Why does salty water make hard boiled egg shells more fragile? Boiling eggs in salty water rather than regular water makes the salt react with egg shells making them more fragile. My guess is that it's either sodium or chloride reacting with the calcium in eggs. Can someone please explain what is going on in there, and why the reaction makes the egg shell more fragile?

Here's an example YouTube video this effect is mentioned.

The eggshell is made of calcium carbonate mostly, but the crystal structure is held together by matrix proteins (more references in Wikipedia). Salt creates ionic interactions with the protein molecules, breaking protein-protein interactions, and in this case, probably protein-mineral interactions as well, debilitating the eggshell structure.


Wavelength and Colour in Coordination complexes My textbook said that coordination complexes with weak field ligands require a larger wavelength for obtaining enough energy for electron transfer(and hence, the manifestation of a colour), but if E = hv(E is the energy, h is Planck's Constant and v is the frequency), then shouldn't the wave of higher frequency be needed?

Low spin (formed from strong field ligands) complexes result in a larger difference in energy between the $t_{2}g$ and $e_{g}$ orbitals. High spin (caused by weak field ligands) complexes are, evidently, the opposite.
This means thats high spin complexes require lower energy photons than low spin. Since wavelength and frequency are inversely proportional, high spin complexes require lower frequency (longer wavelength) photons to promote transitions.


Using molecular orbital for determining leaving group trend and nucleophilicity Here is the question.

  Which one has better leaving group, cyclohexanol or cyclohexane that has $\ce{OH_{2}^{+}}$ substituent? Explain.

I know the answer is the latter because if both compounds are subjected in substitution reaction, the latter will yield a weak conjugate base ($\ce{H2O}$). There are some interesting facts from there which wants me to prove the answer with molecular orbital approach.

Here, I cite the part from it.

  The leaving group itself does play a slight role here in determining the energy of the $\unicode{x3c3}^*$ orbital and thus whether it is energetically favourable for the nucleophile to attack.

Based on the cited part, looks the orbital plays the role in nucleophilicity although the iodide example gives counterexample of nucleophilicity vs leaving group.

Does HOMO orbital play role in determining nucleophilicity? Based on molecular orbital of $\ce{OH-}$ and $\ce{H2O}$, the HOMO of $\ce{H2O}$ is higher than the former, thus is it correct to interpret $\ce{H2O}$ is more nucleophilic than $\ce{OH2+}$?
If nucleophilicity criteria is solely based on energy of HOMO orbital, is there a bound where the energy of HOMO orbital becomes electrophilic (or the LUMO orbital becomes nucleophilic)?
If leaving group is a thermodynamic factor (as noted by first source), would molecular orbital alone (including frontier molecular orbital) suffice to predict the leaving group order?
Can molecular orbital describes the iodide case (good leaving group but good nucleophile)?

The leaving group trend is supposed to be for similar weak bases (halides, $\ce{OTs}$, and $\ce{OMs}$).

MO theory can certainly be used to explain substitution reactions from a mechanistic perspective, and to help rationalize trends of relative reactivity. However, I think that the best way to build a mental model that gives you practical predictive power is to also incorporate the simpler ideas that mostly reduce to Lewis acid/base theory that most introductory textbooks teach.

Remember that "nucleophile" and "leaving group" are roles that molecules take when they participate in a specific type of reaction. They are not intrinsic molecular properties. MO theory can fill in part of the picture, but other factors (such as steric hindrance in the nucleophile or electrophile), solvent selection, heat, concentration, etc. can all affect reaction rates and how well a given compound takes up the role of being a nucleophile or electrophile in a given reaction context.

So, to answer your questions:

1) The energy of the HOMO does play a role in determining nucleophilicity since, from a mechanistic perspective, a substitution reaction occurs when the HOMO of a nucleophile overlaps with (and populates electrons into) the LUMO of the electrophile. However, HOMO energy levels are difficult to predict off-hand. A better explanation for the hypothetical difference in nucleophilicity between H2O and H2O+ is that H2O+ is strongly electron-deficient, with the oxygen lacking an octet. Therefore, it is extremely unlikely that the oxygen will act as a nucleophile. This is just Lewis acid/base theory.

2) The LUMO is, by definition, unoccupied. There are no electrons in the LUMO that could act in a nucleophilic manner towards another orbital. On the flip side, if the HOMO is occupied then that orbital is "full" and there is no way that it could act as an electrophile by accepting more electrons.

3) No - see the second paragraph of my answer. The energy of the electrophile's LUMO in question -- the sigma-antibonding orbital -- is certainly relevant, but many other factors are at play: sterics, the bond length between the leaving group and the electrophilic carbon, solvent effects, etc. The most useful, generalizable approach to predicting LG ability is Lewis basicity; the weaker the base, the better the leaving group. Thus, H2O is a better leaving group than OH-, Br- is better than Cl-, etc.

4) From a Lewis/acid base standpoint, one would predict I- to be a good leaving group since it is a very weak base. MO theory predicts that I- will be a better nucleophile than the other halides because its valence electrons reside in 5p orbitals, which are much higher in energy and more close in energy to the sigma-antibonding LUMO of the electrophile.


Separate precious metals from catalytic converter I am looking to separate the small quantities of platinum and palladium in a catalytic converter. However since these are very inert metals I am unsure how to proceed about doing so. Melting the converter is an option I have considered but I am unsure whether I will be able to reach high enough temperatures in order to obtain a decent amount of metal, so I am leaning more towards performing reaction.

Let me try to talk you out of it ;-)

The precious metals in a catalytic converter sit in a honeycomb ceramic that contains a lot of synthetic Cordierite. This material isn't only heat-resistant, but also quite hard.

Industrial recycling of catalytic converters typically starts with

dismantling (decanning) the ceramic monolith using guillotine shears
breaking down the ceramic to smaller pieces in hammer mills connected by a closed tubing system
milling the material to dust in a ball-mill blender

You might want to wear a filter mask during these steps.

Industrial recycling almost exclusively uses pyrometallurgical tretment (melting the metals) then.

Having ruled that out, there are two other options:

hydrometallurgical treatment
In order to separate the metals from the ceramic, you can either try to

dissolve the ceramic in hot alkaline solution in a closed reactor under pressure (to reach temperatures above 100 °C). This leaves the metals intact.
leach the metals out by using strong oxidizing acids, such as mixtures of concentrated nitric and hydrochlorid acid. You have the metals in solution now.

What are your plans for personal protection during these steps and for the waste water management?
biomobilization
There is some ongoing research to replace the unpleasant conditions of hydrothermal treatment with microorganisms that can leach the solid metals and store them. As an example, have a look at an article by Helmut Brandl, Stefan Lehmann, Mohammad A. Faramarzi, and Daniel Martinelli, published in  Hydrometallurgy, 2008, 94, 14-17. The authors examined, whether metals could be converted to cyanido complexes, such as $\ce{[Pt(CN)4]^{2-}}$, by cyanide-producing microorganisms. A PDF of their article, titled Biomobilization of silver, gold, and platinum from solid waste materials by HCN-forming microorganisms is available at no cost from ZORA (Zurich Open Repository and Archive). In the case of platinum, however, the results were rather disappointing.


What is the correct Lewis structure of NH3O? Choose the most appropriate Lewis structure for $\ce{NH3O}$ among the two:
 and

The oxygen in the second case has 3 nonbonding pairs around it.
I chose the first as an answer, yet the second seems to be the correct one. Can anyone please explain why?

Technically, both the structures $\ce{H2N-OH}$ and $\ce{H3N\bond{->}O}$ may exist. However, in reality hydrogen atom is rather prone to migration and the second structure is not favorable. So, for a compound with composition $\ce{NH3O}$, the correct structure would be $\ce{H2N-OH}$. The structure of second type is stable for compound $\ce{NOF3}$ and may be observed for amine oxides like $\ce{(C2H5)3N\bond{->}O}$ or pyridine N-oxide.

Why $\ce{H2N-OH}$  and not $\ce{H3N\bond{->}O}$? One  can come with several explanations for that, but I would like to focus on one: charge distribution. In $\ce{H3N\bond{->}O}$, the nitrogen has to carry a formal positive (and actually very real) charge, while for $\ce{H2N-OH}$, the structure has no formal charges. Still, according to some sources[1] up to 20% of hydroxylamine in water exists as ammonia oxide $\ce{H3N\bond{->}O}$, probably, due to stabilization by hydrogen bonds.

Somewhat similar uncertainty may be found for phosphorus compounds (like hypophosphorous acid) and sulfur compounds (like sulfinic acids).

Kirby, A. J.; Davies, J. E.; Fox, D. J.; Hodgson, D. R. W.; Goeta, A. E.; Lima, M. F.; Priebe, J. P.; Santaballa, J. A.; Nome, F. Ammonia oxide makes up some 20% of an aqueous solution of hydroxylamine. Chem. Commun. 2010, 46 (8), 1302. DOI: 10.1039/b923742a.


What is the rule for the punctuation marks for a bicyclo compound? What is the rule for the punctuation marks within the bracketed figures for a bicyclo compound? Is it a 'dot' or 'comma'? In some books there are commas whereas dots in some others.

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-16.2.2 Full stops
Full stops are used to separate numbers that denote lengths of bridges in polyalicyclic names constructed according to the von Baeyer system (…) and of chain lengths between spiro atoms in von Baeyer spiro names (…).

Examples:
bicyclo[3.2.1]octane
6-oxaspiro[4.5]decane


Maximum required order for the Boys function I am planning to pretabulate a lookup table in order to evaluate the Boys function. We can expand the Boys function as:

$$F_n (x_t + \Delta x) = \sum _{k=0}^\infty \frac{F_{n+k}(x_t)(-\Delta x)^{k}}{k!} \hspace{16pt} \tag{9.8.12}$$

We can then use downward recursion (upward is numerically unstable) in order to evaluate the function for any value of $n$ (order):

$$F_n (x) = \frac{2xF_{n+1}(x)+\exp(-x)}{2n+1} \tag{9.8.14}$$

I would like somebody to explain how one chooses which value of $n$ the lookup table will take? My working/logic so far as:

The Hermite integral has the following recursion (McMurchie-Davidson scheme, $x$ direction):

$$R^{n}_{t+1,u,v} = tR^{n+1}_{t-1,u,v} + X_\mathrm{PC}R^{n+1}_{t,u,v} \tag{9.9.18}$$

$\hspace{16pt}$ with:

$$\hspace{16pt} R^{n}_{0,0,0}=(-2p)^{n} F_{n}(p R_\text{PC}^{2}) \tag{9.9.14}$$

$\hspace{16pt}$ and so the recursion should limit $n$ to $n\leq t+u+v$.

In 2-electron integrals, the Hermite integrals are bound by $t \leq l_\mu + l_\lambda + l_\nu + l_\sigma $. Therefore $n$ is bound by the sum of the total angular momenta of the basis functions, i.e. $n \leq L_\mu + L_\lambda + L_\nu + L_\sigma$ where $L_\mu = l_\mu + m_\mu + n_\mu$.
Therefore, considering the extreme case of four $L=3$ basis functions, $n$ should be bound by $n \leq 12$.

Does this mean that a lookup table should start at $n=12$ and always recur down from this, or should I make different lookup tables and assess which one to use depending on the largest angular momentum functions in each individual computation? Are there any flaws or holes in my above logic?

All equations taken from Molecular Electronic Structure Theory, Helgaker et al. (2000).

I think your logic seems fine.

When you make your look up tables, you want to make a table for each $n$ up to some value of $n_\text{max}$. Consider you want to calculate an integral that only requires $n=2$, then it would be very inefficient to calculate $F_{12}$ and then down recur.

In your equation (9.8.12) we can notice that in the sum we have the Boys function as $F_{n+k}$. According to "Molecular Electronic Structure Theory", we might need the first $6$ terms of the sum. I.e. when doing your tabulation you will need to calculate all $n$ up to $n_\text{max}+6$.

Also when you do your tabulation you need a maximum range of your tabulated Boys function, i.e. a maximum value for where you can start using the long range approximation afterwards (else your tables will be HUGE). The long range approximation is given as:

$$F_n(x)\leq\frac{(2n-1)!!}{2^{n+1}}\sqrt{\frac{\pi}{x^{2n+1}}}\tag{9.8.10}$$

For which values of $x$ to start using this approximation for different $n$ you can take a look at Table II in Efficient recursive computation of molecular integrals over Cartesian
Gaussian functions. I am sure newer litterature on this topic exist, but I haven't encountered it.

As a side note, you can also calculate the Boys function with out tabulation (this might be less effective). By using the following formula:

$$F_n(x)=\exp(-x)\sum_{i=0}^\infty\frac{(2n-1)!!(2x)^i}{(2n+2i+1)!!}$$

This formula makes an implementation of the Boys functions much easier. If you are making an integral code from scratch, I would suggest to implement the above formula. You can always speed up the code later. Note with the above formula you can also just calculate the Boys function for the maximum $n$ you need and then use downwards recursion. If your goal is to implement the Boys function via. tabulation, then you can also use the above formula to generate the values for your look-up tables.


Thermodynamics of chemical reactions "at constant pressure" Chemical reactions occur at constant temperature and pressure.

Consider a gaseous, equilibrium reaction: $\ce{2NO2(g) <=> N2O4(g)}$. Most questions/textbooks formulate such questions by stating: The reaction happens at ($T$) temperature and ($P$) pressure.
This gives the impression that the surrounding Temperature and pressure are constant.

The first confusion that then arises is: For thermodynamic quantities such as enthalpy that involve pressure, which pressure should we use, internal or external?

After reading some chem.SE answers, it seems to me that $P_{sys}=P_{surr}$ is a basic assumption. However this doesn't seem to go well with the example I mentioned: Since the pressure of the system will be given by $(n_1)RT/V + (n_2)RT/v$, where $n_1$ and $n_2$ are the moles of $\ce{NO2}$ and $\ce{N2O4}$. Clearly, as the reaction progresses, ($n_1+n_2$) changes, and thus,the pressure of the system changes. If the surrounding pressure is kept constant, this beaks the "assumption": $P_{sys}=P_{surr}$.
Am I missing something?

You have correctly stated the pressure of the system is $$p = \frac {(n_1 + n_2)RT}{V}$$
But at the end of your question you incorrectly imply the volume is constant and the pressure changes. That is not true. Constant pressure means the external constant pressure ( like the atmospheric one ) keeps the system pressure constant. Imagine there is a massless, frictionless piston, ensuring $p_\mathrm{ext} = p_\mathrm{sys}$. It is similar as a thermostatic water bath keeps the system temperature constant.
So less confusing is to rewrite the ideal gas state equation :
$$V = \frac {(n_1 + n_2)RT}{p}$$
There are 4 variables: $p, V, T, n_\mathrm{tot}=n_1 + n_2$.
But there are just 3 degress of freedom = 3 independent variables.
Either $V, n_\mathrm{tot}, T$ are given, and $p=f(V, n_\mathrm{tot},T)$,
either $p, V, T$  are given, and $n_\mathrm{tot}=f(p,V,T)$,
either $p, n_\mathrm{tot}, T$  are given, and $V=f(p,n_\mathrm{tot},T)$,
either $p, n_\mathrm{tot}, V$  are given, and $T=f(p,n_\mathrm{tot},V)$.
We have already given 2 variables: $p, T$, with $n_\mathrm{tot},V$ remaining free.
But there is the equilibrium relation
$$K_p = \frac {p_{\ce{N2O4}}} {({p_{\ce{NO2}} )}^2}=\frac {p \cdot ( 1 - x_{\ce{NO2}}) } {({p \cdot x_{\ce{NO2}} )}^2}$$
$$K_p \cdot p \cdot {({x_{\ce{NO2}} )}^2} + x_{\ce{NO2}} - 1 = 0$$
Solving the quadratic equation we would finally get $n_\mathrm{tot}$, so all 3 degrees of freedom are saturated with given values for $p, T, n_\mathrm{tot}$.
So in our case, the following applies:
$p, n_\mathrm{tot}, T$  are given, and $V=f(p,n_\mathrm{tot},T)=\dfrac {n_\mathrm{tot}RT}{p}$


Why do buffers need to be composed of equal amounts of the acid and salt? Say you have $\ce{NaOH}$ and $\ce{CH3COOH}$ and want to make a buffer.
Theory teaches 6 mols of $\ce{CH3COOH}$ and half, 3 mols, of $\ce{NaOH}$ would be needed.
Then you'd be left with 3 mols of $\ce{H2O}$, 3 mols of $\ce{NaCH3COO}$ and 3 mols of $\ce{CH3COOH}$.

However, what if you mixed 3 mols of $\ce{NaOH}$ with 12 mols of $\ce{CH3COOH}$? According to my calculations, you'd be left with 3 mols of $\ce{H2O}$, 3 mols of $\ce{NaCH3COO}$, and 9 mols of $\ce{CH3COOH}$.

In the latter example there is a clear imbalance in the salt:acid ratio. However, won't this still act as a buffer, while not as strong as an equally concentrated 1:1 buffer? So it will work, it's just not ideal?

They don't need to be comprised of equal amounts.  However, when you are near the 50:50 point, the buffer is "unbiased" (a contrived term), and it can equally accommodate assaults to its pH going in either direction.

If, as in your first example, you're sitting at equal concentrations of sodium acetate and acetic acid, dropping in a 1 mole of either NaOH or HCl won't overwhelm the buffer capacity (pH will change, but not as quickly as it would when you exceed capacity).

Conversely, in your second example, if you add 4 moles of NaOH you'll still be buffered, (acetate will rise to 7 moles, acid fall to 5), but if you add 4 moles of HCl, you'll protonate all the acetate back to acetic acid and the buffering effect will collapse.

So while they don't need to be at the 50:50 point, it's generally a good idea to stay in that order of magnitude (which is why recommended buffering ranges are roughly 1 pH unit); 10:90 is iffy (used appropriately, it works fine), but 1:99 is nearly pointless.  The pKa is the theoretical pH when at a 50:50 mix, so different chemicals are used for different pH buffers (Good's buffers shown below).


Problem with the solubility and Ksp of zinc hydroxide My book gave me that solubility (g/L) is $4.2\times 10^{-4}$, so solubility (mol/L) is $4.2\times10^{-6}$. Then, using the chemical formula $\ce{Zn(OH)2}$, I know that the solubility product is $K_\mathrm{sp} = \ce{[Zn^{2+}][OH-]^2}$, so:
$$K_\mathrm{sp} = (4.2\times10^{-6})(2\times(4.2\times10^{-6}))^2 = 3.0\times10^{-16}$$

However, the book says that the $K_\mathrm{sp} = 3\times10^{-17}$. Where did I go wrong?

There are no simple relation between the solubility and the solubility product when a doubly charged ion is involved. The measured solubility is always much bigger than the value obtained from the solubility product. This is due to the doubly charged ion.

Doubly charged ions like $\ce{Zn^{2+}}$ are usually hydrolyzed in water, and are partly transformed into basic ions like $\ce{[Zn(OH)]^+}$ by the following reaction : $$\ce{Zn^{2+} + H2O -> [Zn(OH)]^+ + H+}$$ If $\ce{Zn(OH)_2}$ is dissolved in water, it will not only produce $\ce{Zn^{2+}}$ ions but also basic ions by the following equation :
$$\ce{Zn(OH)_2 -> [Zn(OH)]^+ + OH^-}$$
As a consequence, the concentration of non complexed $\ce{Zn^{2+}}$ may be calculated from the solubility product of $\ce{Zn(OH)_2}$ but this $\ce{[Zn^{2+}]}$ is much smaller than the measured solubility of $\ce{Zn(OH)_2}$, because the measured solubility of $\ce{Zn(OH)_2}$ is the sum of $\ce{[Zn^{2+}]}$ and $\ce{[Zn(OH)]^+}$. It is even possible that ion pairs like  $\ce{Zn^{2+}(OH^-)_2}$ are dissolved in water without being dissociated.

In the case of $\ce{Zn(OH)_2}$, the measured solubility is $4.2 ·10^{-6}$ M. And the concentration of $\ce{Zn^{2+}}$ taken from $\ce{K_{sp} = 3·10^{-16}}$ is $\ce{[Zn^{2+}] = 1.95·10^{-6}}$ M. This is much less than the measured solubility. It means that the total concentration of the dissolved ion pairs $\ce{Zn^{2+}(OH^-)_2}$ and of the basic ions $\ce{Zn(OH)^+}$ is equal to the difference $(4.2 - 1.95)·10^{-6} = 2.25·10^{-6}$ M. Only $46.4$% of all the dissolved $\ce{Zn(OH)_2}$ produce  simple $\ce{Zn^{2+}}$ ions (which may be $\ce{[Zn(H_2O)_6]^{2+}}$ by the way).


Name of these parts in CoQ10? Am giving presentation over Coenzyme Q10 Wednesday for my final project, but we haven't studied any biochem this year.  So questions have arisen.

CoQ10 has a Quinone head, an isoprenyl tail, and these additional oxygens on the left and three methyl groups not a part of the Quinone head or tail.

Is there a name for these two oxygens and three methyl groups?  Or a category?  Like the benzene ring and two oxygens is the Quinone head, what are the additional oxygens and methyl groups called, if anything?

CoQ10 diagram: http://www.mysupplementstore.com/media/wysiwyg/structure-of-CoQ10.jpg

The additional groups are referred as methyl and methoxy groups. There is no category for these groups. It appears that the methoxy groups stabilizes the intermediates of the oxidation or reduction reactions.


How does hyperconjugation explain stability of partial positive charges? Recently I was reading this answer to the question Why does boron add to the less substituted carbon in the hydroboration of an alkene?.
See the transition state given in the above answer, and the reason given to explain it's stability.

As far as I know, hyperconjugation is a result of σ-p interaction in carbocations.
If partial charges are developed(as in the case of linked answer), there is no empty p orbital in them.

So how can hyperconjugation be used to explain their stability?

The solution is referring to the partial positive charge on the carbon opposite the developing C-B bond. It is stabilized by substituents similar to how a full carbocation is stabilized by substituents, e.g., in the tert-butyl carbocation. This doesn't need to be a completely empty p-orbital—electron density will be dynamically influenced by any imbalance in charge distribution—but if it's easier to conceptualize, you could imagine it as the pi-bond electrons being fully transferred to the C-B bond, leaving an empty p-orbital on the opposite carbon, which is stabilized by hyperconjugation.

The reality should be closer to a dynamic process at both sites, though. The pi-bond transfers electron density into the C-B bond, while the substituents stabilize the rising positive charge on the other carbon. As the C-B bond develops, the B-H bond simultaneously weakens, and begins moving its electron density toward the partial-positive carbon.


Why not "Oxygate" OO4(2-)? So many atoms can be surrounded by oxygen atoms to make anions - chromate, phosphate, etc.   In particular there's Tellurate $\ce{TeO4^{-2}}$, Selenate $\ce{SeO4^{-2}}$ and Sulfate $\ce{SO4^{-2}}$ - why not go one step more and have "Oxygate" $\ce{OO4^{-2}}$?

I would guess that such a thing, if it could be created at all, would break up into an molecular oxygen and ozone $\ce{O2 + O3}$ with the electrons doing whatever they please.  But sulfate and the heavier VI central atoms don't break up.  I've never encountered a good clear explanation about this.

It's not necessary that an element shares some chemical property with all the others in its group. If it did share all chemical properties with the rest of the group, then we could just as well call the elements (something akin to) isotopes of each other. You'd have a pretty boring periodic table then.

What's happening here is that the most striking difference between elements in a group is at play. That difference being the number of electrons.

Our electron shells are broken into "orbitals". Every orbital can hold 2 electrons. $s$ orbitals come in sets of one, $p$ in sets of 3, $d$ in sets of 5, etc. Not every shell is made up of all orbitals--the first shell has only one $s$ orbital, the second shell has $s$ and $p$, and so on. Electrons fill into these orbitals by a triad of rules (which I won't go into here)-- the Aufbau principle, the Pauli exclusion principle (which is implicit in my statement that each orbital holds two electrons), and Hund's rule.

Now, an interesting thing is that bonding can only occur with orbitals with close energies. Thus, if the $4s$ orbital is involved in bonding, the $1s$ orbital will not be.

OK. Now, let's first look at Sulfur's outer electron configuration. There are 2 electrons in a $3s$ orbital, 4 in the $3p$ orbitals (so there is enough space for 2 more electrons here), and you have some "vacant space" in the $3d$ orbitals.

In the sulfate ion:

sulfur is "accepting" 6 electrons in this bond. There's some nifty mixing of orbitals that goes on, but basically, those electrons need some place in sulfur's outer shell to hole up--two electrons per orbital as usual.

Since we have 6 extra electrons, we need 3 orbitals. We have one "left over" $p$ orbital, and 5 $d$ orbitals, so we have enough space to accommodate those electrons and form a bond.

OK. Now, let's try the same thing for Oxygen. It has, in its outer shell, a full $2s$ orbital, 2 full $2p$ orbitals, leaving an "extra" $2p$ orbital empty. As mentioned before, the second shell only has $s$ and $p$ orbitals, so there obviously aren't any $d$ orbitals. You've probably noticed the issue here-- we have six electrons and only one extra orbital to stuff them in. This isn't possible, so oxygen never forms $\ce{OO4-}$. The other chalcogens($\ce{Se, Te}$, etc) all have $d$ orbitals in their outer shells, so they have no trouble forming such ions.

Note that I've grossly oversimplified the bonding process, deliberately leaving out a few crucial aspects (which would otherwise take extremely long to explain). If you want to know about the mechanism behind bonding, you may want to read up on orbitals first, then VSEPR theory, and then MO theory.


Reduction potential and electron affinity Why is it that fluorine has a much greater reduction potential than chlorine, despite the fact fluorine has a lower electron affinity than chlorine?

I understand that fluorine has a lower electron affinity than chlorine due to fluorine's compact size and higher negative charge density.

So what makes fluorine a good oxidizing agent - one better than chlorine?

Could it be that the reduction potentials are referring to diatomic fluorine rather than elemental fluorine? And similarly diatomic chlorine rather than elemental chlorine?

Oxidising power , i.e the ability to accept eelectrons for any element depends on its electron gain enthalpy as also 2 other factors:

1) Hydration energy - greater the better, which basically means that water molecules easily overcome attractive forces in the solute particles to break chemical bonds and that easy separation of solvent molecules to accommodate the solute occurs in a solution and that formation of new attractive interactions between solute and solvent particles is not that common.

Rem: sum of delta H 1, 2, and 3 is the overall enthalpy of the solution process. Rem also that H1, H2 are positive, H3 are negative...........read this up if you have forgotten.

Due to its small size Fluorine's hydration energy is very high.

2) Enthalpy of dissociation (i.e. bond dissociation energy) - lower the better

Fluoride has a greater electron-electron repulsion among the lone pairs in the small sized F2 molecule where they are much closer to each other than in case of Cl2, hence the enthalpy of dissociation of F2 is lower than Cl2.

These two results in fluorine to have a greater positive electrode potential than chlorine, which results in greater oxidising power for F2.


Trioxane fuel tablet recrystalization I had a bunch of old military trioxane fuel tablets.  I decided to open all of the packets, maybe a dozen, and place the contents in an empty tin coffee can with a plastic lid.  I put the can in a dark closet.  I let it sit undisturbed for several months.  Upon opening the can, I saw crystal structures growing off of the sides of the tin can.  They appear to be of a lightweight somewhat flexible plastic composition and are flammable in the same way that the rest of the fuel is.  If I disturb them by damaging them or removing them, more will grow over time, if I let the container sit.  What is this reaction?  Has anyone witnessed this before?

It's subliming (aided in this case by decomposition) and re-depositing.

Substances that turn directly from solid to vapor (e.g. naphthalene, water ice) are said to sublime, rather than evaporate. The inverse, changing from vapor to solid, is termed deposition.

When water vapor deposits on very cold surfaces, such as "dry ice", it grows into amazing dendritic crystals:


Gattermann-Koch reaction - Role of cuprous chloride I am unable to figure out what the role of $\ce{CuCl}$ in Gatterman-Koch reaction is. Does it provide $\ce{Cl-}$ ions by dissociating into $\ce{CuCl2}$ and $\ce{2Cl-}$, but that can also be achieved by using more $\ce{HCl}$? So why do we add $\ce{CuCl}$ in the reaction?

According to this site:

  The formylation of aromatic compound can be accomplished by treatment with $\ce{CO}$ and $\ce{HCl}$ in the presence of Lewis acid. The reaction is performed either under pressure or using copper(I) chloride or anhydrous aluminum chloride whose role may be to aid the reaction between $\ce{CO}$ and $\ce{HCl}$ via the complex which it forms with $\ce{CO}$.

As you can see, it does not involve in the mechanism (provide $\ce{Cl-}$ via dissociation of $\ce{CuCl}$) but act as a promoter to speed up the reaction.


Bromination of an alkene in presence of water Bromination of an alkene in presence of water results in its mechanism in a cyclic bromonium ion and "afterwards $\ce{H2O}$ binds to the carbon atom surrounded by more electron donating groups."
This is what we learned in class and I couldn't understand why. Shouldn't an electron donating group decrease the "strength" of an electrophile?

Also to my surprise, in an exercise, $\ce{H2O}$ was added to the formerly $sp2$ carbon bonded to $\ce{-OCH3}$ which donates electrons by resonace in an alkene.
But $\ce{H2O}$ was added to the cyclic bromonium and not to the alkene directly! So why to care about  the resonance effect of $\ce{-OCH3}$?

While there is no formal carbocation intermediate, the cyclic bromonium intermediate has a few different resonance forms (reference):

As you can see on the right, in the structure representing the resonance hybrid, partial positive charges build up on the electrophilic carbons connected to the bromine. The carbon that is more substituted with electron donating groups has a greater partial positive charge (and therefore more electrophilic character) because of the stabilizing effect of the substituents. That's why the nucleophile attacks at the more substituted carbon.


Reaction of 2-ethyl-5-methyl-3-phenylfuran under acidic conditions I was asked:

  Does protonation occur to the electronegative oxygen?
  What is the function of the heat?

I think that there will be protonation at C-3 which will break the aromaticity and the ring will open to form a diketone product.

Your answer looks perfectly reasonable. Acid hydrolysis of furans is a well established route to 1,4 diketones. The heat is required to overcome the resonance stabilisation of the furan ring.


How isotopes atomic mass was defined? So we say that 1 Amu (u) = 1/12 of the mass of a Carbon 12 atom.

Im not sure about this, but I believe that first we measured  the mass of 1 carbon 12 atom (This includes the mass defects that occur during the formation of nucleus) and then defined that 1 atomic mass unit has a value of 1/12 of the carbon 12 atom mass... Therefore one Carbon 12 atom has exactly 12 atomic mass units.

Until here everything is fine.

Now my question is... To define all the other elements isotopes atomic mass did we first measured each isotope atom mass and then said " If 1/12 of the Carbon 12 atom is equal to X grams and this X grams are equal to 1 atomic unit the then my isotope atom which has Y grams must have Y atomic mass units " did we work it out this way or did we use calculation to define other isotopes mass based on carbon 12 atomic mass ?

Basically what I want to know is, do we need to have the value of an atoms mass in grams or kilograms before we know how many Amu´s it has, or can we calculate it ?

Historically, the relative masses of the isotopes were measured relative to O=16.

There is a good historical explanation in A New Mass-Spectrograph and the Whole Number Rule (1927).

At this time (1927), though it was understood that many elements had more than one isotope, only one isotope of oxygen was known.


light Intensity and its effects on energy and emitted electron on metal surface we have a metal surface with electrons in a specific energy state say X ,now we shine light on it of specific wavelength and intensity.
now what if i double the intensity of light , what would happen to the number of electrons emitted and the average energy of the electrons ?

my guess is that since intensity is doubled, then number of photons emitted also double and the energy also doubles .... but i am not really sure about it

what if i double the intensity of light , what would happen to the
  number of electrons emitted and the average energy of the electrons?

This is known as the "Photoelectric Effect" and is described by the following equation

$$\ce{E_{max}=h\nu- WF_{M}}$$

where $\ce{E_{max}}$ is the maximum kinetic energy of the electron escaping from the metal surface, $\nu$ is the frequency of the incoming photon and $\ce{WF_{M}}$ is the workfunction for the particular metal. Your guess is part right, doubling the intensity of the incoming beam doubles the number of electrons ejected from the metal surface, but their kinetic energy distribution remains unchanged.  The maximum kinetic energy of an emitted electron is still determined by the above equation and therefor still capped by $\ce{E_{max}}$ and the kinetic energies of all electrons emitted are still distributed from 0 to $\ce{E_{max}}$.  If you'd like more information, here's a good, concise reference.


Reduction and re-oxidation of TiO₂ I'm having a hard time finding data about an experiment that I know has been done a bunch of times:

I have a sheet of single crystal rutile $\ce{TiO_{2}}$ (110) that I want to reduce to various $\ce{TiO_{2-x}}$ forms where the x is what I want to vary. I want to try to avoid conversion to the other Magnéli phases if possible $(\ce{Ti_{n}O_{2n-1}})$. To do this reduction, I have a high temp oven that I can flow with 95% $\ce{N_2}$ and 5% $\ce{H_2}$ or simply evacuate. I can set the flow rate to cubic centimeters per second. I need a chart that will tell me what temperature, flow rate, and duration I need to convert my sheets surface to a desired reduced value (i.e., I want a specific x).

I then want to reoxidize the surface by using lower temperatures in a furnace that I can fill with oxygen at different pressures. I need a chart that will tell me what temp, partial pressure of $\ce{O_2}$, and duration I need to re-Oxidize the surface.

The goal here is not to simply get the surface, but to test other properties. I was hoping this step (finding the proper charts) would have been fairly straightforward, but my literature searches haven't turned up quite what I need. I was hoping someone here can point me to a resource that has this data.

I can't post a comment, so this is not a true answer but I think can help. If I've understand what you mean you want to perform a Temperature-programmed reduction (TPR) followed by a temperature-programmed oxidation (TPO). In this case I think this old book can help: Temperature-programmed Reduction for Solid Materials Characterization. I don't know much about single crystal but I think that many things from TiO2 studies of nanomaterials can be consider valid for your case, this online book can be helpful to understand all the variables.


Comparison of rate of reaction in bromination and chlorination I did research, about this one but the data and other things prove chlorination is faster than bromination, because chlorine is more reactive. But, some data says reaction of chlorine is more exothermic then bromine, that is the reason. But, if this is the case, then according to me bromination should be faster. Because, less energy is needed to start bromination and less energy will be given out in formation hydrogen bromide.
And how reaction being more exothermic or less exothermic determines the rate of reaction?

According to the Hammond postulate, a highly exothermic reaction is more likely to have a low transition state barrier than a less exothermic reaction.

Lower barriers lead to faster reactions.

This is a multistep mechanism, so for this to be relevant the rate-determining step.  According to this lecture the rate limiting step is the hydrogen abstraction from the alkane, forming a hydrogen-halide bond.

The bond dissociation energy for HCl is 431.8 kJ/mol and 365.7 kJ/mol for H-Br.  The formation of the H-Cl is this much more exothermic, and the reaction will be faster.


Determination of a chemical compound in a non-homogeneous sample I want to know if it is possible to determine a chemical compound in a non-homogeneous sample. I am asking this because we are aiming to create a device that will detect histamine level in fish without homogenizing the samples. In all the journals we've read (which are a lot), fish samples are homogenized (blended + chemically-treated) before histamine is determined.

I am a computer engineering student with basic knowledge on biochemistry. I'm researching on this because this is our thesis topic (which our adviser has forced recommended to us).

Any help/recommendation regarding this is appreciated. Thank you!

If you analyze non-homogeneous samples, the results can't be interpreted as being representative of the whole organism (or population). So, for example, if histamine is unevenly distributed within the body of a fish, your results would vary widely depending on what part of the fish you analyzed.


How does liquid gallium interact with a gold leaf I would like to know what happens when gold leaf comes into contact with liquid gallium. Will the gold dissolve into (alloy with) the gallium, wet the gallium or have no reaction with gallium?
The reason I am asking this question is to figure out if I could gild some gold leaf onto a base metal (that is compatible with gallium) and use the gallium as an 'adhesive' between the gold and base metal. I am planning on using this as a gold plated cathode in electrolysis experiments.

Broadly speaking, looking at the Au-Ga phase diagram one gets:

(Now, this is based on a paper from 2011, a more recent 2017 paper is open access).
Several things to notice (or get from the paper):

The solubility of gold in solid gallium can be considered to be zero.

The solubility of gallium in gold is not zero, but is on the order of a few atomic percent.

Slightly above the melting point of gallium, the solubility of gold in the liquid is also very small.

There are multiple intermetallic phases across the diagram, 4 of which are stable at (and below) room temperature.

Formation of an amalgam can be ruled out.

So, it is a pretty normal binary phase diagram. Now the phase diagram does not provide kinetics information, so the time needed to actually form any intermetallic compounds (and if those formed near room temperature would even be the equilibrium phases) is not known.
Now, if we look at how gold leaf is used, there is no glue used - the leaf is place, smoothed and hammered into place, and it just sticks. So try just sticking it.


Litmus paper - turning red, blue and even bleached I have blue litmus paper, and if I put it in an acid it turns red. I also have red litmus paper, and if I put it in a base, it turns blue.
I know the question about litmus was asked and answered and I like the answer:
How does the litmus pH indicator work?
However, damp litmus paper also becomes bleached in presence of chlorine gas. I believe it has nothing to do with the mechanism described in aforementioned question. I wonder what the mechanism is. Wikipedia actually suggests:

  For instance, chlorine gas turns blue litmus paper white – the litmus dye is bleached, because of presence of hypochlorite ions. This reaction is irreversible, so the litmus is not acting as an indicator in this situation.

I also found an equation which describes formation of hypochloric acid (source):
$$\ce{Cl2 (g) + H2O (l) -> HClO (aq) + HCl (aq)}$$

The problem that I have now is as follows: We know that litmus paper gets discolored because of contact with hypochloric acid. But what is actually the chemical reaction for that?

I can supply some details now, and hopefully this ought to qualify as an answer.

As I mentioned earlier litmus is a mixture of 10-12 dyes (CAS number: 1393-92-6).

The acid-base indicator properties of litmus are primarily due to 7-hydroxyphenoxazone chromophore (pictured below)

The answer you linked to discusses the acid-base indication mechanism in some detail, so I shall skip over that.

Anyway, what this serves to establishing that it is indeed a extended $\pi$-conjugated system that we are dealing with in the chromophore.

Now, $\ce{HOCl}$ would bring about halohydrination (basically an electrophilic addition) across the $\pi$ bonds, thus disrupting the conjugated system.

Halohydrins are compounds that contain an $\ce{-OH}$ and $\ce{-X}$ groups on adjacent carbons.

This image describes the general mechanistic scheme in a simpler case:

Anyway, after said addition reaction has taken place in the chromophore, the $\pi$-conjugated system is disrupted, and thus, we see an absence of colour.


Baking Soda + Citric Acid + Water = CO2, how much though? Baking soda (sodium bicarbonate) + citric acid react when mixed with water to form some amount of carbon dioxide gas ($\ce{CO2}$). I am trying to determine what percentage of the products is carbon dioxide: in other words, how much does reaction make?

I'm not sure what you mean by 90% and 10%. If you mean 90% yield (i.e. 90% or more of the bicarbonate ions in sodium bicarbonate will become carbon dioxide), then yes, that will happen if there is enough acid. How much is enough acid? We'll get to that in a minute. If you mean something else, like 90% of the mass of the components will become carbon dioxide, then that probably will not happen.

Let's look at why 90% of the mass of products is not carbon dioxide.

Bicarbonate anions react with protons from acid to produce carbonic acid, which decomposes to form carbon dioxide:
$$\ce{H+ + HCO3- -> H2CO3 -> H2O + CO2}$$
Let's start with a gram of sodium bicarbonate. How many bicarbonate ions are present? The molar mass of sodium bicarbonate is $84.0066\mathrm{\ g/mol}$, where a mole (mol) is a quantity equal to the number of carbon-12 atoms in 12 grams of carbon-12 or an Avogadro's number $(6.022\times 10^{23})$ of things. Atoms and molecules are very small, and the numbers of them in macro measurements are large.
So we have
$$\mathrm{1.00\ g \times \dfrac{mol}{84.0066\ g}= 0.0119\ mol}$$
$$\mathrm{0.0119\ mol \times \dfrac{6.022\times 10^{23}}{mol}=7.17\times 10^{21}\ molecules}$$
Let's stick with the mole unit since it gives us easier to conceptualize numbers.
The molar mass of carbon dioxide is $\mathrm{44.01\ g/mol}$, so if all of the bicarbonate ion in sodium bicarbonate is converted into carbon dioxide, we will get $\mathrm{0.0119\ mol}$ of carbon dioxide which is
$$\mathrm{0.0119\ mol \times \dfrac{44.01\ g}{mol}=0.524 \ g}$$
So 1.00 gram of sodium bicarbonate will product 0.524 grams of carbon dioxide. We have not even accounted for the citric acid yet. Citric acid is $\ce{H3C6H5O7}$, and it is tempting to think that some of those carbon and oxygen atoms in that molecule might become molecules of carbon dioxide. They can, but not in a reaction with sodium bicarbonate (acid-base). If you react citric acid with oxygen (combustion), then you can produce carbon dioxide.

How much acid?

Citric acid serves as a proton source - in fact it has three acidic protons. That means in theory, each molecule of citric acid can donate three protons to convert three bicarbonate anions into carbon dioxide:
$$\ce{H3C6H5O7 + 3HCO3- -> C6H5O7^3- + 3H2O + 3CO2}$$
However, those three acidic protons are not equal (and the reason for that would make a great follow-up question). Two of the protons are more acidic than carbonic acid, and the third one is slightly less acidic than carbonic acid. In theory, since carbonic acid decomposes into carbon dioxide, which escapes from solution, we can count on Le Châtelier's principle to pull even that third proton. However, in practice, we might not want to wait around. Let's use one equivalent of citric acid per two equivalents of sodium bicarbonate:
$$\ce{H3C6H5O7 + 2HCO3- -> HC6H5O7^2- + 2H2O + 2CO2}$$
So, to figure out how much citric acid (molar mass $\mathrm{192.12\ g/mol}$) we need, we have to do some calculations. Remember that 1 gram of sodium bicarboncate is 0.0119 moles, and we need 0.5 moles of citric acid for every mole of bicarbonate:
$$\mathrm{0.0119 \ mol\ bicarb\times \dfrac{1\ mol\ CA}{2\ mol\ bicarb}\times\dfrac{192.12\ g\ CA}{1\ mol\ CA}=1.14\ g\ citric\ acid}$$


Are there any non-Jahn-Teller type distortions? My question is essentially, if ANY gradual type of deviation from a higher possible symmetric arrangement of the nuclei in a molecule is covered by the Jahn-Teller model.

More specifically I have in mind some lets say "ionic complexes" between a very small cation and some larger (possibly weakly coordinating) anions.

Very specifically, I remember I have seen for example a calculated equilibrium structure of a beryllium bis-2-fluorobenzoate $\ce{(Be[o-C6H4FCOO]2}$, isomer with $\ce{F\bond{...}Be-O}$ coordination), which showed a distinct deviation from $\ce{C2}$ symmetry. The only type of JT-distortion this would qualify for is 2nd order one. But then since its largely an ionic species I suppose the HOMO-LUMO gap should be so large that 2nd order JT is very unlikely (I am aware that $\ce{C2}$ is not Jahn-Teller active but through vibronic coupling of close-lying $\ce{^2A}$ and $\ce{^2B}$ also JT-like distortions can occure).

I imagine that if the cation is very small so that it "dangles" in the gap between the anions it might be energetically more favourable if the cation decides to stick on one than to balance in the centre. Just a naive picture. But would such a thing - if it exists - be subsumed under JT as well?

So, is distortion possible for ionic aggregates and would such a thing also be described by JT?

Apparently there is some theorem e.g. to be found in an article from Isaak B Bersuker with title "SPONTANEOUS SYMMETRY BREAKING IN MATTER INDUCED BY DEGENERACIES AND PSEUDODEGENERACIES" in Advances in Chemical Physics, Volume 160, First Edition. Stuart A. Rice and Aaron R. Dinner. © 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc; p 187-

The theorem reads like:

"DEGENERACY AND PSEUDODEGENERACY ARE THE ONLY SOURCE OF SPONTANEOUS SYMMETRY BREAKING (SSB) IN ATOMIC SYSTEMS"

And can be reformulated by:

" [...] can we get structural instability of high‐symmetry configurations in any electronic state without involving interaction with other states [...]? The answer to these questions is no: any instability and SSB in polyatomic systems leading to their formation, deformation, and transformation involves at least two electronic states, degenerate or pseudodegenerate. This is the two‐state paradigm in structural instability."

So in essence that means all types of molecular distortions can be traced back to some JT effect.


Full Valence Shell (Octet Rule) vs Formal Charge When does formal charge have more precedence over the octet rule / an atom having a full valence shell when drawing structures?

When you draw $\ce{BCl3}$ for example, there are no double bonds - just three single bonds. Because of this $\ce{B}$ does not end up with 8 valence electrons. The explanation I have seen for this is that a double bond makes it so that the formal charge is undesirable. When drawing structures which has a higher precedence: formal charge or ensuring the atoms have full valence shells?

First, keep in mind that Lewis-structures and formal charges are huge simplifications.  In reality, at the quantum level, charges and bonds are spread out and it's somewhat unrealistic to say "Atom X has a charge of Q" or "Atoms X and Y have an Nth order bond between them".  There's merely a cloud of electron density, and it's somewhat arbitrary to say which parts of that cloud belong to a given atom or bond.

It's perfectly valid to draw a structure with a double bond.  And in reality, I'm sure there is hyperconjugation that leads to a small amount of double bond behavior.  However, such a resonance structure puts a negative charge on an electropositive element, and a positive charge on a very electronegative element, and thus such a structure has a smaller contribution to the overall structure.


How does one find the probability of tunneling along a reaction coordinate computationally? The idea for this question comes from Jan's comment over here. I realized I don't know how I would compute the probability of tunneling in the process of a reaction.

When I say "compute" I don't mean something rather trivial like finding the probability of a quantum harmonic oscillator being past the classical turning points which can just be done analytically.

For instance, if I were studying some reaction computationally (to estimate rate constants or something which depends on tunneling) and constructed a potential energy surface for a bond-breaking process, let's say the transfer of a hydrogen, how do I account for the possibility of tunneling through the barrier along this reaction coordinate?

I could speculate, but I imagine I would be wrong. How would one do this using common computational software and techniques?

If the reaction coordinate $\xi$ can be expressed as the change in the relative position between two groups of the molecule (e.g. for instance as a change in a bond length so that $\xi$ is given in units of length) the problem may be treated as one dimensional and one may use computational software to calculate the potential energy $V(\xi)$ along this path. You may then proceed solving the time-independent Schroedinger equation using numerical techniques such as Numerov's integration method. However, in many cases WKB theory provides an accurate estimate.

In terms of WKB, the tunneling probability is given by
$$
T=e^{-2G}
$$
where $G$ is the Gamow factor given by
$$
G=\frac{2}{\hbar}\int_{\xi_a}^{\xi_b}{ \sqrt{2\mu_\text{red}\left [V(\xi) - E \right ]}}d\xi
$$
where $\mu_\text{red}$ is the reduced mass along the reaction coordinate and $E$ is the energy of the incoming particles. The integral runs over the clasically forbidden region from $\xi_a$ to $\xi_b$. The Gamow factor thus represents the square root of the integral over the "reaction barrier" multiplied with the effective mass that tunnels. Of course, you should take zero point energy into account and in the case of light reactants also the vibrational and rotational structure of the products as this may lead to resonances.


What do numbers in parenthesis after elements respresent? I have seen elements with a number in parentheses after them in literature, but cannot find what it means. What does the number in parentheses after an element represent, such as in W(100) or Mo(110)?

https://aip.scitation.org/doi/abs/10.1063/1.1727919
https://www.sciencedirect.com/science/article/pii/0039602880900886

These are called Miller indices and they identify a particular surface of a crystal.  (You can form different faces by cutting the crystal in different ways.)  When you see something like "Mo(110)" it means "the 110 surface of a crystal of molybdenum."


Question on deflection of cathode rays in a magnetic field My book states that :

  When only electric field is applied, the electrons deviate from their
  path and hit the cathode ray tube at point A. Similarly when only
  magnetic field is applied , the electron strikes the cathode ray tube
  at point C.

I tried to apply the Fleming's Left hand rule used to find the direction of force on a current carrying wire where the direction of current is taken opposite to the direction of flow of electrons.

The rule states that :

  Stretch the thumb, forefinger and middle finger of your left hand such
  that they are mutually perpendicular. If the first finger points in
  the direction of magnetic field and the second finger in the direction
  of current, then the thumb will point in the direction of motion or
  the force acting on the conductor.

Applying this rule I found that the electrons should hit the cathode ray tube at A in a magnetic field . But in the book it is given that they strike it at C.

Please explain me where I went wrong . Thank you.

1)  for individual charges Lorentz force F =q[v*B], so curl your right hand from velocity vector of positive charge towards magnetic field so that your thumb points at the force acting on the charge. Here we have considered positive charge

2)It is better use this mnemonic to find out the direction of force to avoid confusion

Stretch the thumb, forefinger and middle finger of your right hand such that they are mutually perpendicular

fore finger denote direction of electron moving
middle finger denotes the direction of magnetic field (north to south on outside the magnet)
finally force direction can be found out using your the direction where your thumb is pointing

The book is correct

Tip- If current is mentioned then Use your left hand, to avoid twisting your hand in weird ways! .


In vibrational spectroscopy, what affect do overtones have on the spectra? I've been in a lecture about vibrational overtones in anharmonic oscillators. How would these affect an IR spectrum?

Since a spectrum, including its overtones, is essentially a property of a molecule, one cannot really say how the overtones affect a spectrum. So, I will answer this question by first describing what an overtone is, giving an example of an effect that overtones sometimes participate in, and then an example of how overtones can be useful.

Overtones are charatceristic vibrational frequencies of a molecule which take place at approximately integer multiples of the $v=0$ vibrational modes. The existence of these overtones demonstrate that molecular vibrations are anharmonic in nature. This follows from the fact that for a purely harmonic vibration, the only non-zero transitions are those for which $\Delta v=\pm 1$. That is, harmonic oscillators can only move on state at a time. Thus, the fact that we observe these overtones tell us that the vibrations are not harmonic or anharmonic.

Now, in IR and Raman spectroscopy, there is an effect called Fermi Resonance. This effect basically allows for the mixing of multiple vibrational states. If I'm not mistaken, this also only happens for anharmonic vibrations because energy needs to be allowed to be shared between the two states which is not possible if the vibrations are pure harmonic, normal modes. I bring this up because it is quite common that one observes a Fermi resonance between an overtone and a higher frequency ground state mode. To quote the Wikipedia page,

  High resolution IR spectra of most ketones reveal that the "carbonyl band" is split into a doublet. The peak separation is usually only a few cm$^{−1}$. This splitting arises from the mixing of νCO and the overtone of HCH bending modes.

Thus, if you wish, one effect that overtones have on IR spectra is their ability to participate in Fermi resonance, which leaves the appearance of a peak being split.

Sometimes it can be difficult to know if a peak being observed is in fact an overtone. Generally, overtones have low intensity, but if they participate in a Fermi resonance, they gain intensity and will look like a normal ground state vibration (depending on what mode they come from).

One place I have seen something likely similar to this is in the following paper[1]. They compute the IR spectrum of a water cluster, including the bending overtone, and show that it is likely the experimental spectrum was mis-assigned. In this example, being able to compute a spectrum independent of the bending overtone is quite useful.

Experimentally, overtones can be quite useful if there is a large number of peaks around the same frequency as a particular mode you are interested in studying. For instance, it is thought that the water dimer may play a significant role in absorption of radiation in the atmosphere and thus contribute to global warming. It is very hard, however, to determine the abundance of water dimers in the atmosphere because of the large amount of water vapor which covers up its entire infrared spectrum. Thus, ref. [2] used the fourth overtone of the $\ce{O-H}$ stretch to experimentally measure water dimers in the atmosphere.

Another example of the same use of overtones is in ref. [3]. There, the authors are interested in studying a possible intramolecular hydrogen bond, but one cannot just look at the lowest energy vibrations because they are swamped by a regular $\ce{O-H}$ stretch. Thus, the authors record infrared spectra in the regime where one would expect to find the first, second, third, and fourth overtones. It is confirmed that these are the peaks observed by simply computing the spectra from first principles. Thus, one also learns which geometric configurations cause which overtone peak.

In summary, overtones are not so much something that "affect" a spectrum, but are a necessary part of the spectrum because we live in a world where molecules vibrate anharmonically. I have also given examples of when overtones are quite a useful tool.

Just to get you thinking a bit, if and when you learn about two-photon or two-dimensional spectroscopies, imagine how overtones can then be used to gain even deeper insight into the molecules of interest.

[1]: Wang, Y., & Bowman, J. M. (2013). IR spectra of the water hexamer: Theory, with inclusion of the monomer bend overtone, and experiment are in agreement. The journal of physical chemistry letters, 4(7), 1104-1108.

[2]: Pfeilsticker, K., Lotter, A., Peters, C., & Bösch, H. (2003). Atmospheric detection of water dimers via near-infrared absorption. Science, 300(5628), 2078-2080.

[3]: Howard, D. L., Jørgensen, P., & Kjaergaard, H. G. (2005). Weak Intramolecular Interactions in Ethylene Glycol Identified by Vapor Phase OH− Stretching Overtone Spectroscopy. Journal of the American Chemical Society, 127(48), 17096-17103.


What is a 'triplet biradical transition state'? My text book just referred to this term. I have absolutely no idea what it is. Articles on the internet are too difficult for me to comprehend.

In some alkenes, steric crowding in the planar state may be so severe that the latter is no longer an energy minimum but becomes an energy barrier with the ground state becoming twisted. In this context the compounds in figure are of interest. The drop in rotational barrier from stilbene ($\pu{42.8 kcal mol-1}$, $\pu{179 kJ mol-1}$)to compound $\ce{P}$ ($\pu{21.1 kcal mol-1}$, $\pu{88.3 kJ mol-1}$) to compound $\ce{Q}$ (“small barrier”, cf.) may be explained by increasing stabilization of the triplet biradical transition state. However, the “negative” barrier in $\ce{R}$, that is, the fact that the stable ground state of $\ce{R}$ is non-planar and that the planar conformation represents the transition state for rotation requires a different explanation; the difference between $\ce{Q}$ and $\ce{R}$ is presumably due to the steric interaction of the four chlorine substituents in the planar conformation of $\ce{R}$. This situation resembles that in o,o'- tetrasubstituted biphenyls to alkenes with low or “negative” barriers.

This is the question given: for which the above is the solution

Source: "Advanced Problems in Organic Chemistry by MS. Chouhan 9th Edition", Chapter GOC, page 53

A triplet biradical state simply implies that the bonding pair will be cleaved homolytically .  For example the triplet biradical transition state for

Will be


Why are combustion reactions spontaneous? I learned that spontaneous reactions are reactions whose activation energies can already be overcome with the temperature they're already in. Since the temperature they're in already provides enough energy to overcome the activation energy, additional input of energy is not necessary and so it is a spontaneous reaction. So if this is true, why are some reactions like combustion reactions considered spontaneous if the temperature alone wasn't even enough to get over their activation energy? Is this definition of spontaneous wrong?

Clarification on the definition of spontaneity must be considered first, as it's meaning is often misinterpreted (and through colloquialisms is further confused). In order to do so, we must separate reaction concepts into thermodynamics (where free energy and spontaneity are discussed) and kinetics (where activation energy and rate are discussed).

Thermodynamics: looking only at the reactants' and products' Gibbs free energies, and the difference between them ( &Delta G ), a reaction is defined as (a) spontaneous if the difference is negative, (b) at equilibrium if zero, and (c) non-spontaneous (a sign that it is spontaneous in reverse) if positive. Thus, thermodynamics is interested only in which form is more stable energetically: the reactants or products. No details about the path (see kinetics below) is included. Check out Khan Academy's "Gibbs free energy and spontaneity" page for some great figures and further detail.

Kinetics: interested in the pathway between reactants and products, and how much energy change is required to obtain an arrangement that leads to one or the other. In other words, this is where the activation energy becomes important, as it is the energy barrier that must be overcome before a reactant could even think about becoming more stable. This barrier impacts the rate at which reactants and products can be formed. A "spontaneous" reaction in kinetics would be a "fast" reaction, one with a nearly zero or negative activation barrier (no representative "hump" in those internal reaction coordinate diagrams).

In summary, thermodynamics is interested in stability of reactants or products, and is not directly discussing the path by which it would arrive at one or the other. Hence, a reaction can be "spontaneous" and still not occur without further prompting because it is kinetically too difficult (read "high activation energy") to get there. However, a "spontaneous" kinetic reaction implies the path is "downhill" (so to speak) and thus is also a "spontaneous" thermodynamic reaction.

Following that understanding, you mention temperature, which is where I believe the confusion builds as it is mixed with the "spark" in combustion. Temperature does effect the Gibbs free energy by the following relationship:

ΔG = ΔH−TΔS

Thus, reactants can be "pushed" with temperature to become products by adding the necessary energy to a non-spontaneous or equilibrium system to breach the activation energy "hump," and be "spontaneous" at that temperature. In other words, they can indeed be triggered just by increasing the temperature. Examples of this are found in spontaneous combustion events as temperatures are slowly raised to a point at which the energy barrier is breached, and reactants are becoming products while releasing heat (because these spontaneous reactions are exothermic, recall negative &Delta G). Heat release then causes surrounding reactants to overcome the same barrier energetically, and result in what we know as fire/burning events.

What about a spark, then? A spark contains the energy needed to begin the combustion process (e.g. from gasoline and oxygen to carbon dioxide and water) but in a sudden shock that triggers this cascade of heat release much more rapidly. A CDC study illustrated the impact of this latent heat given off by the initial sparked reaction by observing a stoppering of combustion by removing the heat with conduction (i.e. blowing out a candle).

So, all in all, "spontaneity" as a definition perhaps isn't wrong, but is used in various circumstances to describe different processes; here, either stability between states or rate at which the states will change.

Be sure to check out the other related stack exchange page mentioned by Mithoron in the comments, with a great answer regarding spontaneity by Brian and Melanie Shebel. Covers more detail about the thermo/kinetic descriptions, whereas here we cover a more broad scope to address your questions about "spontaneity" in general.


What does higher electron gain enthalpy mean here? While talking about ionic bonds my book states :

  Higher the value of electron gain enthalpy of the atom, greater the ease of formation of the anion from it, i.e., other atom should have high value of electron gain enthalpy.

Now I think it should say more negative than just higher (as negative enthalpy means release  of energy and hence more stability). So what should it be? Why the chemistry books (I have to read)  seem to always avoid the term magnitude(Is there some hidden assumption which are generally in the field of inorganic chemistry  ?

I take for granted that what you call "electron gain enthalpy" is generally described as "electron affinity". This is the energy gained (or lost) when an electron is added to an atom, ion or molecule. In general the electron affinity is a measure for the tendency of elements to form anions. The trouble is that you may find tables giving these numerical values as positive or negative values. For example, for fluorine $\ce{F}$, the electron affinity is $\pu{338 kJ/mol}$. But you may find books where the same affinity is given as $\pu{-338 kJ/mol}$. It depends on the sense you consider the reaction $\ce{A + e- -> A-}$.


How do primary alcohols react with dihydropyrans Can anyone also explain me the mechanism of this reaction?

I think it might be formation of acetal or hemiacteal reaction but I am not sure about it.

Please have a closer look at your starting materials and the reaction conditions.

The reaction is performed under protic, ahydrous conditions.
One of the starting materials is a primary alcohol.
The other starting material apparently looks like a cyclohexene where one carbon adjacent to the $\ce{C=C}$ double bond has been replaced by an oxygen atom. This is a dihydropyrane.

Now, let's try a mind game:

Supposed that your second starting material is cyclohexene, what would happen under anhydrous, protic conditions?

   The proton would probably add to the $\ce{C=C}$ double bond to form a cyclohexyl cation. Since the starting material is symmetric, there is only one secondary cation formed. Note that the situation would be different if you would start with methylcyclohexene! Then, a secondary or a tertiary cation could be formed. Remember inductive effects to judge which one is favoured.

How would the primary alcohol come into play here?

   It would probably add to the cyclohexyl cation and yield an ether.

In reality, your second starting material is a dihydropyrane, think cyclic vinyl ether. How will the lone pairs on the oxygen atom effect the protonation?

   If in doubt, draw! Your $\ce{C=C}$ double bond now has two markedly different carbon atoms. Protonation can result in two different cations. I suggest to draw some resonance structures to figure out which one is favoured. Remember the lone pairs ;-)

What happens in the presence of a primary alcohol now?

   Again, it will add to the cation.

How does the final structure look like?

   I'm certain that you have figured this out by drawing and once you have, you will easily identify the structure class :)

Note that the reaction in question is quite useful to reversibly protect alcohols in sequences of chemical transformations.


Why is Ionization Energy definition restricted to gaseous atoms? Here's the definition of ionization energy that I am learning at this level:

  energy required to remove 1 mole of electrons from 1 mole of gaseous atoms to form 1 mole of uni positively charged gaseous ions.

I would like to know, why did they specify that the removal of electrons have to be at gaseous state and the ions that form are gaseous as well? Does it have to do with the amount of energy supplied that causes the elements to vapourize?

In fact, it is not restricted at all. You may compare the energies needed to remove one electron out of a solid; these are also meaningful, albeit in a different way, and known for a wide range of substances. But when you are talking about atoms, you want to measure atoms, and the only way to have an undisturbed lone atom is to put it in a gaseous state. And of course, the resulting ion should be in that state, too; after all, you can't make a solid out of positive ions alone, and if you want to add negative ions to counterbalance them, an ambiguity arises as to which ions should you use.


What is the correct name for this lab technique in crystallization? Take a glass rod and rub vigorously the wall of the flask, the substance will crystallize out of the solution.

Take a fire polished stirring rod and etch (scratch) the glass of your beaker. The small pieces of glass that are etched off of the beaker serve as nuclei for crystal formation

But reaching back to my halcyon years of organic synthesis and crystallization, recalling vivid memories of scratching the sides of Erlenmeyer flasks with a glass rod to induce crystallization

It's a common lab technique. But does it have a name?

I think the most specific you can get on this phenomenon is nucleation , specifically heterogenous nucleation / surface nucleation.

The rough surface formed creates scope for nucleation to occur, and  this creates the crystals.

I don't think there is a specific name for this technique as a lab tenchique .


On the mechanism of the nitrogen-removal step of the Wharton olefin synthesis In today’s reaction seminar, the Wharton transposition made an appearance. In our practice case it started from the epoxide as in scheme 1 below as on Wikipedia; Kürti/Czakó (page 482) begin from the enone which is oxidised by hydrogen peroxide to give the same epoxide.

Scheme 1: General mechanism of the Wharton olefin synthesis starting from an α,β-epoxy ketone to give the allylic alcohol.
The first steps are clear to me, I am interested in the mechanism of the final one. Our problem set’s solution key (taken from a Japanese book that I cannot reference for lack of understanding) draws a concerted [1,3] migration as depicted in scheme 2a. Kürti/Czakó depict a sequential release with an intermediate vinyl anion. Wikipedia makes no note on that step and does not discuss mechanistic proposals.

Scheme 2: Mechanisms given for the final step. (a) concerted pathway given by our problem set. (b) stepwise pathway given by Kürti/Czakó.
Kürti/Czakó also point out that:

Alternatively, the formation of a vinyl radical has been proposed.6
6: G. Stork, P. G. Willard, J. Am. Chem. Soc. 1977, 99, 7067–7068.

The concerted mechanism seems slightly unreasonable due to the strained four-membered ring that would be required in the transition state. Likewise, the stepwise mechanism requires the formation of a vinyl anion which — while possible — is generally not very stable. In its defence, the step would be rendered irreversible by the liberation of nitrogen gas and the subsequent anion immediately captured by any acidic proton around.
Which mechanism does the reaction most probably follow?

Reference:
L. Kürti and B. Czakó: Strategic Applications of Named Reactions in Organic Synthesis. Background and Detailed Mechanisms, Elsevier Academic Press, Burlington, MA, USA, 2005, page 482.

The Wharton rearrangement presents an opportunity to convert one α,β-unsaturated ketone to its regioisomeric allylic alcohol. To this end Ohloff, et al.,1,2 explored the transformation of α-ionone 1 into α-damascone 2. Epoxide 3 was exposed to hydrazine in methanol to afford bicyclic allylic alcohol 4 (30%) and the uncyclized allylic alcohols 5 and 6 (20%, ~1:1). The proposed mechanism (3 $\rightarrow$ 7 $\rightarrow$ 8 $\rightarrow$ 9) goes through the vinyldiazene 8, which loses nitrogen to form reactive species. Ohloff proposed a mechanism invoking the vinyl anion 9, which is either protonated by methanol to afford acyclic allylic alcohols 5 and 6 or undergoes cyclization with the endocyclic double bond followed protonation of the resultant secondary anion. The bothersome aspect of this mechanism is, can a vinyl anion in a sea of methanol survive long enough to avoid protonation and undergo cyclization? The same point can be raised about  the traditional Wolff-Kishner reaction run at elevated temperature in protic ethylene glycol.

Stork and Williard3 explored the scope of the cyclization under Ohloff conditions finding that both 5- and 6-membered rings can be formed. Cyclization of epoxyketones 10a and 10b formed 11a (85%) and 11b (60%), respectively, as mixture of stereoisomers. In the former example, little of the uncyclized product was formed. These authors suggested a radical-based mechanism because in a typical reaction, the ratio of cyclized-to-uncyclization product was not markedly affected by employing the more acidic β,β,β-trifluoroethanol or the less acidic tert-butyl alcohol as a solvent. The authors argue that if a vinyl anion were involved, the more acidic solvent would be expected to increase the amount of uncyclized product at the expense of cyclized material. An intense yellow color developed which faded with time upon mixing the epoxide and hydrazine independent of whether or not cyclization occurred. A strong UV absorption at $\pu{232 nm}$ and a weak one at $\pu{409 nm}$ were attributed to the vinyldiazene. A concerted vinyldiazene cyclization was considered but with less enthusiasm.

Taber and Stachel4 designed a set of experiments to resolve this dichotomy based on the known stereochemistry of radical and anionic cyclizations. Tri-n-butylstannane-initiated cyclization of bromide 12a gave an excess of cis-cyclopentane 13a over trans-14a in accord with the cyclization of bromide 12b to 13b and 14b (13b>14b).5 Alternatively, metal-halogen exchange of iodide 15a with t-BuLi gave more of trans-cyclopentane 14a over the cis-isomer 13a in an anionic cyclization in accord with the results of Bailey6 for iodide 15b. The Wolff-Kishner reduction of ketone 16a in triethyleneglycol (TEG) afforded an 11:1 ratio of trans-14a over cis-13a. This result led the authors to conclude that the Wolff-Kishner cyclization is anionic as opposed to radical in nature.

While it is not necessary that the Wharton rearrangement and the Wolff-Kishner reduction must proceed via the same mechanism, in my opinion there is ample opportunity for more definitive results. Should an occasion arise where only the radical or anionic mechanism would apply, then the definitive experiment has been conducted.

References:

G. Ohloff, G. Uhde, "Über eine ungewöhnliche Cyclisationsreaktion bei der Umsetzung von (+)‐Epoxy‐α‐dihydrojonon mit Hydrazinhydrat (English translation: About an unusual cyclization reaction in the reaction of (+)-epoxy‐α-dihydrojonone with hydrazine hydrate)," Helv. Chim. Acta, 1970, 53(3), 531-541 (DOI:  https://doi.org/10.1002/hlca.19700530309).
K. H. Schulte-Elte, V. Rautenstrauch and G. Ohloff, "Synthesen der stereoisomeren (±)‐γ‐Damascone (English translation: Syntheses of the stereoisomeric (±)‐γ-damascones)," Helv. Chim. Acta, 1971, 54(7), 1805-1812 (DOI:  https://doi.org/10.1002/hlca.19710540709).
Gilbert Stork, Paul G. Williard, "Five- and six-membered-ring formation from olefinic $\alpha, \beta$-epoxy ketones and hydrazine," J. Am. Chem. Soc., 1977, 99(21), 7067-7068 (DOI: https://doi.org/10.1021/ja00463a053).
Douglass F. Taber, Shawn J. Stachel, "On the mechanism of the Wolff-Kishner reduction," Tetrahedron Lett. 1992, 33(7), 903-906 (DOI: https://doi.org/10.1016/S0040-4039(00)91571-5).
A. L. J. Beckwith, I. Blair, G. Phillipou, "Preferential cis cyclization of 6-hepten-2-yl and related radicals. Example of orbital symmetry control," J. Am. Chem. Soc., 1974, 96(5), 1613-1614 (DOI: https://doi.org/10.1021/ja00812a066).
William F. Bailey, Timo T. Nurmi, Jeffrey J. Patricia, Wei Wang, "Preparation and regiospecific cyclization of alkenyllithiums," J. Am. Chem. Soc., 1987, 109(8), 2442-2448 (DOI: https://doi.org/10.1021/ja00242a032).


Mechanism for conversion of ketone to dichloride with phosphorus pentachloride I would like to know the mechanism of this reaction leading to the formation of a geminal dichloride. I am unable to predict how the $\ce{C=O}$ breaks and how the 2 $\ce{Cl}$ atoms replace the oxygen atom. What kind of a reaction is it?

Note: This reaction is mentioned on page 310 of Solomon and Fryhle's  Organic Chemistry in the chapter "Alkenes and Alkynes". It's the precursor of the preparation of cyclohexylacetylene through E2 mechanism using 3 equivalents of $\ce{NaNH2}$

I would encourage you to think about how these molecules are likely to interact with each other and try outlining a possible mechanism.

Step 1: The carbonyl oxygen carries a partial negative charge due to its higher electroneegativity. It also has a high  affinity towards bonding with the $\text{III}^{\text{rd}}$ row atoms which have $\ce{d}$-orbitals, such as $\ce{Si}$, $\ce{P}$ and $\ce{S}$.

$\ce{PCl5}$ exists as $\ce{[PCl4+][PCl6-]}$. The lone pairs on the oxygen attack the central phosphorus atom.

Step 2: The negatively charged phosphonate complex has high energy (nucleophilic) $\ce{P-Cl}$ bonds. They attack the electrophilic carbonyl carbon. Note that these bonds are more nucleophilic than basic and would prefer substitution over elimination.

Step 3: The lone pairs on the oxygen are next to the $\ce{d}$-orbitals of the phosphorus. They can push out one of the chlorines attached to the phosphorus.

Step 4: The chloride pushes the $\ce{-OPCl3+}$ leaving group out giving the dichloride. As orthocresol pointed out, this step is likely to be an SN1 type of substitution owing to the excellent leaving group and the rather weak nucleophile $\ce{Cl-}$.

Note: Steps do not represent elementary reaction steps. The $\ce{PCl5}$ formed after $\ce{PCl6-}$ gives its $\ce{Cl-}$ is ignored. I have seen/heard Step 3 and Step 4 being described as concerted, but I have shown them here separately as I am not sure of those "rumours".


The products of a chain reaction beginning with acetone and sodium amide I encountered the following question in an exam:

  $\ce{CH3COCH3 + NaNH2}$  giving out $\ce{A}$.
  $\ce{A}$ then reacts with $\ce{C2H2}$ giving out $\ce{B}$.
  $\ce{B}$ then reacts with $\ce{H+}$ to give $\ce{C}$.
  $\ce{C}$ then reacts with 1 equivalent of $\ce{H2}$ in presence of $\ce{Pd}$ to give $\ce{D}$.
  $\ce{D}$ then is heated to $\pu{400 ^{\circ}C}$ in presence of $\ce{Al2O3}$ to give $\ce{E}$.

I reasoned it as the first reaction could be an acid-base reaction, but that didn't help much. Then I tried a nucleophilic substitution, but there aren't any good leaving groups and still we get stuck after the first step.

One of the teachers reasoned it as the first reaction being reversible and it goes back to $\ce{CH3COCH3}$ which then reacts with all other and it ends up at isoprene which is the correct answer according to the answer key, but this explanation seems very flawed to me as then what was the purpose of the first step? Can someone please check and write the steps and products in each reaction?

Acetone and sodamide give acetone sodium enolate A, nothing further happens until acetylene is introduced. The pKas of acetone and acetylene are both around 25 so an exchange occurs reforming acetone and creating sodium acetylide. This seems an unusual way of making sodium acetylide, but I don't think this is a real world example (unless someone can find a reference).

Sodium acetylide adds to acetone giving the sodium alkoxide of tertiary alcohol B

Work up with dilute acid gives the tertiary alcohol C.

Hydrogenation over Pd catalyst reduces the triple bond to a double bond giving D.

Passing D over alumina at 400C dehydrates the tertiary alcohol to give isoprene.


Acid-catalyzed hydrolysis of diphenyl malonate What are the complete, acid-catalyzed hydrolysis products (with application of heat) of diphenyl malonate?

According to me, they are phenol, carbon dioxide, and acetic acid.

The aqueous acid and heat conditions indicate to me that we are returning from the carboxylic acid derivative (ester) to the carboxylic acid. The mechanism involves protonation of the carbonyl oxygen and nucleophilic attack by water. This ejects alcohol (well, in this case, phenol).

The heat induces the malonic acid intermediate to decarboxylate and this results in the production of carbon dioxide and acetic acid.

Is my analysis correct?

If the question is "What are the products of hydrolysis of diphenyl malonate?", then my answer would be 2 equivalents of phenol and one equivalent of malonic acid.

If the question is "What are the products of diphenyl malonate when heated in an acid aqueous solution?", then my answer would be 2 equivalents of phenol, one equivalent of acetic acid, and one equivalent of carbon dioxide.

I see it sort of as a semantic question, since the hydrolysis of an ester gives an alcohol and a carboxylic acid, as you suggest. I don't consider decarboxylation to be a hydrolysis reaction, literally defined as cleavage of bonds by addition of water. Water isn't incorporated into the products of the decarboxylation step.


How to remove epoxy on clothes? How to remove epoxy on clothes in the perspective of chemistry?
More precisely in my case, how to remove the hardener of epoxy, which consists of two components: hardener and resin.
The clothes is made in nylon, please take the least damage to the cloth into high priorities of consideration.
Does alcohol help?
Thank you very much for your reply.

Lestoil, PineSol or other organic-solvent cleaner should get out the unset hardener. Once set, epoxy is hard to remove; you might try methylene chloride-based paint removers [toxic!]. Some hardeners are organic peroxides, which may bleach spots on the clothing. –


How do I convert salinity measured in mg/L Cl into parts per thousand as NaCl? I have collected water data using a refractometer that records salinity in ‰ (parts per thousand) as $\ce{NaCl}$. I want to compare my data to other data but every journal article I read  has the results in $\pu{mg/L }\ce{Cl-}$

How do I convert my $\pu{ppt}$ data into $\pu{mg/L}$ or vice versa so I am comparing apples to apples?


What does a comma signify in inorganic chemistry? $\ce{(Ca{,}Co)CO3}$

I have never ever seen this before. What does this mean exactly?

Quoting [1]:

  Comma. In accord with general usage, the comma "," should be used to
  indicate continuous substitution, partial as well as complete, if
  there is no change of CN. Vacancies (symbol $\square$) are treated here just
  like atoms. This usage is restricted to chemical formulas for atoms on
  a specific site. The comma is not appropriate for structure site
  formulas because different symbols indicate a change in the
  coordination number. Examples for proper use of the comma include
  forsterite-fayalite $\ce{[(Mg{,}Fe)2(SiO4)]}$ and hornblende $\ce{\{(Na{,}\square)Ca2(Mg{,}Fe{,}Al)_5[(Si{,}Al)4O11]2(OH)2\}}$. In structure type formulas, the
  comma is used to mean "or" where substitution may occur. An example is
  in the zeolites where H2O and M units may both occur.

Comma representation are often used in writing chemical compositions of minerals as the elements and its concentration vary to a certain extent. See minerals.net:

  A number of minerals contain a varying amount of two or more elements like Aurichalcite, which has a chemical formula of
  $\ce{(Zn{,}Cu)5(CO3)2(OH)6}$ contains an unspecific varying amount of
  zinc (Zn) and copper (Cu). This is indicated by comma separating the
  Zn from the Cu. If a chemical formula with two elements in
  parenthesis is separated by a comma, the number of those elements
  vary. Aurichalcite has a variable amount of zinc of copper where the
  combination of both these elements totals five. The more dominant
  element is usually listed first. (Emphasis added.)

For more information, see this previous Chemistry.SE question and see here to find steps regarding how to write such type of formulae using chemical markups.

References:

Smith, D. K.; Roberts, A. C.; Bayliss, P.; Liebau, F. A systematic approach to general and structure-type formulas for minerals and other inorganic phases. Am. Mineral. 1998, 83 (1-2), 126–132. DOI: 10.2138/am-1998-1-212. Online version: http://database.iem.ac.ru/mincryst/fst.html


why does a pot of boiling water quiet down after it reaches boiling point but is still roiling? I've noticed that when water boils, it makes a lot of noise. After the water reaches boiling point, even though the water is still ""rolling", the water quiets down. What is all the noise about and why does it go away?

Oh, you've come to the right place.  Many years ago I asked this question as a kid on a TV write-in programme called Don't Ask Me.  The TV people invited me on the show and built a series of demonstrations to explain.

The first thing that happens when you warm up a cold pot of water is that the pan bottom gets to boiling temperature.  The water starts to boil.  Liquids boil when their vapour pressure equals atmospheric pressure.  Otherwise bubbles could not form in the the liquid.  So, the pan bottom gets hot, bubbles form, but then they move up into cooler water and implode as the vapour pressure drops.  This makes a sharp snapping sound.  Lots and lots of these imploding bubbles make a roaring sound.

As convection currents then start to mix up the water, it becomes more uniform in temperature, and the bubbles actually rise to the top of the liquid and burst.  The implosions stop and you get the softer bubbling sound.

Hope this answers your question!


Is this a spontaneous or a non-spontaneous process? An insulated tank initially contains a liquid-vapor mixture of water. Then, an electric heater is turned on and kept inside the tank until all the liquid becomes saturated vapor.

According to the 2nd Law of Thermodynamics, a spontaneous process occurs when the change in entropy of the universe is greater than zero.

In this case, since the tank is insulated, $Q=0$ and the change in entropy of the surroundings is also equal to $0$; hence, $\Delta S_\mathrm{universe}$ = $\Delta S_\mathrm{system}$. Since adding work to the system (via the electric heater) results in an increase of the thermal energy of water and its evaporation, $\Delta S_\mathrm{system}>0$. Therefore, $\Delta S_\mathrm{universe} > 0$, and the process is spontaneous.

But wouldn't the necessity of adding work to the system (in order for the water to evaporate) mean the process is non-spontaneous?

Think like this, if the process occurs it is spontaneous.

Without the heating the process is not spontaneous because it does not occur.

With the warm up, you've modified the system. It is this system that is spontaneous (with heating).

Your last statement is right for the unheated system.

Notice, there are two distinct situations.


Weak base titrated with a strong acid If you have a weak base XOH titrated with a strong monotonic acid, where do you start with that? Would you first have the base react with water and break up and determine how much of it does? Or just it breaking up alone, not reacting with water? Reacting just with the acid right away? I'm not sure where to start for this problem type...

I need to find the pH at the equivalence point. I have the concentrations of the acid and base, which are both equal, no volume given.

Equivalence point means equal numbers of moles of acid and base have been added. If $x$ moles of the weak base $\ce{B}$ and the strong acid $\ce{HA}$ have been added, then the $\ce{H^+}$ ions will react away $\ce{OH^-}$ until all the weak base has associated into $\ce{BH^+}$ and $\ce{OH^-}$  (which of course has been reacted with $\ce{H^+}$ forming water). This leaves us with water and $x$ moles of the salt $\ce{BHA}$. This salt is ionic and so soluble in water, so it completely ionises to form $\ce{BH^+}$ and $\ce{A^-}$ ions in solution.

As $\ce{BH^+}$ is the conjugate acid of the weak base we had originally, it can react with water in an equilibrium reaction as follows:

$\ce{BH^+} + \ce{H2O} \rightleftharpoons \ce{B} + \ce{H3O^+}$

As $\ce{B}$ is a weak base, the above equilibrium will lie to the right, and hence there will be a considerable concentration of $\ce{H3O^+}$ in the solution at equivalence, which leads to an acidic pH.

Calculating the pH depends on the fact that $K_b K_a = K_w$ for a base and its conjugate acid. This is derived here.

For the equilibrium

$\ce{BH^+} + \ce{H2O} \rightleftharpoons \ce{B} + \ce{H3O^+}$

the acid dissociation constant has the form

$K_a = \dfrac{[\ce{B}][\ce{H3O^+}]}{[\ce{BH^+}]}$

As we know $K_b K_a = K_w$, we can rewrite this equilibrium expression as

$[\ce{H3O^+}]=\sqrt{[\ce{BH^+}]\dfrac{K_w}{K_b}}$

To find pH you can $-\log$ this. (Note $[\ce{B}]=[\ce{H3O^+}]$ as one molecule of each is produced for each molecule of $\ce{BH^+}$ that reacts with a water molecule).

So in order to find the pH at the equivalence point, you need to know the $K_b$ of the base you're using. And as everything is in the same volume, all the volumes will cancel from the concentration terms.

I hope this helps, and please correct me if I've made any errors.


Do pipettes always outclass glass syringes? I was previously working in a lab preparing lots of samples for gas chromatography that exclusively used glass pipettes and was able to get really high precision between replicates. Now I'm working on my own project doing a lot of liquid dilutions that need to be very accurate with an older pipette that I am looking to potentially replace for better results. I'm specifically looking at 1 mL volumes.
Tolerances for glass syringes seem to be about +/-1% (not as much info), where pipettes are +/-0.6% accuracy and +/-0.2% precision. There were definitely some other reasons to use syringes vs pipettes at my old lab, but are syringes really always outclassed by pipettes? Can an experienced syringe operator beat a pipette?
I realize tip choice, needle, calibration, and ambient temperature controls probably have larger effects than these differences (though seemingly favor syringes), so I am wondering more about the relative optimal performance of each.

Glass syringes and glass pipettes are high precision and high accuracy tools of an analytical chemist. Like any tool at home, such as a cloth cutting scissor and a plant pruning clipper do the same job of cutting but their applications are different.
As an analytical chemist, I have not seen anyone using microsyringes for standard preparations in volumetric flasks. One uses a class A pipette for volumetric transfers and to make different standards.
Now you wish inject known microliter or nanoliter volumes in a GC injection port or an HPLC rheodyne, can we use glass pipettes? You would choose a glass syringe to pierce the GC septum.
The tip is that you would use the highest quality glass pipette (class A) and highest quality syringes in each case. class A 1 mL pipettes are commercially available.
If you really wish to compare, you would have to do a calibration yourself on a analytical balance under tightly controlled conditions.


Would Schweizer's reagent dissolve nitrocellulose? Schweizer's reagent, or tetraamminediaquacopper dihydroxide, is used in the production of rayon. It does this by dissolving the cellulose which can be acidified to form strands of cellulose. I have had some success with this.

So, seeing as cellulose and nitrocellulose are rather similar, would it be able to dissolve it too? If not, why? The nitrocellulose would have to be able to survive the addition of acid, which should not be a problem since it is synthesized using sulfuric and nitric acids. So, as long as I use sulfuric acid it should work fine.

Something else that might affect it is how nitrated the cellulose is? Cellulose dinitrate and cellulose hexanitrate might act differently.

And now, assuming that it can dissolve it in some way, it could be extracted from that from celluloid plastic, like that found in ping pong (table tennis) balls?

So, by adding some ping pong ball to Schweizer's reagent, then letting it dissolve and adding it to a solution of sulfuric acid, the camphor in the celluloid would be left behind and you should end up with some sort of "nitro-rayon". This is a bit of a stretch, but to me seems feasible.

I will try this when I can, but for now I hope this is enough to prompt speculation.

Schweizer's reagent dissolves cellulose because it forms a complex with hydroxyl groups of glucose moieties. Therefore, a completely nitrated cellulose will not dissolve in Schweizer's reagent. However, the reagent is quite basic and it will cause a hydrolysis of the nitrocelulose, so it may dissolve eventually, but after precipitation it will be normal cellulose again. But there is also a good information for you: nitrocellulose easily dissolves in many organic solvents e.g., acetone, ethyl acetate, diethyl ether and so on. So once you have a nitrocellulose, you do not need to play with Schweizer's reagent.


Increasing elasticity of microcrystalline wax I'm doing sculpture and I've been playing around with some microcrystalline wax. The pliability of it once warm is great, but I'm running into the problem of it tearing / ripping when I stretch it. The same issue causes problems when smoothing it out, as the surface "gives way" under my finger if I press hard enough.

I've tried mixing in petrolatum and mineral oil in various different quantities but with no success; both just make the wax stickier and softer with no elasticity benefits. Is there another method which will increase the stretchiness?

The wax (presumably a mixture of mostly n-alkanes?) constantly re-crystallises under strain, making it brittle in the same direction. Petroleum acts as a plasticiser, easing the flow of the amorphous phase, but also easing the recrystallisation even more. In effect it doesn't really help with your problem.

You could try to find another, harder wax with longer alkyl chains, and try softening that by mixing in your present material.

And generally take your time when working it. When the material becomes less opaque, i.e. whiter, then it is already hardening. Stay below that threshold. High strain rates let the crystallites grow, which are stiff, so at the edges of the crystallites, the strain rates are even higher.