Why water cools from outside to inside? It is a well-known fact that

Water kept in a container, on cooling, results in freezing with initially at the circumference of surface
(1D), then on the surface (2D) and then throughout the volume (3D).

Why does it happen that way?
My attempt:
I thought that this may be due to their exposure levels.The outer surface is more exposed to cooling than the inner ones and hence cools first.However, I am not satisfied with this explanation.
I think there is some relation between this and degrees of freedom.I feel as there are more degrees of freedom in 3-D, there's more movement you should suppress to lower its energy, so maybe they freeze slower.
Is the above reasoning correct? I would also like to know different ways to reason out this.

This has everything to do with heat conduction, Setting aside freezing or melting for the time being, the two things that are happening in combination are (1) heat conduction and  (2) the capacity of a parcel of solid to store thermal energy by its temperature changing.  Heat conduction is driven by a spatial gradient in temperature, with heat flowing from a hotter region to a colder region.
So if the outside surface temperature of an object is suddenly increased, the interior temperatures will not change at all initially, and the temperature gradient will be zero everywhere, except at the surface, at which it will be very high. This will cause heat to begin flowing into the very outer layer of the object.  The heat that flows into this outer layer will cause its temperature to begin rising (to store at least part of the heat).  This temperature rise will then create a temperature gradient for heat to begin flowing into the next layer inward, where its temperature will begin rising.  This kind of sequence continues until, later, the temperature at the very center of the object begins to change.  So, as a result of conductive heat transfer, the temperature rise propagates like a wave traveling inward into the object.
This can all be described mathematically and modeled by the so-called transient heat conduction equation.


Does hydrogen peroxide exhibit intramolecular hydrogen bonding interactions? We know that intermolecular hydrogen bonding is seen in hydrogen peroxide $(\ce{H2O2})$. But is intramolecular H-bonding, i.e. hydrogen bond between an oxygen atom and a hydrogen atom attached to the adjacent oxygen atom possible? How strong will it be, compared to intermolecular H-bonding and why?

Preamble:

I will conveniently avoid entering into a debate as to what a hydrogen bond is, given the numerous theories which attempt to do so from various perspectives.
I will, however, attempt to use the relatively cheap and easy natural-bond orbital (NBO) analysis method to examine the possible existence of an intramolecular hydrogen bond (HBond) in the ground-state, global minimum configuration of H$_2$O$_2$.  In doing so, we will assume that the mere existence of an $n\rightarrow\sigma^{*}$ overlap indicates that an intramolecular HBond is indeed present (a commonly accepted metric).  For comparison, I will look at the planar form of H$_2$O$_2$ (containing a trans-type bond configuration) in order to minimize any overlap between the oxygen lone pair and the corresponding $\sigma^{*}$ orbital.  I note here that the NBO analysis is purely a qualitative method, and caution should be used when attempting to make quantitative comparisons with this method.

Methods:

The H$_2$O$_2$ molecule was constructed in $C_2$ symmetry (Structure A). A full optimization and corresponding harmonic vibrational frequency computation was employed at the MP2/heavy-aug-cc-pVTZ level of theory, where non-hydrogen (i.e. heavy) atoms are augmented with diffuse functions.  The basis set is denoted henceforth as haTZ.  The magnitudes of the residual Cartesian gradient components of the optimized structure were less than $4.0\times10^{-5}$ $E_{\textrm{h}}$ $a_0^{-1}$.  The frozen-core approximation was invoked for all computations.
A subsequent NBO analysis of the fully optimized H$_2$O$_2$ molecule was performed at the MP2/haTZ level of theory using the SCF density.
This procedure was also applied to a H$_2$O$_2$ molecule constructed in $C_{2h}$ symmetry (Structure B).
All computations were performed with the G09 (Rev. D01) software package.

Results:

Structure A is a minimum on the MP2 potential energy surface with zero imaginary modes of vibration whereas structure B is a transition state with one imaginary mode of vibration.

The NBO analysis of structure A reveals two $n\rightarrow\sigma^{*}$ interactions (where $n$ is an electron lone pair on oxygen and $\sigma^{*}$ is the $\sigma$ antibonding orbital of the neighboring O$-$H entity.  The existence of exactly two of these interactions is due to the symmetry of the molecule, and only one will be reported here for simplicity.

The following figure illustrates the overlap of a lone electron pair on O with the adjacent $\sigma^{*}$ orbital in structure A.  The occupation number for this antibond, 0.00351, indicates a very small overlap.  The corresponding second-order perturbation energy E$^{(2)}$(donor$\rightarrow$acceptor) is a modest 1.74 kcal mol$^{-1}$.  We note that this is merely an estimated measure of the strength of the interaction.

Interestingly, upon rotation of the torsion angle, $\tau$(H-O-O-H), to 180$^\circ$ (structure B), the $n\rightarrow\sigma^{*}$ interaction disappears.  The intramolecular hydrogen bonding  may play an important role in stabilizing the lowest-energy conformer in H$_2$O$_2$ (structure A) as the transition state, structure B, does not contain this interaction.

Comparing the strength of an intramolecular HBond to an intermolecular HBond is not straightforward.  Intermolecular interactions are frequenly and efficiently computed by using the supermolecular approach by taking difference of the electronic energy of a complex and the energies of each non-interacting body which make up the complex.  Intramolecular interactions cannot be determined in this way without breaking bonds, thereby skewing the resulting number.  To facilitate comparison to the intramolecular HBond in H$_2$O$_2$, an NBO analysis was performed on the water dimer (H$_2$O)$_2$ (in $C_S$ symmetry) due to the presence of a strong, intermolecular hydrogen bond.   The corresponding E$^{(2)}$ for an $n\rightarrow\sigma^{*}$ interaction is a much larger 6.82 kcal mol$^{-1}$, nearly 5 kcal mol$^{-1}$ stronger than that seen in H$_2$O$_2$

Conclusions:

An NBO analysis of the lowest energy conformer H$_2$O$_2$ indicates the possibility of the presence of weak, intramolecular hydrogen bonds.  These intramolecular hydrogen bond-like interactions could play a role in stabilizing the H$_2$O$_2$ complex.
The strength of the interaction is much smaller than that computed for the water dimer by about 5 kcal mol$^{-1}$.

(Note: I strongly recommend performing a Bader analysis on H$_2$O$_2$ to compare with the NBO analysis before reaching any solid conclusions.)

ADDENDUM:

I have tried to avoid calling this intramolecular interaction a hydrogen bond.  Based on the metric imposed in the preamble, it has a property of a hydrogen bond, however, the strength of this intramolecular interaction is so weak, it really does not reflect a true hydrogen bond (e.g. A typical hydrogen bond is about 10 kcal mol$^{-1}$ and a weak hydrogen bond is usually considered to be on the order of 2-3 kcal mol$^{-1}$.  If we take the NBO results at face value, the interaction is just too weak to be considered as a hydrogen bond.  Coupled with the fact that the geometry itself does not facilitate reasonable HBonding configurations.

Upon reflection of these issues, I conclude that there is, in fact, no intramolecular hydrogen bonding in H$_2$O$_2$.


How to prove that entropy is a state function? Consider the following steps:
\begin{align}
\mathrm{d}U &= \mathrm{d}q - p\,\mathrm{d}V\\
\mathrm{d}U &= C_V\,\mathrm{d}T\tag1\\
\end{align}
Side question: Is equation $(1)$ only true for a perfect gas or all substances?
\begin{align}
C_V\mathrm{d}T &= \mathrm{d}q - p\,\mathrm{d}V\\
\mathrm{d}q &= C_V\,\mathrm{d}T + p\,\mathrm{d}V\\
\end{align}
Divide by T:
$$\mathrm{d}S = \frac{C_V}{T}\,\mathrm{d}T + \frac pT\,\mathrm{d}V$$

The proof requires a substitution of $\frac pT=\frac{nR}V$ because when it is then differentiated with respect to $T$ it equates to zero and so does $\frac{C_V}{T}$ when it is differentiated with respect to $V$ which shows it is an exact differential. However, if you do not make the substitution you do not get zero so the two differentials do not equal each other. This would suggest that entropy is not a state function. Am I going wrong somewhere?

I don't understand what you mean by saying "if you don't make the substitution you do not get zero".  The relevant point is that, for ideal gases, $$p = nRT/V,$$ so $p/T$ is constant with respect to $T$.  Therefore $$\frac \partial{\partial T}\left(\frac pT\right)=0$$ for an ideal gas, whether or not you use the substitution to simplify the process.

If you don't use the substitution, you need to use the quotient rule to do the partial derivative properly, and you should still end up getting zero if you do it correctly.  Could that be where your confusion is coming from?


Why exactly does atomic radius increase down a group? I realize this should be a fairly basic question, but I'm still not quite satisfied with what I've been told from numerous sources. The general explanation seems to state that as we go down a group, there are more electron shells present to contribute to a "shielding" phenomenon, where inner electrons cancel out part of the inward force from the nucleus with a repulsive force. Thus with more energy levels present and more shielding, valence electrons begin to reside further and further away from the nucleus.

But at the same time, textbooks make a big deal about the concept of effective nuclear charge, which states that the "effective" force a given valence electron feels is a function of the net charge of the nucleus and non-valence electrons combined—the effective nuclear charge. I find this contradictory, since as we go down a group, the effective nuclear charge remains constant (equal numbers of protons and non-valence electrons are added). So what am I missing here? Why does shielding matter? Is the force on a single valence electron a function of the net nuclear charge or do we have to look at it from a more piecewise level, considering how other electrons are in between it and the nucleus and how it interacts with those individual electrons?

The description of $Z_\mathrm{eff}$ you gave is a bit too simplistic. As you correctly said, that would predict the same value of $Z_\mathrm{eff}$ for all elements in a group, which is not true.

In general we have $Z_\mathrm{eff} = Z - \sigma$ where $Z$ is the nuclear charge (that solely depends on the number of protons) and $\sigma$ is the shielding constant, which reflects the electron-electron repulsions but is not a simple function of the number of core electrons.

One of the earliest models to determine $\sigma$ was described by Slater in 1930.1 These are easily found online as "Slater's rules". These are quite simplistic, and so people tried to find better ways to calculate $\sigma$. Nowadays, one very popular source for values of $Z_\mathrm{eff}$ are the Clementi values.2 That's still quite long ago, so you can imagine that since then, we have come up with even more complicated ways to calculate it.

A peculiarity is that the values of $Z_\mathrm{eff}$ actually increase down the group. At the very least, that should dispel the myth that $Z_\mathrm{eff}$ only depends on the number of core electrons.

(Source: Keeler & Wothers, Chemical Structure & Reactivity: An Integrated Approach, p 264)

"Surely that should mean that the atomic radii decrease down the group!" you say.

I don't fault you for thinking that at all. However, that's not what we see: the atomic radii clearly increase going down the group. The second factor is the principal quantum number, $n$.

In the hydrogen atom, an electron in the 1s orbital has a much smaller radius3 than an electron in the 2s orbital. That's something that comes right out of the quantum mechanical description of the hydrogen atom.4 The same can be said for other atoms: a 2s orbital in lithium is smaller than a 3s orbital in lithium.

Note that I didn't compare a 2s orbital in lithium with a 3s orbital in sodium. Why? That's because sodium has a larger effective nuclear charge, and this serves to contract all the orbitals in sodium relative to those in lithium. Essentially, the orbitals are pulled inwards by the larger $Z_\mathrm{eff}$.

So, as you can see, going from lithium to sodium, there are two competing factors:

Larger $n$ suggests that the outermost electron should be further away from the nucleus.
Larger $Z_\mathrm{eff}$ suggests that the outermost electron should be nearer to the nucleus.

Unsurprisingly, it turns out that the variation of atomic radius is dependent on a combination of the two. We know that the atomic radius increases going down the group, so that must mean that the increase in $n$ (going from 2 to 3) outweighs the increase in $Z_\mathrm{eff}$ (going from 1.28 to 2.51), although it's sadly difficult to find an exact mathematical formulation in any textbook. It's likely that the exact dependency is very complicated and nowadays, such atomic properties are mostly calculated computationally anyway, which removes the need for such a mathematical formula.

Notes and references

1 Phys. Rev. 1930, 36 (1), 57

2 J. Chem. Phys. 1963, 38, 2686

3 I'm talking about $\langle r \rangle$, not the radius of a Bohr orbit, although that does also increase going from 1s to 2s.

4 I'm avoiding the circular logic that "higher energy means further away". It is further away because that is just the form of the radial wavefunction obtained by solving the Schrodinger equation - period.


Why does acetylene, having an acidic proton, not react with strong bases like NaOH and KOH? In acetylene (ethyne, HC≡CH), the carbons are sp-hybridised, which make the attached protons more acidic than those in alkenes or alkanes. (See also: Relative acidities of alkanes, alkenes, and alkynes)

Why doesn't acetylene react with strong bases like NaOH or KOH?

The $\mathrm{p}K_\mathrm{a}$ of acetylene is $25$. It is very weakly acidic, and hence, $\ce{KOH}$ is not sufficiently alkaline to react with it. Even methanol, with a $\mathrm{p}K_\mathrm{a}$ of $15.5$ does not react much with hydroxide ion. On other hand, $\ce{NaH}$ or $\ce{NaNH2}$ have more basic anions than hydroxide, and weak acids such as acetylene or alcohols can react with them.

Predicting the relative basicities can be done qualitatively by invoking the Brønsted conjugate acid-base theory. The rule to remember is, a base gives its corresponding conjugate acid in solution, and stronger the conjugate acid, weaker its conjugate base (and vice versa). This can be understood from the following

$$\ce{HA<=>H^+ + A-}$$

Here, $\ce{A-}$ is the conjugate base of the acid $\ce{HA}$. Now, if $\ce{A-}$ is highly unstable, it will have a stronger tendency to react with a proton and get converted to its conjugate acid which is relatively more stable. Since the equilibrium lies more to the left in that case, we say that the conjugate base is strong (and consequently the conjugate acid is weak). Now consider the following anions for the sake of example.

$$\ce{HC#C-} > \ce{NH2-} > \ce{OH-} > \ce{F-}$$

These are listed in decreasing order of stability, and hence also their strength as bases. This is because as the electronegativity increases from carbon to fluorine, the negative charge becomes increasingly stable on the atom. More electronegative elements are increasingly stable with the extra electron due to higher effective nuclear force of attraction. Correspondingly, by the dictum "stronger base is weaker conjugate acid", the order of the acidity is:

$$\ce{HC#CH} < \ce{NH3} < \ce{H2O} < \ce{HF}$$


Hydrogen vs neon in terms of entropy I know that both molecular size and complexity increase overall entropy, but between the two I'm not sure which increases it more. Would 1 mole of hydrogen have the greater entropy because it is two atoms instead of one, or would 1 mole of neon have the greater entropy because it is a larger atom?

Assuming you have equal moles of two different perfect gases, according to the Sackur–Tetrode equation the difference in their molar translational entropy is:

$$\frac{\Delta S_\mathrm{trans}}{R} = \frac{3}{2}\ln\left(\frac{M_2}{M_1}\right)$$

where the $M_i$ are the molecular masses. For instance, since $\ce{D_2}$ and $\ce{He}$ have identical masses this term would vanish, i.e. $ \Delta S_\mathrm{trans} = 0$.

If the atoms are in their electronic ground states then the electronic state makes no contribution to the entropy. On the other hand, a diatomic molecule has rotational and vibrational degrees of freedom. Assuming it's a rigid rotor, then in the high temperature limit (in which many rotational quantum levels are significantly occupied),

$$\frac{S_\mathrm{rot}}{R} = 1 + \ln\left(\frac{8\pi^2IkT}{h^2\sigma}\right)$$

where $I$ is the moment of inertia, $\sigma$ is a symmetry number.

Assuming the diatomic molecule is a harmonic oscillator, then

$$\frac{S_\mathrm{vib}}{R} =  x(e^x - 1)^{-1} - \ln(1 - e^{-x})$$

where $x = hν/kT$ and $T$ is around room temperature. However, the contribution of the vibrational entropy is tiny compared to the rotational component, so it can be ignored.

That would wrap up the story, in any case, if nuclear spin didn't have something to say about it. Nuclear spin contributes additionally to the entropy of $\ce{H2}$ and $\ce{D2}$ as explained in this post, resulting in an altered rotational entropy term. This rotational/spin contribution lifts the entropy of these molecules above those of Ne and He.

Therefore at room temperature and $\pu{1 atm}$ pressure, the molar entropy of $\ce{D2}$ is expected to be larger than that of helium due to the rotational contribution of $\pu{31.3 J mol-1 K-1}$.

Similarly, the translational entropy of neon is $\pu{3.5 J mol-1 K-1}$ higher than that of hydrogen gas, but $\ce{H2}$ has a rotational entropy of $\pu{18.7 J mol-1 K-1}$ so the entropy of neon is expected to be smaller by about $\pu{15.2 J mol-1 K-1}$.


Does a certain "Badecker reaction" actually exist? A book I read, mentions a certain Badecker reaction under coordination chemistry. It goes like this:

$$\ce{Na2[Fe (CN)5 NO] + Na2SO3 -> Na4[Fe (CN)5 (NOSO3)]}$$

I googled up the name, but I didn't get anything even remotely related to chemistry.

That particular book does have a pretty long history of typos and a couple of factual errors here and there (but I still use it because its got quite a few useful tips …)

So the question is; is there such a thing called the Badecker reaction (I even tried googling it up with an umlaut, Bädecker reaction, because, well … Badecker looks a little German, but then I was directed to a Wikipedia page on cadmium oxide. So what is the reaction’s name?

Also even if it doesn't go by that name, does the reaction actually even exist?

The reaction in question is actually called the Boedeker reaction. How the book came from Boedeker to Badeker I do not know, but maybe there was one instance of Boedeker being written Bödeker which was later misparsed by a human or a robot and mistook for an a. (Replacing oe for ö is a common mistake with German names since the latter is typically used in orthography but names don’t adhere to orthography.)

The reaction is well-known and often used as a qualitative analysis for sulphite ions in solution, much like the same nitroprusside ion can also be used to detect sulphide ions. In chemical formulae, it is:

$$\ce{[Fe(CN)5(NO)]^2- + SO3^2- <=> [Fe(CN)5(NOSO3)]^4-}$$

The reaction in itself is also rather interesting since for over 100 years the structure of the addition complex was unknown. Bottomley et al. were finally able to obtain crystal structures of the probably structurally similar ruthenium complex $\ce{[RuCl(bpy)2(NOSO3)]}$.[1] These show that the addition product is — very much like in the sulphide case — formed by nucleophilic attack of the sulphite sulphur’s lone pair to nitrosyl nitrogen creating a nitrosylsulphite anion $\ce{O3SNO-}$. Thus, the complex formed should be named pentacyanidonitrosylsulphito-κN-ferrate(II).

Reference:
[1]: F. Bottomley, W. V. F. Brooks, D. E. Paéz, P. S. White, M. Mukida, J Chem. Soc., Dalton Trans. 1983, 2465. DOI: 10.1039/DT9830002465.


How to make spectra smooth by using gaussian or lorentzian function I want to use some origin spectrum to fit to my experimental spectra. However, my experimental spectra is not so smooth because of finite resolution of spectrometer. I used to use 'origin' to make it smooth, but it seems not reasonable because I don't know how 'origin' makes it. Then I was suggested using Gaussian or Lorentzian function.
A lot of nonsense, in one word, please teach how to use gaussian or lorentzian function.

please teach how to use gaussian or lorentzian function.

It will take several hours of discussion or 4-5 lectures. You are mixing several things.
a) Briefly, your collected spectrum is noisy. You want to remove the noise. What type of spectra is this?
b) The finite resolution of your spectrometer has nothing to do with noise. Increasing the resolution mathematically or so-called deconvolution is another story.
c) The easiest solution is to "fit" Lorentzian or Gaussians to your experimental spectrum after denoising it (step a). Experimental spectra are neither pure Gaussians nor pure Lorentzians.
Here are the steps. I am mentioning the keywords so that you can search the right terms.

Can you get hold of PeakFit software? It has a better function library than Origin.
Denoise your spectrum, using Savitsky-Golay filter.
Once excessive noise has been filtered, use a trial function such as Gaussian or Lorentzian and see which one fits better. This is checked by residuals.
In an experimental spectrum, not all peaks have the same width so you will have to use multiple Gaussians or Lorentzians to fit a segment of a spectrum.

Search "Pragmatic Introduction to Signal Processing Applications in scientific measurement 2019 edition" by Prof. Tom O'Haver. It is a free book and contains all the practical stuff which would be very useful to you.


Quick way to find magnetic nature of compounds or ions According to the Molecular Orbital Theory, the method to check whether a given compound is paramagnetic or diamagnetic is to draw the molecular orbitals and then to see whether we have all the electrons paired or is there some unpaired electron. However, I want to know, is there a quick way to find the magnetic nature without drawing the orbital diagrams?

Well, for some compounds it's easy, because there are an odd number of electrons - thus there's no way to pair them all.

Otherwise, there is no quick-fire way without doing the MO diagram.

Consider $O_2$ which has an even number of electrons, but is a ground-state triplet with two unpaired electrons. The only way to know is to look at the MO diagram.

If you practice, you'll realize certain patterns and develop more of an intuition.


Clausius-Clapeyron - Slope of the best-fit-line I'm using the equation $\frac{y_2-y_1}{x_2-x_1}$ to find the slope of the best fit line for this problem but I can't get the figure shown in the book (-3773x K).  I've been at this a while and I really need to know so I can move on.  What am I doing wrong?  I've looked at half a dozen general chemistry texts and none of them make this step explicit.

Using your formula, $\mathrm{slope}=\frac{y_2-y_1}{x_2-x_1}$, and eyeballing the most extreme points on the chart, $\mathrm{slope} = \frac{0-6.2}{0.005 - 0.0033}$, I get a slope of 3647 K, pretty close to the "official" answer.

If you want to get more fancy, you could use a software program to fit this data.

my_data <- data.frame(vapor_pressure = c(1.1, 6, 28, 95, 293, 521),
                      temperature = c(200, 220, 240, 260, 280, 300))

# linear regression on transformed variables
my_data$ln_vp <- log(my_data$vapor_pressure)
my_data$inverse_T <- 1/(my_data$temperature)
linear_fit <- lm(data = my_data, formula = ln_vp ~ inverse_T)
linear_fit$coefficients['inverse_T']

inverse_T: -3787.46829114867

This answer is even closer to the "official" answer.  The remaining difference is probably due to rounding errors or differences in fitting algorithms used.  It only affects the answer by 0.3%.


Predicting the generation of singlet oxygen On Wikipedia, I saw the reaction $\ce{H2O2 + NaOCl -> O2 + NaCl + H2O}$, where the $\ce{O2}$ is in the $^1\Delta_g$ (singlet) state. Can anyone offer an explanation for why it isn't in the usual triplet state, maybe using the theory of molecular orbitals?

Molecular orbitals are actually, to the best of my knowledge, not really helpful in explaining this observation. Instead, a much more fundamental concept does a much better job: the conversion of (electronic) spin.

The reacion you mentioned can be analysed in terms of spin states. Nobody will disagree that hydrogen peroxide, sodium hypochlorite, sodium chloride and water are singlet compounds. Therefore, using generic Greek letters as placeholders for the different compound to illustrate the point that their true doesn’t matter much, we can write the equation as:

$$\ce{^1\unicode{x391}\ + ^1\unicode{x392}\ -> ^?\Gamma\ + ^1\Delta\ + ^1\unicode{x395}}\tag{1}$$

We see that we are reacting two singlet compounds two three compounds and two of the products are also singlets. This means, every spin on the left-hand side of the equation is paired and most of the spins on the right-hand side, too. To generate a triplet product, we would need to flip a few spins but that is a slow process on the timescale of a chemical reaction. (Note that the conversion from singlet oxygen back to triplet can be observed as a faint red glow demonstrating that it is not instant but delayed.) Thus, in the moment of the particle collision, generating all-singlet products from all-singlet reactants (if only a single product potentially can have an unpaired spin) is the only option. Thus:

$$\ce{^1\unicode{x391}\ + ^1\unicode{x392}\ -> ^1\Gamma\ + ^1\Delta\ + ^1\unicode{x395}}\tag{2}$$


Making aspirin without acetic anhydride I want to make aspirin. Normal method is to react salicylic acid with acetic anhydride, but it is extremely hard to find acetic anhydride. How may I synthesize aspirin with acetic acid and salicylic acid only?

I have watched a video where someone makes it with sulphuric acid, salicylic acid and acetic acid. Can I use phosphoric acid insted of sulphuric acid as a catalyst?

I just looked in the literature, and it has been done even without catalyst in a "reactive distillation column" at high temperatures (patent US201113194873). If you can remove the water formed, you can technically drive the reaction forward, even if it is slow. A catalyst will just make that process easier, requiring less heat and time.

Phosphoric acid is a much weaker acid than sulfuric acid, and it does not have the same dehydrating properties. However, if you have a proper setup, it should do the trick. A big problem with phosphoric acid is that the most concentrated you can normally get is 85%, which means you still have to remove 15% of water somehow. If you don't, you'll likely hydrolyse any aspirin formed and get your starting material back.


What changes does zinc chloride undergo when ‘fusing’ on an atomic level? How does this influence reactivity? I have been using a reaction with an organozincate starting material. This zincate is freshly prepared from zinc chloride and a Grignard reagent. The experimental procedure ‘handed down’ from more senior members of the lab include fusing the zinc chloride before use. ‘Fusing’ means:

heating under high vacuum with a heat gun until the salt melted
cooling the liquid down rapidly with the cold air stream of another heat gun to receive a glass-like solid

This is then dissolved in THF and the Grignard added.

In a recent group seminar I spoke about the trouble I had in achieving consistency with that reaction. Part of the resulting discussion revolved around the method of fusing and what I had been doing wrong. The PI noted how important it is for the fusing to be performed correctly as it influences the reactivity.

What actually happens to the zinc chloride during this process? My little understanding of inorganic solid-state chemistry leads me to believe that the initial crystals break down but the resolidified glass should essentially have the same structure. I am also at a loss how to explain any differences in reactivity following this method.

While it is not the most easy task to search for such references, the concept of fused zinc chloride does exist in the literature; the following two references use it in their experimental sections:

P. Dotta, A. Magistrato, U. Rothlisberger, P. S. Pregosin, A. Albinati, Organometallics 2002, 21, 3033–3041. DOI: 10.1021/om020314q. Synthesis of the ligand 4 on page 3040.
P. G. Gassman, J. G. Schaffhausen, J. Org. Chem. 1978, 43, 3214–3223. DOI: 10.1021/jo00410a024. Synthesis of compound 28 on page 3220.

One purpose is to eliminate water as @Waylander has already pointed out.

Another is to have a clean crystal surface. Aging is a common process of crystals. The surface is transformed slowly. The crystalline structure changes since it is not the same to be surrounded with other atoms (molecules, ions) of the same composition than with nothing or a gas.

Finally, another purpose is to have a crystal size as small as possible. The smaller the crystals the bigger the surface area, so more is available and the reaction proceeds faster.
 The small crystals are obtained in the rapid cooling. The faster, the smaller.

The zinc chloride solidifies in one lump, but there are still crystals.
They only have grown together.
If it were a real glass, the reactivity would be the highest since there would be no packing.
This can only be achieved with extreme rapid cooling.
When cooling is fast but still takes some time, crystals grow.
When the cooling is slower, the crystals grow bigger.
The smaller they are, the more surface the lump has.

Melting must be done for the other reasons, but when you cool it must be done fast because, if not, you will get a less reactive mass.
It is not a comparison between molten and non molten product, but between fast or slow cooling of the molten salt.


Do Acids/Bases undergo dilution/bio-degradation when exposed to air? I spilled some cleaning fluid/drain opener in my trash can, a long time ago(almost 3 months ago), I didn't tried to clean it up. The cleaning fluid was diluted, and I spilled a small amount not more than 30 ml, I think.

The bottle does not say, what the cleaning fluid contains, but it probably has Sodium Hydroxide(if it is an alkali-based cleaner) or Hydrochloric acid(if it is an acid-based cleaner).

Now, what I have heard about Sodium Hydroxide, is that it reacts with carbon dioxide in the atmosphere, and lose its corrosiveness over time.

So, is it possible that NaOH got completely carboxylated in 3 months and nothing corrosive remains?

Now, about strong acid such as Hydrochloric acid, since they are hygroscopic, do they become dilute over time?

What you heard about NaOH is right: it surely got completely carboxylated in 3 months, maybe even in 3 days. Not that the resulting compounds are completely neutral to the environment, but yes, they lack that corrosive ability.

As to the more general question of what happens to inorganic substances once they are out in the open: dilution is quite possible, biodegradation is not so likely. Some of them, like hydrochloric acid (and most other acids), would pretty soon end up reacting with some random compounds out there to yield relatively harmless products which would become normal part of the environment. Others (think of heavy metals) may have long-lasting adverse effects.


Hydrates and melting points Basalt melting point: wet vs dry

I had a student ask me this in class the other day; I thought about it, but I still can't work it out.

Many mineral and rock hydrates (take basalt, for instance) have much lower melting points than their anhydrous counterparts.  I could not craft a plausible explanation for this.

I realize that hydrates have different crystal structures due to the presence of water, but I am unsure how this will affect the melting points.  If anything, the possibility of hydrogen bonding would make the melting point higher, no?

It's been so long you probably won't even see this, but in case someone else is searching for an answer, here are some interesting answers I've found: (The websites are in French, so I tried to translate the intersting bits)

  Igneous rock samples can be melted in the laboratory by heating them at elevated temperatures (1000 ° C to 1300 ° C). If one operates under pressure and in the presence of volatile elements, the melting temperature of the rock and its minerals is much lower (650 ° C to 950 ° C). It is known that magmas contain varying amounts of volatile elements that lower the melting point of many silicates, thus allowing the magma to remain liquid even at temperatures below those required to melt rocks in the laboratory. Indeed, the temperatures that exist in the different levels of the earth's crust are not sufficient to melt rocks in the absence of volatile elements.
  The main volatile elements found in magmas can be commonly observed in the gaseous state at the vents of active volcanoes. It is essentially water vapor, carbon dioxide, hydrogen, chlorine, fluorine (in the form of acids) and many other gases [...].

Taken from http://www.fossiliraptor.be/rochesetphenomenesignes4.htm

  In the case of hydrated paragneiss, during the warming of this rock, the melting begins as soon as it reaches the first reaction ((4) Qz + Ab + Gold + H2O ⇄ Liquid), around 650 ° C. Indeed, this gneiss contains all the reagents necessary for this reaction to take place.
  For a dry paragneiss, on the other hand, nothing happens on the curve corresponding to reaction 4; indeed, one of the reagents, water, is lacking. However, when the temperature exceeds 850 ° C, this rock crosses the curve corresponding to reaction 5 ((5) Bt + Ab + Qz + AlS ⇄ Liquid + Cord / Gt + KF, which can be decomposed as follows: (6) Bt + AlS ⇄ Gt / Cord + KF + H2O, then 7) Qz + Ab (Pg) + Gold (KF) + H2O ⇄ Liquid) and the two coupled reactions described above take place: the biotite decomposes to liberate water; therefore, we are in a situation where all the necessary products are present: the merger can take place.
  In other words, a hydrated system will melt (cross its solidus) at a lower temperature than a system without free water: the two reactions correspond respectively to "hydrate solidus" and "dry solidus"!

Taken from: https://planet-terre.ens-lyon.fr/article/reactions-de-fusion.xml#equation4


Stability of diboron(B2) and dilithium(Li2) Why is $\ce{B2}$ more stable than $\ce{Li2}$?

Both $\ce{B2}$ and $\ce{Li2}$ have bond order equal to 1. Moreover, $\ce{Li2}$ has only 2 electrons in the antibonding molecular orbital while $\ce{B2}$ has 4 electrons in the antibonding molecular orbital. Thus, $\ce{Li2}$ should have been more stable than $\ce{B2}$ if we consider the number of electrons in the antibonding molecular orbital as the parameter of stability. Does the fact that $\ce{B2}$ is smaller than $\ce{Li2}$ makes it more stable?

Firstly, the use of molecular orbital diagrams allows you to predict the electronic configuration, if a molecule would exist or not (as you say, evaluating the bond order), and some magnetic properties. Moreover, if you have "the same molecule" as O$_2$ and its ion  O$_2^-$, etc, you can say the higher bond order, the bigger stability.

But if the bond order is the same (your case), you can not talk about stability using those diagrams.

Coming back to the question, looking the dissociation energy (the energy relative to separated atoms) for B$_2$, you will see it is bigger in module.
So it will be more stable.
I do not have a qualitative way to estimate which of those molecules is more stable.

The way to know it theoretically is using electronic structure programmes.


Stereoisomers of compounds named tribromocyclobutane Here is the link to the page in which I don't understand the answer to the last question (Question 13). The question states:

  Draw all of the constitutional isomers and stereoisomers of compounds named
  tribromocyclobutane (all structures must contain a cyclobutane ring).

The following image exists as a part of the answer.

But why is the following image not part of the answer?

Are the above two images not diastereomers (hence stereoisomers)?

The second configuration shown cannot exist simply because of the geometry of the orbitals at the carbon center. The angle between the bromine's (those on the same carbon) is much less than 108° even though there is no forced strain.


Why are there free electrons in a metal? Recently, we covered metallic bonding in chemistry, and frankly, I understood little.

I understand that:

Metals bond to each other via metallic bonding
Electricity can flow via free or delocalized electrons

But, I do not understand why the metal atoms turn into ions and delocalize the electrons, why don't the metal atoms stay as atoms?

When electricity flows, the electrons are considered "free" only because there are more electrons than there should be, and because the transition metals, such as iron, copper, lead, zinc, aluminum, gold etc. are willing to transiently accept and give up electrons from the d-orbitals of their valence shell.

Transition metals are defined in part by their stability in a wide range of "oxidation states"; that is, in several combinations of having too many or too few electrons compared to protons. This is thought to be because of the d orbital in their valence shells. Compared to the s and p orbitals at a particular energy level, electrons in the d shell are in a relatively high energy state, and by that token they have a relatively "loose" connection with their parent atom; it doesn't take much additional energy for these electrons to be ejected from one atom and go zooming through the material, usually to be captured by another atom in the material (though it is possible for the electron to leave the wire entirely). This impetus can be caused by many things, from mechanical impact to chemical reactions to electromagnetic radiation (aka light, though not all of it visible); antennas work to capture radio frequencies, because the light at those frequencies induces an electric current in the wire of the antenna. Now, in the absence of a continuous force keeping the electron in this higher energy state, the electron (and the metal atoms) will naturally settle into a state of equilibrium. Electricity is generated when just such a force is acting on the metal, giving energy to the electrons in the d orbital and forcing them to move in a certain direction.

This impetus can come from many sources, as discussed, be it the movement of a magnet within a coil of wire, or a chemical redox reaction in a battery creating a relative imbalance of electrons at each of two electrodes. The end result is that the electrons, given additional energy from this voltage source, are ejected from their "parent" atom and are captured by another. The "holes" left behind by these electrons are filled by other electrons coming in behind them from further back in the circuit. Thus, the energy provided by the voltage source is carried along the wire by the transfer of electrons.

The analogy typically made is to the flow of water, and it generally holds in many circumstances; the "voltage source" can be thought of as being like a pump or a reservoir, from which water flows through pipes, and the amount of water and the pressure it's placed under (by the pump or by gravity) can be harnessed to do work, before draining back to a lower reservoir. The pipes are similar to wires in many ways; the larger the diameter, and the smoother the inside of the pipe, the more and the faster water can flow through it (equivalent in many ways to the thickness and conductivity of the metal wire), and when under enough pressure (high enough voltage), the pipes will actually expand slightly and hold more water than they would at low pressure (this is a property of wires and other electrical conductors called "capacitance"; the ability to store a charge while under voltage and to discharge it after the voltage is released).


homogeneous mixture - Are they truly uniform in composition and proportion? If 5ml alcohol is dissolved in 1 L of water ( the purpose is to highlight the low quantity of one of the liquids) , how is it possible for 5 ml to evenly dissolve in 1 L of water ? There is not enough of alcohol to sit between water molecules , isn't it ? So how is the mixture's proportion/composition is uniform ?

Or is there anything that I miss to understand in the term - composition?

Maybe it is just a question of what you mean by "evenly dissolved". True enough, in your example, the proportion of alcohol to water is not "even" - rather the solution is indeed mostly water containing only a little bit of alcohol. But for a true solution such as this, the alcohol is dispersed evenly (uniformly) throughout the bulk of the solution, so that if you repeatedly withdrew a sample and analyzed it, you would get the same result no matter which part of the liquid you sampled.

That would not be the case for oil and water - they would exist in separate phases and which phase you sampled would give a different analysis result.


Term symbols for nitrogen: determining J I'm trying to figure out what terms are possible for nitrogen with the electron configuration $\ce{[He] 2s^2 2p^3}$. There is an old question on StackExchange and the corresponding answer was a great help already but doesn't mention $J$ or specify what terms exactly are possible. I'll try to explain how I would determine the possible terms:

The s-electrons needn't be considered for they complete the s subshell, this means I'm left with three p-electrons. I do get the same results for the possible values of $L$ and $S$ as user orthocresol provided with his answer namely:

$L=3,2,1,0$

$S=3/2, 1/2$

Now the possible values of J range from $L+S$ to $|L-S|$:

For $L=0$ there are two S-Terms*: $^4\!S$ and $^2\!S$ giving

$$^4\!S_{3/2} \quad ^2\!S_{1/2}$$

$L=1$ gives two terms*: $^4\!P$ and $^2\!P$, the former term has the following possible J values: $J = 5/2, 3/2, 1/2$ and the latter: $J=3/2, 1/2$. The P-Terms of this electron configuration are:

$$^4\!P_{5/2} \quad ^4\!P_{3/2} \quad ^4\!P_{1/2} \quad ^2\!P_{3/2}\quad ^2\!P_{1/2}$$

Doing the same for $L=2$ there are two D-Terms*: $^4\!D$ and $^2\!D$, possible J values for the former: $J = 7/2, 5/2, 3/2, 1/2$ and for the latter: $J = 5/2, 3/2$. This gives six different terms.

I'm stopping at this point because I think I'm not doing it right. Using the NIST ASD I only see five terms with this electron configuration in the Grotrian diagram:

$$^4\!S_{3/2}, ^2\!D_{5/2}, ^2\!D_{3/2}, ^2\!P_{1/2}, ^2\!P_{3/2}$$

and according to the book "Physical Chemistry: A Molecular Approach" Table 8.4 for three p-electrons only $^2\!P,^2\!D,^4\!S$ Terms are possible. What's wrong with my approach that I'm that much off?

*Question on the side: I don't think it's correct to use the term "term" here because what I'm talking about is split into actual terms. Is there a way to refer to, for example, the $^4\!D$-Terms in general, maybe term system?

Firstly, regarding the extra question: Atkins' Molecular Quantum Mechanics (5th ed.) uses "term" for $^2\!S$, and "level" for $^2\!S_{1/2}$.

Back to the main question. It's been a long time since I did term symbols, so I am happy to be corrected, but if I am not wrong, your $^4\!P$, $^4\!D$, and $^2\!S$ terms are not allowed because of the Pauli exclusion principle. The argument is pretty similar to the one in the comments on my answer, which you're already aware of.

A term with $S = 3/2$ must have a set of states with $M_S = +3/2, +1/2, -1/2, -3/2$. Likewise, a $P$ term with $L = 1$ must also have a set of states with $M_L = +1, 0, -1$. Consequently, the total number of states associated with a single term $^{2S+1}\!L$ is $(2S+1)(2L+1)$. (The total number of states in a level, $^{2S+1}\!L_J$, is $2J+1$, and you can also show that the sum of this over the allowed values of $J$ is equal to $(2S+1)(2L+1)$.)

One of those twelve states in the purported $^4\!P$ term must have $(M_S, M_L) = (+3/2, +1)$. However, $M_S = +3/2$ can only be achieved if all three electrons have the same spin (spin up in this case), which in turn necessitates that the three electrons are all in different p-orbitals, such that $M_L = \sum m_l = +1 + 0 + -1 = 0$.

Ergo, a state with $(M_S, M_L) = (+3/2, +1)$ cannot exist, and the entire term $^4\!P$ cannot be possible. The same argument also applies to the $^4\!D$ term.

I suspect that showing that the $^2\!S$ term is forbidden is somewhat trickier. I had a similar issue before with carbon, which is an easier system to understand (only two electrons to consider instead of three), so maybe you will find this useful: Pauli-forbidden term symbols for atomic carbon.

Out of the three surviving terms ($^2\!P$, $^2\!D$, $^4\!S$) you already see from NIST that all the possible $J$ values (from $|L-S|$ to $L+S$) are allowed, so there is no issue with your calculation of $J$.


Why add water first then acid? From school, I remember a very important rule: first you need to pour the water and then the acid (when you need to mix them) not vice-versa. This is because otherwise the aсid becomes very hot and splashing may happen.

So, why does it get hotter when water is poured into it? What reaction takes place?

This is mostly the case for sulfuric acid. Commercially available sulfuric acid is dense (~1.8 g/ml) and when water is added, it may not mix. In this case a layer of hot weak acid solution is formed, which boils and sprays around. When acid is poured into water, it flows down the flask and mixes much better, so no boiling occurs.

The reason this occurs is due to the large amount of energy released in the hydration reaction of sulfuric acid ions. Do not believe that heat comes from dissociation, as the dissociation of acids, bases, and salts always consumes energy. The energy is released from subsequent hydration, and the release may be high, especially if $\ce{H+}$ or $\ce{OH-}$ ions are hydrated.


find the pH where is maximum the concentration of a amphiprotic substance Considering a general amphiprotic substance NaHA. I know that the maximum concentration of the specie $HA^-$ is found at $pH = \dfrac{pKa1+pKa2}{2}$. Why? How i can confirm this statment?

The equations are:

$$ \ce{HA^- + H_2O <=> A^{2-} + H_3O^{+}} \quad Ka2$$
and
$$ \ce{HA^- + H_2O <=> H_2A + OH^{-}} \quad Kb=\dfrac{Kw}{Ka1}$$

Let's take an aqueous solution of a salt $\ce{NaHA}$ with the initial concentration $C$ when added to water. It will completely dissociate  according to the eaquation: $\ce{NaHA(s) \rightarrow Na^+ +HA^-}$.

$\ce{HA^-}$ will participate in three equilibria:

$\ce{2HA^- \leftrightarrows H2A +A^{2-}\quad \quad \quad  }$ ${K_1^0=K_{A2}/K_{A1}}$

$\ce{HA^- +H2O\leftrightarrows H3O^+ +A^{2-}\quad }$ $K_2^0=K_{A2}$

$\ce{HA^- +H2O\leftrightarrows OH^- +H_2A^\quad }$ $K_3^0=K_{B1}={K_w}/{K_{A1}}$

In most cases, $K_1^0$ is far bigger than $K_2^0$ and $K_3^0$. So, the first equilibrium is the preponderant reaction, and this reaction will impose the pH of the solution.

Let's now calculate the product $K_{A2}\times K_{A1}$:

$K_{A2}\times K_{A1}= \frac{[\ce{}A^{2-}].[\ce{H3O+}]}{\ce{[HA^-]}}.\frac{[\ce{}HA^{-}].[\ce{H3O+}]}{\ce{[H_2A]}}$

According to the stoichiometry  of the preponderant reaction, we have $\ce{[A^{2-}]=[H_2A^]}$. So the product
$K_{A2}\times K_{A1}= \ce{[H3O+}]^2}$ .
i.e. $\ce{pH}=0.5(\ce{p}K_{A2} +\ce{p} K_{A1})$


Which electron is knocked off a Helium atom? When a He atom is ionized, which electron is removed first, the one with the spin up or down? or is it just random?

An interesting question! Does the polarization of ionizing radiation interact with electron spin? Apparently, that has been investigated to some extent, and some spin information can be obtained from an entangled electron and photon.

So the answer to your question is not a clear-cut "yes" or "no", but depends on the parameters of the incident photon.


Space in naming Acetophenone The IUPAC name for Acetophenone is : 1-Phenylethanone.
An other name is : Methyl phenyl ketone.
My question here why there's a space in "Methyl phenyl ketone" although there's no space in "1-Phenylethanone"?

The two mentioned systematic names correspond to different types of nomenclature. The name “1-phenylethan-1-one” is formed according to substitutive nomenclature, and the name “methyl phenyl ketone” is formed according to functional class nomeclature. Both systematic names are in accordance with current IUPAC recommendations. However, the preferred IUPAC name (PIN) for such acyclic ketones is generated using substitutive nomenclature. Note that the omission of the locant “1” for the suffix “one” in “1-phenylethanone”, while permissible in general usage, is not allowed in PINs, thus the name “1-phenylethan-1-one” is the PIN. Furthermore, the name “acetophenone” is retained, but only for general nomenclature.

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

  P-64.2.2.1 Acyclic ketones

  Unsubstituted acyclic ketones are systematically named in two ways:

  (1) substitutively, using the suffix ‘one’ and the prefix ‘oxo’; the presence of several ‘one’ characteristic groups is denoted by the numerical multiplying prefixes ‘di’, ‘tri’, etc.; the final letter ‘a’ of a numerical multiplying prefix is elided before the suffix ‘-one’, for example, ‘tetrone’;

  (2) by functional class nomenclature using the class names ‘ketone’, ‘diketone’ etc.; substituent groups are placed, as separate words, in alphanumerical order before the class name.

  Method (1) generates preferred IUPAC names.

Note that the rule for functional class nomeclature (method 2) stipulates that the groups shall be written as separate words.


The definition of density Could density be defined as the measure of matter per unit volume instead of mass per unit volume because mass is a measurement of matter?

It depends, there are different kinds of densities: mass density, particle density, charge density, probability density ...

Depending on how your measure your matter, mass, number of particles, charge (e.g. for electrons), you could use the corresponding density.

"Matter per unit volume" would only work in a broader sense, since it does not specifically state what kind of measure you want to use. Most of the time one simply defaults on mass, but this depends on the specific field.


Why does sodium require energy to ionize? If you look at $\ce{Na}$, it has a very low ionization energy, but the question is, why does it have an ionization energy at all? If it can get to the electronic configuration of $\ce{Ne}$, by losing an electron, which is a very stable configuration, then it should release energy. Consider the analogy to gravitational potential energy, you can consider $\ce{Na}$ to be analogous to standing on a very high ledge. And $\ce{Na^+}$, is the position at the ground. Obviously, getting to the ground is much more stable. And if I jump down from the ledge I convert my potential energy to kinetic energy and ultimately to heat, thus releasing energy. Why does sodium require energy to ionize?

And, consider ferric chloride $\ce{FeCl_3}$.

Why does iron form this bond? I mean, it makes sense from chlorine's point of view to share an electron with iron. All the chlorines have a full outer shell. But what about iron? It had $2$ valence electrons, now it has $5$. It would make much more sense for iron to bond with chlorine to form $\ce{FeCl_6}$, since then it, too would have a full outer shell. Or, $\ce{Fe}$ could give away its $2$ electrons to $2$ chlorines and form ionic bonds. Why does this not happen?

There are two very different questions here so let's break things down a bit
Why does sodium have an ionization energy?
Since electrons are negatively charged and nuclei (owing to the presence of protons) are positively charged, electrons will typically be drawn to the nucleus.  The situation gets a bit more complicated by (a) the quantum nature of energy and (b) the presence of other electrons.  Complication (a) is the reason why electrons do not behave like planets orbiting the sun because if they did the electron would eventually be pulled in to the nucleus.  (More details probably requires a different question.)  The second complication is more relevant to your question, because additional electrons will both repel the electron we are trying to eject through ionization and also shield some of the positive charge of the nucleus.  (Think about wearing sunglasses; they don't make the sun less bright, they just make less light reach your eyes.)
Your comparison of Na to that of an object on a cliff is not correct; the sunglasses analogy I used above is better (although it's flawed as well).  The reason why sodium, or any other atom/ion for that matter, has an ionization energy is because it requires energy to separate two oppositely charged particles.  The attraction may be weak, as it is in sodium, but it is still non-zero.
Why does $\ce{FeCl3}$ bond the way it does?
First, it is a good idea to look at the wikipedia page for iron(III) chloride, which shows that the molecule actually binds in an octahedral fashion, where two additional chlorides from neighboring molecules bind in a bridging fashion.  This is a type of bonding that is typically not discussed until one's third year of chemistry education.  Valence shell electron counting works really well for identifying "stable" electron configurations of ions and covalent molecules made out of lighter elements (C, N, O, for example).  The presence of d-orbitals, plus the ionic nature of the interaction between Fe(III) and chloride ions, plus the Lewis acid/base concepts that do a better job at describing and rationalizing bonding in transition metal complexes, makes the concept of valence electrons not the best model to explain bonding in $\ce{FeCl3}$.
Without access to molecular orbital theory, it is perhaps best to explain the bonding in iron(III) chloride from a strictly ionic bond nature.  Fe(III) is a +3 cation and therefore requires three -1 anions to balance the charge.


Oxidizing 2-Methylpropan-2-ol using potassium dichromate. Why is heating required? In Chemistry class, we had to combine $\ce{K2Cr2O7}$ with 2-Methylpropan-2-ol and heat it for the oxidation to occur.

But why do we have to heat the mixture when, for example, butanol, can be oxidized via use of the same reagent but without heating it?

Tertiary alcohols (such as 2-methyl-2-propanol, in your specific case) cannot be directly oxidized by $\ce{Cr(VI)}$ salts like $\ce{K2Cr2O7}$ because they have no $\alpha$-hydrogen. On the other hand, a primary alcohol (such as 1-butanol) will be readily oxidized, even under mild conditions and with reagents much gentler than $\ce{K2Cr2O7}$.

With heating and sufficiently acidic conditions, a tertiary alcohol can be eliminated to form an alkene, which can then be oxidatively cleaved to yield two molecules with carbonyl groups. In the case of 2-methyl-2-propanol, one would presumably obtain acetone and either formaldehyde or formic acid, depending on how far the oxidation is permitted to go. (Though, under conditions vigorous enough to effect the oxidative cleavage of the alkene, it's highly unlikely the aldehyde could be isolated. I would also note that carbon dioxide is a product of oxidative cleavage of terminal alkenes when $\ce{KMnO4}$ is the oxidizer, but whether this occurs with $\ce{Cr(VI)}$ salts is unknown to me.) This is really the only reaction I can imagine to have plausibly occurred, if any.

To the best of my knowledge, $\ce{K2Cr2O7}$ is not very efficient for this purpose, and the procedure itself strikes me as pretty odd if the oxidative cleavage described above is the intended reaction. It may be that no reaction was intended to happen, since oxidation reactions of this type can be used as chemical tests to differentiate alcohols by degree of substitution.


Can I use the mono ammonium phosphate found in fire extinguishers to make crystals? I recently opened up a old decompressed fire extinguisher out of curiosity. After some research I discovered that the dry powder inside, mono ammonium phosphate, can be used to make beautiful crystals. However, I found this powder to be extremely hydrophobic and could never get its dissolve in water at all. I mean this stuff is amazingly hydrophobic. Its fun to see how it reacts in water. But why cant it dissolve and what kind of preparation of mono ammonium phosphate will dissolve and become crystals?

If you google for MAP, you'll find out that it is quite soluble in water, ranging from 370 to 401 g/litre of H2O at temperatures around 25 C. Thus, what you report contradicts literature data on MAP's solubility. So, maybe the powder you found is not MAP, it's something else (CaCO3 comes to my mind, because it is insoluble in water and yields CO2 when heated, but I may be wrong).


Nomenclature of a poly-nitrile For this compound, I think the ethyl group should be in the parent chain and the numbering should start from the right-most carbon. Hence, the compound should be named 2-methylhept-1-ene-1,3,4,5-tetracarbonitrile, instead of the name given above. Am I wrong somewhere?

Your approach is correct.

The given compound includes the characteristic groups $\ce{-CN}$, which are expressed as suffix ‘carbonitrile’. Since there is only one kind of suffix, the order of seniority of suffixes is not relevant in this case.

Functional suffixes are always attached to the name of the parent structure, which is modified by ‘ene’ or ‘yne’ endings if necessary.

When there is a choice among acyclic parent structures, a principal chain must be chosen. The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) stipulates various criteria for the selection of the principal chain. These criteria are applied, in the order listed, until a decision is reached.

For the given compound, there is a choice between two different principal chains:

    and

The first relevant criterion in this case provides that the principal chain has the maximum number of substituents corresponding to the principal characteristic group (suffix).

  P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).

Therefore, the principal chain of the given structure has all four substituents $\ce{-CN}$ corresponding to the suffix ‘carbonitrile’.

Furthermore, the current IUPAC recommendations stipulate various criteria for the assignment of locants when several structural features appear in a compound.

In particular, the order of seniority provides that low locants are given first to principal characteristic groups (here: carbonitrile) before low locants are given to multiple bonds (here: the double bond) and before low locants are given to detachable alphabetized prefixes (here: methyl).

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (c) principal characteristic groups and free valences (suffixes);

  (…)

  (e) saturation/unsaturation:

    (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;

    (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (…)

Therefore, the resulting name is (1E)-2-methylhept-1-ene-1,3,4,5-tetracarbonitrile (not (6E)-6-methylhept-6-ene-3,4,5,7-tetracarbonitrile) since the locant set ‘1,3,4,5’ for the principal characteristic groups is lower than ‘3,4,5,7’.


Is using melting/boiling point good to measure volatility? I wanted to figure out a way to compare volatility in organic compounds, so I read on many forums on the differing ways to measure volatility, which included enthalpy of vaporization and melting/boiling points.

I am planning on using melting/boiling point to prove that branching of chains in organic compounds causes weaker volatility.

According to this Wikipedia page:

  In chemistry and physics, volatility is the tendency of a substance to
  vaporize. Volatility is directly related to a substance's vapor
  pressure. At a given temperature, a substance with higher vapor
  pressure vaporizes more readily than a substance with a lower vapor
  pressure.

So, I'm not sure how much information on volatility you will get from melting point measurements.  Ideally, the enthalpy of vaporization would be the best measurement, but, if it fits your needs, the boiling point at ambient pressure is a simple way to determine whether there is a significant difference in the volatility of various compounds.


Ligand exchange reactions... are they one way or reversible? Are ligand exchange reactions one-way  reactions or reversible? I know this is a very silly question but its not said outright in any place...

For example, in my high school chemistry book, these two ligand exchange reactions are shown:

$$\ce{[Cu(H2O)6]^{2+} (aq) + 4Cl- (aq) <=> [CuCl_4]^{2-} (aq) + 6H2O (l)}$$

$$\ce{[Cu(H2O)6]^{2+} (aq) + 4NH3 (aq)-> [Cu(NH3)4(H2O)2]^{2+} (aq) + 4H2O (l)}$$

The first is reversible but the second is one-way

Also, my teacher stated that ligand exchange reactions are irreversible... this confused me further

Can someone please tell me what is right...

Let's take the binding of $\ce{O2}$ and $\ce{CO}$ to hemoglobin as example. $\ce{CO}$ does bind much better than $\ce{O2}$!

We can consider the reaction of oxygen-loaded hemoglobin with $\ce{CO}$ a ligand exchange reaction. If that would be completely irreversible, the logical consequences in the case of an intoxication would be: [...] six feet under ;)

Instead, intoxicated patients are treated with oxygen at higher pressure.

To me, that sounds like Le Chatelier in the ICU and it only makes sense in the case of equilibria.


Why do biphasic systems of fluorous and organic solvents form? It's possible to select a perfluorinated hydrocarbon that is immiscible in another organic solvent, forming a distinct phase boundary. Such systems may become miscible at elevated temperature which is exploited in the design of some catalytic processes1.

Why is it that many fluorous solvents are immiscible with organic solvents? Are their polarities sufficiently different or is there another reason?

(1) Housecroft, C.E.; Sharpe, A.G.; Inorganic Chemistry 2e; Pearson Prentice Hall; 2005, pp. 798-799

Fluorous solvents have the odd property of being both hydrophobic and lipophobic (Ref) and thus are not miscible with either aqueous or many organic solvents as you've noted. Fluorine, as the most electronegative element, does odd things to a molecule (and is often used in pharmaceutical compounds just because of some of these odd properties).

In the perfluorinated hydrocarbons, the carbon-fluorine bonds are quite polar with the electron density higher toward the fluorine. The solvents themselves, however, are not polar due to free rotation about the C-C bonds, so they are not miscible with polar solvents and are hydrophobic.

Because the F is so electronegative, the electrons are quite tightly held and the molecules have a low polarizability and unusually weak London dispersion forces, so they aren't miscible with non-polar solvents either. These weak London forces are illustrated by the fact that the perfluorinated hydrocarbons have nearly the same boiling point as their corresponding normal hydrocarbon, despite having more mass. (Ref)

You mentioned heat as one way to render the organic and fluorous solvents miscible. Adding gaseous $\ce{CO2}$ to the biphasic system also makes the phases miscible in many cases due to the high solubility of $\ce{CO2}$ in both fluorous and organic solvents. (Ref)

This reference suggest that shape may also be a consideration when considering flourous solvent's miscibility with alkanes, also through poor/weak interaction of London dispersion forces in the two liquids.


Which melting point would indicate a metallic structure? Question
  If you have three substances with three different melting points;

  445 °C
  1907 °C
  -7 °C

  Which of these would most likely have a metallic structure?

I understand that the third option would most likely be covalent. But between the first and second option, there are many bonds which have a similar melting point to both answers. Which of these would most likely indicate a metallic strucure to be involved.

Now, before we get to the "expected theoretical" answer, I should make one point clear: predicting the solid state just based on the melting point data can not be perfect! Browsing the data on melting points on Wikipedia, you'll find that vanadium has a melting point of $\pu{1910 ^\circ C}$, while zinc has a melting point of $\pu{420^\circ C}$. Both these values are very close to the options (1) and (2), hence, one can very well argue that both the options represent a metallic structure.

But this is definitely not the answer your teacher expects. A theoretical (and again, practically wrong) approach instead is to realize that the values of intermolecular forces increase in the order of molecular solid < metallic solid < ionic solid < covalent networked solid. Thus, the option (3) is probably a molecular solid, option (1) can be either an ionic solid or a covalent networked solid, while option (2) must be the metallic solid as expected.

  I understand that the third option would most likely be covalent.

Just get rid that misconception already. Covalently bonded networked solids have higher melting points than several metals and even ionic solids! Diamond melts at $\ce{3800 K}$, while boron melts at $\ce{2349K}$. Our beloved ionic solid sodium chloride melts at a poor $\ce{1073K}$, while all the 3d series metals melt under $\ce{2200K}$! Hence, the assumption, that low melting point indicates a covalent solid, is false.


Skeletal formula of methyl chloride Recently in The Periodic Table, there was a short discussion regarding this structure:

At first sight, I couldn't figure out what is the structure. After a quick google search, I came to know it is trichloromethane or chloroform (chemspider).
Now, I thought of the structure of dichloromethane and found out to be:

From the structures, It is observed that these are the skeletal formula of tri- and dichloromethane and generally in these structures, the letter C is not denoted, the C-H bond in not denoted and C-Me is denoted by a dash.

Now, following these rules, I was now curious of the skeletal formula of methyl chloride. Is it just a chlorine protruding a bond??

I couldn't find out this structure on the internet and this means that this formula is not valid although, this formula follow the basic rules. Am I wrong in my argument or are the rules vague?

Further, hydrocarbons are the basic skeleton of organic compounds. To draw really complex organic compound, we first have to figure what are the hydrocarbon chains. But still, methane and ethane cannot be drawn using skeletal formula. From propane onwards, you can represent them in skeletal formula.

So, there are certain ambiguities regarding skeletal formula (some simple and basic compounds cannot be drawn) but they are widely preferred. Why is this so?

I searched hard and eventually found out the structural formula of ethane. It is a dash as expected but still, it is a dash. Who would mind that it is ethane?

It is a valid generalisation of the skeletal formula principle to draw chloromethane as you did; you may also see it drawn in the following way:

$$\ce{-Cl}$$

In a similar vein, ethane can be drawn as a simple horizontal (or $60^\circ$ ascending or descending) line, ethene as an equality sign and ethyne as a triple line.

However, the concept of skeletal formulae, which was primarily invented to simplify drawing large, complex molecules, does not extend well to these very small ones. A chlorine atom with a line can be mistaken as $\ce{Cl-}$, a horizontal line as a dash, a double line as an equality sign and so on. Thus, while formally valid, you will not see this representation used often for very small molecules. Typically, it would be used from three explicit atoms upwards — as is the case in your second example, dichloromethane.


Deriving the equation for oxygen estimation by Winkler’s method We were doing $\ce{O2}$ estimation by Winkler's method

The procedure and reactions involved in this experiment are:

Carefully fill a 300-mL glass stoppered bottle brim-full with sample water.
Immediately add 2mL of manganese sulfate to the collection bottle.
Add 2 mL of alkali-iodide reagent (NaOH + KI) in the same manner.
Stopper the bottle with care to be sure no air is introduced. Mix the sample by inverting several times.
Add 2 mL of concentrated sulfuric acid.Then mix well.
In a glass flask, titrate 100 mL of the sample with sodium thiosulfate to a pale straw color. Titrate by slowly dropping titrant solution from a calibrated pipette into the flask and continually stirring or swirling the sample water.
Add 2 mL of starch solution so a blue color forms.
Continue slowly titrating until the sample turns clear.

$$\begin{align}\ce{MnSO4 + 2NaOH &-> \underset{{white ppt.}}{Mn(OH)2} + Na2SO4}\\[0.6em]
%
\ce{2Mn(OH)2 + O2 &-> \underset{{orange-brown cloudy ppt.}}{2MnO(OH)2}}\\[0.6em]
%
\ce{2Na2SO4 + 2H2SO4 &-> 4 NaHSO4}\\[0.6em]
%
\ce{MnO(OH)2 + NaHSO4 + 2KI &-> \underset{{golden-brown colour}}{I2} + MnSO4 + K2SO4 \\&\quad\quad + 2Na2SO4 + 3H2O}\\[0.6em]
%
\ce{I2 + 2Na2S2O3 &-> 2NaI + Na2S4O6}\end{align}$$

And at the end of the experiment we needed to calculate the $\ce{O2}$ concentration in mg/l by the following equation:

$$c(\ce{O2})\ \mathrm{(mg/L)} = \frac{(V_1 \times N \times E \times 1000)
}{[V_4 {(V_2-V_3)/V_2}]}$$

Here,

$V_1=$ Volume of titrant $\ce{Na2S2O3}$ (ml)

$V_2=$ Total volume of water sample (ml)

$V_3=$ Volume of ($\ce{MnSO4 + KI}$ solution) added (ml)

$V_4=$ Volume of analyte taken in titration

$E=$ Equivalent weight of $\ce{O2}$

$N=$ Normality of $\ce{Na2S2O3}$ solution

So how can I derive this equation?

Reference followed: Ghosh, K. C. & Manna, B. (2005) Practical Zoology

A different book though had a different formula :

Conc. of O2 (mg/l)= 0.2 x (Volume of sample water in ml) x (Strength of $\ce{Na2S2O3}$)

[Concepts]

Normality, N: The normality is the relation between the sustance equivalence and the liters of a solution. The sustance equivalents are referred to the charges per mole of a sustance. The number of equivalents (NOE) of a sustance is obtained as follows:

Acid equal to number of hydrogens.
Hidroxide equal to number of hydroxils.
Salt equal to number of cations.

To determine the normality of a solution, these are the steps you must follow:

Weight Equivalent Gram, WEG:
\begin{equation}
    WEG=\frac{Molecular\ weight\ of\ solute}{NOE}
   \end{equation}
Weight Equivalent Number, WEN:
\begin{equation}
    WEN=\frac{Mass\ of\ solute}{WEG}
   \end{equation}
Normality, N:
\begin{equation}
    N = \frac{WEN}{L}
   \end{equation}
Doing some basic algebra, you find that:
\begin{equation}
   N=M\cdot NOE
  \end{equation}
Where M stands for Molarity.

Dilution factors, DF: It is employed when the solution concetration of your sample is too concentrate. In order to determine the concentration of the substance you want to find, you have to dilute the initial solution. This is done through the dilution factors. I.e. you have 25 ml of a solution 10 M and you are asked to dilute it 100 times. To do so, you have to take an aliquot (A portion of a large whole) of 0.250 ml, pour into a volumetric flask of 25 ml and then fill it with destilled water to capacity. Then, you have a diluted solution of 0.1 M. In general, a dilution factor is expressed as:
\begin{align}
   Dilution\ factor=\frac{Total\ volumen}{Sample\ volume}
  \end{align}
Dilution: Dilution is the process of decreasing the concentration of a solute in solution, usually simply by mixing with more solvent. To dilute a solution means to add more solvent without the addition of more solute. This can be represented as:
\begin{align}
   V_1\cdot M_1 &= V_2\cdot M_2\\
   M_2&=\frac{V_1}{V_2}\cdot M_1
  \end{align}

[Answer]

Now, if you put all together:

If you look at the first parenthesis at the right of the equa, having in mind that E is WEG and doing some basic algebra:
\begin{align*}
   V_1\cdot N\cdot E\cdot 1000&=V_1(L)\cdot\frac{WEN}{L}\cdot WEG\cdot 1000\\
   &=V_1(L)\cdot\frac{\frac{Mass\ of\ solute}{WEG}}{L}\cdot WEG\cdot 1000\\
   &=V_1(L)\cdot\frac{Mass\ of\ solute}{WEG\cdot L}\cdot WEG\cdot 1000\\
   &=V_1(L)\cdot\frac{Mass\ of\ solute}{L}\cdot 1000\\
   &=V_1(mL)\cdot\frac{Mass\ of\ solute}{mL}\\
   &=Mass\ of\ solute\cdot\frac{V_1(mL)}{mL}\\
   &=Mass\ of\ solute
  \end{align*}
The square bracket is just a dilution factor.
If you put together the last two steps with the dilution section, you find a similarity.

[History]

If you are interested in the origins of this method, I recommend you the following paper:

Winkler, L. "Die Bestimmung des in Wasser Gelösten Sauerstoffes". Berichte der Deutschen Chemischen Gesellschaft. (1888) 21, 2, 2843–2855.

I cannot give you a brief handout of it, my german is basic. But, I want to point out that in page 2845, you will find an equation similar than the indicate in dilution section.

[How to understand the Winkler method]

  The Winkler method for the determination of dissolved $\ce{O2}$ involves adding an excess of ‘white’ $\ce{Mn(OH)2}$ which contains $\ce{Mn^{2+}}$ ions.

  For every molecule of $\ce{O2}$ two molecules of $\ce{I2}$ are formed. The quantity of iodine present can be found though titration of the sample with sodium thiosulphate.

[Bibliography]

Shashi,  B.P. A complete review on: effluent testing and treatment in pharmaceutical industry.


What would cause a sudden drop in humidity in my home? I have a ventless propane heater newly installed in my home, which has been an increasingly source of paranoia for me (constant condensation on the windows, worries about carbon monoxide poisoning).  It's been a interesting few weeks of trying to find the right balance of windows to crack; needless to say I'm looking for a new heater.
Today, I noticed that NONE of my windows had condensation on them, even though the heater had been running for hours and the relative humidity outside was 100% (~30F) with fog and sleet.  I would expect it to feel like a swamp in here.

Now for the chemistry question- I know that the incomplete combustion of propane in a low oxygen environment would produce carbon monoxide, but is it typical for the air to become more dry as well?  Does the lack of oxygen also produce less water, and maybe more H2 or something like that?  What would cause the unusual drop (or perceived drop?) in humidity in my house?

the humidity is caused mainly from sources of water in your home such as plants, pets and of course humans. As you breath out water is coming out of you. And the condensation of water on windows is caused by different temperature outside/inside especially at the interface glass/air.
There is no way that heater can reduce humidity in sealed room (in means of decomposition of water as you suggested). Imho if you have CO alarm you have nothing to be afraid of.


Interpretation of TD-DFT results For the first time I'm doing TD-DFT calculations (wB97XD functional) in Gaussian 09 for an open-shell system and the results look like the hell of a mess for me. The molecule is rather big, so I started with just 10 excited states to see how it would work.

First off, the ground electronic state is $^2A_2$, orbital 453 is HOMO and 454 is SOMO. Now to the excited states. In the output I found, for example, the following piece that looks strange to my eyes

Excited State   3:  2.139-B2     0.9431 eV 1314.65 nm  f=0.0816 <S**2>=0.894
453B -> 454B       0.98895
453B <- 454B       0.17190

The first line is fine: it says that the third excited state is $^2B_2$ (taking spin contamination into account).
The second line is also fine: it tells that the determinant corresponding to excitation from HOMO to SOMO has the largest coefficients in the CI expansion.
But what is the meaning of the third line?

Time-dependent DFT can be used to predict excitation energies through a linear-response formulation.

In this Gaussian result, beyond the first line, you are looking at the largest coefficients in the configuration-interaction (CI) style expansion.

(It's not strictly CI, but the implementation of time-dependent HF or RPA is essentially the same for TDDFT or Tamm-Dancoff Approximation TDDFT.)

Anyway, since you're performing a calculation on an open-shell system, the formulation for the state will include all one-electron excitations AND all one-electron de-excitations. That state has a de-excitation from 454B into 453B (i.e., the SOMO will fill the hole left by the excitation). Now you may think "but that doesn't accomplish anything." Remember that Gaussian only prints the dominant coefficients.

For this reason, Rich Martin created the Natural Transition Orbitals method as a nice way to express the excitation, rather than a set of coefficients and the ground-state orbitals.

  Oftentimes, there is no dominant configuration in the list of excitation amplitudes, thereby making a straightforward interpretation of the excited state difficult. This is particularly unsatisfactory when attempting to determine the qualitative nature of an excited state. An additional complication in the DFT case is that in principle all orbitals in DFT but the HOMO are devoid of physical significance. However, chemical intuition is built on the orbital construct, and a simple orbital interpretation of "what got excited to where" is important.

This is available in Gaussian from the Population keyword and in other codes.

Incidentally, there's a nice review of TDDFT in: "Progress in Time-Dependent Density-Functional Theory" Annual Review of Physical Chemistry
Vol. 63: 287-323.


What's the H-C-H bond angle in ethene? The carbon is $\mathrm{sp^2}$ hybridised and is therefore planar and should also, theoretically be $120^\circ$. However, VSEPR theory suggests that the π bond would "need more space" due to greater electron repulsion. As a consequence the $\ce{H-C-H}$ bond angle would be smaller. However, since the π bond is out of the plane of the molecule does this actually happen?

The H-C-H bond angle in ethene is ca. 117 degrees and the H-C-C angle is ca. 121.5 degrees.  There are two reasons that combine to explain this angular deformation in ethene.

First, from these bond angles and Coulson's Theorem (ref_1, ref_2) we can determine that the C-H sigma bonds are $\ce{sp^{2.2}}$ hybridized and the C-C sigma bond is $\ce{sp^{1.7}}$ hybridized.

From these hybridization indices (the index is the exponent "n" in the $\ce{sp^{n}}$ expression) we see that the C-C sigma bond has higher s-character content (1 part s to 1.7 parts p - 37% s) than the C-H bonds (1 part s to 2.2 parts p - 31% s). Since there is more s character in the C-C bond, it is lower in energy and the carbon sigma  electrons will tend to flow  towards this lower energy C-C bond.  Consequently, the C-C sigma bond will contain more electron density than the C-H bonds. Therefore, the electron repulsion between the C-C sigma bond and C-H sigma bonds will be greater than the electron repulsion between the two C-H bonds.  Hence the H-C-C bond angle will open up slightly from the $\ce{sp^{2}}$ ideal of 120 degrees and the H-C-H angle will close down slightly in order to minimize the bond-bond electrostatic repulsions.

Second, steric factors (which are also really just another way of describing electron-electron repulsion) may also come into play. To whatever extent the cis H-C-C-H hydrogen-hydrogen repulsion is more destabilizing than the geminal H-C-H hydrogen-hydrogen repulsion, it will also serve to increase the C-C-H bond angle and shrink the H-C-H bond angle.


Definition of NMR peak In NMR a spectrum consists of noise and peaks. While steps, such as repeated scans or smoothing of the spectra can reduce noise some still remains. As such one step usually taken is to bin the data so that only peaks are chosen, usually with a threshold and finding maxima/minima.
This raises the question however, which is how do you define a peak. If you use a threshold a poor baseline can result in this failing, for instance if the baseline is above the threshold set.

Is there a widely used method/rule of thumb which is used to decide if a region contains a peak, or is just noise?

A general term used in all analytical chemistry is the lower limit of detection (LoD), which is generally defined as 3 times the standard deviation of the baseline/blank. So if you calculate the $\sigma$ standard deviation of the baseline in question, and you have a signal that is over $3 \sigma$, you can generally accept that as a peak. This comes down to the so-called 68-95-99.7 rule, according to which there is only a 0.3% probability that a peak higher than the LoD is an artifact.


Are there any full worked examples of DFT calculations? I just started learning DFT and now I am totally confused.

Assuming I want to use B3LYP:
\begin{align}
v_s\left(\textbf{r}\right) &= v_\text{ext}\left(\textbf{r}\right) + \int d^3r\frac{n\left( \textbf{r}\right) }{\left|\textbf{r} - \textbf{r}^\prime\right|} + v_{\text{XC}}\left[ n \right]\left(\textbf{r}\right)\\
v_{\text{XC}}\left( r \right) &= \frac{\delta E_\text{XC}}{\delta n \left(\textbf{r}\right) }\\
\end{align}

I know that we use B3LYP for approximation of $v_{\text{XC}}$ - but I have no idea about how - and then we use some basis sets to run an SCF calculation to minimize the energy.
I downloaded a bunch of articles and bought some books and read them, but all of them give very detailed discussion about all the basics (e.g. what a functional is, or Kohn and Sham theorem, etc.), but no one talks about how to use them.

I am looking for a reading material or video that explains all the calculation, step by step, for example, some material that shows all the calculations for a CO molecule, from the beginning to the end of the SCF calculation.

The XC potential for DFT actually consists of two terms: $V_x$ and $V_c$. Depending on which XC functional you choose, the exchange part is either computed exactly (using Hartree-Fock) or using fitted parameters or in a combination of both. In the case of B3LYP the exchange is 20% Hartree + 8% Slater + 72% Becke88. The correlation functional consists of three (3) different terms (LYP) in a different form. If you take a different functional, say M06-2X, it has different functional type (54% Hartree exchange, 200+ parameters!).

Coming to the next part: You use a basis set only to compute the kinetic energy and exchange terms in the system. Let me walk you through a HF routine first:

The Hamiltonian consists of four terms(nuclear kinetic energy is zero because of Born-Oppenheimer Approximation) : $T_{elec} + V_{nn} + V_{en} + V_{ee}$. The nuclear-nuclear potential energy is constant (because they are 'clamped'). The first and third terms are fairly trivial, and are computed with relative ease. The electron-electron potential term consists of Pauli's exchange term, and the coulomb repulsion term. They are the computationally expensive part of the HF calculation, and the eigenvalues (energies) are solved self-consistently.

So, when you obtain the HF energy of a system, it is obtained variationally with respect to all the energy components in an interacting system of electrons and nuclei. Coming to DFT:

The first term you mentioned $V_{ext}$ is the external potential, which is usually the nuclear potential for an unperturbed system. The second term is equivalent to Coulomb repulsion. The third term is evaluated using the XC potential of your choice(every DFT method differs only at this part). You evaluate the energies for each and every orbital using the effective potential $V_s$ which you mentioned. Note: You evaluate energies for every orbital separately.

This is because of the KS theorem that equates the density created by non-interacting particles to be the same as that of interacting particles. So, we solve Kohn-Sham equation instead of Schrodinger equation, under a constraint that the sum of square of all the orbitals gives the density:

$$\rho(\mathbf r)=\sum_i^N |\phi_{i}(\mathbf r)|^2$$

This means that the energy computed using DFT is minimum with respect to a non-interacting system of electrons. This is the biggest assumption of DFT, which is quite valid too. As a result, DFT provides a good account of correlation energy at a very minimal computational expense (HF accounts for ZERO correlation)!!!

I took a computational chemistry course, where DFT was a small part, so I am not able to explain more in detail. I used Prof. Kieron Burke's ABC of DFT book, which he has uploaded online in his site for free use. It is a good source to start. There is also a video lecture of Prof. David Sherrill on Youtube: https://www.youtube.com/watch?v=5orzn-XA29M .

I missed the last question: For any molecule:

Compute the kinetic energy (using the KE functional)
Compute the external potential ($V_{ext}$)
Compute the Coulomb repulsion using an initial guess density ($n(r)$)
Compute the XC potential from the XC functional
Use Kohn Sham equations to find out orbital energies (under the constraint that the density ($n(r)$) is equal to the density of interacting particles (electrons). In this case, the density must be equal to the total number of electrons.
Using the newly obtained density, recompute the effective potential ($V_s$)
Perform until self-consistency is achieved. Then total energy is the sum of occupied orbital energies times the occupancy number.

Hope this helps!


Which of the following procedures will allow the student to determine the rate constant, for the reaction? Is the answer A or B because C and D won't give me the rate constant?

From the graph it is clear that the reaction follows first order kinetics ($\ce{[X] = [X_0] e^{-kt}}$) where $ [X_0]$ is initial concentration. Taking $ln$ on both sides we get
$\ce{ln[X] = ln[X_0] +(-kt)}$
So if you plot a graph between $ln[X]$ and $t$ then you have the magnitude of slope as k ie the equilibrium constant.
So correct answer is A.


Why does rdkit's `AllChem.ReactionFromSmarts.RunReactants` return a tuple of tuples? In rdkit, the function AllChem.ReactionFromSmarts.RunReactants returns a tuple of tuples. Does anyone know what the two dimensions are for? From the (admittedly few) examples I've seen, the inner tuples are always just one molecule.

The docstring says,

RunReactants( (ChemicalReaction)arg1, (tuple)arg2) -> object :
apply the reaction to a sequence of reactant molecules and
return the products as a tuple of tuples

example:

from rdkit import rdBase, Chem
from rdkit.Chem import AllChem, Draw
rxn = AllChem.ReactionFromSmarts('[c:1][c:2][c:3]>>[c:1][c:2]([F])[c:3]')
m = Chem.MolFromSmiles('c1ccccc1')
ps = rxn.RunReactants((m,))

((<rdkit.Chem.rdchem.Mol at 0x10be03818>,),
 (<rdkit.Chem.rdchem.Mol at 0x10bef92f0>,),
 (<rdkit.Chem.rdchem.Mol at 0x10bef9a28>,),
 (<rdkit.Chem.rdchem.Mol at 0x10bef99d0>,),
 (<rdkit.Chem.rdchem.Mol at 0x10bef9768>,),
 (<rdkit.Chem.rdchem.Mol at 0x10bef9920>,),
 (<rdkit.Chem.rdchem.Mol at 0x10bef96b8>,),
 (<rdkit.Chem.rdchem.Mol at 0x10bef98c8>,),
 (<rdkit.Chem.rdchem.Mol at 0x10bef95b0>,),
 (<rdkit.Chem.rdchem.Mol at 0x10bef9710>,),
 (<rdkit.Chem.rdchem.Mol at 0x10bef9558>,),
 (<rdkit.Chem.rdchem.Mol at 0x10bef9ad8>,))

The reason I ask is that I want to put the products into a pandas dataframe (already accomplished) and label the columns and indices according to the intended scheme.

From a tutorial I wrote on SMARTS reactions in rdkit:

  The output is a tuple of tuples. The inner tuples are there because even reactions that take only a single input molecule can result in multiple output molecules (e.g. hydrolysis). The outer tuples are there because rdkit applies your reaction incidence of the appropriate motifs in the input molecule, even if it occurs more than once.

The examples in the tutorial illustrate both cases where a single application of the reaction rule creates multiple molecules (the inner tuple(s) would have length > 1), and also cases where a reaction rule can be applied more than once to a molecule, which would lead to the outer tuples having length > 1.


What is the IUPAC name of fullerene like molecules? I wonder what the IUPAC names of fullerene-like molecules, I mean all those hydrocarbons with 3-D structure. Do we have another name of these compounds or are they classified as aromatic?

I think I've found relevent sources here:

https://old.iupac.org/publications/pac/1997/pdf/6907x1411.pdf
http://www.sbcs.qmul.ac.uk/iupac/fullerene/

However, I still wonder if we have another name for all those hydrocarbons with 3-dimensional structures. Should they be regarded as a subclass of "aromatic" or do we have another name for them?


Reaction of furan with chloroform in base This reaction looks like the Riemer-Tiemann reaction. The only difference is the reactant. First step is the formation of carbene. Second step involves the attack of π-electrons of furan on the vacant orbital of carbene which results in two possible type of products:

The answer given is of type (B) (That is, dichlorocarbene is attached like product B). But both results in the same position of carbocations. Aren't both products produced in equal quantity?

Also, in the case of pyrrole, the opposite happens. That is, the major products are:

The key to this question lies in realising that the Reimer–Tiemann reaction is not an electrophilic aromatic substitution but it is in fact

  addition of a carbanion to a strongly electron-deficient carbene.

This was discussed by @KlausWarzecha and @ron in answers to a previous question.

Therefore the major product will be the one where the attack comes from the carbon with the greatest electron density. Using MolCalc (partly developed by @JanJensen) I generated an electron density plot of furan which clearly shows that there is more electron density on the 3 and 4 carbons than on 2 and 5. This explains the given major product.


Is there any method to synthesize salicylaldehyde from salicylic acid? I know salicyclic acid can be produced from salicaldehyde by treatment with $\ce{KMnO4}$.

Is there any way to reverse this, i.e- produce salicaldehyde from salicyclic acid?

Salicylic acid can be converted to salicylaldehyde using a triangular or square wave alternating voltage. The process of this conversion is explained in this paper$\ce{^{1}}$but it is behind paywall so here's the abstract:

  The effect of superimposing sinusoidal, triangular, and square‐wave
  alternating voltage (aV) on the potentiostatic reduction of salicylic
  acid to salicylaldehyde has been studied. Polarization curves with aV
  modulation were measured with an amalgamated copper rotating disk
  electrode. Enhanced dc current densities were observed with increasing
  magnitude of superimposed aV. However, aV frequency and type did not
  have significant effect on the resulting dc current densities. Batch
  cell electrolysis experiments were conducted with a H‐type cell. The
  conversion of salicylic acid and yield of salicylaldehyde were
  measured as functions of electrolysis time, aV magnitude, type, and
  frequency. The superimposed aV significantly enhanced the reaction
  rate, conversion of salicylic acid, and the yield of salicylaldehyde.
  The energy consumption of ac electrolysis was higher than that of dc
  electrolysis.

doi: 10.1149/1.2100342 J. Electrochem. Soc. 1987 volume 134,issue 12, 3074-3077


Reduction of alkyl halides to alkanes For all the three parts, only methane has been given as the product in my textbook.

In 1) and 2), according to the mechanism I wrote, I get both ethane and methane as products. I think both these products should be formed in good yield. Then, why my products do not match?

In 3) , I got methane as the only product as given in the textbook. Can you please check the mechanism?

Your answer for question #3 is correct. However, there is a fundamental misconception you used in #1 and #2 that led to an incorrect answer.

In your notes, it seems that electrons from the metal ($\ce{Zn}$ and $\ce{Mg}$) somehow "attack" the alkyl halide and generate the methine ion ($\ce{CH3-}$). This does not happen for two reasons: Firstly, electrons are not free species in solution (like ions and molecules), they are only transferred from one molecule (reductant) to another (oxidant), so they cannot attack anything by themselves. Secondly, the methine ion is an extremely strong base (the pKa for methane is around 50, so the methine ion, its conjugate base, has to be VERY basic), which could only be produced by using an even stronger base to abstract a proton from it.

What actually happens in both cases is what is called metallic insertion. The metal ($\ce{Zn}$ or $\ce{Mg}$) is inserted between the carbon-halide bond, forcing the carbon atom to hold a partial negative charge (the author of this picture forgot to include the $\ce X$ atom bound to $\ce M$ in the organometal compound, but it is not that relevant here).

https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/alhalrx4.htm

When it happens, you end up having a nucleophilic carbon, and it could indeed react with a $\ce{CH3Cl}$ molecule to yield $\ce{CH3-CH3}$ under a unimolecular substitution mechanism ($\ce{S_N1}$). However, such substitution reactions are very slow compared to acid-base reactions, and since you have acidic hydrogens available in both cases ($\ce{HCl}$ in question #1 and $\ce{H2O}$ in #2) the acid-base reaction (abstraction of a H atom from $\ce{HCl}$ or water) would happen fast enough to prevent the formation of $\ce{CH3-CH3}$ adduct.


Do light and heavy water form an azeotrope? (Interested in the general principle too, not just for water and its variants, but with particular emphasis on it.)

When you distill a weak ethanol-in-water solution, you get a condensate that's ethanol-enriched, but that only takes you asymptotically up to 95%, because ethanol and water form an azeotrope with that composition. Is there a similar "enrichment limit" one can achieve in enriching water with deuterium, using only distillation at a fixed pressure? (I know one can "break the azeotrope, but I don't want to get into this.) Or could one, given enough stages in a cascade, approach 100% enrichment arbitrarily closely?

This is a good question, but no, there is no azeotrope in any isotopic mixture, as an azeotrope requires non-ideal mixing of two components to form a  minimum or maximum critical point on the plot of total vapor pressure as a function of composition as shown in Figure 1 below. Isotopic Mixtures have a near-zero enthalpy of mixing, making isotopic mixtures the closest thing to an ideal liquid mixture. Because of this ideality of mixing the boiling of protium-deuterium oxide mixtures looks much like the phase diagram in Figure 2 which follows Raoult's law, though this is actually quite exaggerated compared to semi-heavy water mixtures.  Even if there was a theoretical point at which an azeotrope could exist for the negligible enthalpy of mixing it would occur at extreme concentrations of $\ce{D2O}$ or $\ce{H2O}$ and be unobservable due to the "doped" species existing as semi-heavy water ($\ce{DHO}$).  This is further evidenced by the fact that other physical properties of isotopic mixtures follow a linear curve when measured as a function of concentration.

Figure 1. Generic Plot of vapor pressure as a function of composition

Figure 2. Generic Boiling phase diagram for ideal solutions

To answer your question about distillation only enrichment, it is possible though impractical.  Commercial enrichment of $\ce{D2O}$ first uses hydrogen sulfide in the Gridler-Sulfide process, once enriched to around 30%, distillation is used to finish the process to the desired isotopic purity.


Adhesion of polymerized fats to cast iron I'm trying to make an informed decision as to which oil is best to use for cast iron seasoning. I've read all kinds of conflicting information online and the only consistency seems to be that seasoning involves polymerization of fatty chains.

This explains the formation of the coating itself, but I'm interesting in choosing the oil that is least likely to separate from the cast iron surface. So my question is this:

What is the chemical (or physical) reaction (or process) which is responsible for the adhesion between the polymerized oil and the surface of a cast iron pan? Is it different from that of unpolymerized oil?

Thank you for asking this question. As a French, I am 200+% obsessed with food, cooking and every dish my grandma was cooking for the family. I am also an organic chemist so I don't buy the "everyday bullshit" (if that makes sense).

My first (personal) remark would be that if there were a way to quickly season a cast iron pan, every manufacture would use it and build its reputation on that.

This brings me to a more physical/chemical remark: (1) seasoning cast iron takes time ; (2) many different fats are recommended, from pork fat (low in unsaturated fat) to olive oil (high in unsaturated fat). But only unsaturated fat can polymerize.

Overall, that points to something else, which is that burnt fat would play a major role in seasoning. Burning fat will create carbon (which iron cast contains) and aromatic compounds (from Maillard's reactions) which have affinity for carbon.

As a whole, I would suggest you to cook in this/those pan(s) as often as you can and deglaze it to make some sauce as we do in French cuisine.