Why do stabilised ylids lead to trans alkenes in the Wittig reaction? Why do stabilised ylids (such as $\ce{Ph3P=CHCO2Me}$) lead to (E)-alkenes in the Wittig reaction, in contrast to unstabilised ylids which are (Z)-selective?

The (E)-alkene is formed from stereospecific collapse of the more stable trans oxaphosphetane. Does the observed preference have anything to do with thermodynamic control over formation of the intermediate oxaphosphetane?

TL;DR: With stabilised ylids, the trans oxaphosphetane is formed, under kinetic control. This is due to minimisation of dipole-dipole repulsions in the transition state for oxaphosphetane formation:

There is no equilibrium established between cis and trans oxaphosphetanes. The formation of the trans oxaphosphetane is irreversible, and its collapse leads to the formation of the (E)-alkene.

Long answer

Part 1: Oxaphosphetane reversibility?

The mechanism of the Wittig reaction has historically been controversial. Pioneering research, most notably by Edwin Vedejs and his group, has led to our current understanding of the mechanism as being a concerted [2+2] cycloaddition to form the oxaphosphetane directly. For more information, see: Which is the currently accepted mechanism of a Wittig reaction?

Within this framework, the formation of (Z)-alkenes from unstabilised ylids is understood to be because of a minimisation of steric repulsion in the cycloaddition transition state, as was proposed by Vedejs in 1988.1 For more information, see: Why does the unstabilised Wittig reaction selectively form cis alkenes?

For a long time, it was believed that oxaphosphetane formation from stabilised ylids was reversible. Consequently, the trans oxaphosphetane was preferentially formed (due to minimisation of steric repulsions), under thermodynamic control. The collapse of this then led to (E)-alkene formation.

However, this has been proven to be all but incorrect (and is sadly still propagated on widely used websites such as Wikipedia). A major problem with this theory was that oxaphosphetanes formed from stabilised ylids decompose so fast that they are not detectable via NMR spectroscopy.2 It is therefore implausible that the reversal of oxaphosphetane to starting materials can occur before forward reaction to form products.

The nail in the coffin was provided by Vedejs.2 If oxaphosphetane formation (from stabilised ylids) is indeed reversible, then it would mean that if the cis oxaphosphetane is synthesised, it should decompose to form the (E)-alkene via isomerisation to the trans oxaphosphetane. While the oxaphosphetanes themselves cannot be synthesised (they simply decompose to products), the corresponding betaines can be synthesised by deprotonation of β-hydroxyphosphonium salts (their conjugate acids).

The work by Vedejs proved that the syn betaine - which necessarily cyclises to form the cis oxaphosphetane - decomposed stereospecifically to the (Z)-alkene, hence disproving the idea of reversibility in the Wittig reaction. First, the relevant β-hydroxyphosphonium salts were synthesised:

Deprotonation of this formed a betaine, which would immediately cyclise to form a cis oxaphosphetane (this was well-established by then; see the question linked earlier on the mechanism).

If any of the (E)-alkene was formed, then it would mean that there was a mechanism by which either the betaines, or the oxaphosphetanes could interconvert. This would therefore not be conclusive evidence as to whether oxaphosphetane formation was reversible. However, if none of the (E)-alkene was formed, then it would necessarily mean that betaine and oxaphosphetane formation were both irreversible.

In their experiments, an E:Z ratio of > 98:2 was observed for all cases except for R = Ph. This indicated that in fact oxaphosphetane formation was not reversible, and that the stereoselectivity of the Wittig reaction cannot be based on a thermodynamic argument.

For R = Ph, 8% of the (E)-isomer of ethyl cinnamate was observed. Experiments with added 4-chlorobenzaldehyde ruled out the possibility of reversal to the ylid + carbonyl, as any ylid formed would preferentially react with the more electrophilic 4-chlorobenzaldehyde, leading to ethyl 4'-chlorocinnamate - but this was not seen. Furthermore, ethyl cinnamate was shown to be configurationally stable under the experimental conditions. It was finally found out that this occurred because, instead of deprotonating on oxygen, DBU deprotonated the carbon atom α to the ester, which led to loss of stereochemical integrity. This was further supported by deuterium labelling studies.

As a consequence of this, Vedejs concluded that oxaphosphetane formation was irreversible, and noted that for reactions with a stabilised ylid, the Hammond postulate implied a late transition state, which would "feel" stabilising or destabilising influences of the oxaphosphetane. Vedejs therefore proposed an almost-planar transition state, with the stereoselectivity rationalised by steric repulsions between the $\ce{CO2Me}$ and $\ce{R}$ substituents in the TS:

Part 2: Steric repulsions or dipole alignment?

In 2003, Varinder Aggarwal's research group managed to synthesise stabilised ylids with a $\ce{P(OMe)3}$ group instead of the usual $\ce{PPh3}$, and discovered that these ylids led to substantially lower (E)-selectivity.3 Vedejs' model could not be used to rationalise these observations:

To explain these, Aggarwal and Harvey put forth a different model, based on dipole-dipole interactions and supported by DFT studies:4

For the usual $\ce{PPh3}$ ylids, there is a strong dipole pointing from ylid carbon to ester group; in the trans TS, this dipole points opposite to the dipole of the C=O group in the carbonyl component. Hence, there is electrostatic attraction in, and stabilisation of, the trans TS. This explains the usual high (E)-selectivity.

However, for the new $\ce{P(OMe)3}$ ylids, the dipole moment of the ylid is much smaller. This is because the negative charge on the ylid carbon is delocalised into not only the ester group, but also the σ* orbital of the P–O bonds, which are energetically low-lying. This accounts for the lower (E)-selectivity of the phosphite-derived ylid.

As far as I know, this is the currently accepted model for why stabilised ylids lead to (E)-alkenes. Our mechanistic understanding of the Wittig reaction is very nicely summed up by Aggarwal and Harvey,5 in a grand computational study of all types of ylids. In it, they provided a full potential energy surface for the reaction of $\ce{Ph3P=CHCO2Me}$ with benzaldehyde (note that there is a typo in the ylid starting material in the diagram):

Ending with a quote from their paper, which perfectly summarises this answer:

  In the case of the triphenylphosphonium stabilized ylide, TS2 is very low in energy, showing that reversal cannot occur in these reactions, which is in agreement with experiment [...] E/Z selectivity is thereby determined solely by the relative energy of cis and trans TS1s. These energies, as well as the geometry of the respective TS1s, are strongly affected by a dipole-dipole interaction between the two reactants [...] The outcome for stabilized ylides is thus a high E selectivity.

References

Vedejs, E.; Marth, C.F. Mechanism of the Wittig reaction: the role of substituents at phosphorus. J. Am. Chem. Soc. 1988, 110 (12), 3948–3958. DOI: 10.1021/ja00220a037.
Vedejs, E.; Fleck, T. J. Kinetic (not equilibrium) factors are dominant in Wittig reactions of conjugated ylides. J. Am. Chem. Soc. 1989, 111 (15), 5861–5871. DOI: 10.1021/ja00197a055.
Aggarwal, V. K.; Fulton, J. R.; Sheldon, C. G.; de Vicente, J. Generation of Phosphoranes Derived from Phosphites. A New Class of Phosphorus Ylides Leading to High E Selectivity with Semi-stabilizing Groups in Wittig Olefinations. J. Am. Chem. Soc. 2003, 125 (20), 6034–6035. DOI: 10.1021/ja029573x.
Robiette, R.; Richardson, J.; Aggarwal, V. K.; Harvey, J. N. On the Origin of High E Selectivity in the Wittig Reaction of Stabilized Ylides: Importance of Dipole−Dipole Interactions. J. Am. Chem. Soc. 2005, 127 (39), 13468–13469. DOI: 10.1021/ja0539589.
Robiette, R.; Richardson, J.; Aggarwal, V. K.; Harvey, J. N. Reactivity and Selectivity in the Wittig Reaction: A Computational Study. J. Am. Chem. Soc. 2006, 128 (7), 2394–2409. DOI: 10.1021/ja056650q.


H2SO4 with ethanoic acid reaction? Earlier today I was preparing an ester (propyl ethanoate) and I noticed that instead of putting the propanol, ethanoic acid and $\ce{H2SO4}$ in the test tube and warming it (taking time and external energy) I could instead mix sulfuric acid with the ethanoic acid, which made the test tube go very hot, and then by adding the propanol and shaking it split into two layers. On pouring this into water I ended up with an immiscible layer which appeared to be an ester (it was sweet smelling and considering the reaction it would be likely to be an ester).

I cannot think of which reaction was causing the heating up initially. The test tube was clean so the only reactants were the sulfuric acid and ethanoic acid, and it appeared that a gas was being given off (as shaking it with a rubber cap on caused it to pop out of the tube, but no effervescence could easily be seen). Can anyone help? Furthermore, would the eventual product be propyl ethanoate or something else (it was suggested that it was dehydrogenating the ethanoic acid so the actual product would be something like propyl ethenoate, but from a quick look on the internet this would seem to require an iron catalyst (or another catalyst) and a high temperature, neither of which was present)?

I suspect that the ethanoic acid contains a residual amount of water and that the concentrated sulphuric acid is reacting with the water. Since this is a very exothermic reaction the test tube gets heated up which evaporates some of the ethanoic acid, causing the bung to pop off. If it's just slight evaporation this would also explain why you couldn't see any effervescence.


Radius of hydride (H⁻) ion What is the radius of the hydride $(\ce{H-})$ ion? There are several sources with conflicting reports on the same.

Inorganic Chemistry: Principles of structure and Reactivity cites Pauling’s 1960 study of the hydride ion, stating that

  it has a radius of 208 pm compared to 216 pm for the iodide ion

This would imply that it is larger than the fluoride, chloride and even the bromide ion! This article in nature disproves the above and states:

  the ionic radius of an $\ce{H-}$ ion is 134±2 pm, which is slightly larger than the radius of $\ce{F^-(II)}$ (128.5 pm) and almost the same as the radius of $\ce{O^2-(II)}$ (135 pm). This evaluation agrees with the generally accepted trend that the radius of an $\ce{H-}$ ion is nearly the same or slightly larger than that of an $\ce{F-}$ ion.

Which one of these is correct? Also, could additional citations be given to tell the exact size of the hydride ion?

Unfortunately, there is no caliper-like tool to measure ionic radii directly.
They are determined based on experimental data such as crystal structure or crystal lattice energy using empirical relationships.
The exact value would inevitably deviate from one method to another.

Pauling [1, pp. 150–152] used a hard-sphere model which doesn't account for overlap in crystal lattice and was predicting higher interatomic distances compared to the crystallographic data, hence an overestimated value of the anionic radii.

Several limitations of the hard-sphere model can be resolved by applying a soft-sphere model.
According to the recent work by Lang and Smith [2], who also compared applicability of both hard- and soft-sphere models, $r_\mathrm{i}(\ce{H-}) = \pu{139.9 pm}.$

CRC Handbook of Chemistry and Physics [3, p. 12-22] lists the value of thermochemical radius determined using Kapustinskii equation from a set of known lattice energies: $r_\mathrm{i}(\ce{H-}) = \pu{(148 ± 19) pm}.$

References

Pauling, L. The Chemical Bond; Cornell University: Ithaca NY, 1960.
Lang, P. F.; Smith, B. C. Ionic Radii for Group 1 and Group 2 Halide, Hydride, Fluoride, Oxide, Sulfide, Selenide and Telluride Crystals. Dalton Trans. 2010, 39 (33), 7786. DOI: 10.1039/c0dt00401d.
Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data, 97th ed.; Taylor & Francis Group (CRC Press): Boca Raton, FL, 2016. ISBN 978-1-4987-5429-3.


NMR splitting of protons in a ring I was presented with the following compound and asked to comment on the splitting of the proton $\ce{H_a}$ in $\ce{^1H}$ NMR.

The question presented today was:

  What is the chemical shift (below) and splitting pattern for $\ce{H_a}$ up to and
  including $\ce{^4J}$?

I relabelled the molecule as shown:

but I was left stuck here. I know that $\ce{H_a}$ will be split by $\ce{H_d}$ and $\ce{H_e}$ to give a doublet of doublets (dd) but I was wondering, since the molecule is asymmetric. Now

are the protons $\ce{H_c}$ and $\ce{H_d}$ chemically equivalent but magnetically inequivalent

to yield the final splitting pattern as dddd?

The two protons Hb and Hc are also chemically inequivalent, because they have a different chemical environment with the methyl group, i.e. they don´t have the same chemical shift.

In a chiral molecule, the protons in a methylene group are rarely really chemically equivalent. There is usually a preferred orientation.


Solubility, pressure and Henry's law Determine the solubility of $\ce{N2}$ in water exposed to air at $\pu{25^\circ C}$, if the atmospheric pressure is $\pu{1.2 atm}$, assume that the mole fraction of nitrogen is $0.78$ in air and the Henry's Law constant for nitrogen in water at this temperature is $\pu{6.1 \times 10^{-4} M/atm}$.

I have the answer, but i want to know how to solve for it. I've been reading through my notes and the book trying to find a reference through which to approach this problem and i have found none. If anyone can guide me in the right direction that would suffice.

What are the units of your answer? You need a solubility, which is an amount per volume. If your amount is moles, the your solubility is in moles/liter: $\text{M}$.

That unit appears in the Henry's Law constant! $6.1\times 10^{-4} \frac{\text{M}}{\text{atm}}$. What can you do to get rid of the atmospheres in the denominator? You probably need the pressure and the mole fraction.


How to accurately calculate the electromotive force for various conditions in a salt water battery? This question is on the same topic as this one but more complex.
So, let's consider the cell with $\ce{Zn}$ and $\ce{Cu}$ electrodes inside the $\ce{NaCl}$ solution.
On the $\ce{Zn}$ electrode we have
$$\ce{Zn + 2Cl^- -> ZnCl2  + 2e^-},$$ and on the copper one
$$\ce{H3+O + e^- -> 1/2 H2 + H2O}.$$

For both reactions we can find standard electrode potentials, but they are given for normal conditions, and $1~\mathrm{M}$ concentration of $\ce{Zn^{2+}}$ and $\ce{H3+O}$. To calculate the EMF for non-$1~\mathrm{M}$ concentration we use the Nernst equation. In this case we have something like
$$E=E_0 - \frac{\mathcal{R}T}{n\mathcal{F}}\ln{\frac{[\ce{Zn^{2+}}]}{[\ce{H3+O}]}}.$$
But really in the initial $\ce{NaCl}$ solution we do not have any $\ce{Zn}$ ions (so their concentration is very close to zero and unknown to us). Of course, in few microseconds some $\ce{Zn}$ from electrode will be dissolved creating some finite potential. But in the experiment when putting $\ce{Zn}$ electrode into the same solution we always observe the approximately same EMF value, $\approx 0.8~\mathrm{V}$. How to predict theoretically this value?

Furthermore, my question is how to calculate the EMF of such cell as precisely as possible for different experimental conditions, e.g. different temperature?

I have made an experiment showing that the EMF of a 'cell' with $\ce{Zn}$ and $\ce{Cu}$ electrodes into $\ce{NaCl}$ solution is decreasing when heating the $\ce{NaCl}$ solution.

It's strange, let me explain why. The Nernst equation tells us that EMF is increasing with $T$ when $[\ce{Zn^{2+}}]$ is less than $[\ce{H3+O}]$ (because $\log$ is negative), and otherwise EMF is decreasing with $T$ ($\log$ is positive). But in my opinion the concentration of $\ce{Zn}$ ions is always less than $[\ce{H3+O}]$ because we do not have $\ce{Zn}$ in the initial solution. So then logarithm is always negative so EMF is growing with $T$.
In this case, why do I observe the EMF decreasing for $\ce{Zn-Cu}$ pair?

And the last question. When we talk about $\ce{Zn-Cu}$ pair we assume that zinc dissolves; and the hydrogen gas is created on the copper electrode.
But why do I have the EMF when putting into a $\ce{NaCl}$ solution the $\ce{Al}$ ($\ce{Fe, Cr, etc}$) electrodes instead of $\ce{Cu}$, and remain instant the $\ce{Zn}$ electrode? I am not sure that hydrogen is created in such pairs: all these metals have a negative standard electrode potential. Measured EMFs of all these ($\ce{Zn-Al, Zn-Fe, Zn-Cr}$) pairs are significantly different, so the second metal is important in these cases. What reactions are going on in such pairs?

I suspect that the reason you see a decay in the cell potential of $\ce{Zn/1M~ NaCl/Cu}$ with increasing temperature is due to increased corrosion of zinc. In $\ce{NaCl}$ solutions the anodic corrosion reaction is dissolution of $\ce{Zn}$ as $\ce{Zn^{2+}}$ and the cathodic reaction is reduction of dissolved oxygen to form an insoluble zinc hydroxide (Ichiro Suzuki, Corrosion Science 1985,25 (11), 1029-1034). The insoluble hydroxide passivates the metal surface. So, you have two effects which would contribute to EMF decay: 1) increased $\ce{Zn^{2+}}$ concentration and 2) passivation of the electrode.

One last thing which may or may not be helpful to you - if you are planning on utilizing the EMF from this cell to drive any kind of current you must account for the hydrogen overpotential on the copper electrode. It takes a lot of extra voltage to make hydrogen on pretty much any metal except platinum. You may find that you have a cell that looks good at open circuit (i.e. high potential with no current flow) only to see it crumble to nothing when you try connecting it to a load.


Why do Zr, Sn, and Ti enolates form syn aldols selectively? In Organic Chemistry (2nd ed.) by Clayden, Greeves, and Warren, it is stated that (p 869)

  [...] the enolates of some metals [Sn(II), Zr, Ti] give syn aldols regardless of enolate geometry.

There are also literature examples of syn-selective Zr aldol reactions, e.g. Tetrahedron Lett. 1980, 21 (48), 4607–4610.

If they were to obey a simple Zimmerman–Traxler chair-like transition state model, as is applicable for Li or B enolates, then (Z)-enolates should give syn aldols and (E)-enolates anti aldols. How does the mechanism or stereochemical model differ for these enolates?

For most aldol reactions, the relative 1,2-diastereoselectivity is a consequence of the enolate geometry.*

Both tin and titanium enolates are easily formed, and have been shown to exist predominantly in the (Z) configuration.

When reacting a cyclic (Zimmerman Traxler) TS, its commonly observed that (Z) enolates afford 1,2-syn aldol adducts, due to the orientation of the enolate   (the (Z) enolate always has an axial R-group).

The following examples are instructive. In both cases (tin or titanium), we first form a (Z) enolate using the conditions shown. When the aldehyde (R-CHO) is introduced, chelation using the metal organises the Zimmerman Traxler TS. The lowest energy TS possible is the one in which the aldehyde is arranged with its bulky R group equatorial, however the enolate cannot put the methyl group (in this case) equatorial, as the enolate geometry enforces its axial positioning. Following both transition states through, we get the 1,2-syn aldol adduct in both cases.#

  Image taken from Andy Myers' CHEM115 Handout: Stereoselective, directed aldol reaction

The description given in Carey is fundamentally incorrect for the reactions shown (even if it does predict the correct product). Looking at the references he gives, they're all fairly old, before more recent experimental and computation studies have validated the cyclic TS model for these kinds of metal enolates.

*: Mukaiyama aldol reactions do not conform to this, since they occur via a fundamentally different transition state (an open TS) to the Zimmerman Traxler model commonly found for other aldol reactions.

#: In this case, we also get enantioselectivity via the use of a chiral auxiliary, and depending on the metal used the enantioselectivity can be switched around due to chelation or lack thereof between the auxiliary and the metal


Why is the melting point of BaCl2 > BeCl2 while CsCl<NaCl I'd think that increasing ionic character by increasing electronegativity difference would result in strong bonds but in the case of NaCl and CsCl(and the other chloride of the group) the case is different.

Sodium chloride and caesium chloride are both primarily ionic compounds. Within ionic compounds, a greater melting point can be caused by a higher ionic charge or a higher charge density/better interaction. Since both have the same charge while sodium has a smaller ionic radius, the sodium-based crystal is held together better due to more interaction than the caesium based one.

When comparing barium chloride and beryllium chloride, you are essentially comparing apples and oranges since beryllium chloride is a primarily molecular compound. In its gas phase, it forms $\ce{BeCl2}$ monomers and $\ce{Be2Cl4}$ dimers. The intramolecular attractions in molecular compounds are typically rather strong with the intermolecular ones are weaker. For ionic compounds such as barium chloride, we cannot distinguish between the two since we cannot define molecular entities.

When comparing ionic compounds to molecules, the ionic compounds typically have a much higher melting point (with exceptions; both low-melting ionic compounds and high-melting molecular compounds).


Electroplating to create a scratch-free surface? Can electroplating improve the texture of a steel ball bearing so that it will have a smooth surface free of scratches even a few microns wide? I'm trying to find a way to remove scratches (about 10 microns wide) from carbon-steel electrodes and have recently started looking into electroplating. I have already ruled out electropolishing and using rotary/vibratory tumblers.

In general, this doesn't really work. To level a surface by electrodeposition, material would have to deposit in recesses in the surface faster than it does onto surface features that stick out. Normally, the opposite occurs because things that stick out are more accessible for diffusion. Electropolishing works because it's doing the opposite process in solutions viscous enough to enhance the differences in diffusion.

Not to say that it's impossible to level surfaces by electrodeposition. It's been known for a long time that certain additives to plating baths can change the deposition selectivity to this end. However, the mechanisms behind these effects are not well known and it seems to be pretty difficult to achieve consistently. Also, you won't be able to electrodeposit carbon steel, so you would have some other metal on the surface, which may or may not adhere well. I wasn't able to find any specific method for carbon steel—electropolishing seems to be much more common.


Is hydrazine a better nucleophile than ammonia? Which is a better nucleophile, $\ce{NH3}$ or $\ce{NH2NH2}$? For the sake of limiting scope, I want an answer for behavior in a protic solvent though non-protic solvent answers are welcomed.

My logic is that since $\ce{NH2NH2}$  has 2 lone pairs it will be a better nucleophile. I would like to know if this conclusion and reasoning is correct, or if I'm correct for the wrong reason why, or if I'm wrong, why is ammonia a better nucleophile?

Yes, hydrazine is a better nucleophile than ammonia. This is the case in both water (a protic solvent) as well as acetonitrile (an aprotic solvent). A quantitative approach may be found in the Mayr nucleophilicity parameters (Ctrl-F "ammonia" and "hydrazine"):

$$\begin{array}{ccc} \hline
\text{Nucleophile} & N\text{ parameter in }\ce{H2O} & N\text{ parameter in }\ce{MeCN} \\ \hline
\ce{N2H4} & 13.46 & 16.45 \\
\ce{NH3}  & 9.48 & 11.39 \\ \hline
\end{array}$$

The larger, the more nucleophilic. It also makes sense that the protic solvent water reduces the nucleophilicity of both species.

For more details I suggest reading the articles cited in the database entries. The most relevant one is: J. Org. Chem. 2012, 77 (18), 8142–8155. Inside there is a good discussion of the methodology.

As for why hydrazine is a better nucleophile, the jury is still out. In general there is a phenomenon whereby nucleophiles with lone pairs on two adjacent atoms are more nucleophilic than expected, which is called the α-effect. [Note that I deliberately avoid saying that hydrazine is more nucleophilic "because of the α-effect", because that merely begs the question. The α-effect is not (yet) a theoretical explanation, but rather an empirical observation.]

I don't have the interest to summarise reviews on the α-effect (at this point in time, at least) so I will quote Mayr from the article above:

  The α-effect, which has been investigated for various reactions of nucleophiles including acylations, Michael additions, SN2 reactions, and nucleophilic aromatic substitutions has been the topic of several reviews, but its origin and extent are still discussed controversially. [...] Several theories on the origin of the α-effect have emerged, which include the destabilization of the ground state by electron repulsions, the stabilization of the transition state, thermodynamic stabilization of the products, and solvent effects.

See also: Is there a general consensus on the causes of the alpha-effect? Unfortunately, that question does not have a proper answer, and I believe the direct answer to that question is "no". As Jan's comment indicates, too, the explanation must go beyond "hydrazine has two lone pairs but ammonia has one", because that alone doesn't actually explain anything.


How can the Bonding in IF7 be explained using LCAO method? I know that the VSEPR theory explains $\ce{IF_7}$ (iodine heptafluoride)'s structure as a pentagonal bi-pyramidal one. The valence bond theory can be used to say that it has $\mathrm{sp^3d^3}$ hybridisation (I think). How can the bonding in it be explained using the LCAO (Linear Combination of atomic orbitals) method in molecular orbital theory?

I got this doubt because there are only two "axial" d-orbitals ($\mathrm{d}_{z^2}$ and $\mathrm{d}_{x^2-y^2}$) among the 5 present, which can have head-on overlap with p-orbitals (non-axial orbitals can only have sideways overlap). In the excited state, the iodine atom has 7 half filled atomic orbitals (an s-orbital, three p-orbitals, and 3 d-orbitals) and according to LCAO each of them has to combine with one p-orbital of each fluorine atom to form molecular orbitals by σ and π bonds. But, the formation of only 6 σ bonds can happen between an iodine and 6 fluorine atoms, as σ bonds involve head-on overlap of orbitals, and there are only 6 orbitals capable of doing that (the s and p orbitals, and the two 'axial' d-orbitals). Therefore, the seventh fluorine atom must be in pπ-dπ bond with fluorine.

This doesn't explain "five equivalent bonds and two bonds of equal length (all σ bonds)" as predicted by VSEPR theory. How many σ bonds does it really have?

No 6-centre-10-electron bond.

The bonding situation in $\ce{IF7}$ is explained by Christe et. al. (see ssavec's answer) as a 6-centre-10-electron bond. In the article itself I found no real evidence that supports this statement and I believe it to be completely wrong. The following section is taken from Karl O. Christe, E. C. Curtis, David A. Dixon, J. Am. Chem. Soc. 1993, 115 (4), 1520–1526. and might as well be the culprit of this monstrosity.

  The bonding in the square-planar $\ce{XeF4}$, can be described by two semi-ionic, 3-center 4-electron (3c-4e) bonds for the four $\ce{Xe-F}$ bonds and two lone valence electron pairs on $\ce{Xe}$ ($\mathrm{s^2p}_z^2$ hybrids). The 3c-4e bonds involve the $\mathrm{p}_x^2$ and $\mathrm{p}_y^2$ orbitals of xenon. Addition of an $\ce{F-}$ ion to the equatorial plane in $\ce{XeF4}$, results in pentagonal-planar $\ce{XeF5-}$ and the formation of a semi-ionic, 6-center 10-electron (6c-10e) bond involving the delocalized $\mathrm{p}_x^2\mathrm{p}_y^2$ hybrid orbitals of $\ce{Xe}$ and six electrons on the five $\ce{F}$ ligands. The two lone valence electron pairs on $\ce{Xe}$ in $\ce{XeF5-}$ are analogous to those in $\ce{XeF}$.
  The planar $\ce{IF5}$ fragment of $\ce{IF7}$, has essentially the same bonding as $\ce{XeF5-}$, as shown by the atomic population calculations given in Table VII. [...]

They continue to compare the populations of the valence orbitals in question, without stating the what methods they used to determine that.

  The electronic structure calculations were done at the ab initio molecular orbital (MO) and local density functional (LDF) levels. [...]

A more accurate description of the bonding in in $\ce{IF7}$

Iodine heptafluoride is one of the classic examples when it comes to describing hypervalency and octet-expansion. Early attempts included d-Orbitals into describing that phenomenon (VSEPR), resulting in a description that yields the correct geometry. We now know that this theory is obsolete and more often wrong than correct.
A very principle problem in describing the bonding in this molecule is assuming that it is completely covalent bound. Christe et. al. acknowledge that this is wrong by stating "semi-ionic" bonds. This is - as we will see - already a stretch.

A more accurate terminology for these kind of molecules is hyper-coordinated; i.e. the central atom has a higher coordination number than common.

The following analysis utilises a calculation on the local density approximation level of theory. The used basis set is of triple zeta quality and employs a small effective core potential for iodine. The density fitting approximation was used to somewhat speed up the calculations. This is often denoted as DF-BP86/def2-TZVPP. The program used is Gaussian 09 rev. D.

Geometry

The performance of the method is not overwhelming. The bond lengths are overestimated by quite a bit. As cited by Christe, observed is $r_\mathrm{obs}(\ce{I-F_{ax}})=178.1~\mathrm{pm}$ and $r_\mathrm{obs}(\ce{I-F_{eq}})=185.7~\mathrm{pm}$. I calculated $r_\mathrm{BP86}(\ce{I-F_{ax}})=190.4~\mathrm{pm}$ and $r_\mathrm{BP86}(\ce{I-F_{eq}})=195.6~\mathrm{pm}$. Other methods, that I could still reasonably perform did not do much better. The MP2 result is $r_\mathrm{MP2}(\ce{I-F_{ax}})=186.6~\mathrm{pm}$ and $r_\mathrm{MP2}(\ce{I-F_{eq}})=191.9~\mathrm{pm}$.
For the purpose of the analysis the DFT method should be enough. (I had a look at the other results, too; they don't differ significantly.)

Quantum theory of atoms in molecules (QTAIM)

I would usually start with looking at the orbitals, but in this case it is more appropriate to look at the electron density itself. Better yet, we look at the Laplacian of the electron density. The depicted QTAIM plots are the Laplacian distribution within the molecule. Solid blue lines indicate charge depletion $\nabla^2\rho>0$, dashed red lines indicate charge accumulation $\nabla^2\rho<0$, red spheres are bond critical points, solid black lines are bond paths, dark red lines are zero flux surfaces (these separate the atoms). See an earlier answer of me for some more information on QTAIM.
We have two bond critical points to consider, one for the axial and one for the equatorial bond. We find low values of electron density at both; $\rho(\ce{I-F_{eq}})=0.125~\mathrm{a.\!u.}$; $\rho(\ce{I-F_{ax}})=0.137~\mathrm{a.\!u.}$. The Laplacian at both points is positive; $\nabla^2\rho(\ce{I-F_{eq}})=0.33~\mathrm{a.\!u.}$; $\nabla^2\rho(\ce{I-F_{ax}})=0.42~\mathrm{a.\!u.}$.
These values indicate predominantly ionic bonding. The graphs nicely show that the atoms appear as separate entities.

With that background it seems wrong arguing the bonding situation by forming covalent bonds, like they are assumed in MO or VB theory.
However, as wrong as it is to neglect ionic bonding, it is equally wrong to neglect covalent bonds.

The Orbitals of $\ce{IF7}$

There are a total of four orbitals with contributions from iodine and fluorine. There are not more than eight bonding electrons. I cannot imagine that Christe et. al. found a different solution, as his level of theory was not better (maybe worse). They either did not know how to interpret these results or simply overlooked it.
Here are the relevant lines from the output.

 Alpha occ 17 OE=-1.196 is I1-s=0.23 F7-s=0.18 F8-s=0.18
 Alpha occ 25 OE=-0.587 is I1-p=0.30 F7-p=0.28 F8-p=0.28
 Alpha occ 26 OE=-0.568 is I1-p=0.31 F4-p=0.19 F3-p=0.19
 Alpha occ 27 OE=-0.568 is I1-p=0.31 F2-p=0.21 F6-p=0.14 F5-p=0.14

(OE is the orbital eigenvalue in Hartree, the contributions missing from 1.00 are at the other atoms, but to small to trigger printing.)

In the LCAO bonding picture we have a totally symmetric bonding orbital from all atoms, an orbital referring to the axial bonds, and two degenerate orbitals for the equatorial plane.

Charges and Wiberg Bond indices

The ionic character of the bonds can also be seen in the charge distribution. While the iodine has a high positive charge $q(\ce{I})=3.24$, the fluorines have a negative charge $q(\ce{F})=-0.46$. These are AIM charges calculated with the MultiWFN program package.
The Analysis with natural bond orbitals give a similar result, $q(\ce{I})=3.20$, $q(\ce{F})=-0.46$.
Looking at the Wiberg bond indices matrix formed from natural atomic orbitals, we find a bond index of $0.54$ for the equatorial bonds and a slightly higher value of $0.56$ for the axial bonds.

Conclusions

It is not possible to determine a number for how many sigma bonds are in the molecule, since this is strictly only possible for covalent bonds. The bonds are here predominantly ionic, with small covalent contributions. The number of bonding electrons does not exceed eight and therefore obeys the octet rule.
VSEPR fails, as it was expected. The term hyper-valency should be avoided and replaced by the term hyper-coordination.


Why does everything go in the direction of positive entropy change? I read about entropy from different sources. Still don’t get why (universal) entropy change has to be greater than zero for irreversible process, i.e. $\mathrm{d}S_\mathrm{Univ}>0$. Is it a result of pure observation or is there any theoretical background of this phenomena?

Unfortunately the linked question "Does adsorption violate thermodynamics?" does not answer this, as it only shows that overall entropy change is greater than zero. But doesn’t provide a reason where it comes from.

I am looking for a simple explanation in classical thermodynamics.

This answer us quite long, but at each stage should be straight forward and will hopefully show you exactly where the second law comes from in classical thermodynamics. If you are going to spend the time learning something you might as well do it properly right?

Introduction

The first law of thermodynamics tells us that in any process energy is conserved, it can be transformed and converted into different forms, but the total amount is unchanged.

The second law imposes limits on the efficiency of energy conversion processes. For a robust understanding of the second law it is necessary first to understand Carnot cycles. As such I will divide my answer into sections. Many of the points in the following are intricate in their application and a good knowledge of the first law is required. At the time classical thermodynamics was developed the main interest of the research was into the efficiency of heat engines. Historically these are quite interesting, but that is why you need to understand a Carnot cycle first which in essence is just a really efficient engine (the most efficient).

Keep in mind that in any heat engine a supply of heat $Q$ can either be converted to work $W$ or internal energy $\Delta U$. Since an isothermal process means $\Delta U=0$ then the most efficient cycle will have two isotherms. However two isotherms do not constitute a cycle. The most efficient way to connect these two isotherms is with adiabatic processes (no heat exchange with the surroundings). With two adiabatic lines we finally have our cycle. Graphically to maximise the work you want to maximise the area under the cycle.

By cycle we mean that a system returns to exactly the same state as before the change or process occurred.

Carnot Cycles

Work can be obtained from an engine of there are heat sources at different temperatures, and heat will flow from a hot body to a cold body without work being performed until thermal equilibrium is attained. Carnot realised therefore that any temperature difference either produces work, or is wasted as spontaneous flows of heat.

Lets use an ideal gas as our substance. The figure shows a Carnot cycle. It is reversible and has two reservoirs. Where $ab$ is an isotherm at temperature $T_1$ and heat $Q_1$ enters from a reservoir at $T_1$. The line $cd$ is an isotherm at a lower temperature $T_2$ where heat is given to another reservoir at $T_2$. The line $bc$ and $da$ are adiabatic processes. The work done $W$ is the area enclosed by the cycle.

We might choose to represent the Carnot engine according to the figure below. Where $W$ is the work done by the engine. The efficiency $\eta$ is a measure of "what we get out for what we put in" so in this case the work output for the heat input.
\begin{equation}
\eta =\frac {W}{Q_1}
\end{equation}
If we remember the first law and that $\Delta U=0$ then we may write $W=Q_1-Q_2$ denoting the sign convention for heat into the system and work done on the system to be positive and vice versa.
\begin{equation}
\eta =1-\frac{Q_2}{Q_1}
\end{equation}
Kelvin-Planck statment

It is impossible to construct a device that, operating in a cycle, will produce no effect other than the extraction of heat from a single body at a uniform temperature and the performance of an equivalent amount of work.

This basically says that some heat needs to be given to a body of colder temperature otherwise by the above equation we would have 100% efficiency. If you were to violate this you could just have perpetual motion machines or cars that just absorbed heat and produced work. It may be summarised as a process whose only effect is the complete conversion of heat into work is impossible. However the opposite process is allowed  where all the work is transferred to heat completely

The Clausius statement

*It is impossible to construct a device that, operating in a cycle, produces no effect other than the transfer of heat from a colder to a hotter body. *

What this means is that work must be performed if heat is to be transferred form a colder to a hotter body. We might be more familiar with this statement if we called this engine a fridge. An engine that extracts heat from a cold body and delivers it to a hot body when work is performed.

The kelvin-Planck statement is entirely equivalent to the Clausius statement. We will skip any proof or mathematics of why this is so.

Carnot's Theorem

No engine operating between two reservoirs can be more efficient that a Carnot engine operating between those same two reservoirs.

The proof is as follows. Assume that such an engine did exist $E'$ with efficiency $\eta '$ that takes heat $Q_1'$ from the hot body and does work $W'$ delivering heat $Q_2'=(Q_1'-W')$ to the cold bath.

Let us now imagine a Carnot engine $C$ with heats $Q_1$ and $Q_2$, efficiency $\eta$ adjusted in such a way that it performs the exact amount of work $W$ as the "more efficient engine $E'$ so that $W=W'$, and $Q_2=Q_1-W$.

Iff,
\begin{equation}
\frac{W'}{Q_1'}>\frac{W}{Q_1}\ \ \ \ \ \ \ \ \ then \ \ \ \ \ \ \ \ Q_1>Q_1'
\end{equation}
The Carnot engine is reversible so it may be driven backwards, acting as a composite device with $E'$. It extracts positive heat $(Q_1-Q_1')$ form the cold reservoir and delivers the same heat to the hot reservoir with no external work being required.

However since the reservoir is large the temperature is unchanged with the addition of this heat. This means that the Clausius statement is violated. Therefore $\eta'>\eta $ is not true, and $E'$ can never exist. If $\eta '=\eta$ then $Q_1'=Q_1$ and no net heat transferred for no work. Therefore we say,
\begin{equation}
\eta' \leq \eta
\end{equation}
Entropy

Consider a substance that can do work under going a cycle so that at the end of its cycle its state is unchanged at $T_1$. Lets say we change the first state to infinitesimally close neighbouring state 2 at $T_2$ by adding a small amount of heat $\delta Q_1$ from . If we do this with a Carnot engine with the temperature baths of state 1 $T_1$ and some reservoir at $\bar T$ then $T_1$ supplies $\delta Q_q$ to the working substance while the Carnot engine supplies the same amount to the $T_1$ reservoir to leave it unchanged, in turn taking heat $\{\bar T/T_1\}\delta Q_1$ from a principle reservoir.

So in short we have performed a change from states 1 to 2 with external work $\delta W_1$ and  extraction of heat $\{\bar T/T_1\}\delta Q_1$ from the principle reservoir. This is one system of a composite process of identical systems. The diagram really helps here!

The composite system is assumed that $\Delta U=0$ and that the heat supplied to it is;
\begin{equation}
Q=\sum _i\frac{\bar T}{T_i}\delta Q_i
\end{equation}
Where the sum is over all the little systems making up the composite. The external work performed is;
\begin{equation}
W=\sum _i\delta W_i
\end{equation}
The first law is then $0=Q-W$. Looking at the second diagram below for the composite system we can see that this violates the Kelvin-Planck statement. The only way that this works is if work is done on the system and an equal quantity of work flows out. Or it works if both $W$ and $Q$ are zero.
\begin{equation}
W=Q\leq 0
\end{equation}
So
\begin{equation}
\bar T\sum _i\frac{\delta Q_i}{T_i}\leq 0 \ \ \ \ \ \ \ or \ \ \ \ \ \ \ \sum _i \frac{\delta Q_i}{T_i}\leq 0
\end{equation}
If you take the limit of this then over one complete cycle we arrive at the Clausius inequality!
\begin{equation}
\oint \frac{dQ}{T}\leq 0
\end{equation}
It is important to know that the $T$ appearing in the integral is the temperature of the heat supplying bath!

If the cycle is reversible then there is no reason why we couldn't do the proof in the other direction and hence obtain the inequality sign the other way around. This only holds for reversible cycles. The only way to satisfy both inequalities physically is if for reversible cycles we equal the integral to zero exactly. This means that for reversible cycles the integral is equal to zero!

The final step is to now use this Clausius inequality to prove the second law. Imagine a cycle that is completely reversible between two states $i$ and $f$ with two reversible paths $r_1$ and $r_2$ connecting them.
\begin{equation}
\oint \frac{dQ_r}{T}=_{r_1}\int^{f}_{i} \frac{dQ_r}{T}+_{r_2}\int_{f}^{i}\frac{dQ_r}{T}=0
\end{equation}
Thus,
\begin{equation}
_{r_1}\int ^{f}_{i}\frac{dQ_r}{T}=_{r_2}\int^{f}_{i}\frac{dQ_r}{T}\end{equation}
What all this means is that the integral is path independent SO THERE MUST BE A PATH INDEPENDENT FUNCTION !! We call this state function entropy.
\begin{equation}
S_f-S_i=_r\int ^f_i \frac{dQ_r}{T}
\end{equation}
For this equality to hold the heat exchanged must be defined over a reversible path! That is to say if the entire cycle is not reversible then we have to use the Clausius stamen again. Lets suppose that one off the paths is irreversible now.
\begin{equation}
 \int^f_i\frac{dQ}{T_0}+_r\int ^i_f\frac{dQ_r}{T}\leq 0
\end{equation}
 With a bit of interval switching,
\begin{equation}
\int ^f_i\frac{DQ}{T_0}\leq -_r\int ^i_f\frac{dQ_r}{T}=_r\int ^f_{i}\frac{dQ_r}{T}=S_f-S_i
\end{equation}
And finally the way we normally write this is in an infinitesimal form,
\begin{equation}
\frac{dQ}{T_0}\leq dS
\end{equation}
This is more general than before since it includes the previous equation too, equality holds if the cycle is reversible. So , in an infinitesimal irreversible process between a pair of states the entropy change is $dS$, if the system is thermally isolated then $dQ=0$ and we generate the principle of increasing or "maximum" entropy.
\begin{equation}
S_f-S_i \geq 0
\end{equation}
Or if you prefer $dS\geq 0$. The entropy of a thermally isolated system increases in any irreversible process and is unaltered in a reversible one.


What is crystallographic equivalence? I’ve learnt that in crystallography, there are many crystallographic directions that are “equivalent.” For example, in a cubic crystal system, the [011] and [110] and [101] directions are “crystallographically equivalent.” See the image below:

I cannot begin to fathom how they are equivalent. These are vectors after all, and from what I can tell with my rudimentary knowledge of maths, these vectors are not equivalent in direction. Yet all materials science textbooks and powerpoints insist that, for cubic systems, any permutation of positive and negative indices will yield a same equivalent direction. That is, the [011], [110], and [101] vectors are… the same vector?

I share the same confusion with planes, even if they aren’t vectors. For cubic systems, just as for directions, any arrangement of positive and negative indices produces an equivalent plane. For example, (100) and (010) are the same plane, when to me they clearly look different.

So what exactly is “crystallographic equivalence”?

Directions & sites are said to be crystallographically equivalent, because there are symmetry operators that relate them.  So

learn about simpler symmetry operations (proper axes, mirror planes, centres of inversion) or advanced ones (like screw axes, glide planes, etc.)
learn how to recognize them in every-day's life.  It may be helpful for crystallography, and in other fields too: IR / Raman spectroscopy, for example, not only for understanding (descriptive), but for forcasting (predictive).  Tutorials like this represent an entry.


Chemistry: A Chemical Dilemma We accept that, via observations that an electron possesses a negative charge and a proton a positive charge, both of which attracting each another. My dilemma lies here, what is the reason for the force of attraction or repulsion? If it may be due to the desire to become more stable and is logically comprehensible via mathematics of potential energy – but why does a body seek the desire to become more stable at all in the first place?

Though this question may be silly, and some might say this goes into metaphysics, but does anyone know why, or do we assume that everything – or the universe to be more precise, acts as it is, accepting it as a fact, and construct our calculations and laws above it? Are all chemical and physical knowledge based on taking the mechanics of the universe for granted?

The question was bothering me for sometime and finally decided to ask in search of an answer.

You have it a little backwards.  We don't observe charge, and then infer the existence of a force.  It's the other way around.  We observe forces between objects -- they repel or attract each other -- and then build models which explain which objects are attracted or repelled by which other objects, and by how much.  That leads us to invent the idea of charge and assign it to objects, and later still to connect charge to other properties of the objects.

If your question is why is there an electromagnetic force at all, the answer is probably that it is easier to imagine a universe in which there are forces between degrees of freedom than one in which there are not.  Our current understanding of the universe is that can best be described as a set of fields, like the surface of the sea, and everything we see is some kind of ripple or wave in that field.  It's easy to see why ripples on the surface of the ocean should interact with each other -- exert "forces" of some kind on each other -- and hard to see how they could just totally ignore each other.  That's how it is with nature.

Granted, it looks like there exist wholly separate bodies (particles, objects, things) and we wonder why they should interact with similar things far away.  But that may be because our perception is severely limited, and the underlying fields are much more complex than the surface of an ocean.  If you want to explore the topic further, start by googling "wave-particle duality in quantum mechanics."


Designing a theoretical synthesis of methamphetamine from phenylpropyne I am asked to synthesize "methamphetamine." Note that there is no configuration designated for this molecule, and methamphetamine refers to a very specific configuration. In any case, here are the steps I took:

Reduction with hydrogen gas and Lindlar's catalyst. This transforms the alkyne into a cis alkene.
Epoxide formation with m-CPBA.
Ring opening with sodium methylamide. This is a base, and it should attack the less sterically hindered side of the epoxide - the side without the phenyl substituent. This opens the ring and gives me almost what I want. Now I just have to remove the hydroxyl group.
Removal of hydroxyl group: I decided to use TsCl and heat. The tosylate group should leave upon heating and a secondary benzylic carbocation that'd form. Sodium hydride was added as to add to the carbocation formed when the TsCl leaves. However, a possible issue is that the hydride anion, being a good base, would give me the elimination product as well.

Alternate pathway below

I think the synthetic route you've proposed is quite good for the most part. However, I think your use of $\ce{NaH}$ to remove the tosylate is incorrect. $\ce{NaH}$ is a very strong base, but not a good reductive hydride donor in most instances. Here, it will almost certainly just react by eliminating the tosylate to give a double bond. However, LAH (and various similar reagents) should achieve the desired effect. Alternatively, one of the more popular dehydroxylation reactions is the Barton-McCombie deoxygenation.

An alternative route might begin with hydroboration-oxidation of the alkyne directly. Generally, when the two ends of the alkyne are equally substituted, one obtains a mixture of isomers, and electronic factors are usually more decisive than sterics. That being said, in this case, I suspect that it might be possible to control the regiochemistry by selecting a sterically hindered borane, perhaps 9-BBN or catecholborane. I'm not certain, however. Assuming one can indeed install the ketone at the end of the alkyne further from the phenyl group, then reductive amination should give the desired product. Direct reductive amination should be possible in this case (e.g., with $\ce{NH2Me}$, catalytic acid, and $\ce{NaBH3CN}$), allowing for a one-pot reduction.

Hence, the proposed reaction scheme would be:
\begin{align}\ce{
Ph-C#C-CH3 &->[\mathrm{1)~9-BBN}][\mathrm{2)}~\ce{H2O2,~NaOH}] Ph-CH2-C(O)CH3\\\\
           &->[\ce{NH2Me,~[H+]}][\ce{NaBH3CN}] Ph-CH2-CH(CH3)NHMe
}\end{align}

An additional point in favor of this approach is that it can be made highly enantioselective by use of asymmetric catalysts (q.v., Noyori asymmetric hydrogenation).

Addendum

In response to your second reaction scheme, jerepierre's comment above is definitely correct about a competing elimination reaction in the last step being a problem. It's especially likely because the proton transfer is fast, $\ce{NaNHMe}$ is extremely basic, the elimination is highly entropically favored, and the resulting $\pi$-bond is conjugated to the ring, giving an especially stable product (the major elimination product by a large margin should be the Zaitsev one). In sum, I think both kinetic and thermodynamic factors are heavily against you here. If you really want to go the route of nucleophilic substitution to attach your amine, better approaches are probably the Delépine reaction, any of the azide reduction methods, or the Gabriel synthesis (approximately in that order, I think). All of those suffer from the fact that you would then need to methylate the resulting primary amine, and avoiding extensive polyalkylation is more or less impossible, as far as I know.

There's also an issue before that stage, in that hydroboration-oxidation will not predominantly install the alcohol at the correct position, giving mainly the benzylic alcohol instead, unless you use a suitably bulky borane to achieve the desired regioselectivity. Since you only want to replace the alcohol with a bromine, I think $\ce{HBr}$ in the presence of peroxide (or some other radical initiator) should work, with no intermediate step and a reasonably low proportion of unwanted side-products.


What happens on heating an ester? While solving a few questions a came across a problem which asked me what would happen when sec-butyl ethanoate was heated.
I had never read about this reaction before and took to researching through a few of my textbooks, but could not find an explanation as to what would happen.
I also thought along the lines free radical formations and tried to make an attempt.
However according to the answer the reaction was a syn elimination and it would give a Hoffmann product. But I couldn't get any more information out of it.
But now I was stumped. So please help me out of you know anything about this.

Edit:
After googling for sometime I came across a paper which said that this particular compound would decompose to give cis butene.
However I could not understand the reason for this. Here is where I saw it:
https://pubs.acs.org/doi/pdf/10.1021/i360047a014

This is an example of a Thermal syn elimination. The Wikipedia article is a good place to start: here

image from here
From ref 2

Unsymmetrical Acetate and xanthate esters yield a mixture of all the possible alkenes, but there is usually a predominance of the more highly substituted alkene.


How to determine sulfate ions in a solution by precipitation? For example I am given an unidentified solution and am told that it may or may not contain sulfate ions. How can I confirm or deny the presence of sulfate in this solution?

Acquaint yourself with the so-called solubility rules. Depending on which cation sulfate is paired with, the sulfate compound may or may not be soluble. Generally, compounds containing sulfate ($\ce{SO_4^2-}$) compounds are soluble. (For essentially the same reason that sulfuric acid $\ce{H2SO4}$ is a strong acid, if you're interested.)

Well-known insoluble sulfate compounds are $\ce{PbSO4, SrSO4, CaSO4}$, and $\ce{BaSO4}$. Note that all of these cations have a somewhat high (+2) charge and are large.

If sulfate anions are present in your aqueous solution and you a a soluble compound containing $\ce{Pb^2+, Sr^2+}$, etc, then those cations will form an insoluble precipitate with the sulfate.


Properties of steel and aluminum alloys I have been trying to compare 304 stainless steel with 7075 aluminum for some personal research. This steel alloy has a specific heat capacity of 500 J/kg-C, while for aluminum the value is 960 J/kg-C. Does this mean it is harder to heat up aluminum than steel? Then, steel has a thermal conductivity of 130 W/m-K, while for aluminum the value is 16.2 W/m-K. Does this mean that steel conducts heat away better than aluminum? Thank you.

Heat capacity (specific heat) varies inversely with atomic mass, the Dulong–Petit law. Al is about 27 amu, and Fe is about 56 amu, so as you noted, it would be expected that aluminum stores more heat. An extreme example is lead solder, which has such low specific heat that a calloused plumber's hand can wipe a solder joint with little discomfort.

As for thermal conductivity, that property is associated with the rigidity of crystal lattices, where sound travels as phonons, as in diamond, and with electrical conductivity, where "freely moving valence electrons transfer not only electric current but also heat energy." Aluminum and copper are among the best metallic conductors, so are used for cooking-pan bottoms to spread heat evenly. Since stainless steel has many additions to Fe, such as Ni and Cr, these inclusions provide discontinuities at grain boundaries that further impede heat transfer. Dewar flasks are made of stainless steel or glass, rather than Al, because they are poor conductors of heat.


Electrolysis of aqueous solutions Water competes with any ions in solution for both oxidation and reduction. There are $1000\ \mathrm{g\ l^{-1}}/18\ \mathrm{g\ mol^{-1}}\approx55.56\ \mathrm{mol\  l^{-1}}$. That is truly huge in comparison to any amount of salt we can dissolve. Why doesn't water dominate any electrolysis reaction since even if a small percentage of the collisions of water can accomplish the reaction? The sheer number of collisions on its side will ensure it wins. Also, if hydrogen ions and hydroxide ions alone participated, a concentration of $10^{-7}$ means that it's several orders of magnitude away from the dissolved salt's ion concentration, which means that water can never dominate. Yet, we use a comparison of the thermodynamic feasibility in the form of reduction potentials to decide which competing ion gives the product. How is it that the concentration effects become comparable enough to allow this in most cases?

It is good question given the fact that this idea is not commonly discussed. The main source of your confusion is the following idea

Why doesn't water dominate any electrolysis reaction since even if a small percentage of the collisions of water can accomplish the reaction? The sheer number of collisions on its side will ensure it wins.

Unlike chemical reactions, electrode reaction do not solely require collisions. It requires an additional process: electron transfer. No matter how much collisions are occurring, the electrode is unable to transfer or withdraw electron from water in high salt concentration. Thermodynamics tells us (from the Nernst equation) how easy it would be to remove or add an electron to a water molecule at a given salt concentration. It is silent as to why this happens and what is the mechanism.
Electrode processes are very complicated. Marcus got a Nobel Prize on electron transfer.


Dissociation of benzoic acid Benzoic acid is a weak acid, so it does not completely dissociate in water. I know that the dissociation equation of benzoic acid is:

$$\ce{C6H5CO2H <=> H+ + C6H5CO2-}$$

The question I'm having trouble with is:

  Would the dissociated form be more or less likely to dissolve in water?

Is this basically asking if the ions on the right are more likely to dissolve when compared to $\ce{C6H5CO2H}$?

If this is what the question is asking, I think yes, the dissociated form will be more likely to dissolve as they are very polar when compared to $\ce{C6H5CO2H}$ and they can be surrounded by polar water molecules. But I'm not sure.

Solubility of benzoic acid(mw 122.12):

1.7 g/L (0 °C) - 0.00139 moles/100 ml
2.7 g/L (18 °C)
3.44 g/L (25 °C)
5.51 g/L (40 °C)
21.45 g/L (75 °C)
56.31 g/L (100 °C) - 0.04611 moles/100 ml

Solubility of sodium benzoate (mw 144.10):

62.69 g/100 mL (0 °C) - 0.4350 moles/100 ml
62.78 g/100 mL (15 °C)
62.87 g/100 mL (30 °C)
71.11 g/100 mL (100 °C) - 0.4935 moles/100 ml

So yes, sodium benzoate is more soluble than benzoic acid, but consider calcium benzoate which is less soluble.

Calcium benzoate (mw 282.31):

2.32 g/100 mL (0 °C) - 0.00822 moles/100 ml
2.72 g/100 mL (20 °C)
8.7 g/100 mL (100 °C) - 0.03081 moles/100 ml

So the "real" answer is - it depends...


Term symbol for atomic nitrogen: why can't S be -1/2? The spectroscopic notation for an element is given by $^{2S+1}L_{J}$. I want to use this notation for nitrogen with configuration

$$\mathrm{[1s^22s^22p^3]}$$

This is what I did. $l=1$ for p orbital. There are three electrons in p orbital, two electrons: spin up and spin down, and the remaining electron in spin up position, so that my $$S = \frac 12 + \left(-\frac{1}{2}\right) + \frac12 = \frac12$$ which gives

$$J=l+s=1+1/2=3/2$$

My query is can't the remaining electron in p orbital be in spin down position so that my $S$ becomes $-1/2$ and so that $J=l+s=1/2$.

Why not go one step further and make all the electrons have spin down? Then you should get $S = -3/2$, no?

In fact, what's happening is: you are confusing the projection of the spin quantum number, $M_S$, with its magnitude, represented by the spin quantum number $S$.

The $\pm1/2$ that you are summing are the individual projections of the electron spins, $m_s$. When you add up projections, you get a projection $M_S$. This projection can indeed be $-1/2$, and it can indeed be $-3/2$ as well, but the term symbol does not depend on $M_S$; it depends on $S$.

To obtain the total spin quantum number $S$ you have to use the Clebsch-Gordan series to couple different sources of angular momenta together. In this case we would first couple the spins of two electrons

$$S_{12} = s_1 + s_2, s_1 + s_2 - 1, \cdots, |s_1 - s_2|$$

Note that for an electron, it has spin $s = 1/2$. Again, the projection of the spin can be $+1/2$ or $-1/2$, but that is not what we are interested in. We are interested in the magnitude of its spin, which is only $1/2$, not $-1/2$.1 If we plug $s_1 = s_2 = 1/2$ into that equation above we get

$$S_{12} = 1, 0$$

and now if we couple this to the spin of the third electron we get

$$S = S_{12} + s_3, S_{12} + s_3 - 1, \cdots |S_{12} - s_3|$$

Again, $s_3 = 1/2$ only. But $S_{12}$ can take on two values so we will have to evaluate this sum two times, one time for each possibility:

$$\begin{align}
S_{12} = 1 &\Longrightarrow S = \frac 32, \frac 12 \\
S_{12} = 0 &\Longrightarrow S = \frac 12
\end{align}$$

Note that the absolute value in the Clebsch-Gordan series forbids the possibility that $S < 0$. This makes sense since a magnitude of a spin cannot be negative.

In the end your two possibilities are $S = 3/2$ or $S = 1/2$. Due to Hund's first rule the $S = 3/2$ term is the ground state and the $S = 1/2$ terms are excited states.

Next, the $L$ in the term symbol is not equal to 1 just because the p orbitals have $l = 1$. Note that one is a small $l$ and the other is a big $L$. The big $L$ in the term symbol is obtained again by coupling the individual values of $l$ for each orbital. In this case all three p orbitals have $l =1$ and you would have to work through the Clebsch-Gordan series again.

$$L_{12} = l_1 + l_2 , l_1 + l_2 - 1, \cdots , |l_1 - l_2|$$

$$L = L_{12} + l_3, L_{12} + l_3 - 1, \cdots, |L_{12} - l_3|$$

and if you set $l_1 = l_2 = l_3$, should reach the conclusion that $L$ can take on the values of $3,2,1,0$. However you have to be careful about this, because the $L = 3$ value is forbidden by the Pauli principle, which is not taken into account by the Clebsch-Gordan series. A term with $L = 3$ would necessitate the existence of a microstate with $m_L = +3$, which would imply that all three electrons are in the p orbital with $m_l = +1$, which is not possible.

1 Technically, the spin is $\displaystyle \sqrt{\frac{1}{2}\left(\frac{1}{2} + 1\right)}\hbar$, but whatever. The correct phrasing is that the quantum number representing the magnitude of the spin can only be $1/2$.


Why macromolecule ligands such as Crown ethers form more stable complexes than open chain ligands? This was in J. Huheey's book and no explanation was given

The reason for crown ethers being more stable than their equally long open chain derivatives is the difference in Gibbs Free energy. (If you are not used to these terms, the Wikipedia article will help I think, though it is quite, long and thorough.)

The main point to take from the use of the Gibbs free energy in this case is this: it is made up of two parts, enthalpy ($\Delta{}H$) and entropy ($\Delta{}S$).

$$
\Delta{}G = \Delta{}H - T \cdot{} \Delta{}S
$$

If only the binding affinity of the (crown) ether chain is considered, only the enthalpy is relevant. The strength of the individual ether moieties is as strong in a crown as it is in a chain. The difference then, lies with the entropy. To have a crown ether form a structure around a ligand 'costs some entropy'. The same entropy difference should be considered for the chain ether, with added to that the cost of pulling the ends of the chain together! That difference in entropy results in the difference in $\Delta{}G$ and therefore yields a more stable complex. (The ends of the crown ether are also pulled together, this cost of entropy however, has been paid in an earlier reaction when making the crown ether, and therefore has no influence here.)

You can see this also from the relation between the Gibbs free energy and the equilibrium constant of the complex (In the imaginary reaction $\ce{A + B -> AB})$:

$$
\Delta{}G = - RT \cdot{} Ln (K)
$$

where,

$$
K = \frac{[\text{AB}]}{[\text{A}] \cdot{} [\text{B}]}
$$

A higher $\Delta{}G$ leads to a higher equilibrium constant.

Hope this helped, and I'd be happy to clarify anything further.


Identification of unknowns with varying solubilities and states of matter White solid A water insoluble cleaved hot to solid gas B and C. B is insoluble in water and in $\ce{HCl}$ but soluble to $\ce{HNO3}$ the react gives solution D and E colorless gas which becomes red on contact with air. The solution D gave white precipitate F with $\ce{HCl}$. The gas C is reacted by heating with magnesium G yielding a white solid which reacts with water yielding a white solid and the gas J characteristic suffocating odor. J is a basic , H is soluble in dilute $\ce{H2SO4}$. The gas J oxidized to C while the gas under specific conditions ($\ce{Pt}$ catalyst and $800^\circ\mathrm{C}$) is oxidized to the gas E.

My attempt:
C: $\ce{N2}$,
G: $\ce{Mg3N2}$,
H: $\ce{Mg(OH)2}$,
J: $\ce{NH3}$,
E: $\ce{NO}$
\begin{aligned}\ce{
N2 + 3Mg &-> Mg3N2\\
 Mg3N2 + 6H2O &-> 3Mg(OH)2 + 2NH3\\
 Mg(OH)2 + H2SO4 &<=> 2 H2O + MgSO4\\
 4 NH3 + 5 O2 &->[\ce{Pt~$800^\circ\mathrm{C}$}] 6 H2O + 4 NO\\
 4 NH3 + 3 O2 &-> 6 H2O + 2 N2\\
}\end{aligned}

I cannot really find anything else.

The solutions which you have already found look good so far. For solid A, we need to find a compound which thermally decomposes into $\ce{N2}$ and an element which is only soluble in oxidizing acids like $\ce{HNO3}$, and whose chloride is insoluble in water or acids. Silver azide could possibly be a good fit, as it can be cleaved into $\ce{Ag}$ and $\ce{N2}$ by heating, and silver chloride has a very low solubility in water. Care would have to be taken with the temperature when actually performing the decomposition of $\ce{AgN3}$, as it is explosive.

$$\ce{2AgN3 -> 3N2\uparrow + 2Ag}$$

Dissolving silver (solid B) in nitric acid gives a silver nitrate solution D and $\ce{NO}$ as colorless gas E.

$$\ce{3Ag + 4HNO3 -> 3AgNO3 + 2H2O + NO\uparrow}$$

$\ce{NO}$ turns red on contact with air when it is oxidized to nitrogen dioxide by atmospheric oxygen.

$$\ce{2NO + O2 <=> 2NO2}$$

From solution D, insoluble $\ce{AgCl}$ (F) precipitates when $\ce{HCl}$ is added.

$$\ce{Ag+ + Cl- -> AgCl\downarrow}$$


How many moles of aspartame are present in 4.00mg of aspartame? (3 sig figs) This is a part of my online chemistry homework (mastering chemistry)

I know how to do it but when I enter $\pu{1.36E-5}$ into the program, it says that it's incorrect and that the correct answer does not depend on $\times10$. I'm not sure what it wants from me. The answer should be given in moles with $3$ sig figs.

Try putting in 0.0000136. Remember that zeroes between the decimal point and the last numbers do not contribute to significant figures. Good luck.


What is the E/Z configuration of cyclic compounds? What are the configurations of 1-bromo-3-methylcyclopent-1-ene and 1-nitro-2-(propan-2-yl)benzene? Are they E or Z? Are they cis or trans?

What are the rules for these types of isomerism?

What is the meaning of cis and trans?
cis means on the same side (or face), trans means on the opposite side (or face).
Can I use cis and trans for cylic compounds?
Yes, you can describe the relative orientation of substituents on a ring using cis and trans, such as in the case of 1,3-dibromcyclohexane.

Can I use cis and trans for alkenes?
Yes, in priciple. But you should rather use E and Z. Note that these descriptors are taken from German.

E means entgegen and refers to opposite directions (or sites), while Z means zusammen, which is together, that is the same direction (or site).
What about the compounds in my question?
E/Z or cis/trans are not applicable here.
In the first case, the $\ce{Br}$ atom is in the ring plane, the methyl group can be above or below it. You could describe the absolute stereochemistry at the centre bearing the $\ce{CH3}$ substituent using R or S.
In the second case, there's no alkene at all but a disubstituted benzene. The relative orientation of the substituents on these systems can be described by assigning the proper numbers or by using the ortho- meta- or para-prefix. For this compounds, 2-isopropylnitrobenzene or o-isopropylnitrobenzene would be ok.


How to display ligands as ball and stick and macromolecules as cartoons (if protein) / rings (if nucleic acids) in Avogadro? Avogadro has many good qualities but I can't seem to find a way to display ligands differently to macromolecules in the same view window. Frankly I can't seem to find a way to display any two things differently (like one as ball and stick, another as ribbons, or w/e) to one another in the same view. It seems like in a given view window everything has to be displayed the same way, there's no room for making one part displayed one way and another part displayed another way.

This was also asked on the Avogadro discussion forum

It's possible in Avogadro 1.x and discussed in the documentation

Select part of the molecule
Click on the "options" button for a particular display type (e.g. "ball and stick")
Click on the button to only show the selected atoms


Why does Xe have a higher boiling point than Ne? I am certain that the reason involves intermolecular forces, but since both $Xe$ and $Ne$ are noble and non-polar gases, shouldn't these forces have a much smaller effect or negligible on boiling point?

The main interaction between noble gases is Dispersion. (inert gas that's where the name comes from)
Dispersion is larger in heavier atoms. (Considerable larger, 18+18+8 more electrons difference is massive)

an example:
When Interaction energy passes zero point and thermal energy you can have the solid phase. This is why you will not have solid helium.


Extracting elemental ratios from X-ray photoelectron spectroscopy (XPS) Is it possible to extract information on elemental ratios from X-ray photoelectron spectroscopy (XPS)?  For example, I am reading a paper and in it they say that the carbon to oxygen (C:O) ratios of the four samples are 2.8, 2.9, 5.1, and 23.3.  However, the authors do not provide the actual XPS spectra, not even in a Supporting Information-type addendum.

I am a computational chemist and have no experience with XPS.  I do know that elements, and even particular hybridizations, give characteristic peaks at characteristic binding energies (in units of eV, for example).  This, I know, enables qualitative analysis of compounds.  But is there a way to extract quantitative information from XPS?  Suppose I want to determine the C:O ratio in a compound containing only C, O, and H.  Do I take the ratio of the C and O peak heights, or do I take the ratio of the area under the peaks corresponding to C and O?

Do you know of any good review articles that would help me to understand the basics of XPS?

I found some information in Skoog, Principles of Instrumental Analysis, 6th ed.  On page 597:

  Quantitative Applications.  Once, XPS was not considered to be a very useful quantitative technique. However, there has been increasing
  use of XPS for determining the chemical composition of the surface
  region of solids.  If the solid is homogenous to a depth of several
  electron mean free paths, we can express the number of photoelectrons
  detected each second $I$ as $$I = n \phi \sigma \epsilon \eta A T \ell$$ where $n$ is the number density of atoms ($\text{atoms cm}^{-3}$) of the sample, $\phi$ is the flux of the incident X-ray
  beam ($\text{photons cm}^{-2} \text{ s}^{-1}$), $\sigma$ is the
  photoelectric cross section for the transition
  ($\text{cm}^{2}/\text{atom}$), $\epsilon$ is the angular efficiency
  factor for the instrument, $\eta$ is the efficiency of producing photo
  electrons ($\text{photoelectrons/photon}$), $A$ is the area of the
  sample from which photoelectrons are detected ($\text{cm}^{2}$), $T$
  is the efficiency of detection of the photoelectrons, and $\ell$ is
  the mean free path of the photoelectrons in the sample ($\text{cm}$).

  For a given transition, the last six terms are constant, and we can
  write the atomic sensitivity factor $S$ as $$S = \sigma \epsilon \eta A T \ell$$  For a given spectrometer, a set of relative values of $S$
  can be developed for the elements of interest.  Note that the $I/S$ is
  directly porportional to the concentration $n$ on the surface.  The
  quantity $I$ is usually taken as the peak area, although peak heights
  are also used.  Often, for quantitative work, internal standards are
  used.  Relative precisions of about 5% are typical.  For the analysis
  of solids and liquids, it is necessary to assume that the surface
  composition of the sample is the same as its bulk composition.  For
  many applications this assumption can lead to significant errors.
  Detection of an element by XPS requires that it be present at a level
  of at least 0.1%.  Quantitative analysis can usually be performed if
  5% of the element is present.


What is the pOH of a 0.0100 F solution of dipotassium phtalate? What is the pOH of a 0.0100 F solution of dipotassium phtalate? pKa1 = 2.950 and pKa2 = 5.408 for phtalic acid.

I did it for phthalic acid by calculating pH and then just doing pOH = 14.00 - pH, which gave me pOH = 11.60. Not sure if I did it correctly, I'm only just (re)learning this. I just need a clear and basic explanation, if possible. The two pKa's are what confuse me... which one do I use? I've been searching on google and through my textbook for 2 days now but I'm just confusing myself even more at this point. I know it's probably really easy stuff, but I haven't done chem in 4 years... trying to refresh my knowledge.

So, we know that dipotassium phtalate is a conjugate base of a diprotic acid, so the addition of it into water will remove some $\ce{[H+]}$s from $\ce{[H2O]}$ and increase the level of $\ce{[OH^{-}]}$ with means a lower pOH.

We calculate the pOH by first converting to $\mathrm{K_a}$
$$\mathrm{pK_{a2}=10^{-5.408}=3.908*10^{-06}}$$
$$\mathrm{K_{b1}=\frac{K_w}{K_{a2}}= \frac{1*10^{-7}}{3.908*10^{-06}}=2.559*10^{-2}}$$
$$\ce{\frac{[HA][OH^-]}{[A^-]}=\mathrm{K_{b1}}}$$
$$\mathrm{\frac{x^2}{F-x}=2.559*10^{-2}\rightarrow 1.60*10^{-2}}$$
We did that last one assuming that x was negligibly small.
We now know
$$\ce{[HA]}=\mathrm{x=1.60*10^{-2}}$$
$$\ce{[H+]}=\mathrm{\frac{K_w}{\ce{[OH^-]}}=\frac{K_w}{x}=6.25*10^{-6} \rightarrow pH=5.20_4}$$
$$\mathrm{pOH=14-pH \rightarrow 14-5.20_4 = 8.79_6}$$


Can a low concentration of an acid disolve solids just as well? I'm in the process of removing some hard water stains with vinegar.  It's not working so well.

The stains are in a large fish tank, which I have filled with water and vinegar.  I would expect a lower concentration of acetic acid to cause the acid to dissolve the calcified material slower, but it doesn't seem to be having any effect.

Is there a threshold-like effect on an acid's ability to dissolve such solids, such that a lower concentration would simply be unable to dissolve them (rather than merely dissolving them slower)?

The rate of dissolution depends on $\mathrm{pH}$ and $\mathrm{pH}$ dependent equilibrium reaction chain :

$$\begin{align}
\ce{2 CH3COOH + CaCO3 v &<=> (CH3COO)2Ca + Ca(HCO3)2}\\
\ce{2 CH3COOH + Ca(HCO3)2 &<=> (CH3COO)2Ca + 2 H2CO3}\\
\ce{H2CO3 &<=>   H2O + CO2(aq)}\\
\ce{CO2(aq) &<=>   CO2(g)}\\
\end{align}$$

what can be written in the equivalent ionic form:

$$\begin{align}
\ce{CaCO3 v &<=> Ca^2+ + CO3^2-}\\
\ce{CH3COOH &<=> CH3COO-  + H+}\\
\ce{H+ + CO3^2- &<=> HCO3-}\\
\ce{HCO3-  + H+ &<=> H2CO3}\\
\ce{H2CO3 &<=>   H2O + CO2(aq)}\\
\ce{CO2(aq) &<=>   CO2(g)}\\
\end{align}$$

6% vinegar (1M) acetic acid has $\mathrm{pH=p}K_{\mathrm{a}}/2=2.37$.

Vinegar with 1/11 acid  spent has pH 3.75, with activity of $\ce{H2O+}$  24 times lower, so does is lower the reaction speed.

Vinegar with the acid half spent has pH 4.75, with activity of $\ce{H2O+}$ 240 times lower.

Vinegar with 10/11 acid  spent has pH 5.75, with activity of $\ce{H2O+}$  2400 times lower.

If pH raises enough, equilibrium concentration of $\ce{CO2(aq)}$ does not cross the gas solubility limit anymore and reaction approaches equilibrium, as production of gaseous carbon dioxide ceases..

If initial concentration crosses some minimal threshold, $\mathrm{pH}$ raises to level not able to produce gaseous carbon dioxide before water gets saturated by $\ce{CO2}$.

Some carbonate gets dissolved as bicarbonate, but with low initial acetic acid concentration, this dissolution is slow.

See also carbonic acid Wikipedia article.

Another effect is diffusion limitation. Without stirring, the acid near walls may get spent, while the acid abundance in free volume is intact, without access to limescale. Rate of acid diffusion to walls limits rate of dissolution.

Personally I would suggest rather citric acid, that have much higher acidity.

And/or, instead of filling the whole volume by diluted solution , rather using wall washing by concentrated solution.


What reaction could rubidium undergo in a vacuumed chamber with Macor ceramic and glass? In the center of the chamber on the photo below you can see the hexagonally-shaped Macor  glass-ceramic consisting of

Silicon dioxide $\ce{SiO2}~46\%$
Magnesium dioxide $\ce{MgO}~17\%$
Aluminum oxide $\ce{Al2O3}~16\%$
Potassium oxide $\ce{K2O}~10\%$
Boron oxide $\ce{B2O}~7\%$
Fluorine $\ce{F}~4\%$

The original color of Macor is white as you can see on the second picture on the side. I insert the rubidium into the chamber. It all became black at the temperature of approx. $\pu{150–200 °C}.$ What could rubidium possibly react with?

It was in the vacuum $\pu{E-8 Torr}.$ I have heard that the window material can also react with rubidium. Rubidium is in vapor state (according to optical spectrum at temperature $~\pu{100 °C}$ and $\pu{E-4 Torr}).$ Rubidium tends to condense on the cold part of the chamber, so it is possible to touch the ceramic.

I heat the chamber outside, so rubidium should be in vapor state inside of the chamber, and Macor is connected to the chamber, so it should be close but lower than the chamber temperature.

One candidate for reaction with the rubidium is silica, $\ce{SiO2}$.  Reference [1] reports that black powders are formed when alkali metals are contacted with silica.  Such powders are more stable in dry air than ordinary alkali metals.  The reference suggests that electrons are imparted from the alkali metals into the silica forming an electride complex.

Reference

Shatnawi, M.G., Paglia, Gianluca, Dye, James, Cram, Kevin, Lefenfeld, Michael, Billinge, Simon, "Structures of Alkali Metals in Silica Gel Nanopores: New Materials for Chemical Reductions and Hydrogen Production", Journal of the American Chemical Society 129, 1386-92 (2007).  DOI 10.1021/ja067140e


Which easy-to-acquire clear liquid would be best for testing whether or not a twenty-sided die is balanced? In another question on another stack, I asked if the Saltwater float present in a specific youtube video was capable of determining any differences in mass between the faces of a die in determining the balance of a group of dice, and was met with very good results and a positive affirmation that yes, it was very possible to do so.

Which led me to believe that there was a possibility that denser liquids might have the possibility of testing the balance of the dice even better, with much less mess. This led me to go through and search the densities of many liquids to determine which would be best for the job.

Water came in at ~1g/mL at room temperature, which was the medium for the test initially.
Saltwater was at about ~1.028g/mL. Dice were able to float in this.
H2O2 (Hydrogen peroxide) clocks in at ~1.44g/ML, but I tested this one myself with a commercially available H2O2 and was unable to float dice in it just on its own. It isn't dense enough to float a die on its own.
Originally I thought Galium might be a good choice since it has a high density and is very easy to melt, but it isn't transparent.

So the question is this:

Is there a liquid, that is easy to acquire, that has a higher density than H2O + MgSO4 (Epsom Salt), that it would be easier to test a dies balance in and at what density does a d20 float?

Generally when referring to dice, the most well known manufacturer is probably Chessex. The dice are made out of some kind of plastic polymer of some sort. I don't really know any specifics about their chemical composition, but they could just be injection molded plastic dice.

As for the approximate mass of said dice.. This site states that the dice themselves are ~5mg (That doesn't sound right. I learned in school that a Paper clip is a gram, Dice weigh much more than a paper clip..)

You don't want a denser liquid. You want a liquid with just the right density so the dice would barely float, almost fully submerged. To this end, you mix the solutions of different concentration and so fine-tune the density, as is beautifully described in the accepted answer to your rpg.SE question.

A denser liquid is worse for your purpose. Ultimately, with a very dense liquid (like mercury) the dice would just sit on one face, much like they do on a tabletop.

In the relatively rare case your dice might be made of different kind of plastic and wouldn't float at all. That's when you genuinely do need a denser solution. Chemistry knows a lot of salts; which of these are available to you is another question. Saturated solution of $\ce{MgSO4}$ has the density of 1.3 or so; $\ce{CaCl2}$ will get you to ~1.4, and other salts farther yet, but they are also more difficult to come by.

As for the reference books with some density values, you have to read those really, really careful to see what they actually refer to, and then read again, just in case. 1.028 looks like the density of seawater; it is salty all right, but that's pretty far from saturation. The solution you use is a lot more dense. 1.44 for hydrogen peroxide looks like the density of the pure compound, which you surely can't acquire (which in turn is probably good for your survival); what you had is probably the 30% solution which is a lot less dense.

As for the density at which your d20 will float... well, how are we supposed to guess that? For all we know, your die may be carved out of wood; also, chances are it is cast in bronze. The resulting figures may vary pretty widely. Upd. So it is plastic, after all. Well, that still leaves a pretty wide room for wiggling. You can't tell the density until you measure it. Chemistry is an experimental science, they say.

So it goes.


M/P nomenclature for axial-chirality Synthesis and Chiral Resolution of C3-Symmetric Tribenzotriquinacenes, Eur. J. Org. Chem. 2016, Early View article

  DOI: 10.1002/ejoc.201600890

I came across this graphical abstract with M and P overlayed. After reading the article it became clear that M and P were being used as stereochemical descriptors, presumably of some kind of axial chirality/screw sense.

When I've previously looked, I've often seen either (+)/(-) (where + is the right-handed screw and - is the left) since this is actually describing the physical property of the molecule as measured, in the same way that (+)-proline describes the way in which the molecule interacts with plane polarised light. In many cases (R) and (S) are also used to describe axial chirality (BINAP, for example).

A quick google of the issue seems to suggest that (-) --> M and (+) --> P (cf. Wikipedia, confirmed by the IUPAC goldbook), but I was wondering if there is something more systematic going on here (i.e. a (+)-molecule doesn't always have an (R)-configuration).

In my head, the fact that there are descriptors like M and P (cf. E/Z and R/S) suggests that one should be able to look at the molecule and assign these by some kind of priority rule rather than by physical measurement (and by extension, does M/P actually predict +/- in a way that R/S doesn't).

The (+)/(-) notation refers only to the sign of the experimentally measured optical rotation. This notation alone does not tell you anything about which enantiomer you have.

The M/P notation refers to the spatial arrangement of atoms (see https://en.wikipedia.org/wiki/Atropisomer for an explanation).

What is the relationship between (+)/(-) and M/P? NONE.
You cannot say a priori if a certain compound for which you know the spatial arrangement of atoms will give you a positive or negative optical rotation.
Similarly, you cannot deduce the spatial arrangement of atoms from an optical rotation measurement.

There are methods to establish the relationship between the two descriptors, which means assigning the absolute configuration of a chiral molecule.

People have come up with rules to relate the two (e.g. binaphthyls, biphenyls), and these rules often work. More specifically, I am talking about the so-called Exciton Chirality method by Harada and Nakanishi (see DOI: 10.1021/ar50056a001).

The reasons why it works (often) or fails (sometimes) go probably beyond the scope of the question, but if you are interested I can maybe look for some paper explaining the problems.


Why can't iodide be oxidised to iodate in acidic medium? Even if we use a strong oxidising agent like 'permanganate ion in acidic medium' it doesn't help oxidise iodide to iodate, but a relatively weaker oxidising agent 'permanganate ion in alkaline medium' helps to form iodate from iodide.
I have found one discussion similar to this but it's not convincing.
Even in alkaline medium the potential is negative although significantly lesser than in acidic medium.
Then how does the reaction even proceed in alkaline but not in acidic?

Permanganate oxidizes iodide to iodine $\ce{I2}$ at all pH values. In acidic conditions, the reaction is finished here. But in basic solution, the iodine $\ce{I2}$ is transformed into iodide and iodate according to $$\ce{3I2 + 6 OH^- -> 5 I- + IO3^- + 3 H2O}$$The mixture  iodide + iodate reacts the other way as soon as the solution becomes acidic : $$\ce{5I^- + IO3^- + 6 H+ -> 3 I2 + 3 H2O}$$


How to explain the products of the nucleophilic aromatic substitution of 1-chloro-3-methylbenzene with NaOH? I guess this reaction is a nucleophilic aromatic substitution where $\ce{ONa-}$ is the nucleophile $\ce{Cl-}$ the leaving group, is that right?

Methyl group is a ortho-, para- director. So why should I have a meta product here? Don't we need an electron withdrawing group ($\ce{NO2}$ , carbonyl) in a NAS-reaction which is at ortho or para positions to the leaving group?

I am a bit confused about the shown picture.

What is going on here is a benzyne process. Under the conditions of intense heat the hydroxide removes one of the protons either side of the Cl. Cl- then leaves to create a triple bond, giving a benzyne. As you might imagine this is a highly reactive species and reacts with the OH- with no regioselectivity; either end can get hydroxylated and you have two possible benzyne intermediates. Thus you get the mixture of products.


Solution based measurement of Solvent-Accessible Surface Area of macromolecules The Solvent-Accessible Surface Area (SASA) is a valuable metric for looking at protein folding and protein-protein interactions. However, this measurement is typically done by calculating the SASA from a solved (and generally static) structure.

Chemical probes like diazirine and hydroxyl radicals show some bias regarding where they tend to bind. I'm realizing while I'm writing this question that NMR is a perfectly valid method to determine a solution based structure and then calculate the SASA. Similar strategies has also been used to examine structured RNAs. I'm curious about the variety of these methods and how accurate they are.

As you noted, the solvent-accessible surface area is a geometric concept, and as such is most easily calculated using computational geometry tools from a given molecular (or crystalline) structure. However you obtain this atomistic structure (X-ray, neutrons, molecular simulation, NMR, …) is of course of significant importance to the resulting area, but the calculation method will be the same.

There are, however, a few studies which have attempted to link the solvent-accessible surface area to directly measurable properties, and in particular to spectroscopic properties. A quick search of the literature reveals three examples:

Amide H/2H Exchange MALDI-TOF Mass Spectrometry
UV Raman directly on the exposed residues
Using diazirine as a probe


How does alkaline hydrolysis of peptides occur? I have been learning about zwitterions and the isoelectric point, and have been trying to apply this to hydrolysis of peptides. From the diagram in my textbook, I can see that when acid hydrolysis occurs, the amino acids formed from the hydrolysis act as bases because the pH is below their isoelectric point, so they accept the $\ce{H+}$ ions forming positively charged ions.
Now when it comes to alkaline hydrolysis, I don't understand why the peptide would not just be hydrolysed, and then act as an acid and donate the extra $\ce{H+}$ ion on its amine group to the $\ce{OH-}$ ions to form negative ions and a $\ce{H2O}$ molecule. Instead, in my text book it shows something like this:

$$\ce{-NH-CH(CH2SH)-C(=O)-NH-CH(CH3)-C(=O)- + NaOH \\-> -NH-CH(CH2SH)-C(=O)-O^{-}Na^{+} + H2N-CH(CH3)-C(=O)-}$$

I'm really sorry if this makes no sense whatsoever...

The unprotonated amino group ($\ce{R-NH2}$) does not protonate $\ce{OH-}$ because it is an extremely weak acid. Primary and secondary amines have a $pK_a$ of around 35, while the $pK_a$ of hydroxide is about 13-14 in aqueous solution. Deprotonating $\ce{NH2}$ groups requires much stronger bases in non-aqueous medium. Therefore, the carboxylic acid group of the amino acid is deprotonated in aqueous alkaline solution, as shown in the scheme from your textbook.


Calculating how many grams will I get from a reaction Question:

  $$\ce{2NH3 + CO2 -> (NH2)2CO + H2O}$$

  If we have $637.2\ \mathrm{g}$ of $\ce{NH3}$ and $1142.0\ \mathrm{g}$ of $\ce{CO2}$ how many grams of $\ce{(NH2)2CO}$ will we get?

This is the explanation I was given:

  First, convert reactives to moles:

  $637.2\ \mathrm{g}$ of $\ce{NH3}$ are $37.5$ moles.
  $1142.0\ \mathrm{g}$ of $\ce{CO2}$ are $26$ moles.

  Second, define the stoichiometric proportion between reactives and
  products...

  From $2$ moles of $\ce{NH3}$ we get $1$ mol of $\ce{(NH2)2CO}$
  From $1$ mol of $\ce{CO2}$ we get $1$ mol of $\ce{(NH2)2CO}$

  Third, calculate the number of product moles we would get if each
  reactive was completely consumed:

  Consuming all the $\ce{NH3}$ we would get $18.75$ moles of $\ce{(NH2)2CO}$
  Consuming all the $\ce{CO2}$ we would get $26$ moles of $\ce{(NH2)2CO}$

  The limiting reagent is $\ce{NH3}$, so the most we will get of
  $\ce{(NH2)2CO}$ is $18.75$ moles. Now convert that to grams, we get
  $1125.0\ \mathrm{g}$.

Why are we ignoring the existence of $\ce{H2O}$? Does it not matter?
IF the formula is unbalanced, do I have to balance it first before making this calculation? What if I can't balance it?

Here, $\ce{H2O}$ is a product and you are assuming that the reaction goes to completion, so it does not affect the yield of the reaction in any way. Maybe a suitable analogy is to think of the reaction system as a car - you give the car a certain amount of petrol (say 2 gallons) and it converts it to a certain number of miles (say 70 miles), as well as a lot of exhaust gases. Now the amount of exhaust gases that your car is putting out doesn't affect the fact that if you put in 2 gallons of petrol, you get 70 miles out of it.
If you were to treat the reaction as a reversible one, where you have an equilibrium between reactants and products, then in general, the amount of products present would also matter. But don't worry about this for now, I'm sure you will come across this in your study of chemistry at some point in time.
Good question - if the reaction is not balanced, then yes, you do have to balance it by finding the appropriate stoichiometric coefficients. If you don't know exactly how many miles per gallon (35 in this case) your car consumes, then you won't be able to work out how many miles you can drive if you feed your car 2 gallons of petrol. Not all reactions can be balanced, but if a reaction can't be balanced in one way or another, it can't happen in real life. For example, you can't balance the reaction $\ce{Pb -> Au}$, but as we all know, alchemy was (mostly) a load of rubbish. For this reason, any reaction that you come across in your study of chemistry, as well as in real life, can be balanced.


Melting and boiling point trend in Group II The following picture shows the melting and boiling point trends down group II elements.

I have added question marks where the variability in data was rather disturbing (over two hundred degrees Celsius), or where the value is simply disputed. Nevertheless, this topic is not about the different results from measurements. Instead, it focuses on the "trend" in general.

For group I, we would say

the number of electron shells increases,
the increasing number of protons is unable to compensate for this fact,
therefore, atomic radius increases and
electronegativity, ionisation energies decrease,
hence the metallic bond is weaker $\Rightarrow$ melting and boiling points decrease
atomisation energy decreases.

Here are the atomic radii and ionisation energies of group II elements.

These (apart from radium) do seem to follow the same logic as in group I, however they do not correlate with the melting or boiling points. Some textbooks claim,

  A good measure for the strength of a metallic bond is its atomisation energy.

Let us have a look.

Sure, this does agree better with the boiling and melting points. But note that it is actually not an explanation! Why do the atomisation energies behave in such a way?

This is where most resources either

ignore the problem altogether by not mentioning it,
give an explanation which is claimed by others to be faulty,
few acknowledge the issue and say no simple explanation is possible.

The faulty explanation

Beryllium and magnesium have crystal lattices that are hexagonal closest-packed (close-packed?). Therefore, they have their own trend. Then there are calcium and strontium which have face-centered cubic lattices. Finally, the pair barium and radium have body-centered cubic structures.

Others say this is not enough, because the differences should not be as significant, especially for magnesium but also for barium. Barium has a 8-coordinated structure but has a higher boiling point than 12-coordinated strontium!

It might also be stated that melting points themselves are not the best estimator of the strength of bonds. The bonds are not all broken, but rather rearranged and "made weaker" as a result.

The question

As of now, is there an explanation for melting and boiling point trends of group II elements? Is there consensus?

Its goes down because the attraction between electrons in sea of electrons and cations in the metal lattice becomes weaker therefore less energy is required to interrupt the lattice and easier to melt .


What results from the reaction of pent-3-en-2-ol with hydrogen bromide? I have seen the following reaction in my textbook:

This was quite easy to understand, though I wonder why 2-bromopent-3-ene is not a possible result:

Isn't this how $\ce{-OH^}$ groups and hydrogen halides interact? Any clarification would be appreciated.

It is very rare for a reaction to have no side products, and this reaction is certainly not exceptional in that regard. I'm actually a bit surprised your book chose this reaction to illustrate the regiochemistry of $\ce{HBr}$ addition to an alkene, as this would compete with the formation of an allylic cation, which are generally more stable than alkyl carbocations. Being more stable, these cations will form more quickly:

giving the product you correctly assumed would also be formed. You probably wouldn't obtain this though, as it would continue to react, forming a dibromoalkane (as would the products listed by your book):