay is exponential the spectral line shape is Lorentzian by Fourier Transform theory.

In a medium, the surrounding solvent or gas molecules repeatedly interact with our victim molecule by collisions and by electric and magnetic dipoles etc, and do so  in a random fashion and thus cause fluctuation in its energy levels. These changes are then observed in the transition energy and so broaden the observed transition in the ensemble of molecules. So far this is quite general. This paper examines one model of this, linear + quadratic coupling and the quadratic part is presumably why the peak is split - but I only skimmed the paper so you will have to check this.  The detailed maths is just the technical part, ignore this and try to get to the basic physics first.

(The term 'thermal bath' is just jargon for all the random interactions from solvent/buffer gas surrounding the victim molecule, 'modulation' just means changes, so the first sentence could read 'We describe how random interactions with a solvent or buffer gas changes the observed spectroscopic line shape ....')


What are pharmaceutical uses for Vanillin? What are some pharmaceutics that use vanillin? Is vanillin just an aroma and taste agent to pharmaceutics?

To my knowledge, vanillin nowadays does not have any pharmaceutical uses.

But according to Elisabeth Vaupel in Pharmazeutische Zeitung online, 2002, 38 (SRC), Michael Bernhard Valentini wrote 1714 in his book Museum Museorum, oder vollständige Schau-Bühne aller Materialien und Specereyen, nebst deren natürlichen Beschreibung:

  Sehr erwärmende und zerteilende, anbei aber auch stärkende Kraft, womit sie den Magen stärken, die Winde zerteilen und dem Gehirn, der Mutter und anderen nervösen Gliedern sehr gut tun. Sie treiben den Harn, befördern die monatliche Reinigung, natürliche Geburt und Schwierungen: Treiben auch die Nachgeburt und tote Kinder fort und kommen also dem weiblichen Geschlecht in ihren meisten Krankheiten wohl zu pass. Ingleichen werden sie gegen die erstarrend-machenden giftigen Bisse and andere dergleichen giftige Sachen gebraucht [...] Am meisten aber werden die Vanillen zu Verfertigung der Chocolaten gebraucht, welche sie anmutiger und kräftiger machen. Die Tabaksbrüder brauchen sie auch, den Tabak wohlriechend zu machen.

As a starting material in synthesis, vanillin can be converted to MMDA in 5 steps.


Questions regarding the preparation and recrystallisation of methyl 3-nitrobenzoate I have a procedure for the synthesis of methyl 3-nitrobenzoate (via nitration of methyl benzoate):

  Weigh 2.0 g of methyl benzoate into a dry 50 cm3 conical flask.
  Add 4 cm3 of concentrated sulfuric acid slowly to the methyl benzoate with swirling to ensure thorough mixing. Cool this mixture by partially immersing the flask in an ice-water bath.
  Carefully transfer 1.5 cm3 of concentrated nitric acid into a dry test tube. Cool the nitric acid by partially immersing it in an ice-water bath before slowly adding, with swirling, 1.5 cm3 of concentrated sulfuric acid.
  Allow this mixture to cool. This is the nitrating mixture.
  Using the glass dropping pipette, very slowly add the nitrating mixture (over about 15 min, 1 drop every 10 seconds) to the contents of the conical flask. Stir the reaction mixture as the addition is made. During the addition keep the temperature of the reaction mixture below 6 °C.
  Once addition is complete, allow the flask containing the reaction mixture to stand at room temperature for 15 min.
  Carefully pour the reaction mixture onto a small amount (approx. 20 g) of crushed ice held in a beaker.
  Stir the crushed ice throughout. Solid methyl 3-nitrobenzoate will form.
  Allow the ice to melt and filter under suction. Wash the crude product with a little ice-cold water.

I have two questions:

Firstly, why does the conical flask need to be in an ice bath when adding $\ce{H2SO4}$ to methyl benzoate?
Secondly, why is ethanol used for recrystallisation?

Can't we just heat the mixture of methyl 3-nitrobenzoate after the nitrating mixture has been added?

Answer 1 - you want to keep the whole reaction cool, preferably below 6 °C, because the nitration reaction is exothermic. If it gets too hot you will lower your yield due to side-reactions. There is also the possibility of runaway reaction degrading the starting material and product. This is why you don't heat it.

Answer 2  - someone somewhere has done the comparison crystallisations and determined that EtOH is a good solvent for the recrystallisation. It is also cheap and safe for students to handle.


Does the multiplication and division significant figures rule apply to unit conversions as well? My chemistry teacher told our class that answers to calculations that involve multiplication and division should be given to same number of significant figures as the factor, dividend or divisor with the lowest no. of sig.figs.

For example, to calculate the no.of moles of calcium in a pure 40g sample of calcium, the dividend (the mass) has 1 sig.fig. and the divisor (the molar mass) has 3 sig.figs. Hence the no. of mole sis given to 1 sig.fig.

40/40.1 = 1 mol of Calcium

Does this same rule apply to unit conversions? For example if I wanted to convert 12.4 milligrams to grams, I would need to divide 12.4 milligrams by 1000. 12.4 has 3 sig.figs. whereas 1000 has only 1 sig.fig. Does that mean the figure in grams must be given to 1 sig.fig?

In regards to the calcium situation, it depends on whether you measured 40g of calcium (which would be 2 sig. fig.), or 40g was the stated value (1 sig. fig.). Yes, the number of sig figs in your answer should equal the number of sig figs in the least accurate piece of data.

When converting units, the 1000 is taken as the exact value, i.e. it has infinite accuracy. Here, it is not considered, so the answer should have 3 sig fig from 12.4 (which is also why scientific notation is used to decrease ambiguity).


Determine the geometry of each nitrogen atom in formamidinium ion The geometry at A seems to be straightforward: $\ce{N}$ is surrounded by $\ce{C}$ and two $\ce{H}$'s. By VSEPR, this is trigonal planar.

For the geometry at B: $\ce{N}$ is surrounded by $\ce{C}$, two $\ce{H}$'s, as well as a lone pair. Once again by VSEPR, I determined this was trigonal pyramidal.

However, the answer was given as A, which means that my process for the geometry at B was incorrect. What mistake did I make and how should I fix it?

The hybridisation of any atom is determined by the total count of its sigma bonds and localised lone pairs. This is because, while localised lone pairs occupy hybrid orbitals (as in ammonia), delocalised lone pairs occupy pure $\ce{p}$ orbitals instead. Hence, the latter is not counted in determining hybridisation.

The $\ce{N}$ atom labelled "B" has its lone pair in conjugation with a pi bond. This is perfect for delocalisation/resonance to take place. Hence, that lone pair is not counted in determining hybridisation. Thus, it is also a trigonal planar geometry.

Interestingly, notice that the formamidinium ion has two equivalent canonical resonating structures. The lone pair of the atom B on delocalisation, results in a lone pair on A and a formal positive charge on B, which is just the reverse of the structure shown.


Influence of sodium cyanoborohydride on C=N bonds in DNA bases I would like to crosslink an amino-modified DNA strand with an APTES monolayer (which also has a $\ce{NH2}$ end group) using a glutaraldehyde linker. After the formation of Schiff base linkages between the aldehyde and the amino groups, I would like to reduce the bonds using sodium cyanoborohydride to form secondary amine bonds.

Does the cyanoborohydride also affect the $\ce{C=N}$ bonds in the DNA bases or not?

Sodium cyanoborohydride is a very selective reducing agent. Another (safer) alternative would be sodium triacetoxyborohydride. But this doesn't really matter as far as $\ce{C=N}$ bonds of DNA bases are concerned since they are very stable.


How to find the concentration of an of ammonia solution by titration with nitric acid? I am stuck with this question I am doing:

To determine the ammonia ($\ce{NH3}$) concentration in a cleaning product, a chemist diluted 20 mL of the product to 500 mL. Of the 500 mL solution, 25 mL was titrated with 0.151 M $\ce{HNO3}$. The endpoint occurred at 26.2 mL. Calculate the concentration of the cleaning fluid.

[Ans: 3.956 M]

What I have done so far:

I worked out the moles of $\ce{HNO3}$ that the $\ce{NH3}$ must have reacted with, using the concentration and volume they gave me. I got 0.0039562 moles.
I then wrote the balanced equation and worked out that the stoichiometric ratio of $\ce{HNO3:NH3}$ must be 1:1. Therefore, I worked out that the amount of moles of $\ce{NH3}$ reacting with the $\ce{HNO3}$, must also be 0.0039562 moles.
From here, I initially worked out the concentration of $\ce{NH3}$ reacting with $\ce{HCl}$ and got 0.158 M, but then I realised this wasn't what the question was asking. I tried using 0.02 mL as the volume instead (as this was the very initial volume of $\ce{NH3}$) but still got the wrong answer.

Can someone please tell me what I have done wrong/ guide me in the right direction.

Edit: I have just tried it again, and this time I used the concentration I previously found (0.158 M) and a volume of 0.5 L to work out the number of moles in the 500 mL dilution of the product. I used this value which was 0.079 moles and a volume of 0.02 mL which was the original volume of cleaning liquid, to work out the concentration. I seem to be getting the right answer, but is my working correct?

Yes, although your presentation is a bit confusing.
To be clear,

You determined the quantity of $\ce{HNO3}$ consumed during the titration
$$0.151\ \mathrm{M} \times 26.2\ \mathrm{mL} = 3.956 \  \mathrm{mmol}$$
Then given this titration equation:
$$\ce{HNO3 + NH3 -> NO3- + NH4+}$$
You deduced that the $25~\mathrm{ml}$ sample contained $3.956 \  \mathrm{mmol}$ of $\ce{NH3}$, hence a concentration of
$$\frac{3.956\ \mathrm{mmol}}{25\ \mathrm{mL}}=0.1582\ \mathrm{M}$$
From which you calculated the amount of $\ce{NH3}$ in the $500\ \mathrm{mL}$ sample:
$$0.1582\ \mathrm{M} \times 500\ \mathrm{mL} = 79.12 \  \mathrm{mmol}$$
And finally since the original $20\ \mathrm{mL}$ sample also contains $79.12\ \mathrm{mmol}$ of $\ce{NH3}$, that gives you the original concentration:
$$\frac{79.12\ \mathrm{mmol}}{20\ \mathrm{mL}}=3.956\ \mathrm{M}$$

What I think you should have done in place of 4 and 5 is:

4'. Since the $25\ \mathrm{mL}$ sample is taken from a solution diluted from $20\ \mathrm{mL}$ to $500\ \mathrm{mL}$, the original concentration is:
$$0.1582 \ \mathrm{M} \times \frac{500\ \mathrm{mL}}{20\ \mathrm{mL}}=3.956\ \mathrm{M}$$

Which spares you one calculation's work and potential errors.


Why do atomic orbitals have their unique shapes? Is there a scientific explanation to why p orbitals are shaped like two balloons, etc. I think it has got to do with electron repulsions. Wikipedia says they are 'characterised by unique values of quantum numbers', of which I don't understand.

The pictures that we typically use to represent orbitals are really just graphs showing the 3-dimensional probability of finding an electron (that occupies the orbital being examined) in space.  $\ce{\Psi}$ is the wavefunction that describes the electron with a set of specific quantum numbers.  $\ce{\Psi^2}$, gives rise to the electron density distribution (the shape of an orbital) plotted in the graph.


Calcium Phosphate Dissolution For what reason does Calcium Phosphate dissolute at low pH? What makes it pH dependent? This question arised after reading about the calcium phosphate bridges in casein micelles and how they 'break apart' at a pH below 6, such that the micelle destabilises and essentially falls apart.

The solubility of calcium salts is highly dependent on pH. For example, let’s look at tricalcium phosphate, which presents the solubility equilibrium:
$$\ce{Ca3(PO4)2(s) + H2O <=> 3Ca^2+(aq) + 2PO4^3-(aq)}$$
The solubility product constant for this equilibrium ($K_\mathrm{sp}$) is vastly varied from source to source. Some listed it as $\pu{2.0 × 10^–29 mol^5L{–5}}$, and another listed it as $\pu{2.07 × 10^−33 mol^5L^{–5}}$ at $\pu{25 ^{\circ}C}$ and $\mathrm{pH}~7.00$. However, I found a reliable $K_\mathrm{sp}$ value in literature, which reported that $K_\mathrm{sp}$ of $\beta$-$\ce{Ca3(PO4)2}$ in the $\mathrm{pH}$ range $6.0-7.5$ is $\pu{1.2 × 10^–29 mol^5L^{–5}}$ at $\pu{25 ^{\circ}C}$ and $\pu{2.83 × 10^–30 mol^5L^{–5}}$ at $\pu{37 ^{\circ}C}$ (a negative thermal coefficient of solubility; Ref.1). Therefore, the quantitative analysis of $\ce{Ca3(PO4)2}$ is near impossible.

Still, we can qualitatively look at the problem: If you decrease the $\mathrm{pH}$ of the solution in above equilibrium by adding acid to the solution, some of the phosphate ions would protonate and transform into $\ce{HPO4^2–}$ ions, as what happens in phosphate buffer solutions:
$$\ce{PO4^3-(aq) + H3O+ (aq) <=> HPO4^2-(aq) + H2O}$$
As a result, the concentration of $\ce{PO4^3-}$ ion would be reduced. The system would respond to this reduction by producing more $\ce{PO4^3-}$ ions, according to the Le Chatelier principle. To do so, some solid $\ce{Ca3(PO4)2}$ would dissolve in first equilibrium, and the equilibrium will be shifted to the right.

Keep in mind that in first equilibrium, $\ce{Ca/P}$ ratio is 1.5. The solubility of calcium phosphates is also depend on $\ce{Ca/P}$ (Ref.2). For example, both monocalcium phosphate ($\ce{Ca(H2PO4)2}$) and monocalcium phosphate monohydrate ($\ce{Ca(H2PO4)2.H2O}$) are highly soluble in water at $\pu{25 ^{\circ}C}$ ($\ce{Ca/P}$ is 0.5). On the other hand, $K_\mathrm{sp}$s of dicalcium phosphate ($\ce{CaHPO4}$) and dicalcium phosphate dihydrate ($\ce{CaHPO4.2H2O}$) are $1.26 × 10^{–7}$ and $\pu{2.57 × 10^{–7} mol^2L^{–2}}$, respectively at $\pu{25 ^{\circ}C}$ (Ref.2). Based on these values, when $\mathrm{pH}$ is gradually reduced to a point such that all calcium phosphates would dissolve.

References:

T. M. Gregory, E. C. Moreno, J. M. Patel, W. E. Brown, “Solubility of $\beta$-$\ce{Ca3(PO4)2}$ in the system $\ce{Ca(OH)2-H3PO4-H2O}$ at $5, 15, 25,$ and $\pu{37 ^{\circ}C}$,” Journal of Research of the National Bureau of Standards–A. Physics and Chemistry 1974, 78A(6), 667–674 (http://dx.doi.org/10.6028/jres.078A.042).
L.C. Chow, “Solubility of Calcium Phosphate,” In Octacalcium Phosplrate; L. C. Chow, E. D. Eanes, Vol. Eds., Volume 18 of Monographs in Oral Science (G. M. Whitford, Ed.) Karger Publishers: Basel, Switzerland, 2001, pp. 94-111 (DOI:10.1159/000061650).


Why isn't standard Gibbs energy zero at equilibrium? At equilibrium, everything that $G$ depends on is constant.
The Gibbs' free energy may have been given to you without really emphasizing that it is actually a function of temperature, pressure, and the moles of involved reactants and products:
$$G=f(T, p, n_i)$$ When we are at equilibrium, we really mean all of the following are occurring:

Thermal equilibrium
Pressure equilibrium
Dynamic equilibrium

And hence the  moles of reactants and products is steady/constant (but may react if disturbed).
But when all those things are constant, the Gibbs free energy simply isn't changing at all. If the variables on which $G$ is dependent are constant, $G$ (for a single state) is constant.

Standard Gibbs energy is basically Gibbs energy but at standard states i.e. $\pu{1 bar}$ pressure and substances at their standard states.
So this standard Gibbs energy is also a function of $T$, $p$, and $n$, which at equilibrium are constants. But still $$\Delta G^\circ=-2.303RT\log K$$ at equilibrium which clearly isn't zero.
Why isn't change in standard Gibbs free energy $\Delta G^\circ$ zero at equilibrium while $\Delta G_r$ is?

METHOD 1
Suppose we were to start off with pure reactants in stoichiometric proportions in the standard state of 1 bar, and end up with pure products in corresponding proportions in the standard state of 1 bar.  The change in Gibbs free energy for this process is $\Delta G^0$, as you have answered in your comment to me.  To bring about this change, we use whatever equipment is necessary, typically using a van't Hoff equilibrium box and constant temperature compressions and expansions in cylinders with piston.  If we had assumed different values of the product and reactant pressures in the initial and final states than 1 bar, then the change in free energy would not be $\Delta G^0$, but rather $\Delta G$, with $$\Delta G=\Delta G^0+RT\ln{Q}$$.  If the values of the pressures in this final state happened to be the same as the partial pressures of the reactants and products in an equilibrium reaction mixture, the $\Delta G$ would be 0, and the Q would be equal to the equilibrium constant K.  So we would have $$0=\Delta G^0+RT\ln{K} $$
METHOD 2
Here, we start out with a mixture of various amounts of reactants and products that are not at equilibrium, and we analyze the variation in the free energy of the mixture as the reaction proceeds.  So, let $n_A$, etc. represent the number of moles of reactants and products at any time as the reaction proceeds, let $G^0_A$ etc. represent the free energies of formation of the reactants and products, and let $p_A$ etc. represent the partial pressures of the reactants and products at any time as the reaction proceeds.  So at any time, the Gibbs free energy of the reaction mixture will be:
$$G=n_A(G^0_A+RT\ln{p_A})+n_B(G^0_B+RT\ln{p_B})+n_C(G^0_C+RT\ln{p_C}+n_D(G^0_D+RT\ln{p_D})$$ with $p_A=\frac{n_A}{n_A+n_B+n_C+n_D}P$ , etc., and  $n_A=n_{Ao}-a\xi$, $n_B=n_{Bo}-b\xi$, $n_C=n_{Co}+c\xi$, $n_C=n_{Do}+d\xi$, with $\xi$ representing the extent of the reaction and P representing the total pressure.
If we use this equation to make a plot of G vs $\xi$, the equilibrium of the mixture will be determined when G passes through a minimum.  If we differentiate  this equation with respect to $\xi$ and set the derivative equal to zero, we find algebraically that this happens when $$RT\ln{\left(\frac{p_A^cp_B^d}{p_Ap_B}\right)}=RT\ln{K}=-\Delta G^0$$Work it out yourself and see.
So both very different methods give the same result for the final equilibrium.


Hydrogen bond comparison How do you compare the strength of H bonds when it's the same atom on both sides of the H? Like $\ce{N-H\bond{...}N}$ vs $\ce{F-H\bond{...}F}$? Is one still stronger than the other, or are they both the same since in both cases the pull on both sides is equal?

In some cases, such as [F-H-F]- (biflouride), there is a symmetric hydrogen bond, with both H-F bonds equally strong.

However, just because the two atoms bonding to the H are the same doesn't mean it is a symmetric hydrogen bond.  For example water dimer does not have a symmetrical hydrogen bond, the H is closer to one O atom than the other.

There can be cases where the hydrogen being exactly in the middle is a potential energy minimum, and other cases where the middle is not a local minimum but there are two minima offset from the middle.

See Symmetry of hydrogen bonds in solution for more information.


Unknown yellow substance from mixing ascorbic acid, baking soda, and rubbing alcohol, then boiling So lately I've been using ascorbic acid (very very tiny amounts) to neutralize the chlorine and chloramine in my tap water (temporary solution to improve flavor in lieu of a proper filter).

Just for fun, I decided to react a solution of ascorbic acid with some sodium bicarbonate in an attempt to make a sodium ascorbate solution. Then I got to thinking, how can I get the ascorbate out of the solution?

I figured boiling would do the trick, but for some reason, I thought that adding isopropyl alcohol to the solution might do something, maybe precipitate the crystals out? That probably sounds completely stupid, but forgive me, it's been years since I studied any kind of chemistry (and I wasn't planning on ingesting the solution, so I figured it couldn't hurt to try).

Anyway, the alcohol did nothing visible, so I figured I'd go ahead and book off the solution (figuring the alcohol would dissipate quickly with its low boiling temp).

All went well for some minutes, the alcohol seemed to have boiled off (the smell was gone), and the liquid level was going down so I expected to see some white precipitate forming after a while. Instead what happened surprised me, the solution started turning yellow, then darker as more water boiled off. I did eventually start seeing some cloudiness at the bottom which I assume was the sodium ascorbate, but I didn't boil all of the water off because the color intrigued me.

Here's what it looked like after boiling about 2/3 of the water off:

Any ideas what this could be? My first thought was some contamination, but the pan I used was washed and rinsed very thoroughly.

Some more details on the solution:

about 1/2 teaspoon ascorbic acid
about 1/4 teaspoon sodium bicarbonate
mixed in roughly 2oz water and added a bit more sodium bicarbonate until everything stopped bubbling
when the solution was completely clear and not bubbling, added about 1oz isopropyl alcohol (maybe less)
boiled off about 2/3 of the solution

I eyeballed a lot of this and added stuff as I went (wasn't sure exactly how much ascorbic acid would dissolve) so those amounts may be quite a bit off, but the solution was crystal clear and not bubbling at all after I was done adding everything.

Hope someone with more knowledge than I can offer some insight!

That's sort of the colour that vitamin-C tablets go if they are kept long beyond their expiry date. I'd go with it being oxidation of some sort but chemistry.oregonstate.edu/courses/ch130/old/VITCTEXT.htm indicated that the simple oxidation of ascorbic acid is colourless... so what you've got is a more complex mix of oxidation products.

Instead of boiling off in an open pan, boiling off in a stable atmosphere might work better. You'd probably need some distillation glassware and a source of non-oxidising gas to flush out the oxygen dissolved in the water / keep an oxygen free atmosphere. You could probably get away with any arc welding gas or even use carbonated water as a source of CO2 gas to flush the air out of your equipment. Or of course you could go the whole chemistry-department route and boil the water off in a low pressure system without adding much heat.


What is the effect of substituting oxygen with sulfur in drug molecules? What sort of changes in the properties of organic or drug molecules can be anticipated if you substitute some or all of the oxygen atoms with sulfur atoms, and vice versa.

My interest in this question is for chemical graph theory perspective. There are numerous parameters calculated on molecular graphs (in trying to relate to chemical or biological activity) which are indifferent to the above substitution. I would like to make an appropriate adjustment if the effects are really significant.

In asking this question - Why do chalcogens (Group VI) stink so badly? - I received a couple answers and comments along the lines of "it's not quite so simple". The same would generally apply here; there isn't one change in behavior, or even a finite set of behavioral changes, that could be predicted by an oxygen-to-sulfur exchange.

Just as a basic example, compare methanol ($\ce{CH_3(OH)}$) to methyl mercaptan, ($\ce{CH_3(SH)}$). Methanol is a colorless liquid at RT with a sweetish odor. Methyl mercaptan, with a higher molecular mass due to the sulfur switch, is nevertheless a gas at RT and smells like rotting cabbage. Methanol metabolizes into formic acid and formaldehyde which in sufficient doses causes nerve damage, specifically to the optic nerve. Methyl mercaptan is pretty much its own metabolite; while also toxic in sufficient doses, it's readily filtered out by the liver and kidneys, as it's produced by bacterial breakdown of many of the foods we eat (especially highly-fibrous vegetables, legumes etc).

Perhaps the most common occurrence of sulfur in otherwise organic molecules is a sulfhydryl group (R-SH), forming a thiol. The alcohol-to-thiol switch is relatively common in human biochemistry and as such is also of interest in pharmacology. The Wikipedia page on thiols lists the following basic changes between an alcohol and its thiol analog:

More easily oxidized, producing disulfides, esters, ethers, etc
As thiols, more acidic than alcohols
Contrasting this, thiols dehydrogenate easily to form thiolate bases
Thiolates complex strongly to most metals

In general, you can expect the sulfur analog of pretty much any organic functional group (alcohols->thiols, aldehydes->thials, ketones->thioketones, esters->thioesters, ethers->thioethers, etc) to be more reactive. However, exactly how the sulfur switch will affect the actual reactions the compound will undergo is best studied case by case, as there is no one way that a particular sulfur-bearing functional group will behave when introduced into the complex chemical stew of human biology.

One near-constant; sulfur in small doses can smell pleasant, because sulfur is an essential not-so-trace element in humans (oranges and grapefruit, for instance, have some volatile sulfur compounds that contribute to their smell and taste). Sulfur in high concentrations stinks, because for the same reasons it is essential in biochemistry, too much of it is toxic (which is a pretty good illustration of the First Rule of Toxicology - every substance, in sufficient dosage, is toxic).


What kind of drugs can be absorbed through skin? Some drugs such as nicotine can be administered through skin. I thought the layers of skin are designed to prevent in-flow of any chemical/germs. Not all drugs get absorbed in this fashion. So do drugs like nicotine have any specialities(such as a special chemical group) that fool the barriers of skin? A link to an article/site would be very helpful, or even a book recommendation.

The skin is able to prevent some things from passing in and out of a person but it is not a perfect barrier.

The lipophilic things like nicotine, hydrogen flouride, VX, hexane and iodine vapour can pass through undamaged skin. These things are able to dissolve in the fatty parts of skin. Then they diffuse in through the skin into the person. Thus they enter the person.

Some water soluble ionic things can enter / leave a person through the skin, this can occur through the sweat pores.

For example cesium in humans does come out in sweat, in a 1980s case in Brazil where a lot of people were internally contaminated with cesium-137. It was found that cesium kept appearing on the surface of the skin of the contaminated people.


The Acidity of the Hydrated Iron (III) Ion Why is $\ce{Fe(OH_2)_6^{3+}}$ fairly acidic? This iron has six water molecules coordinated to it.

In other words, water itself is a very weak acid. But when water is coordinated to iron, it becomes more acidic. I can think of a few reasons; can you think of more? Also are my reasons valid?

1) Induction. The negative charge generated through loss of the hydrogen proton is stabilized by the coordinated water molecules. Plus the center of the square planar molecule is greatly positively charged (and therefore can handle the negative charge, despite the fact that iron itself is actually electropositive rather than electronegative - at least in iron's ground state).

2) Size of the ferric ion. The ferric ion is rather big (it is, after all, able to coordinate six water molecules, as opposed to only four for some other metal ions). So whatever positive charge it has after being ionized once is spread out over a large surface area and that makes entire compound relatively stable.

3) Oxidation state of the central ferric ion - positive 3. This indicates a high degree of ionic character in the coordination bonds. Withdrawal of electron density from the water molecules makes the hydrogens even more positively charged and thus even more electrophilic.

4) Positive charge density of the central ferric atom. Coulomb's law is in (significant) effect here. Positive and positive. Like charges repel. Kicking off a hydrogen proton - no problem.

When a water molecule coordinates to $\ce{Fe^3+}$, one non-bonded $sp^3$ hybrid orbital (with one of the 2 lone pairs) of oxygen overlaps with an empty $d$ or $p$ orbital of $\ce{Fe^3+}$. Electron density is transferred from oxygen to iron, and due to the electronegativity of oxygen, the polarization of the $\ce{O-H}$ bond in the water ligand is increased (more electron density is "pulled" from $\ce{H}$ towards $\ce{O}$). The hydrogens are more positively polarized than in a free (uncoordinated) water molecule, and this increases their electrophilic character and acidity to the point that they are readily abstracted by solvent (non-coordinated) $\ce{H2O}$ in aqueous solution:

$$\ce{Fe(H2O)6^3+ + H2O \rightleftharpoons  Fe(H2O)5(OH)^2+ + H3O+}$$

$$\ce{Fe(H2O)5(OH)^2+ + H2O \rightleftharpoons  Fe(H2O)4(OH)2^+ + H3O+}$$

Further deprotonation results in the neutral complex $\ce{Fe(H2O)3(OH)3}$, from which $\ce{Fe(OH)3}$ precipitates.

The aqueous solution of $\ce{Fe^3+}$ aquo complexes is fairly acidic, wth a pH of up to 1.5 depending on iron concentration. The first dissociation equilibrium lies mainly on the product side, which can also be seen in the color of the solution: Pure $\ce{Fe(H2O)6^3+}$ is pale violet and found in crystals of some hydrated $\ce{Fe(III)}$ salts, while aqueous solutions get their orange color from $\ce{Fe(H2O)5(OH)^2+}$.

You have mentioned ion charge (oxidation state) and charge density (which is proportional to ion size) of the metal center as potential factors of influence. One can expect that with increased charge and/or charge density, the polarization of the water ligands also increases, and thus the aquo complex should be more acidic. This is true for many aquo complexes; however, there are also some deviations from this rule, like $\ce{Cr(H2O)6^2+}$ and $\ce{V(H2O)6^2+}$ (reference).


What is 'autogenous pressure'? Compound 1 was hydrothermally synthesized under autogenous pressure[1]

I've seen this term used a few times, but with no real explanation of what it means forthcoming. I suspect it refers to the synthesis being performed at equilibrium pressure inside a sealed container (the aforementioned synthesis is performed in an autoclave at 160 C). Is my intuition correct?

[1] Sun, et al., CrystEngComm, (2012), 14, 3982-3988

Autogenous pressure synthesis is a form of solvothermal synthesis (also known as “hydrothermal synthesis” if the solvent is water). In most publications I could find, the two terms (“autogenous pressure” and “—thermal synthesis”) are used together, making it clear what it is: you seal your solution in a stainless steel autoclave, and heat it up. The pressure thus generated is called autogenous pressure because you don't apply it externally, but it simply derives from the closed nature of the container and the $(\rho,P,T)$ relationship for the fluid (its equation of state).


Anomalous electronic configurations Cu has an anomalous electron configuration. $\ce{Cu ~=~ 1s^2~2s^2~2p^6~3s^2~3p^6~4s^1~3d^{10}}$, it does not follow the usual pattern. In this case, the 3d subshell is filled before the 4s, which usually happens in the reverse order ($\ce{1s^2~2s^2~2p^6~3s^2~3p^6~4s^2~3d^{9}})$.

My question is, can you tell by an element's position on the periodic table (Group #, Row #, Block, etc.) that it will have an anomalous electron configuration? Do I have to memorize which elements have this property?

I am noting that Cu is in the 1B Group of Transition metals which implies that it has one valence electron, this is consistent with its configuration. will this be true for all cases?

Also: $\ce{Fe ~=~ 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6}$ and is a transition metal as well (Group 8). This implies it must have 8 valence electrons. But following the definition of valence electrons (The number of electrons in outermost shell), $n = 4$ is the outermost shell and I get a conundrum. Obviously if I add $\ce{4s + 3d}$ electrons everything is good. Why is this?

Many related questions have been asked on this topic before. I suggest you read these previous posts:

Why do elements in columns 6 and 11 assume 'abnormal' electron configurations?

How can one explain Niobium’s weird electronic configuration?

The Aufbau Principle is a general rule of thumb to find the electronic configuration of an element but usually many exceptions arise in the d and f block elements due to electron-electron interactions.

The concept of valence electrons does get a little fuzzy when you consider $d$ block elements since the energies of the $\ce{3d}$ and $\ce{4s}$ subshells are quite similar so I think its better to avoid it. $\ce{d}$ block elements can show a staggeringly large number of oxidation states (for example manganese which shows almost all oxidation states from 0 to +7). This can't be explained using the concept of valence electrons and the octet rule.

$\ce{Fe}$ itself can lose 2 electrons to form $\ce{Fe^{2+}}$ or 3 electrons to form $\ce{Fe^{3+}}$ so now again you can see that you can't explain this by saying that $\ce{Fe}$ has 8 valence electrons.

Other exceptional configurations: (there are quite a few):
niobium, platinum, gold, cerium, gadolinium, etc.


Using MO theory, give an explanation for the C-C bond length in cyanogen 1. The problem statement, all variables and given/known data

2. The attempt at a solution

I know that the MO diagram for CN- is this:

I am unsure how to draw the MO diagram for cyanogen. I know that the higher the bond order, the shorter the bond length. I have tried to draw a MO diagram for cyanogen by combining two of the MO diagram for CN:

Not sure if this is necessary but I have drawn the following resonance structures:

I think that structure (I) contributes the most as structures (II) and (III) have a positive charge on the nitrogen.

I think the C-C bond in cyanogen is shorter than the "usual" C-C bond lengths due to delocalisation across it but I don't know how to explain this using Molecular Orbital theory.

See my answer to your question on methyl vinyl ether for some background on molecular orbital (MO) theory.

In this case, your molecule is analogous to the all-carbon system, butadiene.  In cyanogen you have 4 p orbitals that can align, overlap and mix together to generate 4 molecular orbitals and you have 4 electrons (1 from each p orbital involved in a pi bond) to place in your molecular orbitals.  Since the carbons in cyanogen are $\ce{sp}$ hybridized you actually have a second set of molecular orbitals that are identical to the first in all respects except that the 4 p orbitals in the second set are rotated 90 degrees with respect to the other set of p orbitals.

Let's count electrons. We have 4 electrons in each of these orthogonal molecular orbitals that each contains 4 p orbitals, so we have 8 electrons in our two orthogonal sets of molecular orbitals.  We also have 2 lone pairs remaining on each nitrogen for a total of 4 electrons.  Finally we have the sigma system; we have 2 C-N sigma bonds and 1 C-C sigma bond each containing 2 electrons for a total of 6 sigma electrons.  For the whole molecule we have 8+4+6=18 electrons as expected.

As I mentioned in my other answer, we typically focus on the delocalized molecular orbitals formed by combining the p orbitals and disregard the sigma system (usually) when we create our MOs. Disregarding the sigma system usually simplifies matters.

There are a number of ways to create the MO diagram for butadiene or cyanogen, but all of them will have the common theme of mixing 4 p atomic orbitals and generating 4 molecular orbitals.  Here is one way to do it, let's start by mixing 2 p orbital to create the MO's for ethylene (or nitrile).

image source

Now we can mix 2 ethylenes (or nitriles) to generate butadiene (or cyanogen)

image source

We label molecular orbitals $\ce{\Psi_1 ~- ~\Psi_4}$ in order of increasing energy (so $\ce{\Psi_1}$ is the molecular orbital at the bottom of the figure and $\ce{\Psi_4}$ is at the top).  Don't worry about the "S"'s and "A"'s in the diagram, they are just telling us about the symmetry properties of the molecular orbitals which we don't need to focus on right now.

Notice how $\ce{\Psi_1}$ has overlap between the orbitals on the two middle carbons.  This indicates overlap and a higher bond order between these 2 carbons, just as your resonance structures II and II indicate.  Just like in butadiene, this carbon-carbon bond has some double bond character and, therefore there is a higher barrier to rotation about it then you might otherwise expect.


Sublimation and recrystallization of caffeine I'm looking to do an extraction of caffeine from energy drinks. I want to use both sublimation and recrystallization in my experiment for advanced purity.

My question is which should happen first, recrystallization or sublimation? Does it even matter which happens first?

This sounds like a nice project.

For the purification of caffeine, sublimation in vacuum and recrystallization from ethanol are both conceivable.

But first you have to separate it from the other ingredients, such as carbohydrates.

According to the manufacturer, 100 mL of Red Bull, a popular brand, contain 11 g sugars and 32 mg caffeine.

In order to isolate the caffeine, two methods are possible:

extraction with $\ce{CH2Cl2}$
solid phase extraction on a C18 column

The latter method might be interesting. Imagine that your pour the energy drink through a preconditioned column of this adsorbent. The caffeine will  be retained, whereas everything else passes through.

In a second step, you wash the caffeine off the adsorbent using ethyl acetate. Then, dry the organic solvent with sodium sulfate, filter off, evaporate the ethyl acetate, dry the crude material and check the melting point. If necessary, use any of the above two methods for purification.

If you'd like to read this in more detail, have a look at a procedure provided at Open BU, an open access initiative of the Boston University.

UPDATE

I'd probably use just one of the methods for purification, depending on the equipment available. But if you insist on using both, I'd recrystallize first.


Suitable reagents for nitration of thiophene I know that for nitration of benzene we can use concentrated nitric and sulfuric acid. As furan and pyrrole are less aromatic than benzene, they can be protonated more easily and undergo polymerisation. We can use $\ce{AcONO2}$ and $\ce{AcOH}$ instead.

Give the order of aromaticity: benzene > thiophene > pyrrole > furan, I'm not sure whether thiophene is suitable for use with concentrated nitric and sulfuric acid.

According to this reference by Katritzky et. al (2005) (see below):

  Thiophenes  are  easy  to  nitrate  compared  to  other  five  membered  heterocycles.  They  react  with  mild  nitrating  agents  such  as  copper  nitrate.

The reference further reports the nitration of thiophene to 2-nitrothiophene by nitric acid/Trifluoroacetic anhydride in 78% yield.

The classic $\ce{c HNO3/c H2SO4}$ conditions are too strong and likely to result in substrate degradation.

Reference:

Alan R. Katritzky, Eric F. V. Scriven, Suman Majumder, Rena G. Akhmedova, Novruz G. Akhmedov, Anatoliy V. Vakulenko, "Direct nitration of five membered heterocycles," ARKIVOC 2005, (iii), 179-191 (https://doi.org/10.3998/ark.5550190.0006.320).


Why is the methyl group electron-donating via the inductive effect? I have been reading mechanisms recently, and I find that $\ce{-CH3}$ is a $+I$ effect group. I understand the mechanisms that require this, but can someone explain at the molecular level why is $\ce{-CH3}$ a $+I$ effect group?

According to http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch04/ch4-3-2-1.html:

"Hyperconjugation is the stabilising interaction that results from the interaction of the electrons in a σ-bond (usually C-H or C-C) with an adjacent empty or partially filled p-orbital or a π-orbital to give an extended molecular orbital that increases the stability of the system."

How I see "hyperconjugation" is the formation of partial π bonds. Let's say we have a methyl group bonded to a positively-charged carbon atom with a vacant p orbital.

Ideally, it would be best if we could have an atom with a fully-filled p orbital, which is parallel to the p orbital of the this positively-charged carbon atom, to come and interact with this carbon atom. Such an interaction would stabilise the electron-deficient carbon atom. This is essentially the basis of hyperconjugation.

However, in our case, we do not have such an atom with a fully-filled p orbital around. Instead, what we have is a few C-H σ bonds. These C-H bonds are not parallel to the vacant p orbital of the positively-charged carbon atom but they are very close to parallel, around 19.5 degrees away, assuming that the geometry of the methyl group bonded to that positively-charged carbon atom is tetrahedral.

These C-H bonds are thus able to interact with the electron deficient vacant p orbitals of the positively-charged carbon atom, maybe not maximally, but sufficiently. These interactions would be what I would call "partial π bonds" and they stabilise the electron deficiency of the carbon atom. This interaction is also what is known as "hyperconjugation".

*Please note that the illustration is slightly inaccurate because the hydrogen atom would not pop off the methyl group for the formation of a full π bond between the two carbon atoms. This is only illustrated to better show the interaction of the C-H σ bonds with the p orbital of the positively-charged carbon atom.


Does the free energy of the universe decrease for every process? The second law of thermodynamics states that $dS_{universe}>0$ for all processes.

In addition to the second law, gibbs free energy is used to consider the feasibility of a reaction. $dG=dH_{syst}-T*dS_{syst}$. If $dG_{syst}<0$, this means that the reaction is spontaneous, but a reaction with $dG_{syst}$ can also occur if energy is input into the system (for example electrolysis).

Therefore, $dS_{univ}>0$ does not imply $dG_{syst}<0$.

However, is there such a quantity as $dG_{univ}$, and if so is the relationship $dG_{univ}<=>dS_{univ}$ true?

Didn't you ask the last question about this, too?

The issue here is that when $\Delta G > 0$, the reaction is indeed non-spontaneous as written. If you wish to add energy to the system, for example, via electrolysis, you should include the additional changes in the computation of the free energy change. You will find that after you sum up everything $\Delta G < 0$.

Also, note that $\Delta S_{\mathrm{universe}} < 0$ does not mean that the process will not happen; it only means that it's (fairly) unlikely to happen.


What is the mathematical basis behind the Jahn-Teller effect? Both first-order and second-order Jahn-Teller distortions play a very important role in chemistry.

It is often said that the Jahn-Teller effect is based on symmetry arguments, and hence nothing can be said about the extent or the exact form of the distortion.

What are these "symmetry arguments", and how can they be derived?

1. Perturbations

As already mentioned, the Jahn–Teller effect has its roots in group theory. The essence of the argument is that the energy of the compound is stabilised upon distortion to a lower-symmetry point group. This distortion may be considered to be a normal mode of vibration, with the corresponding vibrational coordinate $q$ labelling the "extent of distortion". There is one condition on the vibrational mode: it cannot transform as the totally symmetric irreducible representation of the molecular point group, as such a vibrational mode cannot bring about any distortion in the molecular geometry (it may lead to a change in equilibrium bond length, but not in the shape of the molecule).
$\require{begingroup} \begingroup
\newcommand{\En}[1]{E_n^{(#1)}}
\newcommand{\ket}[1]{| #1 \rangle}
\newcommand{\n}[1]{n^{(#1)}}
\newcommand{\md}[0]{\mathrm{d}}
\newcommand{\odiff}[2]{\frac{\md #1}{\md #2}}$

In the undistorted geometry (i.e. $q = 0$), the electronic Hamiltonian is denoted $H_0$. The corresponding unperturbed electronic wavefunction is $\ket{\n{0}}$, and the electronic energy is $\En{0}$. We therefore have

$$H_0 \ket{\n{0}} = \En{0}\ket{\n{0}} \tag{1}$$

Here, the Hamiltonian, wavefunction, and energy are all functions of $q$. We can expand them as Taylor series about $q = 0$:

$$\begin{align}
H &= H_0 + q \left(\odiff{H}{q}\right) + \frac{q^2}{2}\left(\frac{\md^2 H}{\md q^2}\right) + \cdots \tag{2} \\
\ket{n} &= \ket{\n{0}} + q\ket{\n{1}} + \frac{q^2}{2}\ket{\n{2}} + \cdots \tag{3} \\
E_n &= \En{0} + q\En{1} + \frac{q^2}{2}\En{2} + \cdots \tag{4}
\end{align}$$

In the new geometry (i.e. $q \neq 0$), the Schrodinger equation must still be obeyed and therefore

$$H\ket{n} = E_n \ket{n} \tag{5}$$

By substituting in equations $(2)$ through $(4)$ into equation $(5)$, one can compare coefficients of $q$ to reach the results:

$$\begin{align}
\En{1} &= \left< \n{0} \middle| \odiff{H}{q} \middle| \n{0} \right> \tag{6} \\
\En{2} &= \left< \n{0} \middle| \frac{\md^2 H}{\md q^2} \middle| \n{0} \right> + 2\sum_{m \neq n}\frac{\left|\left<m^{(0)} \middle|(\md H/\md q)\middle|\n{0} \right>\right|^2}{\En{0} - E_m^{(0)}} \tag{7}
\end{align}$$

The derivation of equations $(6)$ and $(7)$ will not be discussed further here.1

Distortions that arise due to the $\En{1}$ term are called first-order Jahn–Teller distortions, and distortions that arise from the $\En{2}$ term are called second-order Jahn–Teller distortions.

2. The first-order Jahn–Teller effect

Recall that

$$E_n = \En{0} + \color{red}{q\En{1}} + \cdots \tag{8}$$

Therefore, if $\En{1} > 0$, then stabilisation may be attained with a negative value of $q$; if $\En{1} < 0$, then stabilisation may be attained with a positive value of $q$. These simply represent distortions in opposite directions along a vibrational coordinate. A well-known example is the distortion of octahedral $\ce{Cu^2+}$: there are two possible choices, one involving axial compression, and one involving axial elongation. These two distortions arise from movement along the same vibrational coordinate, except that one has $q > 0$ and the other has $q < 0$.

In order for there to be a first-order Jahn–Teller distortion, we therefore require that

$$\En{1} = \left<\n{0}|(\md H/\md q)| \n{0}\right> \neq 0 \tag{9}$$

Within group theory, the condition for the integral to be nonzero is that the integrand must contain a component that transforms as the totally symmetric irreducible representation (TSIR). Mathematically,

$$\Gamma_{\text{TSIR}} \in \Gamma_n \otimes \Gamma_{(\md H/\md q)} \otimes \Gamma_n \tag{10}$$

We can simplify this slightly by noting that the Hamiltonian, $H$, itself transforms as the TSIR. Therefore, $\md H/\md q$ transforms as $\Gamma_q$, and the requirement is that

$$\Gamma_{\text{TSIR}} \in \Gamma_n \otimes \Gamma_q \otimes \Gamma_n \tag{11}$$

In all point groups, for any non-degenerate irrep $\Gamma_n$, we find that $\Gamma_n \otimes \Gamma_n = \Gamma_{\text{TSIR}}$. Therefore, if $\Gamma_n$ is non-degenerate, then

$$\Gamma_n \otimes \Gamma_q \otimes \Gamma_n = \Gamma_q \neq \Gamma_{\text{TSIR}} \tag{12}$$

and the molecule is stable against a first-order Jahn–Teller distortion. Therefore, all closed-shell molecules ($\Gamma_n = \Gamma_{\text{TSIR}}$) do not undergo first-order Jahn–Teller distortions.

However, what will happen if $\Gamma_n$ is degenerate? Now, the product $\Gamma_n \otimes \Gamma_n$ contains other irreps apart from the TSIR.2 If the molecule possesses a vibrational mode that transforms as one of these irreps, then the direct product $\Gamma_n \otimes \Gamma_q \otimes \Gamma_n$ will contain the TSIR.

In a rather inelegant article,3 Hermann Jahn and Edward Teller worked out the direct products for every important point group and found that:

  stability and degeneracy are not possible simultaneously unless the molecule is a linear one...

In other words, if a non-linear molecule has a degenerate ground state, then it is susceptible towards a (first-order) Jahn–Teller distortion.

Take, for example, octahedral $\ce{Cu^2+}$. This has a $\mathrm{^2E_g}$ term symbol (see this question) - which is doubly degenerate. The symmetric direct product $\mathrm{E_g \otimes E_g = A_{1g} \oplus E_g}$. Therefore, if we have a vibrational mode of symmetry $\mathrm{E_g}$, then distortion along this vibrational coordinate will occur to give a more stable compound. Recall that the vibrational mode cannot transform as the TSIR, so we can neglect the $\mathrm{A_{1g}}$ term.

What does an $\mathrm{e_g}$ vibrational mode look like? Here is a diagram:4

It's an axial elongation, which happens to match what we know of Cu(II). However, there is a catch. The vibrational mode is doubly degenerate (the other $\mathrm{e_g}$ mode is not shown), and any linear combination of these two degenerate vibrational modes also transforms as $\mathrm{e_g}$. Therefore, the exact form of the distortion can be any linear combination of these two degenerate modes. It can also involve negative coefficients, i.e. it might feature axial compression instead of elongation; there is no way to find that out using arguments purely based on symmetry.

Therefore, in this case, it is simply a coincidence that the $\mathrm{e_g}$ mode displayed is exactly the same as the form of the distortion seen in Cu(II). Nevertheless, it is encouraging to see that the axial elongation indeed transforms as $\mathrm{e_g}$ - which lends credence to the group theoretical analysis above.

On top of that, there's also no indication of how much distortion there is. That depends on (amongst other things) the value of $\En{1}$, and all we have said is that it is nonzero - we have not said how large it is.

This is what is meant by "impossible to predict the extent or the exact form of the distortion".

3. The second-order Jahn–Teller effect

Pearson has written an article on second-order Jahn–Teller effects.5

For the second-order term, the energy correction is of the form

$$E_n = \En{0} + q\En{1} + \color{red}{\frac{q^2}{2}\En{2}} \cdots \tag{13}$$

Here, the $q^2$ term means that $\En{2}$ has to be negative if we want to see a second-order Jahn–Teller distortion. Unlike the first-order case, if $\En{2} > 0$, there will not be any distortion.

The second-order correction to the energy comprises two terms. The first term, $\left<\n{0}|(\md^2 H/\md q^2)|\n{0}\right>$, is always positive. (The proof of this is left to the reader. Hint: use the fact that $\md H/\md q$ is hermitian.) It may be interpreted as a restoring force that tries to bring the nuclei back to their original positions, and it is related to the fact that if the electronic state remains unperturbed (i.e. $\ket{n} = \ket{\n{0}}$), the unperturbed nuclear positions represent the most stable nuclear configuration.

The second term has the form

$$\sum_{m \neq n}\frac{\left|\left<m^{(0)} \middle|(\md H/\md q)\middle|\n{0} \right>\right|^2}{\En{0} - E_m^{(0)}} \tag{14}$$

which may seem like a slight monstrosity, but it is actually much easier to analyse than it looks. The summation over $m$ indicates that we are going to count every single electronic state $\ket{m}$ that is not the ground state $\ket{n}$. Since the numerator is a square modulus, it is either zero or positive, and since $\En{0} < E_m^{(0)}$, the denominator is always negative; therefore, this term is necessarily either zero or negative.

If $\En{2}$ is to be negative, then we need the second term to dominate the first. For this to occur, there are two prerequisites:

The denominator must be small, i.e. $\ket{m}$ must be a low-lying excited state such that the energy gap $\Delta E = \En{0} - E_m^{(0)}$ is small;
The numerator must not be zero, i.e. there must be a low-lying excited state of the appropriate symmetry $\ket{m}$ such that $\Gamma_m \otimes \Gamma_q \otimes \Gamma_n$ contains $\Gamma_{\text{TSIR}}$.

The first condition usually means that it will suffice to consider the first few excited states. In many of the examples I know of, the excited state that mixes with the ground state is the first excited state. In such cases one can even simply get rid of the sum and set $\ket{m}$ to be the first excited state.

These symmetry requirements are much less restrictive than previously, and second-order Jahn–Teller distortions tend to be much more widely seen than first-order distortions. A small selection of compounds in which second-order Jahn–Teller distortions are important are: p-block hydrides, $\ce{PbO}$, $\ce{Hg^2+}$, $\ce{WMe6}$, $\ce{R2Sn=SnR2}$, and anti-aromatic compounds such as cyclobutadiene. Let us use octahedral $\ce{Hg^2+}$ as an illustration.

In undistorted $O_\mathrm{h}$ symmetry, $\ce{Hg^2+}$ has a closed-shell $\mathrm{d^{10}}$ configuration and therefore its electronic ground state is $\Gamma_n = \mathrm{A_{1g}}$. However, upon excitation of one electron from the 5d orbitals (specifically the $\mathrm{e_g}$ set) to the 6s orbital (which transforms as $\mathrm{a_{1g}}$), the term symbol changes to

$$\Gamma_m = \mathrm{E_g \otimes A_{1g} = E_g} \tag{15}$$

Therefore, a vibrational mode transforming as $\mathrm{E_g}$ will facilitate a second-order distortion, since

$$\Gamma_m \otimes \Gamma_q \otimes \Gamma_n = \mathrm{E_g \otimes E_g \otimes A_{1g}} \tag{16}$$

contains the TSIR. Again, there is no way of knowing the exact form or the extent of the distortion; we only know that it transforms as $\mathrm{E_g}$. In the case of Hg(II), the distortion is manifested as an axial compression to give a "2 short, 4 long" coordination geometry, which is often described as "linear".

A second factor that favours the distortion is the extremely small 5d–6s gap in mercury, due to relativistic 5d destabilisation and 6s stabilisation. To see the importance of the small $\Delta E$, consider $\ce{Zn^2+}$, which has a larger 3d–4s gap; linear Zn(II) compounds are rare, while linear Hg(II) compounds are the norm.

Most of the time, the relevant excited state $\ket{m}$ arises from promotion of an electron from the HOMO to the LUMO. It is easy to show that if this is the case,

$$\Gamma_m \otimes \Gamma_n = \Gamma_{\text{HOMO}} \otimes \Gamma_{\text{LUMO}} \tag{17}$$

The second-order Jahn–Teller distortion can then be viewed as a reduction in symmetry, such that the HOMO and the LUMO, which transformed as different irreps in the undistorted geometry, now transform as the same irrep and therefore mix with each other. In the case of $\ce{Hg^2+}$:

This interpretation using the symmetry of individual orbitals, however, only works when the relevant excited state $\ket{m}$ is derived by excitation of an electron! In some (admittedly very rare) cases, it is possible that both $\ket{n}$ and $\ket{m}$ are derived from the same electronic configuration. This is the case for cyclobutadiene, and the Jahn–Teller effect in cyclobutadiene cannot be rationalised using orbital mixing. $\endgroup$

Notes and references

(1) For more details, look up perturbation theory in your quantum mechanics book of choice. In such treatments, the perturbation is usually formulated slightly differently: e.g. $H$ is taken as $H_0 + \lambda V$, and the eigenstates and eigenvalues are expanded as a power series instead of a Taylor series. Notwithstanding that, the principles remain the same.

(2) There is a subtlety in that the symmetric direct product must be taken. For example, in the $D_\mathrm{\infty h}$ point group, we have $\Pi \otimes \Pi = \Sigma^+ + [\Sigma^-] + \Delta$. The antisymmetric direct product $\Sigma^-$ has to be discarded.

(3) Jahn, H. A.; Teller, E. Stability of Polyatomic Molecules in Degenerate Electronic States. I. Orbital Degeneracy. Proc. R. Soc. A 1937, 161 (905), 220–235. DOI: 10.1098/rspa.1937.0142. n.b. Considering that I don't have a more elegant proof for it, I don't have much of a right to call it inelegant.

(4) Albright, T. A.; Burdett, J. K.; Whangbo, M.-H. Orbital Interactions in Chemistry, 2nd ed.; Wiley: Hoboken, NJ, 2013.

(5) Pearson, R. G. The second-order Jahn–Teller effect. J. Mol. Struct.: THEOCHEM 1983, 103, 25–34. DOI: 10.1016/0166-1280(83)85006-4.


Can lye be extracted from bamboo ash? I know that the best source of lye is hardwood. A brief search tells me that Bamboo is considered a hardwood of the monocotyledon variety. I've done a lot of searching to find this answer, but I can't seem to find if I can use the same process with bamboo ash as I would with, say, Apple ash, to extract lye.

My question is two-fold:

1) Is it possible to extract lye from bamboo ash?

and if so,

2) Would this lye be a quality lye, capable of being used to create bar soap?

Yes, it's possible but as pointed out in Bamboo: an overlooked biomass resource? Biomass and Bioenergy, Vol. 19, pages 229-244:

  [bamboo] shares a number of desirable fuel characteristics with certain other bioenergy feedstocks, such as low ash content and alkali index

Potassium oxides are between 33-50% of the ash depending upon species and age of the plant.

Potassium oxides were between 0.16-0.58% of dry matter.

Sodium oxides were about a factor of 100 lower.


Specific Heat Derivation The specific heat capacity at constant volume and the specific heat capacity at constant pressure are intensive properties defined for pure, simple compressible substances as partial derivatives of the functions u(T, υ) and h(T, p), respectively,
$$c_v=\left ( \frac{\partial u}{\partial T} \right )_v$$
$$c_p=\left ( \frac{\partial h}{\partial T} \right )_p$$
Can someone explain why this is? The book from which I got this from doesn't derive these expressions.

If a body absorbs a quantity of heat $q$ its temperature will normally rise by a value $\Delta T$. The average heat capacity  over this temperature range is defined as $C_{av}\equiv q/\Delta T$. The instantaneous heat capacity at temperature $T$ is $C\equiv dq/dT$. This definition is not exact enough, however, until the path of heating is specified. From the first law $q$ could be replaced by $dU+dw$ so it is necessary to fix another parameter which is $V$ or $p$ giving rise to the definitions you quote.
Note (a) that at a phase change $C$ becomes infinity because heat is absorbed with no change in temperature. (b) As the heat capacity is the gradient of internal energy with temperature it describes how the internal energy increases. For example when $C$ becomes constant at high temperatures the molecules are absorbing heat a constant and maximum 'rate' with increase in $T$. (c) Using statistical mechanics the heat capacity can be calculated from a knowledge of a molecule's energy levels.


Reason for inability to find the absolute value of internal energy Today in my thermodynamics lecture, my teacher told me that it is not possible to find the absolute value of internal energy and so we have to calculate the change in internal energy.
So, my question is why isn't it possible to find the absolute value of internal energy? Is this statement true?

In order to answer this question, one needs to define what the absolute energy of a system is. Energy can be trapped in a system in ways not yet discovered or fully understood. Think of the energy associated with mass ($E=mc^2$), which is a result that was not known to the early founders of thermodynamics.

We need to define reference points, which we can use to compare our products with. Usually this "zero" point is defined as the energy of the individual atoms in their most energetically favorable state (phosphorus is an exception, since red phosphorus is much more available), such as $\ce{H2}$ and $\ce{He}$.


Is there a difference between ‘ammonium hydroxide’ and ‘aqueous ammonia’? Water ($\ce{H2O}$) can be added to ammonia ($\ce{NH3}$) in order to form ammonium hydroxide ($\ce{NH4OH}$):
$$\ce{NH3 + H2O -> NH4OH}$$

But we know that ammonia is gaseous at STP so writing $\ce{NH3 + H2O}$ equals writing $\ce{NH3(aq)}$ right?

If yes, then the equation will look like:
$$\ce{NH3(aq) -> NH4OH}$$

So my question is: is ammonium hydroxide $\ce{NH4OH}$ simply aqueous ammonia $\ce{NH3(aq)}$? Or am I missing something?
Can I resume the first equation like this?
$$\ce{NH3(g) + H2O(l) -> NH3(aq)}$$

Ammonia is a weak base, ammonium is a weak acid. Water is a very weak acid and hydroxide is a very strong base. Taken together, this tells us that equilibrium $(1)$ is pushed towards the reactant side.

$$\ce{NH3 (aq) + H2O (l) <<=> NH4+ (aq) + OH- (aq)}\tag{1}$$

Thus, it is practically impossible to form stoichiometric ammonium hydroxide; the $\mathrm{p}K_\mathrm{a}$ values tell us that ammonium ions and hydroxide ions put together will generate water and ammonia.

However, it is also true that equation $(1)$ is an equilibrium — everybody has seen a solution of ammonia in water display basic properties so hydroxide ions (and thus ammonium ions) are unquestionably there. The predominant species is dissolved ammonia, though.

$$\ce{NH3 (g) + H2O (l) <=>> NH3 (aq) + H2O (l)}\tag{2}$$

It is correct that the most important process when dissolving ammonia in water is $(2)$ — $(1)$ plays a more minor role. Thus, it is also more correct to speak of solutions of ammonia — but at the same time ammonium hydroxide is commonly understood even if not formally correct.

By the way beware: occasionally students are sent off to fetch ‘solid ammonium hydroxide’ as a joke — similar to students getting sent off to fetch ‘vacuum gas’, ‘three phenol rings’ etc.


Ionization energy of C2-, C2, and C2+ My textbook says that $\ce{C_2}^-$ has a lower ionization energy than both $\ce{C_2}$ and $\ce{C_2}^+$. I calculated that the bond orders of $\ce{C_2}^+$, $\ce{C_2}$, and $\ce{C_2}^-$ are 1.5, 2, and 2.5, respectively.

Why doesn't $\ce{C_2}^-$ have a higher ionization than the other two molecules? Shouldn't the molecule with the larger bond order have the larger ionization energy, because it would be more difficult to remove an electron from the stable bond?

Bond order has to do nothing with ionization energy, it is the bond dissociation energy which has to be dealt with bond order. For minimum ionization energy, the electron has to be lost from the highest energetic orbital (outer).

In $\ce{C2^-}$, the electron is lost from high energetic $\sigma(\ce{2p_x})$ (if you take the bonding along x-axis) orbital, whereas, in the other two molecules electrons are lost from low energetic  $\pi(\ce{2p_y})$ or $\pi (\ce{2p_z})$ orbitals. So $E_{\infty} - E_\text{initial}$ is higher for the latter case, which makes $\ce {C2^-}$ having lower ionization energy.


Circular dichroism of a hydrazone boron complex Last year, I came across a quite puzzling article, in which the authors claim to have found a photophysical explanation for aggregation induced emission (doi:10.1038/nchem.2612). The model proposed implied that the lowest-lying excited state ($\ce{S1}$) was a non-emissive ("dark") state, whereas $\ce{S2}$ (or $\ce{S3}$) was emissive, and depending on the solvent, the molecule could be more or less emissive.

They claim to prove their model by observing the "dark state", and among other concerns, one point that particularly retained my attention was that one of their proofs was a CD spectrum (supporting information page 43) of this molecule:

Now, simply put, I don't understand anything about how it is possible. To precise my trouble, this can be broken into different subquestions (please feel free to answer partially if you wish).

I see strictly no reason why this molecule should show such a CD spectrum in solution (77K in Glycerol/Ethylene glycol (8:2 v/v)). Even if it is chiral, it should be a racemate. Why does this molecule show CD?
The maximum of the band is 150 mdeg. Why is it so high?
Absorption and CD spectra are not completely superimposable. The authors claim that this demonstrates that the band they are looking at consists of two different electronic transitions (hence proving the existence of the "dark state"). I don't find this statement straightforward (especially judging by the signal shape). If you have references which elaborate on this aspect, or other references using this same argument, I would be very pleased to read them.

Am I missing something very important concerning chirality, or CD spectroscopy?

On the origin of chirality, you should read something about atropoisomerism. This sort of chirality arises due to a hindered rotation around a chiral axis.
It is most notably encountered in biaryls, because by introducing substituents you can make rotation more difficult and observe a CD. However, it is also present in very simple molecules. Think about H2O2: every conformation where the dihedral angle is between 0° and 90° is chiral. Probably your teacher didn't point this out when explaining butane conformations, but gauche conformers exist in enantiomeric forms. The problem with these small molecules is that to freeze rotation you should basically go to very low temperatures, and then separate the enantiomers. You would observe CD. If you can do that, you got material for a pretty awesome paper in my opinion!

In this case, think about rotating the phenyl ring around the N-Ph bond. Draw the structures, and you will notice that they are chiral. The claim of the authors is that in a high viscosity solvent this rotation is hindered at room temperature, thus you will observe CD (see more on pg. 46 of the Supp. Info).
Concerning the racemic mixture, I'd say you are right. It is not clear to me why they see CD without separating the enantiomers, or maybe they did but they didn't explain it thoroughly. They refer to some previous work, so maybe you could find more details in other papers.

Why is the maximum band so high?
High with respect to what? 150mdeg really doesn't mean anything if you don't have the concentration and cell length and you can't work out the delta epsilon and the g-factor. CD people never use mdeg because it is not an absolute unit. The authors do not report concentration and cell path, so you can't really say whether it is strong or not, also because absorbance is reported in arbitrary units, so you can't really work out the g-factor. Clearly, they're not CD people, they use it as tool in their research, so the information they provide is not really helpful to answer this question.

Concerning the signals, have a look at figure 64 in the Supp. Info. Think about spectra as a series of lines to which you apply some lineshape function (usually Gaussian or Lorentzian). Now, two isolated bands are exactly superimposable if they have the same intensity and the same broadening. If they are not superimposable, one of these two things is not right, or the bands are not isolated. Since you are speaking about two different spectra, you can always normalise them to the same intensity. Additionally, they are recorded in the same conditions (possibly simultaneously), so there is really no reason why the broadening should be different. The only other possiblity is that there is more than one transition under that signal.
The "dark state" means basically that, in UV such state has a low absorption because of some reason. However, some transition that is dark in UV can have a pretty significant CD signal, because maybe it is only magnetically allowed (think about the n-pi* transition in carbonyl groups, or some charge-transfer state).

I tried my best, however the most interesting point (why do they see CD) still remains obscure to me, mostly because it's weekend and I am too lazy to read the paper! If I remember, I'll maybe read it next week.

Edit:
One thing that I didn't mention is that the sp3 N atom is chiral as well, so actually here, when you put in the axial chirality, you're really speaking about diastereoisomers, if you can fix the chirality of the N atom. This happens in very particular cases (see Cinchona alkaloyds), but since we're in a high viscosity medium, who knows, this might be the case, and explain why only one of the two possible configurations of the chiral axis is preferred.


Does gas chromatography follow the "like interacts with like" rule of thumb? The "like interacts with like" rule of thumb works well when there are polar and non-polar molecules that are mobile (liquid-liquid extraction, membrane vs cytosol, questions of solubility).

Polar molecules interact with polar molecules rather than with non-polar molecules because the interactions are stronger.
Non-polar molecules "hang out" with non-polar molecules because that
way, they don't disrupt the stronger interactions between polar
molecules.

The latter is sometimes called the hydrophobic effect when
   water is present as a solvent. Self-assembly of detergents into
   micelles, folding of proteins, formation of lipid bilayers are all
   rationalized by "like interacts with like".

In a previous post, the question was asked why polar molecules have lower retention times than non-polar molecules when running gas chromatography with a non-polar column. One of the answers invoked "like interacts with like" as rationalization.

I have two questions:

Do polar molecules have lower retention times than non-polar molecules on a non-polar GC column (to make it comparable, the molecules should have similar boiling points)?
Does the "like interacts with like" rule of thumb apply to gas chromatography, given that there are no interactions in the gas phase and there is no hydrophobic effect in the stationary phase (i.e. the non-polar stationary phase interacting with the polar analyte would not disrupt any polar-polar interactions)?

In the post I referred to, the question started with a convincing statement:

  I remembering reading that the interaction between a polar molecule and non-polar molecule is stronger than the interaction between two non-polar molecules.

This would suggest non-polar molecules have the weaker interaction and elute first, even on a non-polar column.

For GC, the most significant factor affecting separation resolution is differences in vapor pressures, boiling points if you will, between analytes.  While column phase selection, oven temperature programming, detector type and temperature, and other factors will have an effect on the efficiency of separation, IMFs are not generally important for the majority of separations.  The "like interacts with like" principle does apply to GC such that in the case you describe, nonpolar analytes will spend more time interacting with the stationary phase while polar will have little to nearly zero interaction and elute first.  The separation of the various nonpolar compounds will, holding all the other column conditions constant, "boil down" to varying vapor pressures.

Edited for clarification:

To begin, I think it beneficial to keep in mind that the question you pose is sufficiently clear we are dealing with polar analytes using a non-polar GC column.  Regarding proprietary column selection, temperature programming, and the choice of various other method parameters, these things are necessarily important in an efficient and successful separation (generally called method development) but they do not do much to answer the principle questions asked about whether 1) polar compounds are less retained on a non-polar column, and 2) does "like with like" apply to GC.  To that end, I will again address these two primary questions and then try to clarify another issue brought into the discussion that does not bear directly on answering the questions as asked.

The most direct answer to your first question is simply, yes, polar analytes do have shorter retention times than non-polar compounds when using a non-polar column because polar compounds have little interaction (partitioning) with the column's stationary phase.  The polar molecules of interest have little affinity for the non-polar stationary phase, i.e. hydrophobicity.  Had it been a polar stationary phase column then IMFs would be significant.  You mention making things comparable by considering compounds of similar boiling points which is precisely why a ramping temperature program is most beneficial for efficiently resolving such a separation.  This too, however, is a method development consideration and thus an example of the applied science of gas chromatography taking advantage of pure science principles.

Considering your second question, this, too, is most directly answered with another yes - the "like with like" concept does apply here, but as you point out, this is a rule of thumb, and as such, a conceptual aid not strict principle.  The qualification about hydrophobic effects in the stationary phase you make in your wording of the question actually explains, as well as I could, what is happening.  Also, as @Blaise points out, the statement with which you end your original question referencing a previous user (I [remember] reading that the interaction between a polar molecule and non-polar molecule is stronger than the interaction between two non-polar molecules.), is in error: the interaction between polar and non-polar molecules is not stronger than between two non-polar molecules.  Perhaps what this person meant is the interaction between two polar molecules is stronger, and relatively much stronger, than interactions between two non-polars.

With regard to your request I clarify why IMFs are not generally important (in this case), it must be remembered that the analytes in GC are not in "pure liquid form" but, rather, in the gas phase and they are completely surrounded by inert carrier gas molecules - little chance for IMFs to be of significance.  The polar molecules to be resolved do not spend a significant time partitioning into the stationary phase so they cannot form immiscible droplets as you brought up.

With respect to @SteffX, your question is about GC specifically and is not a question of what is the best separation method to use.  I take your question to be about using GC (on small sample sizes, right or wrong on my part) to effect a separation for analytical purposes, not as a preparatory separation like distillation.  Also, your original question(s) are with regard to gas chromatography in general and make no reference to column packing, bore, or length (other than being of a non-polar stationary phase) and I tried to answer in general terms.  However, since capillary GC columns were brought into the discussion, I think it deserves clarifying that, in fact, most capillary column separations are not isothermal.  Without going too deeply into something you were not asking about, it should be noted that isothermal temperature separations are a simpler and therefore preferred method development as long as you have good analyte resolution; the more common practice is to program a temperature gradient to resolve similar boiling point compounds - part of your original Question 1.  Lastly, Van der Waals force(s), a subject for much definitional discussion in its own right, have little importance to your polar analytes at GC conditions, and are moot almost before the sample hits the injector.

I hope this has been helpful and, in my opinion, I think you already have a good understanding of this.


What is a "lagged bath"? First question, and this is one I've searched for in vain.

In the context of setting up cooling baths, I've come across the term "lagged" bath.  I cannot extract what this means from any available context.  I assume it has something to do with time passing (i.e. a lag), but I'm not certain.  If I had to guess, I'd assume it means allowing the cooling agent to come to equilibrium temperature, but again, not certain.

Any direct technical experience that can explain this term would be greatly appreciated.

I'm not 100% certain on this, but my best guess on this would be referring to 'lagged' as in 'lagging', as in insulation material. So, a lagged bath is an insulated bath.


How do the strong acids dissociate? In writing net ionic equations involving strong acids, one needs to know how strong acids dissociate. For example, the dissociation of $\ce{H2SO4}$:

\begin{align}\ce{
H2SO4 &-> 2H+ + SO4^{2-}\\
H_SO4 &-> H+ + HSO4^{-}\\
}\end{align}

The reason I ask is because apparently the molecular equation

$$\ce{H2SO4 + 2KOH -> 2H2O + K2SO4}$$

has the net-ionic form $$\ce{H+ + OH- -> H2O}$$

It would be helpful if you could describe in general how I should write the dissociation of the strong acids in the net ionic equation.

This is a little tricky. I would argue that only the first proton dissociates completely and should be considered "strong" - using the rules for writing net ionic equations, where we only dissociate strong electrolytes, this would meant that we would write:

$\ce{H2SO4 + 2KOH -> 2H2O + K2SO4}$

$\ce{H+ + HSO4- + 2K+ + 2OH- -> 2H2O + 2K+ + SO_4^{2-}}$ - here only the first proton dissociates completely

Canceling out spectator ions gives:

$\ce{H+ + HSO4- + 2OH- ->2H2O + SO_4^{2-}}$

Your teacher (or whoever is giving you that answer) is dissociating the second proton as well, to give:

$\ce{2H+ + SO_4^{2-} + 2OH- ->2H2O + SO_4^{2-}}$

Which becomes:

$\ce{H+ + OH- ->H2O}$

There are a few reasons why they might do this, but I would argue that the 1st approach (which only dissociates the first proton) is best because it more accurately reflects what is happening, in the sense that $\ce{HSO4-}$ is a weak electrolyte. This will become important later when you study acid-base equilibria.

In general, you dissociate protons from acids as if they were regular metal cations - just "pull them off" and don't forget the positive charge. For polyprotic acids, the first proton is strongly acidic, or dissociates completely, but the other protons are not. There might be exceptions to this rule but I can't think of any off the top of my head.


Definition of Nucleophilicity I have a rather quick question on the definition of Nucleophilicity. To get straight to the point, am I right in saying that the most appropriate definition of Nucleophilicity, regardless of whether the reaction takes place in solvent or in the gas phase, is the following; Nucleophilicity is a measure of how close in energy the HOMO of a Lewis Base is to the LUMO of a Lewis Acid?

As far as I remember from my lectures nucleophilicity has the same definition problems as electronegativity.
So unlike e.g. ionisation energy you cannot define an absolute scale of ordering which you can verify experimentally
and just as in the case of electronegativity there are several possible definitions (compare https://en.wikipedia.org/wiki/Nucleophile).
Similar to electronegativity the definitions arrive on quite similar results. ($\ce{Br^-}$ is a good nucleophile in polar, protic solvents...)
Other formulations should be "qualitatively consistent" with previous ones.

Your alternative formulation has two big problems:

There is no solvent dependency.
You completely omit steric effects and nucleophilicity is also a kinetic phenomenon.


Boundary lines in phase diagrams and the lever rule My first question is: What exactly happens on a point situated on a line? Or on a point such as E in this figure?

Does point E mean that the $\delta , \gamma + \delta , \delta + \epsilon , \gamma + \epsilon $ phases all coexist here? If so, then why would it be considered a eutectoid invariant point where three phases coexist if in reality there are more than three phases coexisting?

My second question is: How can I apply the tie line and lever rules on a point on a point situated on top of a line, if I want to find Liquid percentage Solid Percentage, and maybe concentrations of one component in the liquid phase and solid phase? Is it undefined on top of the line?

Also, regarding the nature of the lever rule, is it some sort of empirical rule without any real theoretical foundation? At first glance, I can't believe it's actually accurate. Why is it true?

Thanks and sorry if my questions are unclear...

Lots of questions wrapped in one, but they do all come down to understanding what is represented on a phase diagram.  Up front, I'll suggest Porter and Easterling's classic text on 'Phase Transformation in Metals and Alloys' - it has lots of good info for lovers of thermodynamics and phase transitions.

In a system like Cu-Zn that you have pictured above, there are a number of different phases observed.  You could imagine that at a given temperature you could construct a free energy curve for each phase as a function of composition. At each composition, only one will have the lowest free energy, and one might think you could identify the phase diagram that way. However, a straight line across the G vs. composition plot is special - it represents where the chemical potentials of both elements are constant.  So a line that is tangent to two free energy curves represents a range of compositions where the elements in the two phases have the same chemical potential.  This then is how you construct the phase diagram - find the minimum free energy across composition, then find the common tangents (if any). An example is shown here:
The top plot is the free energy curves for all the phases, the bottom plot is the phase boundaries determined by the common tangents.  The dashed vertical lines up top are where the common tangent points are, coming down to their position on the phase diagram.

Now the $\delta$, $\gamma$, $\epsilon$ are specific phases.  In regions with just one of them listed, that is the only phase stable at the temperature and concentration.  Regions such as $\delta + \gamma$ are regions where both are stable. There relative amount has to follow the lever rule, which is just a linear variation across the two-phase region from, say, all $\gamma$ right on the boundary of $\gamma$ with $\gamma + \delta$, to all $\delta$ on the other side.

The horizontal lines are there mostly to emphasize when a new phase appears or disappears.
So, at point E on you diagram above, there is one specific temperature and composition at which $\delta$, $\gamma$, and $\epsilon$ are in equilibrium.  Go up in temperature, and it wants to be all $\delta$; come down in temperature and you enter the two-phase $\gamma$-$\delta$ region with the relative abundance from the lever rule. At P, the horizontal line is to point out that $\epsilon$ disappears into the $\delta$-liquid region.

So, at the eutectoid E, there is one point where the three phases coexist (Gibbs phase rule). Moving away from that point moves into single or two-phase regions. Everywhere along the horizontal line through E is $\gamma$+$\epsilon$, except at exactly point E.

As for the lever rule, that follow directly from coming along the common tangent - the ratios of the two phases have to vary linearly or else you are deviating from the common tangent.


Is this molecule possible? And if so, what is it called? (Cyclic groups) I saw a little chemistry pun with a couple of hydrocarbons included, and it had me thinking if they were actually possible  and if so, what would the name given to it be.

The hydrocarbon I am curious about:

For those wondering what the pun is:

If you want to consider the diene as the main chain, you could call this 2-cyclohexyl-1,3-butadiene. This used to be the proper name for the compound, as double bonds used to take priority over any length of a carbon chain in IUPAC nomenclature. This compound exists, and you could theoretically purchase it at some place like this. In recent years, the proper priorities for IUPAC have changed. Now, the longer chain takes priority over double bonds, and thus this compound should be referred to using the cyclohexane as the main chain. This means that the compound should be called (buta-1,3-dien-2-yl)cyclohexane. For all practical purposes, either name should work, as they both describe the same molecule; however, the second name is consistent with IUPAC standards.


Is the caffeine in coffee an oxidation agent? I heard a friend say:

  You have to watch out for coffee, the caffeine is an oxidation agent.

Now I remember carbon Chemistry from year 12.

You can oxidise ethanol to ethanal, and then to ethanoic acid.

But I don't understand how this would apply to caffeine.

My question is: Is the caffeine in coffee an oxidation agent?

No, caffeine is not known as an oxidizing agent.


How does di-isobutylaluminium hydride ( DiBAl-H ) reduces carboxylic acid $\text{DiBAl-H}$ reduces by electrophilic attack and provides hydrogen by $\text{H}^-$ transfer. Unlike $\text{LiAlH}_4$ it does not directly provide $\text{H}^-$ ion. So it should not undergo acid base reaction with carboxylic acid. But my friend argued that it undergoes acid base reaction. I can't find any reaction mechanism for reduction of carboxylic acid with $\text{DiBAl-H}$, can someone propose a mechanism for it?

Reduction of esters of carboxylic acids directly to aldehydes is of great synthetic interest. Zakharkin and Khorlina in 1962 (Ref.1) has shown that, diisobutylaluminum hydride (DIBAL-H) could be very useful in reducing esters to aldehyde in high yields at low temperatures (about $\pu{-70 ^oC}$). They also found that solvents affect reduction significantly. For example, the reduction reaction in toluene or hexane proceeds to give corresponding aldehydes in yields l0-15% more than those in ether as the solvent. Previously, in 1959, Miller et al. (Ref.2) found that DIBAL-H is a more selective reagent than lithium aluminum hydride in the reduction of nitriles to aldehydes, and it can also be used to reduce benzoic acid to give 72% yields of benzyl alcohol (they also showed esters can be reduced to corresponding alcohols). The conditions and yields of these references are summarized in this page. In Ref.2, during acid reduction, Miller et al. have used 3 equivalents of DIBAL-H and observed that a gas (hydrogen) evaluation upon the addition of the hydride to benzoic acid until one mole of the reducing agent had reacted. Thereafter, the addition of more DIBAL-H gave no gas. Thus, the first reaction is, presumably an acid-base reaction. However, DIBAL-H here should be acting as a Lewis-acid instead of a base, as illustrated in scheme 1:

The ist mono-adduct, then further reacted with second DIBAL-H to give di-adduct, which upon acid hydrolysis would give expected aldehyde (Scheme 2):

References:

L. I. Zakharkin, I. M. Khorlina, “Reduction of esters of carboxylic acids directly into aldehydes with diisobutylaluminum hydride,” Tetrahedron Letters 1962, (14), 619-620 (https://doi.org/10.1016/S0040-4039(00)70918-X).
A. E. G. Miller, J. W. Biss, and L. H. Schwartzman, “Reductions with Dialkylaluminum Hydrides,” J. Org. Chem. 1959, 24(5), 627–630 (DOI: 10.1021/jo01087a013).