How to make an alkaline wax mixture? I want to treat a wax so that it has an alkaline pH of at least 10. I tried just mixing sodium hydroxide with beeswax but the sodium hydroxide reacted with the beeswax and changed its properties. I want the wax to retain its natural properties and not be chemically altered by the pH-increasing component. How can I do this?

If it matters, the ideal wax for my purposes is carnauba wax which is a wax that comes from the leaves of a tree, but beeswax could work as well. It would be nice to have a method that would work for any kind of wax.

The following link from a very old text (pages 279-280) suggests boiling the wax in an alkaline solution. This method may be specific to that wax described as a new vegatable wax from a laurel like shrub from the forests of Para and Bahia. Please explain the application. Why do you want the wax to be alkaline? What are you going to do with the wax?


What is the viscosity of menstruation blood? I am writing my bachelor thesis about alternative materials for feminine hygiene products and for testing I need to know what the viscosity is of menstruation blood. Can I use water as replacement for menstruation blood because this kind of blood exists out of a big part of water? And what is the composition of menstruation blood in percentages?

It is difficult to give a typical value as the composition and hence the physical properties of menstrual blood varies over the course of a woman's period. This reference has a good discussion which you may find useful; you can find on p. 159:

  It is fair to say that a large proportion of menses samples are considerably more viscous than blood or water - about four times as viscous as venous blood and 35 times as viscous as water.


Help with derivation on why antibonding orbitals are more antibonding I have read How can antibonding orbitals be more antibonding than bonding orbitals are bonding?. But I am intersted in a specific derivation. During lecture my professor stated that the overlap $S_{ij}$ integral can be between -1 and 1. In other words,
$$-1<\int\psi^*_i\psi_j\,dx<1$$
He further mentioned that a negative overlap integral means that the orbitals are antibonding. He also mentioned that a positive overlap integral means that the orbitals are bonding, and that an overlap integral equal to 0 means that the orbitals are nonbonding. In other words,

$$S_{ij}<0\Rightarrow\text{antibonding}$$
$$S_{ij}>0\Rightarrow\text{bonding}$$
$$S_{ij}=0\Rightarrow\text{nonbonding}$$

We then solved the Schrodinger equation using the LCAO method. In this case, we were solving for a hydrogen molecule. We only considered the 1s orbitals of each hydrogen atom. In other words,

$$\Psi_{H_2}=c_1\psi_{1s}+c_2\psi_{1s'}$$
Here, $\psi_{1s}$ and $\psi_{1s'}$ are the atomic orbitals of each hydrogen atom.
The derivation that the professor used is similar to the one here: http://www.pci.tu-bs.de/aggericke/PC4e/Kap_II/H2-Ion.htm

In the end he got two wavefunctions with different coefficients
$$\Psi_{H_2}=\frac{1}{\sqrt{1+S_{ij}}}(\psi_{1s}+\psi_{1s'})$$
$$\Psi'_{H_2}=\frac{1}{\sqrt{1-S_{ij}}}(\psi_{1s}-\psi_{1s'})$$
He then mentioned that since $S_{ij}$ is always positive, the coefficient of $\Psi'_{H_2}$ is larger than $\Psi_{H_2}$. In other words,
$$\frac{1}{\sqrt{1-S_{ij}}}>\frac{1}{\sqrt{1+S_{ij}}}$$
And this is why antibonding orbitals are more antibonding than bonding orbitals are bonding.

But, the problem is this: if $S_{ij}$ can be negative (like he mentioned), then we have no guarantee that the coefficient for the antibonding molecular orbital will be larger than that of the bonding molecular orbital. How do we know that the $S_{ij}$ will always be positive in this case? Also, why does this statement not contradict the statement "$S_{ij}$ is negative for antibonding orbitals." What am I missing?

For the 1s orbital in hydrogen, the wavefunction is

$$\psi_\mathrm{1s} = N\mathrm{e}^{-r}$$

where $N$ is some normalisation constant. The normalisation constant is not unique: if $N$ is a satisfactory normalisation constant such that $\left<\psi_\mathrm{1s}\middle|\psi_\mathrm{1s}\right> = 1$, then for any value of $\phi$, the normalisation constant $N\mathrm{e}^{\mathrm i\phi}$ is also valid.

However you can make the assumption, without any loss of generality, that the wavefunctions of both 1s orbitals in dihydrogen ($\ce{H2}$) are of the same phase (i.e. both normalisation constants have the same value of $N$ and $\phi$). As a simple example we can just let $N$ be a positive real number. With this, the wavefunction of the 1s orbitals always have a positive value, and the overlap integral

$$S_\mathrm{AB} = \left<\psi_\mathrm{1s,A} \middle| \psi_\mathrm{1s,B} \right> = \int \psi_\mathrm{1s,A}\psi_\mathrm{1s,B}\,\mathrm{d}\tau$$

always has a positive integrand (no imaginary components, so I dropped the complex conjugate). The integral is therefore positive.

(Note that even though both 1s orbitals have the same exponential form, $\psi_\mathrm{1s,A}$ is not equal to $\psi_\mathrm{1s,B}$ since the orbitals are centred at different points in space.)

In fact, this choice of phase is required so that you can express the bonding and antibonding MOs as the specific linear combinations presented. The "in-phase" linear combination, i.e. the bonding MO, only has the form $N(\psi_\mathrm{1s,A} + \psi_\mathrm{1s,B})$ if you let the two orbitals have the same phase. If you had a different choice of normalisation constant, the bonding MO would be represented by a different linear combination $N(\psi_\mathrm{1s,A} + z\psi_\mathrm{1s,B})$ where $z$ is some complex number determined by the relative phases of the 1s orbitals.

As such, by virtue of expressing the MOs in that form, your overlap integral is necessarily positive.

Note that the above discussion only really applies to 1s + 1s overlap. With other orbitals, which have both positive and negative regions, you can end up with a negative overlap integral depending on the relative phases that you choose.

For example in $\ce{BeH2}$ if you consider the middle Be 2p orbital to overlap with both hydrogen 1s orbitals, one "overlap" will be formed with the positive lobe of the 2p orbital and the other "overlap" will be formed with the negative lobe of the 2p orbital. Assuming that you phase both 1s orbitals to be positive, then the first overlap integral will be positive (positive 1s times positive 2p gives a positive integrand) and the second will be negative (positive 1s times negative 2p gives a negative integrand).

In general your choice of phase is arbitrary; it does not affect the final physical result, and is simply chosen to make the calculations easier. That is to say, the eigenvalues (MO energies) that you get out of it will still be the same. The eigenfunctions may take on a different form, but that is physically unimportant.


What is the mechanism of chlorosulfonation of benzene? The above reaction looks very similar to a conventional sulfonation reaction but if the reaction pathway is analogous to sulfonation I do not see how the $\ce{SO_2Cl^+}$ would form. Surely $\ce{Cl^-}$ would leave rather than $\ce{H_2O}$ in the formation of the electrophile? Please help me find a mechanism.

Also, I would like to approach questions, such as this one, like I would a puzzle rather than calling upon my memory. Is there a series of steps that I should follow? (i.e: 1. Identify electro/nucleophile, 2. Identify acidic protons ...).

Rob, IMO this is a difficult problem, so don't feel bad about not "seeing" the solution right off.

Chlorosulfonic acid (or chlorosulfuric acid) is both difficult and hazardous to work with so it's reactions aren't discussed all that frequently.  It can also give rise to different products at different temperatures.  At higher temperatures it generates $\ce{SO3}$ which will behave as an electrophile and react with aromatic compounds to produce sulfonation products.

image source

At lower temperatures (your case), chlorosulfonic acid reacts by way of the following equilibrium (who knew this?)

$$\ce{3 ClSO2(OH) <=> SO2Cl^{+} + 2 SO3Cl^{-} + H3O^{+}}$$

The equilibrium generates $\ce{SO2Cl^{+}}$ which is the electrophile of interest in your reaction.  $\ce{SO2Cl^{+}}$ will react with an aromatic nucleus to produce the corresponding sulfonyl chloride by way of a typical electrophilic aromatic substitution reaction.

If you'd like to read more about the complexities of reactions involving chlorosulfonic acid see here starting around page 11.

The difficult part here was identifying the source of the electrophile. Looking at the final product you get an idea of what the electrophile is, but how is it produced?  Sometimes you have to do a little googling to sort things out.

Again, IMO this was a tough one.


Electrolysis rust removal - side products - identify lead my kids want to restore an old hand tool made of mostly iron.
I've choosen electrolysis as method for rust removal. We used Natron to enhance the conductivity of the water. The process worked so far and we get as expected some chemical side products swimming on top of the solution. We have brown stuff (iron) and from the wires some blue (silver or copper), and we have back stuff. The solution was pretty black afterwards as well as some scum swimming on top of it.
I don't want my kids to come in contact with (too much) lead. Is there an easy way to determine if the hand tool has lead in it?

Add the information regarding the composition regarding the electrodes. However lead is very heavy it would not like to float on the top. Secondly, lead compounds are mostly white with an exception of oxides (yellow/ red) and iodide (yellowish). Most likely the black floating scum is carbon in iron. Wear gloves, and wash your object with a mild abrasive such as a baking soda, and it should be fine. After all it was meant as a hand tool. It will not become "toxic" after a brief electrolysis.


What is the hybridisation of the carbons in the allyl radical? The example I am talking about is $\ce{CH2CHCH2^.}$.

To find the steric number and hence the hybridization, we use steric number  = No. of lone pairs + number of sigma bonds. In this example, the last C atom has only one electron. Should we count it as a lone pair?

My teacher says that that electron will get delocalised so its hybridization will be $\mathrm{sp^2}$ and not $\mathrm{sp^3}$ (which I think is true). I didn't understand the delocalisation stuff. Is it true that here, there will be delocalisation and if not then what is happening here? If delocalisation is occurring here then please explain it too! Also please tell me in which cases is there delocalisation of electron(s)?

My teacher also said that had it been $\ce{CH3CH2CH2^.}$, then there would be no delocalisation of electrons. Why does this happen?

Radicals are electron-deficient compounds. You should not treat a radical as a lone pair, as they are pretty different.

A good rule of thumb is to assume an orbital that contains the highest possible p-contribution for the single electron. The more s-contribution is mixed in, the ‘closer’ the single electron gets to the nucleus which is less favourable than having an electron pair in a similar orbital. A pure p-orbital fulfils the condition perfectly by having no s-contribution whatsoever. Thus, you should assume radicals to be in p-orbitals whenever possible.

This is similar to carbocations which have an orbital fully devoid of electrons. However, this empty orbital must always be a p-type orbital for electronic reasons. Radicals have the possibility of undergoing hybridisation and e.g. ending up in an $\mathrm{sp}^n$ orbital (examples: radicals on $\mathrm{sp^3}$ hybridised bridgehead atoms; phenyl radical).

Since we just established that the radical will occupy a p-type orbital and since there is a π bond nearby, we can delocalise. In resonance structure terms, this can be visualised in the following way:

$$\ce{H2C=CH-CH2^. <-> H2C^.-CH=CH2}$$

The radical is delocalised across two carbon atoms making the overall system more stable than if it weren’t. In molecular orbital terms, this is basically a stabilisation of the double bond at the expense of the radical-containing orbital which benefits the overall system. The delocalisation can, of course, only occur if there is a π bond nearby; the n-propyl radical does not feature one.


What does negative delta S imply? What does it mean when there is a negative delta $S$ $(-\Delta S)$? One person asked if this implied negative entropy, but I don't see how this could be possible.

Does this imply a negative change in entropy? If the former, how can negative entropy exist?

Negative delta S ($\Delta S <0$) is a decrease in entropy in regard to the system.

For physical processes the entropy of the universe still goes up but within the confines of the system being studied entropy decreases.

One example is a freezer with a cup of liquid water in it.  The freezer will utilize the electrical energy coming in to pump heat from the water until it becomes a solid (ice). At which point the entropy of the system (the contents of the freezer) decreases, however the electrical energy needed to be produced to power the freezer such as coal (burning a solid to a gas) and heat was wasted by the freezer in the process both of which create larger amounts of entropy than was reduced in the system by the freezer.

For chemical processes entropy can be a great driver of many reactions but it is not absolute. A system's favorability to release energy (enthalpy) competes with entropy. For example, an electron of hydrogen may have higher entropy if it drifts from the core proton but the electrostatic forces (and quantum mechanics) energetically keep it bound to the atom.  For isobaric processes, you much determine the change in Gibbs free energy for the reaction to know which way it is driven. For isochoric processes, you must determine the Helmholtz free energy to know which way a reaction is driven.

One example is the oxidation of iron in air.  When the oxygen is in the gas state it has higher entropy but the energy of bonding with iron is so great that at normal pressures oxygen goes from the gas phase and the iron rusts ("enthalpy wins") as delta G is negative.

Now we must consider statistical thermodynamics, this process is pressure dependent. At normal atmospheric pressure, the forward rate of oxygen entering the gas phase is the same as the reverse process. If the iron oxide were held in a sufficiently high vacuum the reverse process would occur and the iron oxide would reduce back to iron like many asteroids ("entropy wins"; note: thermodynamics is equilibrium after an infinite time). Accounting for pressure modifies the Gibbs free energy equation to: $$\Delta G = \Delta G^\circ -RT \ln(P) = \Delta H^\circ -T\Delta S^\circ -RT \ln(P)$$

One thing to note is that for chemical reactions the entropy and enthalpy values are for a standard temperature (such as $298\ \text{K}$). For a spontaneous system with $\Delta S^\circ < 0$ enthalpy must be negative, this heat in reality is absorbed by the system or the environment and produces entropy according to:  $$ \int \mathrm dS \equiv \int \frac{C_v}{T}\,\mathrm dT \equiv \int \frac{\mathrm dQ}{T} $$

This in and of itself produces some entropy in the universe though may not net above zero as the bonding energy is still the major driving force.


butane extraction (microbials) I was wanting to know using a butane extraction method for hemp flower (cannabis sativa) would also extract the microbial DNA present on the flower as well as deriving the cannabis oil (oleoresin)? This extraction process is used to obtain oleoresin(hash) from flower and I want to know if microbial DNA (or living microbes) can also be extracted? The goal of the extraction is to obviously have no microbes present in the end (this is for human consumption). If someone could explain the chemistry as to why it does or doesn't work I would appreciate it.

Thanks,

Kelsey

DNA contamination

DNA is water-soluble and non-volatile (large molecules). You would not expect it to be present in a butane extract. However, traces of DNA are difficult to remove. If you look for DNA (using PCR, for example), you will probably find it. After all, it is a forensic molecule of choice.

Contamination with living microbes

I don't think living cell would survive the extraction. Fungal spores might, though.

Human consumption

There are lots of legal foods where the presence of microbes is desired - yoghurt, beer, cheese are just some examples. All fresh food has its share of microbes, and healthy humans have them too - on the skin and in our digestive tract. Many of these microbes have a role in keeping us healthy, but some make us sick.

As the laws concerning your project differ with location, and the opinions on benefits and risks of your product range widely, I won't comment on those except to say that residual butane would concern me more than traces of microbial DNA.


Synthesis Golf I: Sodium Fluvastatin A full FAQ exists on meta.chem.SE explaining the premise of synthesis golf, why we're doing this, and the ground rules. Please read this before posting an answer.

The target for this first round of synthesis golf is sodium fluvastatin, an artificial statin drug used to treat hypercholesterolemia and in the prevention of cardiovascular disease. You must provide a synthetic route to the target molecule.

In addition, to narrow the scope of this question:

the synthesis must include a way of making the indole (i.e. no buying of the indole core and carrying out multiple functionalisation reactions).
all of the stereo centres must be set. Relative stereochemistry only is fine (the 1,3-syn relationship between the hydroxyls), but if you can think of a way to set them with absolute configuration, then great.

Edit: step count is 15, longest linear sequence of 9 steps

Just to (hopefully) get the ball rolling, here's something I scribbled down. I am sure people here will have better answers. And I have not done a lot of literature research to see whether the steps are plausible, so there might be fatal flaws... but hopefully not; as far as I know/can tell, everything here is relatively sensible.

Retrosynthesis first. I chose a Horner–Wadsworth–Emmons reaction to make the (E)-double bond, and something similar to the Reissert synthesis for the indole. It's not exactly the same because I didn't acylate the benzylic carbon, but the initial disconnection is similar. This leads to the key building blocks:

I chose the C(sp2)-C(sp3) bonds in 2 to disconnect. One can be made by SNAr with the amazing SNAr substrate 1-fluoro-2-nitrobenzene (5), and the other by Pd cross-coupling (I'm guessing cheaper ways probably exist). I've never actually seen SNAr reactions with enolates as nucleophiles, but I guess that if it doesn't work, SRN1-type conditions should still make that step possible:

As for 3, I figured it could be made from a symmetrical intermediate:

Forward synthesis

Synthesis of 3, starting from cyclopentadiene:

  (i) $\ce{BH3.THF}$, then $\ce{H2O2, NaOH}$; (ii) TBSCl, imidazole; (iii) $\ce{O3}$, then $\ce{H2O2}$; (iv) $\ce{TMSCHN2}$; (v) enzymatic desymmetrisation (I did not look up specifically which, but I think it should be possible); (vi) 7, NaH

Next, the synthesis of 2, starting from methylglyoxal:

  (vii) ethylene glycol, TsOH; (viii) Pd cat., ligand, $\ce{NaO^{t}Bu}$, 6; (ix) NaH, 5

Forming the indole:

  (x) Zn, AcOH; (xi) $\ce{NaH, ^{i}PrI}$; (xii) aq. HCl

and finally the HWE, the TBS deprotection, and the last redox step, a diastereoselective boron reduction. I wasn't entirely sure if it would work on an α,β-unsaturated ketone, but Comp Org Synth II, section 8.01 has several examples of it being used on an α,β-unsaturated ketones.

  (xiii) LiHMDS, 3; (xiv) $\ce{TBAF, H2O}$; (xv) $\ce{Et2B(OMe), NaBH4}$


Conversion atom to another One child has claimed to have find a solution to all physical problems. On asking for details, he said that all periodic elements has common components, i.e. electrons, protons, neutrons.
The child has suggested a solution: convert atom to another by adding electron. This way one can get substance like $\ce{H2O, Au, He}$ in abundance.
How can it be done?

I think you are looking for nuclear transmutation which occurs via nuclear reactions or radioactive decay, as explained in the referenced wikipedia page. As you can imagine then, implementation has some important impediments.


Disproportionation of Au(II) $\ce{Cu(II)}$ and $\ce{Ag(II)}$ complexes are known to exist while $\ce{Au(II)}$ is unstable and disproportionates to $\ce{Au(I)}$ and $\ce{Au(III)}$.

My guess as to possible reasons to this effect is that $\ce{Au(II)}$ like $\ce{Cu(II)}$ and $\ce{Ag(II)}$ cannot undergo Jahn-Teller Distortion, the cause of their stability, because of the extremely diffuse nature of $\mathrm{d}$ orbitals of $\ce{Au(II)}$ owing to relativistic effects. Could this be the right reasoning?

Also, what is the cause of the stability of $\ce{Au(I)}$ and $\ce{Au(III)}$ over $\ce{Au(II)}$?

I think this could be a counter answer. In contrast, I found a literature evidence that disproportionation of $\ce{Ag(II)}$ to $\ce{Ag(I)}$ and $\ce{Ag(III)}$ has had happens Ref.1). The abstract states that:

  Interaction of $\ce{Ag+}$ salts in anhydrous liquid hydrogen fluoride, $\ce{aHF}$, with $\ce{AgF4-}$ salts gives amorphous red-brown diamagnetic $\ce{Ag^IAg^{III}F4}$, which transforms exothermally to brown, paramagnetic, microcrystalline $\ce{Ag^{II}F2}$ below $\pu{0 ^\circ C}$. $\ce{Ag^IAu^{III}F4}$ prepared from $\ce{Ag+}$ and $\ce{AuF4-}$ in $\ce{aHF}$ has a tetragonal unit cell and a $\ce{KBrF4}$ type lattice, with $a = \pu{5.788(1) Å}$, $c = \pu{10.806(2) Å}$, and $Z = 4$. Blue-green $\ce{Ag^{II}FAsF6}$ disproportionates in $\ce{aHF}$ (in the absence of $\ce{F-}$ acceptors) to colorless $\ce{Ag^IAsF6}$ and a black pseudotrifluoride, $\ce{(Ag^{II}F+)2Ag^{III}F4^-AsF6-}$. The latter and other $\ce{(AgF)2AgF4MF6}$ salts are also generated by oxidation of $\ce{AgF2}$ or $\ce{AgF+}$ salts in $\ce{aHF}$ with $\ce{F2}$ or in solutions of $\ce{O2+MF6-}$ salts ($\ce{M = As, Sb, Pt, Au, Ru}$). Single crystals of $\ce{(AgF)2AgF4AsF6}$ were grown from an $\ce{AgFAsF6/AsF5}$ solution in $\ce{aHF}$ standing over $\ce{AgF2}$ or $\ce{AgFBF4}$, with $\ce{F2}$ as the oxidant. They are monoclinic, $P2/c$, at $\pu{20 ^\circ C}$, with $a = \pu{5.6045(6) Å}$, $b = \pu{5.2567(6) Å}$, $c = \pu{7.8061(8) Å}$, $β = 96.594(9)°$, and $Z = 1$. The structure consists of $\ce{(AgF)_n^{n+}}$ chains ($\ce{F−Ag−F} = 180^\circ $, $\ce{Ag−F−Ag} = 153.9(11)^\circ $, $\ce{Ag−F} = \pu{2.003(4) Å}$), parallel to $c$, that enclose stacks of alternating $\ce{AgF4-}$ and $\ce{AsF6-}$, each anion making bridging contact with four $\ce{Ag(II)}$ cations of the four surrounding chains “caging” them. There is no registry between the ordered array in one “cage” and that in any neighboring “cage”. The $\ce{F-}$ ligand anion bridges between the anions and, with the $\ce{Ag(II)}$ of the chains, generates a trifluoride-like structure. $\ce{(AgF)_2AgF4AsF6}$ [like other $\ce{(AgF)_n^{n+}}$ salts] is a temperature-independent paramagnet except for a Curie “tail” below $\pu{50 K}$.

However, I cannot find a counter example for $\ce{Cu(II)}$ complexes. Yet, $\ce{Cu(I)}$ is known to disproportionate to $\ce{Cu(II)}$ and $\ce{Cu(0)}$, which has limited $\ce{Cu(I)}$ chemistry. For example, if you react $\ce{Cu(I)}$ oxide ($\ce{Cu2O}$) with dilute sulfuric acid (hot), you'd get a brown precipitate of $\ce{Cu(0)}$ and a blue solution of $\ce{CuSO4}$ (a $\ce{Cu(II)}$ salt), because the disproportionation reaction dominates. You would not get the expected copper(I) sulfate solution from this reaction.

$$\ce{Cu2O + H2SO4 -> Cu + CuSO4 + H2O}$$

However, if you use certain ligands to form $\ce{Cu(I)}$ complexes, they would stabilizes the $\ce{Cu(I)}$ oxidation state. For example, both $\ce{[Cu(NH3)2]+}$ and $\ce{[CuCl2]-}$ are $\ce{Cu(I)}$ complexes, which don't disproportionate.

References:

Ciping Shen, Boris Žemva, George M. Lucier, Oliver Graudejus, John A. Allman, Neil Bartlett, "Disproportionation of$\ce{Ag(II)}$ to $\ce{Ag(I)}$ and $\ce{Ag(III)}$ in Fluoride Systems and Syntheses and Structures of $\ce{(AgF+)_2AgF4^-MF6-}$ Salts ($\ce{M = As, Sb, Pt, Au, Ru}$)," Inorg. Chem. 1999, 38(20), 4570-4577 (https://doi.org/10.1021/ic9905603).
Also read for the Disproportionation of$\ce{Au(II)}$: Khaldoon A. Barakat, Thomas R. Cundari, Hassan Rabaâ, Mohammad A. Omary, "Disproportionation of Gold(II) Complexes. A Density Functional Study of Ligand and Solvent Effects," J. Phys. Chem. B 2006, 110(30), 14645-14651 (https://doi.org/10.1021/jp062501y).


Validating a method to meet accuracy criteria using precision and accuracy? If a method has a precision of 10% and an accuracy of 95 to 105% is required, can this method be validated to meet accuracy criteria? Explain your rationale.

This question was posed by my professor and I'm struggling to come up with an answer.
I understand that precision refers to whether your method gives consistent repeatable results and that accuracy refers to how close your results are to the value you should actually obtain. However, I don't understand how a method's precision and accuracy could validate a method to meet accuracy criteria as the question asks.

Possibly, but the rational is a bit convoluted.

First, as you noted there is a fundamental difference between precision and accuracy. You explained the difference correctly.

Now let's think of a hypothetical example. Suppose that you have a pan balance in the lab and an analytical balance down the hall in a balance room.
So you are going to use the analytical balance (which is accurate and precise) to validate the pan balance.

You weigh a rock specimen on the pan balance and the weights have a precision of 10%. This means +/- 10%. So to get a better precision, you weigh the rock multiple times. Now the average weight will have a better precision than a single weighing. The average will be smaller by a factor of 1/sqrt(n). So 4 weighings will give you +/- 5%.

So you get a bunch of rocks that span whatever weight range that you would use. You weigh the rocks on the pan balance and on the analytical balance (which is assume to give the "true" weight). Now you can establish a calibration curve for the pan balance.

The calibration curve is going to have some error too, so you'll probably have to adjust how many repeat measurements and how many different samples you need to get the desired accuracy. But it should be possible in this example. The real twist here is how you "validate" the calibration.


During the development of aspartame, why was the methyl ester chosen? One frequent argument against the safety of the synthetic sweetener aspartame is that it can be hydrolyzed, with one of the hydrolysis products being methanol, which is known to be toxic. Of course, usually not much aspartame is added in food anyway to yield a sufficient amount of methanol or its oxidation product formaldehyde, but this does beg the question: why wasn't, say, the ethyl ester chosen to be developed as an artificial sweetener? Is there a structure-activity relationship between the alcohol used to esterify the dipeptide and the sweetness of the product?

This question raises more important issues than just the technical "why methyl ester," so I'll address those too.

The easiest explanation for their focus on the methyl ester is that the ethyl ester just isn't nearly as sweet. This report says it is approximately $10\times$ less sweet (see Table VI on page 2689 and the entry '$\ce{Asp-Phe-OEt}$' with "$++$" being defined on pg. 2684). It's an easy economic choice between the two, assuming the only difference is sweetness/g, but you're also worried about the perceived safety concerns of the methyl derivative.

A toxicology class I once took can mostly be summed up by a cliche phrase, "the dose makes the poison." Even too much of a good thing is by definition a bad thing, for example too much oxygen, water, or even the sun can lead to death. We also say that too little of any compound won't cause measurable harm.

It's correct to say that aspartame will be metabolized into amino acids and methanol. Methanol itself is toxic in high doses, in the same manner ethanol is, by acting as a CNS depressant. This acute poisoning can cause respiratory failure or other medical emergencies such as kidney failure. Ethanol poisoning occurs in microorganisms too: it has been proposed that some yeast produce ethanol as a means of eliminating competition. This would have been potentially deadly for a prehistoric mammal too, if it were not for an enzyme, which metabolize ethanol. This enzyme is called alcohol dehydrogenase (ADH), which leads to biochemical pathways that allow the safe removal of ethanol. However, methanol is a different case. ADH instead make formaldehyde, which is then converted to formic acid by ALDH. Formic acid, in large quantities, is used in the animal kingdom as a weapon, most notably in bee venom and from the bites of a fire ant. In low quantities it is fairly benign. Formaldehyde is just a nasty chemical. Both are produced in the body as a natural response to methanol and it's easy to see how too much of either couldn't possibly be desirable. The body also has ways to eliminate these toxic metabolites assuming you don't overwhelm these systems. Hence why a sufficiently small amount of methanol won't make you blind.

Researchers have measured the amount of these undesirable metabolites in the blood after having people, like me, ingest nearly $\pu{1.8 g}$ of aspartame in one sitting. Then they looked at urine and blood concentrations and they couldn't detect any formic acid in the blood but they could clearly see it in their urine for, up to, $\pu{8 hours}$ after a dose. This evidence suggests that acute methanol poisoning isn't going to happen after a ~$\pu{200 lbs}$ person ingests $\pu{1.8 g}$ of aspartame. Let's stop for a second and put this in terms of something we can imagine.

I don't have any reliable numbers, but let's just say that for a $\pu{12 oz}$ diet pop there are $\pu{200 mg}$ of aspartame. That means for every liter of pop, there are approximately $\pu{564 mg}$ of aspartame. That means $\pu{3 L}$ of pop would get me to $\pu{1.8 g}$ of aspartame. That means I would need to drink more than $\pu{3 L}$ in one sitting to overwhelm the excretion systems. This suggests an acute poisoning isn't very plausible.

There is, however, long-term toxicity to consider as well. This is a much more difficult question to probe. "Sure, there is no acute toxicity, but what happens in the long-term with a low dose?" For the answer to this question, we have to turn to toxicologists who have not reported long-term issues. They are looking too. The general consensus is that there is no long-term harm known for the general population. It is hard to quantitate these things, because to perfectly do the experiments necessary would require controlling a bunch of parameters that human experimentation laws prevent, let alone funding constraints. So, researchers make the best with what they have available and using these tools they cannot find any measurable problems with this particular compound in the long-term for the general population. Their analyses are limited by uncertainties in their measurements and this is just the general nature of science.

Tomorrow the majority of scientists in this area of interest could be proven wrong by a very careful experiment, but over forty years of research hasn't shown this chemical to be dangerous at the doses we're likely to be exposed to for any duration of time. We have to make decisions off of the best evidence we have available. I think it's okay for people to apply some level of skepticism to these things, because in science we're continually correcting our accepted knowledge base. Once in a while the general community is proven wrong about something long thought to be true and everybody benefits when that is the case, but we can't say the general community is wrong without substantial evidence. It is for this reason that we have the FDA to evaluate the best available evidence and via a network of reliable strangers, we can enjoy a very cheap/sweet drink that is low in calories. Frankly, I am more worried about the amount of sugar I take in than I'll ever worry about the amount of aspartame I consume -- we have already good evidence that sugar, with a dose too small to cause acute poisoning, is probably killing hundreds of thousands of people due to long-term exposure.


Oxidation of Unsymmetrical Ketones using Selenium Dioxide I have searched the whole internet but I could only see examples where the authors of the article took $\ce{R1}$ and $\ce{R2}$ as the alkyl groups around the carbon. This is regarding the oxidation of carbonyl compounds (I am specifically searching about ketones) using $\ce{SeO2}$. What I understood was $\ce{C=O}$ got introduced on the alpha-carbon and two hydrogens from that carbon were lost. Please tell me the product obtained on reacting pentan-2-one with $\ce{SeO2}$. Will the carbon having 3 alpha-hydrogens become $\ce{C=O}$ or will the other one having 2 alpha hydrogens become $\ce{C=O}$?
References:

Selenium dioxide oxidation of ketones and aldehydes. Evidence for
the intermediacy of .beta.-ketoseleninic, K. Barry Sharpless and
Kenneth M. Gordon Journal of the American Chemical Society,
1976, 98 (1), 300-301 DOI: 10.1021/ja00417a083
https://www.adichemistry.com/organic/organicreagents/seo2/selenium-dioxide-seo2.html

The mechanism of Selenium Dioxide (Riley) oxidation proceeds via the enol tautomer of the ketone (see scheme - image from Wikipedia). If the ketone is assymetric and can enolise to either side then the reaction should give the two di-ketones in proportion to the amount of either enol formed though I cannot find any reference where this has been experimentally demonstrated. In the OP's case of 2-pentanone, a mixture of the two possible products will be formed.


Is there a difference between anti-anxiety and anti-depression drugs? [PREFACE- I've already asked this on Health.SE, but I've drawn a blank there. The place is practically deserted. Also, this is NOT a personal medical question; I'm simply trying to understand the difference between the two classes of drugs.]

I've recently come across two terms "Anxiolytic drugs" and "Anti-depressant drugs". I wanted to know how they differ from each other; if they do.

I mean, sure, "anxiety" isn't the same as "depression" but do the effects of these drugs differ widely?

For example [And this is just an example];

Chlorodiazepoxide is classified as an anxiolytic drug, while Phenelzine is classified as an anti-depressant. Say I had a patient A who's been prescribed Chlorodiazepoxide; patient B has been prescribed Phenelzine by a different physician. Barring possible allergic reactions, or some kind of known sensitivity to either drug, could they possibly take each other's medication and hope for more or less, the same effect? As in; A takes Phenelzine, but B takes Chlorodiazepoxide. [I just cooked up that example, only to see the difference between the two classes of drugs]

To make myself clear, I reiterate:

I would like to know if the two classes of drugs (anti-anxiety and anti-depression) are the same or not. If not, mentioning how the chemistry behind the action of each of the two types of drugs differ would be appreciated.

[Basic]

Receptor: a molecule in a cell membrane that responds specifically to a neurotransmitter, hormone, antigen or other substance.
Agonist: Substance which initiates a physiological response when combined with a receptor.
Antagonist: Substance which inhibits the physiological response when combined with a receptor.
Allosteric: Substance which alters the activity of an enzyme by means of a conformational change induced by a different molecule.

[Response]

GABA receptors are related with anxiety [1]. Anxiety occurs because of low concentration of GABA in the brain. Therefore, Chordiazepoxide is an allosteric agonist.
Dopamine receptors are related with depression [2]. Depression occurs because of low concentration of Dopamine in the brain. Therefore, Phenelzine is a non-selective and irreversible monoamine oxidase inhibitor, which means that the brain does not recycle/degrade the monoamine neurotransmitters (Which dopamine is one of them).


What is the potential to use for non-dispersion interactions? Although liquid argon’s intermolecular potential can be described using a Lennard-Jones (LJ) potential, which takes into account dispersion forces (induced dipole-induced dipole interactions), what potential do you use when also including the effects of dipole-induced dipole and dipole-dipole interactions?

Because all three of these interactions are of the same order in the interelectronic distance ($1/r^{6}$), they are all captured within the Lennard-Jones potential to the extent of the accuracy of the potential. From the "Limits" section:

The term Van der Waals (vdW) forces roughly includes the following intermolecular forces:

  the permanent dipole-permanent dipole force (Keesom force), which is an electrostatic term,
  the permanent dipole-induced dipole force (Debye force), which is an induction or polarization term,
  the induced dipole-induced dipole force (London dispersion force), which is the dispersion term.

the forms of which are

$$
\begin{align}
V_{\text{Keesom}} &= \frac{-m_{1}^{2}m_{2}^{2}}{24\pi^{2}\epsilon_{0}^{2}\epsilon_{r}^{2}k_{b}Tr^{6}} \\
V_{\text{Debye}} &= \frac{-m_{1}^{2}\alpha_{2}}{16\pi^{2}\epsilon_{0}^{2}\epsilon_{r}^{2}r^{6}} \\
V_{\text{LJ}} &= 4\epsilon \left[ \left( \frac{\sigma}{r} \right)^{12} - \left( \frac{\sigma}{r} \right)^{6} \right]
\end{align}
$$

Note that the LJ potential contains both a repulsive ($1/r^{12}$) and attractive ($-1/r^{6}$) component. This is not the form of the dispersion interaction, but is both a suitable mathematical form that encapsulates a few noncovalent (vdW) interactions together and is easy to evaluate on a computer.

You may also be interested in the form of a more complicated potential, such as the AMOEBA polarizable force field, which uses a slightly different form for the pairwise vdW interaction ($i,j$ are atoms)

$$
\begin{align}
U_{\text{vdw}}(ij) &= \epsilon_{ij} \left( \frac{1.07}{\rho_{ij} + 0.07} \right)^{7} \left( \frac{1.12}{\rho_{ij}^{7} + 0.12} - 2 \right) \\
\rho_{ij} &= \frac{R_{ij}}{R_{ij}^{0}} \\
\epsilon_{ij} &= \frac{4\epsilon_{ii}\epsilon_{jj}}{(\epsilon_{ii}^{1/2}+\epsilon_{jj}^{1/2})^{2}} \\
R_{ij}^{0} &= \frac{ (R_{ii}^{0})^{3} + (R_{jj}^{0})^{3} }{ (R_{ii}^{0})^{2} + (R_{jj}^{0})^{2} }
\end{align}
$$

and also considers up to permanent and induced quadrupole moments

$$
\begin{align}
U_{\text{elec}}^{\text{perm}}(r_{ij}) &= M_{i}^{T} T_{ij} M_{j} \\
T_{ij} &= \begin{bmatrix}
1 & \frac{\partial}{\partial x_j} & \frac{\partial}{\partial y_j} & \frac{\partial}{\partial z_j} & \mathrm{L} \\
\frac{\partial}{\partial x_j} & \frac{\partial^2}{\partial x_i \partial x_j} & \frac{\partial^2}{\partial x_i \partial y_j} & \frac{\partial^2}{\partial x_i \partial z_j} & \mathrm{L} \\
\frac{\partial}{\partial y_j} & \frac{\partial^2}{\partial y_i \partial x_j} & \frac{\partial^2}{\partial y_i \partial y_j} & \frac{\partial^2}{\partial y_i \partial z_j} & \mathrm{L} \\
\frac{\partial}{\partial z_j} & \frac{\partial^2}{\partial z_i \partial x_j} & \frac{\partial^2}{\partial z_i \partial y_j} & \frac{\partial^2}{\partial z_i \partial z_j} & \mathrm{L} \\
\mathrm{M} & \mathrm{M} & \mathrm{M} & \mathrm{M} & \mathrm{O}
\end{bmatrix} \left( \frac{1}{r_{ji}} \right)
\end{align}
$$

along with the other typical bonded terms found in modern force fields.

Here is a nice set of notes that describes the interaction between electric multipoles to different orders.


Which of mono-, di-, triphenylmethane is the most acidic? Question

  How would you compare the acidity of the most acidic proton in each of the following molecules?

  diphenylmethane
  toluene
  triphenylmethane

The answer says 3 is the most acidic, followed by 1, and finally 2 is the least acidic. Can someone please tell me how to determine this? I know the benzene group is electron donating group, but am not quite sure how to correlate this knowledge with determining the relative acidities.

You should consider the stability of carbanion that is left behind, after removing $\ce{H+}$.   One could make arguments for resonance stabilization of each of the carbanions. You could "push" the minus charge around the phenyl rings, and there is a general rule that the more "good" resonance structures that one can draw, the more stable the molecule.

It turns out that the tertiary carbanion in 2) is flanked by three phenyl groups (thus many good resonance structures compared to 1) and 3) ).


Does lead carry the negative charge in plumbane? It says in Wikipedia that lead tetrahydride or hydrogen plumbide or plumbane ($\ce{PbH4}$) is supposed to have lead in its -4 state and the hydrogen is in its +1 state because supposedly lead has a higher electronegativity than hydrogen. Is this possible to be synthesized and actually be stable for long enough to at least view it?

$\ce{Pb^{4+}}$ is very unstable due to the inert pair effect, wherein for p-block elements lower oxidation state becomes stabler as we move down the group. Also, the main factor is the electronegativity of lead being more than hydrogen, which you stated in your question (although they are very close). Hence the oxidation states would not be flipped.


Explaining stereochemistry in a sigmatropic 1,3-alkyl migration I have quite a bit of difficulty in explaining sigmatropic rearrangement reactions. Consider the following reaction:

How do I explain the formation of the shown product? When simply moving the electron pair in my head, I come to the other enantiomer, where the $\ce{-CH3}$ group shows in our direction.

I've repeated all my notes on sigmatropic rearrangements but unfortunately, I don't get it. I suppose that I should treat this like an alkyl migration. The latter should happen over a suprafacial intermediate state as we have an uneven number of electron pairs (1) and we have thermal conditions. This results in a suprafacial reaction. Knowing this, I went on and added the HOMO and LUMO orbitals and postulated intermediate steps. Note: Yes, I am aware, that pericyclic reactions happen in a concerted fashion, but somehow I need to see, why the groups on the bridging carbon have this conformation.

So, let's go on to my question...

  Is my proposed explanation okay? If not, please provide some information. I really have trouble with this kind of reactions.

Problem #1 is that if you interchange the methyl and hydrogen groups, you get diastereomers, not enantiomers. But this is a relatively minor point. I assume your question is about the diastereoselectivity, not enantioselectivity (as ron rightly pointed out there is no enantioselectivity in this reaction).

The analysis you have drawn is somewhat similar to a Fukui frontier molecular orbital analysis. In that picture, for a reaction to be thermally allowed, you need to check for constructive bonding overlap between the HOMO and the LUMO, within geometrical reason. This is achieved by connecting lobes of "like" shading: i.e. shaded lobe with shaded lobe, and unshaded with unshaded.

You don't necessarily have to draw it stepwise (although I gather it's your own way of doing it). You can simply draw both interactions in the starting material, like this. The bottom interaction is not an issue. The top interaction, however, is an issue.

Problem #2 is that you tried to connect the large unshaded lobe of the σ* orbital with the white lobe of the π orbital. That is not geometrically reasonable: if you connect these two lobes, then you need to somehow end up with the new σ bond being formed underneath the ring. In other words, you would have to have an antarafacial migration of the alkyl group.

Problem #3 is that you have concluded that the reaction must be suprafacial based on electron counting. Note that in this context, the reaction being suprafacial means that the migrating alkyl group stays on the same face of the ring. In fact, the reaction must occur in a suprafacial manner simply because of geometrical constraints: there's simply no way that the alkyl group is going to migrate to the opposite side. Technically it could, but you would have to break so many bonds in the transition state that it would not be possible.

[Note that this has a different meaning from the suprafacial or antarafacial components in a Woodward–Hoffmann analysis.]

So what is the solution? Well, because the reaction is suprafacial, you need to use the shaded lobe of the π HOMO, which is on the top face of the ring – the same face as the migrating alkyl group. So then for constructive bonding you need to also use the shaded lobe of the σ* LUMO.

Now one might argue that just like how the unshaded lobe of the π HOMO was geometrically inaccessible, so should the shaded lobe of the σ* LUMO. But we already know that the C–C σ bond is going to break in the course of the reaction, so you're going to go through a pathway whereby the bond behind it (with an arrow pointing towards it) can rotate such that the shaded lobe of the σ* can overlap with the shaded lobe of the π* orbital. In fact, that's not too different from what you drew. The transition state is pretty similar to the intermediate in your stepwise reaction – it just has a few more dotted lines.

If you now "join" the shaded lobes then you can see how the correct stereochemistry arises.

In general my preferred way of analysing pericyclic reactions is to use Woodward and Hoffmann's original rules. Here, there is no shading of lobes: you connect the lobes within geometrical reason, then count the number of $4q+2$ suprafacial components, and the number of $4r$ antarafacial components. If the total is odd, the process is thermally allowed; if it is even, the process is thermally forbidden.

I am not going to explain this approach in detail - it is too long for here and it is adequately covered in textbooks which you should have access to - but I will upload the two diagrams I drew:


Determining order of a reaction without isolation I'm writing a computer program to calculate the order of a chemical reaction with respect to each reactant. When I do this by hand, it's easy to isolate which reactant concentration stays relatively the same while the others change. However, when you don't have an isolation method available, is there a non-guess and check way to determine rate orders?

For example, with the sample data of a 1st order reaction with respect to A and B, and second order overall

+--------+-----+-----+------+
| Trial# | [A] | [B] | Rate |
+--------+-----+-----+------+
|      1 |   1 |   1 |   10 |
|      2 |   2 |   3 |   60 |
|      3 |   3 |   2 |   60 |
+--------+-----+-----+------+

There isn't a trial with two constant values, so I'm not sure how you would programmatically determine the orders with respect to each of the reactants without manually trying every possibility until one fit (I realize that this wouldn't take too much computing time to do, but it won't scale well for multiple trials, and I also would prefer a non-brute force solution if possible). If there is no way to determine the rate with respect to each reactant, is there a way to determine the overall rate? This would greatly reduce the number of possibilities of orders when faced with larger rates

Although this is indeed a bit more a math problem than chemistry, I think it is useful to answer it, because this is a commonly encountered problem in reaction kinetics.

First of all, Yes it is possible to get the rates with respect to each reactant separately, as long as you have sufficient experimental data. With sufficient I mean: $1+\text{number of reactants}$. This is because you don't only need the order for each reactant, but also the rate constant for the total reaction

The rate equation

In general a rate equation will have the following structure:
$$ r = k [A]^\alpha [B]^\beta[C]^\gamma...\text{etc}$$
where with etc I mean that you can continue for as many reactants that you have. In your example you have 2 reactants so your rate equation in general will look like this:
$$ r = k [A]^\alpha [B]^\beta $$
To find the reaction order with respect to each reactant we need experimental data on $r$ for given $[A]%$ and $[B]$ like you provided.

The fitting procedure

You will have to write a short program that takes an initial guess for $k$, $\alpha$ and $\beta$ and checks how good or bad that guess is by comparing the calculated rates $R$ with the experimental rates $r$ for each set of $[A]$ and $[B]$. The normal way to make this comparison is based on least-squares:
$$S = \sum_{i=1}^N (r_i-R_i)^2 $$
So you sum up the squared differences between the reaction rate you found experimentally and the calculated rates for all $N$ experimental datasets.

To get the reaction orders and rate constant you want $S$ to be minimal, preferably 0 (although that normally doesn't happen with experimental data due to small measurement errors). Normally $S$ is not immediately optimal on your first guess for $k$, $\alpha$ and $\beta$, so you have to 'guess again'. You could indeed really guess again, but there are much smarter ways of determining the new guess, which will make sure your new guess is a better one than the old one. A common technique for this is called Newton's method. With the new guess you again check the value of $S$. You then repeat the cycle until you cannot find a lower value for $S$, which means that you have found the values of $k$, $\alpha$ and $\beta$ that are optimal for this experimental dataset and thus should be the correct reaction orders for the reaction.


Which d orbitals of sulfur take part in the pi bonds of SO3? In $\ce{SO3}$, 2 $p\pi-d\pi$ bonds are present. But which $d$ orbitals of
  sulfur take part in these $\pi$ bonds?

The answer says $d_{xy}$ and $d_{yz}$. Someone also told me that crystal field theory supports it as while splitting $d_{x^2-y^2}$ and $d_{x^2}$ are raised in energy compared to the other 3, but that theory has nothing to do here.

Unfortunately this is an issue that is, even though thoroughly disproved, still taught in many books and schools.

As Philip already pointed out in the comments: There is little to no contribution of the $d$ orbitals to the bonding in $\ce{SO3}$, as modern quantum chemical calculations prove (compare Wikipedia). This bonding scheme is a relic from a time where quantum theory was not yet widespread available. The little contribution of the $d$ orbitals (usually below 1%) we find is due to polarisation functions, that are added for a more accurate description of the molecule.

There are partial answers to the overall bonding in $\ce{SO3}$ in those two question:

Why is the bond order in the SO₃ molecule 1.33 and not 2?
Is the bond enthalpy of S-O stronger in SO₃ or SO₃²⁻?

But I will go a little further here. In the second post linked, you will find the result of a quantum chemical calculation for $\ce{SO3}$ on the BP86/cc-pVTZ level of theory. Here is the part of the output, containing the major contributions to the molecular orbitals (including the LUMO):

 Atomic contributions to Alpha molecular orbitals:
 Alpha occ 1 OE=-88.394 is S1-s=1.00
 Alpha occ 2 OE=-18.855 is O2-s=0.67 O3-s=0.17 O4-s=0.17
 Alpha occ 3 OE=-18.855 is O3-s=0.50 O4-s=0.50
 Alpha occ 4 OE=-18.855 is O3-s=0.33 O4-s=0.33 O2-s=0.33
 Alpha occ 5 OE=-7.922 is S1-s=1.00
 Alpha occ 6 OE=-5.945 is S1-p=1.00
 Alpha occ 7 OE=-5.941 is S1-p=1.00
 Alpha occ 8 OE=-5.941 is S1-p=1.00
 Alpha occ 9 OE=-1.113 is S1-s=0.39 O2-s=0.17 O3-s=0.17 O4-s=0.17
 Alpha occ 10 OE=-0.993 is O3-s=0.37 O4-s=0.37 S1-p=0.18
 Alpha occ 11 OE=-0.993 is O2-s=0.49 S1-p=0.18 O4-s=0.12 O3-s=0.12
 Alpha occ 12 OE=-0.585 is S1-s=0.24 O3-s=0.14 O4-s=0.14 O2-s=0.14 O2-p=0.11 O4-p=0.11 O3-p=0.11
 Alpha occ 13 OE=-0.502 is S1-p=0.55 O3-p=0.14 O4-p=0.14 O2-p=0.14
 Alpha occ 14 OE=-0.482 is O2-p=0.31 S1-p=0.20 O4-p=0.15 O3-p=0.15 O2-s=0.12
 Alpha occ 15 OE=-0.482 is O4-p=0.26 O3-p=0.26 S1-p=0.20
 Alpha occ 16 OE=-0.374 is O4-p=0.36 O3-p=0.36 O2-p=0.13
 Alpha occ 17 OE=-0.374 is O2-p=0.43 O3-p=0.21 O4-p=0.21
 Alpha occ 18 OE=-0.358 is O2-p=0.60 O4-p=0.15 O3-p=0.15
 Alpha occ 19 OE=-0.358 is O3-p=0.45 O4-p=0.45
 Alpha occ 20 OE=-0.305 is O4-p=0.32 O3-p=0.32 O2-p=0.32
 Alpha vir 21 OE=-0.141 is S1-p=0.45 O2-p=0.18 O4-p=0.18 O3-p=0.18

Now it is important to realise, that these numbers have to add up to 1 (accounting for rounding issues 0.99-1.01). So the reminding contributions are those of polarisation functions. Most notably missing is in orbital 15 the contribution of the O2-p orbital. The missing parts in 16-18 can be attributed to the diffuse and polarisation functions, i.e. higher $p$ and $d$-functions, 0.15; 0.15; 0.10; 0.10. As you can see, the involvement of $d$ orbitals in the total bonding is very little, almost negligible.

If you, however, consider $d$ orbitals to play a role in the bonding, then you have to consider, that all will participate.


E/Z stereoisomers : assigning priorities with a cycloalkane group For the following compound

Right hand side of double bond:
$\ce{CO2CH3}$  is highest priority due to the extra carbon (as we work along both chains)

But for the left hand side oft he double bond I am stuck.

I thought maybe we take the blue path, and as there is 3 carbons as opposed to 2 on the orange path, the blue path is the highest priority :

But my book gives the opposite answer :

There is a similar question here How to assign E/Z configuration according to the Cahn-Ingold-Prelog rules when subsituents differ only by stereochemistry   but in the case of my question the carbons on the cyclo which are directly after the double bond do not have groups attached.

The reasoning you have used is clearly faulty as there is no meaning in assigning more number of carbon to one end in a cyclic system.

Just look at the both the branches you have marked using orange and blue. The logic followed is in accordance to Cahn-Ingold-Prelog(CIP) rules.

The first atom in branches, marked as blue and as orange , have two hydrogen and one carbon each.So, there is nothing to clearly distinguish between the two branches.
So, we look at the only distinguishing feature between the two branches , which is the substituents on the carbon atom(marked as 2) attached to the carbon atom(marked as 1) on both sides.
We can see that the carbon atom marked as 2 on orange marked side has two carbon atoms and one hydrogen atom attached to it. This is of higher priority than the carbon on the blue marked side which has one carbon atom and two hydrogen atom attached to it.

Hence the Z-nomenclature as answered in the book.


Can an aqueous solution conduct electricity forever? We know that pure water does not conduct electricity, but salt water is a decent conductor. This is commonly explained by saying that “the ions carry the current through the solution”, which is an incomplete explanation, because it does not  indicate what happens when all of the ions have migrated to the electrodes.

More complete explanations of conduction through a salty solution (like this one or this one) rationalize the conduction of electricity in terms of a reduction reaction that takes place at the anode and an oxidation reaction that takes place at the cathode. In the case of salt water, chlorine gas ($\ce{Cl2}$) is formed at the anode and hydrogen gas ($\ce{H2}$) is formed at the cathode.

This explanation seems reasonable, but it implies that the conduction of electricity through a solution is fundamentally different than the conduction of electricity through a wire. A copper wire is (usually) unchanged, even after a large amount of electricity passes through it. In contrast, when electricity passes through salt water, two chemical reactions occur (one at each electrode), which fundamentally change the composition material.

This implies that it is not possible for an aqueous solution to conduct electricity forever. Since we are driving a chemical reaction, we are either consuming our salt (forming $\ce{Cl2}$ gas in the case of an NaCl solution or plating it onto the electrodes in other cases), or we are consuming the water by forming $\ce{H2}$ or $\ce{O2}$ gas.

This is surprising to me. So, I’m asking if my thinking is correct: is it possible for an aqueous salt-containing solution to conduct electricity forever, or it will it always eventually consume the reactants and stop conducting electricity as I have surmised?

Since nobody has posted a counter example, I will go ahead and say that my claim is true for direct current (which is what I originally had in mind). The current drives a chemical reaction and, at some point, the reactants will run out and the solution will no longer have enough ions to effectively conduct electricity.

Hopefully someone will post a counter-example and make things more interesting :)

In the case of alternating current, it seems possible to plate material onto one electrode and remove it from the other. When the current switches direction, this process would reverse and the electrodes would remain the same mass. In theory, this could go forever, but in reality I wonder if the electrodes would deteriorate and break due to the many cycles of deposition and erosion.


exchanging H2O in buffer for D2O - pH vs pD? I have TBS pH 7.4 buffer that I am exchanging all the water out for $\ce{D2O}$. How do I find out what the pD is after all the water is gone? I understand that pH meters don't work for deuterium, and it seems like deuterium also changes the spectral properties of the chromophores in pH paper (or in this case pD paper)

Also, if I want to best mimic the alkaline conditions in $\ce{D2O}$ as $\ce{H2O}$, should pD final = pH initial?

The articles by Glasoe,[1] Covington et. al.,[2] and (much more recently) Krężel and Bal[3] suggest in their intros that pD values can be obtained from a pH calibrated electrode by adding a correction constant of about $0.4$. This was determined by comparing the measured pH when the same amount of acid or base was dissolved in $\ce{H2O}$ and $\ce{D2O}$.
References:

P. F. Glasoe, Citation Classic - use of glass electrodes to measure acidities in deuterium-oxide. Current Contents/ Physical Chemical & Earth Sciences 1979, 16, 12-12. A1979HZ34800001 (pdf) Original paper: P. K. Glasoe, F. A. Long, J. Phys. Chem. 1960, 64,:188-190. Doi:10.1021/j100830a521
A. K. Covington, M. Paabo, R. A. Robinson, R. G. Bates, Anal. Chem. 1968, 40 (4), 700–706. DOI: 10.1021/ac60260a013
A. Krȩżel, W. Bal, J. Inorg. Biochem. 2004, 98 (1), 161-166. DOI: 10.1016/j.jinorgbio.2003.10.001 Available at researchgate.net.


Pipetting strong acid with plastic material Can I pipette concentrated sulfuric acid using a plastic pipette tip? Will either the plastic or the acid be affected?

This depends on type of plastic. You definitely can if plastic is perfluoropolyetylene (Teflon). You probably can if plastic is polyethylene or polypropylene, but both will degrade in long term (polypropylene will do it faster). You should not, if plastic is PET or other polyester, or polystyrene.

It is probably most reliable to find which plastic you dealing with and search for 'chemical resistance chart' for this plastic.


What exactly is hydrogen bonding (and does it really need fluorine, oxygen, or nitrogen)? I'm not satisfied with the rationale for the intermolecular attraction known as hydrogen bonding. In my book, it states that

  Hydrogen bonding is a special type of intermolecular attraction between the hydrogen atom in a polar bond (particularly H ¬ F, H ¬ O, and H ¬ N) and non-bonding electron pair on a nearby small electronegative ion or atom usually F, O, or N (in another molecule).

It seems that chemists looked at the data and found they needed a 'fudge factor' to fit the higher boiling points of ammonia, water, and hydrogen fluoride.

Why doesn't hydrogen bonding apply to other atoms like Sulfur or Chlorine? They seem to be electronegative enough IMO. Is there an explanation that explains why this is, rather than saying that it's there? What makes Hydrogen so special?

Note: Chlorine is more electronegative than Nitrogen.

There are two main ways to look at hydrogen bonding.  The first is electrostatic, where the electronegativity of the atoms is used to describe the interaction.  Your argument about chlorine being more electronegative than nitrogen is a good one suggesting that the electrostatic argument is only part of the story, and there is at least one study that suggests that hydrogen bonding does occur with chlorine in polar molecules such as chloroform.

We can increase dramatically the number of atoms considered to undergo hydrogen bonding if we take a molecular orbital approach.  My reasoning here is a summary of what can be found in Inorganic Chemistry by Miessler, Fischer and Tarr.  They, in turn, rely heavily on a new definition of hydrogen bonding recommended by the IUPAC Physical and Biophysical Chemistry Division.

A hydrogen bond is formed when an $\ce{X-H}$ (where X is more electronegative than H) interacts with a donor atom, $\ce{B}$.  The attraction $\ce{X-H...B}$ can be described as consisting of the following components:

An electrostatic contribution based on the polarity of $\ce{X-H}$
A partial covalent character which arises from the donor-acceptor nature of the interaction
Dispersion forces

The first bullet is typically the only phenomenon discussed in General Chemistry classes, and this is not unreasonable since the second bullet requires the introduction of Lewis Acid/Base concepts which may not have been covered.

It is important to note that the new definition includes what I like to call "the proof is in the pudding" where the existence of hydrogen bonding requires experimental evidence, which can be found using a number of methods:

The $\ce{X-H...B}$ bond angle: a bond angle of 180 degrees indicates strong hydrogen bonding and would be accompanied by shord $\ce{H...B}$ bond distances.
A red shift in the IR frequency upon formation of $\ce{X-H...B}$
Hydrogen bonding results in deshielding of the H atom, which can be observed in NMR spectroscopy.
Thermodynamic studies should indicate that $\Delta G$ for the formation of $\ce{X-H...B}$ should be larger than the thermal energy of the system.

While electrostatics is the dominant contributor to hydrogen bonding, an analysis of frontier orbitals can also be insightful.  A thorough answer requires a fair amount of MO theory background, but for those interested, looking at the MO diagram of $\ce{FHF-}$ will be helpful.  In brief, the base ($\ce{F-}$ in this case) has a filled $p_z$ orbital that gains access to a delocalized orbital with lower energy, thus validating the formation of a strong hydrogen bond.


How to find the formula of an binary oxide based on the ionisation energies? I am a little confused about this multiple choice question and what the requirements are to answer it. It is for a year 11 chemistry student that I am helping and I think I am going in a little too much detail to try to solve it to help them. I am sure that I am overthinking this.

  The energies needed to remove electrons in succession from a neutral
  atom of a certain element $\ce{A}$ are given ($\mathrm{kJ\cdot  mol^{-1}}$): $1010$, $2032$, $7056$, $8120$, $9300$.
   The formula  of the oxide of $\ce{A}$ is:
   $\ce{AO}$, $\ce{A2O}$, $\ce{A2O3}$, $\ce{A2O4}$ or $\ce{A3O5}$.

My understanding of the question and what I was trying to do to solve the problem by looking at the ratios between the energies and I was trying to apply
1st ionization energy: $$\ce{X -> X+ + e−},$$
2nd ionization energy: $$\ce{X+ → X^2+ + e−},$$
and 3rd ionization energy: $$\ce{X^2+ -> X^3+ + e−}$$
to the energies below and find a ratio. I thought the ratio was going to be c. I understood that oxide is $2-$, however, I couldn't calculate the $\ce{A}$ value as I didn't know what the element was.

The point of this problem is to test if you can see where there is a jump in ionization energies. The third ionization energy is considerably larger than the others. Although ionization energies will naturally increase in succession, this is the most profound increase, which would suggest that this atom will tend to lose 2 electrons before obtaining a stable octet structure.

This would mean that its ion would be $\ce{A^{2+}}$. Since the oxide ion is $\ce{O^{2-}}$, the charges will be balanced if the ratio of the ions is 1:1, meaning the answer is $\ce{AO}$.


A hint for multiple choice question Give me a hint to solve this question,

what should my first step ?

should convert the grams into moles ?

aluminum= (54g/150g)= 36% ( percent composition of aluminum)
sulphur= 64%

0.36x=20 (36 percentage of what is 20)

x=20/0.36
x=55.5 grams

is there a more easy way ? to do in mind

You might consider writing a correctly balanced equation, and then convert grams to moles.


Do acids really donate a proton? Will acids really donate a proton? In an atom, the proton is inside the nucleus and I don't think a proton will go all the way out of atom dodging all electrons and enter into the nucleus of the base.

It seem impossible but the Brønsted–Lowry theory says acids donate protons.

I am confused—could someone explain what is actually meant by acids donating protons?

Yes, according to the Arrhenius theory, acids dissociate in aqueous solution and release a proton ($\ce{H+}$). The Brønsted–Lowry defines acids  ($\ce{HA}$) and bases ($\ce{B}$) in such a way that their interaction is characterized by the exchange of a proton according to $\ce{HA + B <=> A- + HB}$.

However, this is about the reaction of molecules in which real hydrogen atoms are bound to other atoms.

To illustrate the idea, let's think about the reaction between acetic acid, $\ce{CH3COOH}$, and benzylamine, $\ce{C6H5CH2NH2}$. If we combine both liquids in our imagination, the following will happen:

$\ce{CH3COOH + C6H5CH2NH2 -> CH3COO- + C6H5CH2NH3+}$

The acid donates a proton to the amine, which serves as the base. In the course of this reaction, the acid is converted to its corresponding base, while the base is converted to its corrsponding acid.

You are confusing dissociation and proton exchange with a type of radiocative decay known as proton emission.


Estimating protein binding and disassociation I don't have a background in the area of drugs or pharmacokinetics/Pharmacodynamics but I am trying to understand about protein binding. I was going through this paper.

If $C_b$ is the concentration of the bound drug

$C_f$ is the concentration of the free drug

$C$ is the total drug concentration

$P$ is the concentration of protein binding sites (bound and free)

$K_d$ is the disassociation constant of the drug protein complex

For a one compartmental model the free drug can be found through:

$C_f=\dfrac{-(P+K_d-C)+\sqrt{(P+K_d-C)^2+4 K_d C}}{2}$

The assumptions used in this derivation are:

Binding occurs only to plasma proteins and follows simple
saturation kinetics. The binding process can be described by a single
macroscopic dissociation constant.
Binding equilibrium is achieved virtually instantaneously with
respect to distribution and elimination.
Aside from binding, all other processes (ie., distribution and
elimination) are linear.
Distribution and elimination processes operate only on free
drug.

My question is in how to determine the parameter values of $P$ and $K_d$.
The article has given certain properties of the drug but not a way in how these parameters are found.

The given details on a drug are:

1.The drug has a molecular weight of 150.

Drug binds only to serum albumin, whose concentration is 4.4% and whose molecular weight is 67,000.
At sufficiently low drug concentrations, the drug is 89% bound.
The distribution volume is 50 ml./kg.
Free drug has a half-life of 30 min.

Using these details is there a way to estimate the parameters of $P$ and $K_d$ and if so please explain how they can be derived.

If we call the drug L, and the binding sites S, the equilibrium is the following:

$$\ce{LS <=> L + S}$$

The equilibrium constant expression is:

$$K_d = \frac{[\ce{L}] [\ce{S}]}{[\ce{LS}]}$$

With $[\ce{L}]_\mathrm{tot}$ as the total concentration of ligand and $[\ce{S}]_\mathrm{tot}$ the total concentration of binding sites, we can solve of $[\ce{L}]$, the free ligand concentration to arrive at a quadratic expression like the one in the question.

If the ligand concentration is small compared to the concentration of available binding sites, you can estimate the ratio of free to bound ligand:

$$ \frac{[\ce{L}]}{[\ce{LS}]} \approx \frac{K_d}{[\ce{S}]_\mathrm{tot}}\tag{1}$$

Now, we can look at the statements 1 through 5 and see what information we can extract to determine $K_d$ and $[\ce{S}]_\mathrm{tot}$:

1.The drug has a molecular weight of 150

This information is not directly relevant. I'll get back to it later.

2. Drug binds only to serum albumin, whose concentration is 4.4% and whose molecular weight is 67,000

This allows us to calculate the concentration of protein capable of binding to the drug. 4.4% refers to 0.044 kg of albumin in a solution with a volume of one liter. With a molar mass of M = 67000 g/mol, we get the following concentation:

$$c(\mathrm{albumin}) = \frac{\pu{44 g / L}}{\pu{67000 g/mol}} = \pu{0.66 mM} $$

If we knew the number of binding sites for each molecule of albumin, this would also give us the concentration of sites.

3. At sufficiently low drug concentrations, the drug is 89% bound

If 89% are bound, 11% are free, and the ratio of free to bound is 0.12. "At sufficiently low drug concentration" means the drug concentration is low with respect to the concentration of binding sites, so we can use equation (1):

$$\frac{K_d}{[\ce{S}]_\mathrm{tot}} \approx \frac{[\ce{L}]}{[\ce{LS}]} = 0.12$$

So we can estimate the ratio of dissociation constant and total concentration of sites.

4. The distribution volume is 50 ml/kg

This means for every kg of body weight, the apparent volume in which the drug dissolves is 50 mL. Picture a solvent of unknown volume. You dissolve a solute and then remove some solution to test the solute solution. You should be able to calculate the unknown volume from the amount of solute added and the concentration obtained. If, however, the solute absorbs on the container wall or diffuses through it, your answer is inaccurate.

The human body has an average of 70 mL of blood for each kilogram of body mass. A substance that remains in the blood stream should have a distribution volume of 70 ml/kg if it does not bind to anything.

I'm not sure what to do with this piece of information.

5. Free drug has a half-life of 30 min.

This ensures that the binding reaction has a chance to almost attain equilibrium. If this were not the case, we could not use equilibrium calculations, and would have to resort to non-equilibrium methods such as the ones described in the paper cited by the OP (which is behind a paywall, so I just read the abstract).

  Using these details is there a way to estimate the parameters of P and Kd and if so please explain how they can be derived.

The missing piece of information is the concentration of sites. If we knew the number of sites on each albumin molecule, we could calculate this concentration because we know the concentration of albumin.

Experimentally, what we are missing is data for saturating conditions. As it is, we have one equation for two unknowns, so we only know their ratio. If you assume one binding site per albumin molecule, you would be done. However, looking at the molar masses, it is entirely possible for more than one drug molecule to bind to a single albumin molecule.


What substances do humans consume that are caloric but neither protein, carb, nor fat? On the Medical Sciences StackExchange site I asked Can a calorie be neither protein, carb, nor fat?
 and got a very helpful answer, which was that ethanol (in alcoholic drinks) is caloric but neither a protein, nor a carb, nor a fat.  I then asked a follow-up question as a comment: Where can I find a list of all such substances?  That is, substances that are caloric but neither protein, nor carb, nor fat.  The person who answered my question recommended I post the question to this SE site.

If you go to https://eur-lex.europa.eu/legal-content/EN/TXT/PDF/?uri=CELEX:01990L0496-20081211&qid=1580028914722&from=EN you will see a fairly extensive list of caloric compounds (screenshot pasted below).  This list is probably not comprehensive -- likely there are many compounds we (which includes our gut bacteria) can metabolize for calories.  Rather, it's limited to componds that are caloric and approved for use in food.  [It also doesn't include compounds that are naturally found in foods, and may be caloric, but are only present in very small quantities, e.g., nucleic acids.]

You can see that, in addition to fat, proteins, carbohydrates, and ethanol, the table lists polyols, organic acids, salatrims, and fiber.

Some explanation of the table may be helpful:

Polyols are artifical sweeteners. They include lactitol, maltitol, mannitol, sorbitol, xylitol, erythritol, glycerol, hydrogenated starch hydrolysates,  and isomalt. [Though note that while erythritol is a polyol, it is listed separately as non-caloric.  And while glycerol, aka glycerin or glycerine, is a sweetener, it is also used in food for other purposes.]

Organic acids include acetic acid (the main component of vinegar, other than water), as well as citric acid, ascorbic acid, and malic acid (the latter three are found in citrus fruits).

Fiber refers to dietary fiber, which is the type of fiber we can digest completely.

Salatrims are "short and long chain acyl triglyceride molecules"; they are a type of low-calorie fat substitute.

Source:  Consolidated text: Council Directive of 24 September 1990 on nutrition labelling for foodstuffs (90/496/EEC) Select: 6 CELEX number: 01990L0496-20081211 Author: Council of the European Union Date of document: 11/12/2008


Metal that does not deposit silver (Ag) Hi Chemistry enthusiasts!

I am working on my scholarship exam practice but I am considered myself as a newbie so I would like to hear your opinion :)

  From (1)-(4) below choose the metal that does not deposit silver (Ag)
  on the surface when immersed in aqueous silver nitrate. (1) Cu (2) Fe
  (3) Pt (4) Zn

The answer is Pt but I am not sure how I can solve this problem without any other given information such as electrode potential etc. Please advise.

You do need the electromotive series or reactivity series to answer that question. If you will not have access to it during such exams, you'll need to memorize the order of some common elements, and to understand the arrangement of the Periodic Table.

The most reactive metals are at the bottom left of the table, low in Group I. The "noble" or coinage metals are clustered near the center, with reactivity generally decreasing within a column. E.G., Au is less reactive than Ag, which is less reactive than Cu.

Later, you may learn about the electron configuration of atoms and how that determines reactivity, but at this point, knowing the reactivity series would get you though the test.


Using enthalpies of formation and combustion to find formation enthalpy of Acetylene (Hess' Law) I understand this is a simple question (I’m in high school), but for some reason the numbers just are not working for me.

Given:
$$\Delta H^\circ_\text{c} = -1301\ \mathrm{kJ}\ \text{for}\ \ce{C2H2(g)}$$
$$\Delta H^\circ_\text{f} = -394\ \mathrm{kJ}\ \text{for}\ \ce{CO2(g)}$$
$$\Delta H^\circ_\text{f} = -242\ \mathrm{kJ}\ \text{for}\ \ce{H2O(g)}$$
Calculate $\Delta H_\text{f}$ for $\ce{C2H2(g)}$.

My process:
$$(1) : \ce{C2H2(g) + 5/2 O2(g) -> 2CO2(g) + H2O(g) }\quad \Delta H^\circ_\text{r} = -1301\ \mathrm{kJ}$$
$$(2) : \ce{C(s) + O2(g) -> CO2(g)} \quad \Delta H^\circ_\text{r} = -394\ \mathrm{kJ}$$
$$(3) : \ce{H2(g) + 1/2O2(g) -> H2O(g)} \quad \Delta H^\circ_\text{r} = -242\ \mathrm{kJ}$$

$$\text{Target:}\ \ce{2C(s) + H2(g) -> C2H2(g)}$$

After some experimentation I arrive at:

$$-1\times(1) + 2\times(2) + (3) = \text{Target}$$

As shown through:
$$-1\times(1){:}\ \ce{2CO2(g) + H2O(g) -> C2H2(g) + 5/2 O2(g)} \quad \Delta H^\circ_\text{r} = 1301\ \mathrm{kJ}$$
$$2\times(2){:}\ \ce{2C(s) + 2O2(g) -> 2CO2(g)} \quad \Delta H^\circ_\text{r} = -788\ \mathrm{kJ}$$
$$(3){:}\ \ce{H2(g) + 1/2O2(g) -> H2O(g)} \quad \Delta H^\circ_\text{r} = -242\ \mathrm{kJ}$$
$$\text{Sum:}\ \ce{2C(s) + H2(g) -> C2H2(g)} \quad \Delta H^\circ_\text{f} = 271\ \mathrm{kJ}$$

Now as far as I know, all the givens are accurate empirical values. But the empirical value for $\Delta H^\circ_\text{f}$ for $\ce{C2H2(g)}$ is $227\ \mathrm{kJ}$.

What’s even weirder is that this disparity $(271-227=44)$ is equal to the difference between standard enthalpies of formation for water in gas and liquid states $(-242 - -286 = 44)$.

Have I made an error somewhere, or is water formed in the liquid state during the combustion of acetylene?

Like many past chemistry.se questions, the answer to this one hinges on the presence of the $^\circ$.

$\Delta H^\circ_\text{c}$ is the $\Delta H$ value for a "standard" reaction where acetylene at 1 atm of pressure is combusted isothermally at 298 K with 1 atm of oxygen gas, forming 1 atm of $\ce{CO2}$ and liquid water (which is the most thermodynamically stable form of water at 1 atm and 298 K).

I think it's safe to say that no combusion of acetylene, ever in the history of the world, has occured under those conditions.  But those "standard" conditions are the only conditions where the  $\Delta H^{\circ}_f$ is guaranteed to be -227 kJ/mol.

By the way, in practical and engineering literatures, $\Delta H_c^{\circ}$ of a fuel is commonly called the higher heating value of the fuel.  It corresponds to $\Delta H_{rxn}^{\circ}$ of a reaction like

$\ce{C2H2(g) + 5/2 O2(g) -> 2CO2(g) + H2O(l)}$

The lower heating value, conversely, corresponds to $\Delta H_{rxn}^{\circ}$ of reactions like

$\ce{C2H2(g) + 5/2 O2(g) -> 2CO2(g) + H2O(g)}$


Propagation of uncertainty I was taught that when adding/subtracting measurements, add the absolute uncertainties, and when multiplying/dividing measurements, add the relative uncertainties.

I have a problem in a lab where I need to find the difference between two measurements that are very close together, and are each very precise; however, when I propagate the absolute uncertainties, I get a huge relative uncertainty for the final result (upwards of 50%). Here it is:

(0.050 ± 5%) - (0.044 ± 3%) = ?
= (0.050 - 0.044) ± ((0.050 * 5/100) + (0.044 * 3/100))
= 0.006 ± 0.00382

This ± 0.00382 corresponds to 63.7% relative uncertainty; am I doing something wrong? Both of the original measurements were pretty accurate (<12% RU). In this case can I just add their relative uncertainties instead of converting to absolute, which is very large in comparison with the difference between the measurements?

There is nothing wrong with the ±0.004 uncertainty you obtained. The relative uncertainty of the difference between close results is supposed to be far larger than that of the original measurements. One way of seeing why is stretching your scenario a bit: suppose you got 0.048 rather than 0.044 as the result of the second measurement. The difference, 0.002, would be small enough to be meaningless, as the second result would encroach into the uncertainty range of the first measurement. Your example is not so extreme; still, the large relative uncertainty is justified.

(Aside: an analogous effect can affect computer calculations done with limited precision. Suggestively enough, it is referred to as catastrophic cancellation).


How is combustion speed regulated in solid propellants? My question on Space.SE didn't yield anything I didn't know already: the secondary regulation mechanism – grain (cross-section shape) of the propellant, changing the area of open surface of the propellant as it burns away; the combustion occurs only on the surface, travels inwards only as much as the surface burns away, and produces uniform amount of exhaust gasses per unit of area of open surface; the variance in surface area is what changes the overall thrust.

But I'm currently interested in what is done to composition and/or inner structure of the propellant, so that it burns uniformly on the surface, at predictable (and quite slow) pace, instead of, say, exploding all at once, or at least deflagrating at shocking speed, like common explosives (of very similar composition!) do. What property or reaction keeps the combustion to the surface and makes it progress inwards only so slowly – in case of the Space Shuttle, 290 seconds to burn through about 1.5 m of the propellant.

(I apologize for being too lazy to find all of the citations for my answer.  However, I have a fair amount of experience as an organic chemist, at one point worked in an underground mine with explosives and have built (for entertainment) a solid fuel composite propellant)

Why does propellant "burn" instead of detonating like similar explosives?

A typical solid composite propellant looks like this:  (1) mainly ammonium nitrate (oxidizer) + "fuel" (typically aluminum flake or magnesium) + binder (polymer + perhaps plasticizer) + (maybe) catalysts OR substitute ammonium perchlorate for ammonium nitrate.

The most similar mining explosive is ANFO or variants thereof:  ammonium nitrate + diesel fuel (possible sensitizing substitutions are... some ammonium perchlorate for the ammonium nitrate, some aluminum in addition to fuel, partially oxidized "fuels" such as hexamine nitrate, nitromethane, possibly a small percentage of a secondary explosive such as TNT or RDX, or addition of plastic "bubbles" to simulate the bubbles that sensitize pure nitroglycerine.

The key point is that the latter mining explosive is called a blasting agent.  The taxonomy of high explosives looks like this:  primary explosives can be easily DETONATED (as opposed to deflagration, which is a type of "burning", detonation involves a supersonic thin region of reaction that moves through the explosive, giving it brisance that can shatter rock).  Primary explosives are used in primers, and can be detonated with sparks, physical shocks, etc.  Secondary explosives (dynamites, TNT, PETN) need to be detonate with a primary explosive.  They are relatively insensitive to other means of initiation (one measure of this is the deflagration to detonation distance, which really is the amount of explosive that can burn before detonation takes place.  It is usually quite large).  You can often shoot or burn secondary explosives without them detonating.

Finally, blasting agents are quite hard to detonate.  They require a booster that is a secondary explosive.  Blasting agents are very insensitive and can be burned without detonating, especially if not completely confined.

This answers your first question...  we should not expect solid propellants to detonate without a booster (actually I'd guess that solid propellants are even harder to initiate, as ANFO variants are usually oxygen balanced so that all of the fuel is oxidized, whereas propellants appear to be a bit carbon rich).

Note that blasting agents and solid propellants share the property that the oxidizer and fuel form distinct particles, as opposed to explosives such as nitroglycerine, which are self-oxidizing within the molecule.  The low explosive gunpowder shares the property of having a mixture of discrete particles of oxidizer and fuel, and it merely deflagrates.

TLDR:  The explosives that are most similar to solid composite propellants are difficult to detonate, so it isn't a surprise that they burn instead.

==============

How is the burn rate controlled?  (note, I use burn as burn typically is decomposition at a rate of a few cm per second, while deflagration is at a rate of many meters per second.  Detonation is thousands of meters per second).

A:

All propellant mixtures have (approximately) a burn rate that is a function of pressure.  This approximation is called Vieille's Law.  The formula is

$rate = {rate}_0 + aP^\beta$

where $P$ is the pressure.  Thus, as pressure increases in the rocket motor, burn rate increases.  Ideal values of $\beta$ are between 0.5 and 0.8, otherwise initiation can be difficult or else a slight overpressure could be amplified to cause the pressure to increase dramatically and destroy the rocket.  What seems to affect the burn rate?  I'll list a few:

(1)  Ammonium perchlorate burns much faster than ammonium nitrate.

(2)  Generally, the rate limiting property of a propellant is the particle size of the oxidizer, so more finely divided oxidizer means faster burn rate.

(3)  The geometry of the propellant is important.  The burning tends to occur at the exposed surface, so wide and not very long propellant castings will produce more thrust.

(4)  Metallic catalysts increase burn rate, such as iron oxides and chromates.

(5)  I've heard that a small amount of carbon black increases burn rate by increasing infrared absorption in the casting.  It also makes it look cooler.

(6)  Not so good:  cracks and voids in the casting will increase burn speed, as they lead to more exposed area.  That's why typical binders are rubbers, often including plasticizers, as these are less likely to crack under pressure.

Here's a couple of citations:

A amateur rocketry site

metallic wires as a burn rate catalyst

Rocket engine mathematics


What is meant by "no d-orbital"? Sulfur and oxygen belong to the same group. Sulfur has a vacant d-orbital while oxygen has no vacant d-orbital.

What does having "no d-orbital" mean? Orbitals are just the spaces around the atom. How can you say that there is no space around the atom which makes the d-orbital?

There is a bit of a philosophical debate as to whether orbitals exist only when they're populated, or if they're always there. In both oxygen and sulfur, there is no occupied $d$ orbital in the ground state (so both $3d$ orbitals are vacant), but in sulfur the promotion energy of an electron from a $3s$ or $3p$ orbital to a $3d$ orbital is much less than the promotion energy of an electron in the $2s$ or $2p$ orbitals in oxygen to a $3d$ orbital ($2d$ orbitals don't exist, of course). Arguably, this means that sulfur can access its $3d$ orbitals under the right conditions since the promotion energy required is relatively low and could be supplied in chemically relevant situations (so the sulfur $3d$ orbital is accessible).

It is not expected that an oxygen atom could ever populate its $3d$ orbital in a stable substance. It is possible to occupy an oxygen $3d$ orbital for a short while, however, by exciting the atom with a photon of the proper frequency.


Finding equilibrium constant from standard free energy This question is asking to calculate equilibrium constant for oxidation of glucose from thermodynamics quantities. So the first expression came to my mind was $-\Delta G^\circ=RT\ln K$. I found the value of $\Delta G^\circ=-2\,278.78$ but as I proceed I've got $\ln K=1\,161.94$ which I couldn't reach to the $K$ value. Is there any other ways to solve this question?

$$\ce{C6H12O6(s) + 6O2(g) <=> 6CO2(g) + 6H2O(l)}$$

$\Delta G^\circ_\mathrm r=-2878.78\ \mathrm{kJ}$

At equilibrium, $\Delta G=0$; $Q=K$,

$\Delta G=\Delta G^\circ+RT \ln K$

$2878.78\times10^3\ \mathrm J=\left(8.314\ \mathrm{J/(mol\ K)}\right) (298\ \mathrm K) \ln K$

$\ln K=1161.93596$

Answer from the book is $K=5\times10^{504}$

Your problem seems to be one of calculation using large numbers, so look at it this way:
$$
\ln K = 1161.93596
$$
So
$$
\log_{10}K = \ln K \log_{10}(e) = 504.65\\
\Rightarrow K = 10^{504.65} = 10^{0.65}\times 10^{504} = 4.5 \times 10^{504}
$$
without your calculator exploding.


Why is boiling point of Ammonia greater than Arsine? I have previously read that the boiling point of Stibane(SbH3) is greater than Ammonia(NH3) as ammonia is gas at room temperature and no hydrogen bonding exists in the gaseous form of NH3.
Hence, as SbH3 is greater in size, has greater molecular weight; its boiling pint is higher than NH3.
why does Stibane have a higher boiling point than Ammonia
In the case, of AsH3, even this molecule is gas at room temperature and its boiling point should be greater than NH3, but this is not the case and NH3 has a higher boiling point than AsH3.
I want to know what is the reason behind this trend. (NH3 having a higher boiling point than AsH3)

There is the general trend in the groups 15, 16, 17 of raising of boiling points for the binary compounds with hydrogen, going down the groups.
But the first members of each group - $\ce{NH3, H2O, HF}$ - have anomally with their boiling points being exceptionally high, due hydrogen bonds. The strength of these bonds and the boiling point decreases in the order $\ce{H2O > HF > NH3}$.
See the figure 2 on wordpress page for an illustrative picture:

Hydrogen bonds play role in (not limited to):

evaporation and boiling, as molecules need to gain an extra energy to overcome these bonds and to escape liquid.
condensation, acting as cohesive force for molecules to stick to each other or to a condensation center.
gas pressure, as the inter-molecular cohesive force decreases the gas pressure, compared to the pressure of an ideal gas at the same conditions.

Note that the fact that both NH3 and AsH3 are gases at room temperature is irrelevant for the above.


Resonance and reductions with sodium borohydride I know that sodium borohydride doesn't work particularly well on resonance-stabilized carbonyl groups (i.e. esters) but what about this carbonyl in the center of the above picture? This appears to be somewhat resonance-stabilized ... why would sodium borohydride still work  here? Sodium borohydride is a nucleophilic reagent. On the other hand the second resonance contributor I've drawn is pretty crappy; yes - it's part of an aromatic system - but it's also a carbocation. And the R' group is inductively donating (although I suspect this is a fairly minor point).

Or does it just work in the above synthesis but not with a yield comparable to something such as sodium borohydride and formaldehyde?

Sodium borohydride is definitely strong enough to reduce an acetophenone derivative. It is likely that the resonance stabilization does reduce the reactivity of the ketone. It's not often that a sodium borohydride reduction requires 35 hours. In the cases that I have run, usually less than one hour is sufficient.

I did some quick searching of whether an isolated ketone could be reduced in the presence of an acetophenone, and I did not find an adequate example of a case where only the isolated ketone was reduced. That says to me that the reactivity difference is not that great. Perhaps someone else has come across this though.


Is cis dipole moment is always greater than trans dipole moment? In trans compound the dipole moment always cancels each other!  Then how can it's dipole moment be greater than cis compound's dipole moment?

No.

The dipole moment of trans-cyclooctene ($\pu{0.82 D}$) is greater than cis-cyclooctene ($\pu{0.43 D}$).
See On the Molecular Geometry of trans-Cycloöctene, J. Am. Chem. Soc. 1958, 80 (8), 1953–1955.

  Blomquist, Burge and Sucsy2 have called attention to the fact that while with the cyclodecenes, the cis isomer has the higher density and index of
  refraction as is typical in acyclic systems, with the cyclooctenes, as first noted by Ziegler and Wilms,3 the reverse is true. From these refractive indices, by application of van Arkel’s dipole rule,4 the prediction could be made that the trans-cyclooctene would have a higher dipole moment than the cis.

  [...]

  The dipole moments of the cis and trans isomers of cyclodecene were previously determined,4 and were found to be normal. The conclusion, drawn from the dipole rule, that the trans-cyclooctene would have a higher dipole moment than the cis represents a situation without analogy.

  [...]

  The cis isomer was found to have $\mu_{cis} = \pu{0.43 \pm 0.06 D}$ which may be compared with the value found for cis-cyclodecene of $\pu{0.44 D}$.4 The trans isomer was found to have a dipole moment of $\pu{0.82 \pm  0.03 D}$, compared with the value zero found for trans-cyclodecene.4

(Van Arkel's rule is: The isomer with the higher dipole moment possesses the higher boiling point, refractive index, and density.)


Regarding preservatives in food Can sugars like fructose-glucose or high-fructose corn syrup that are found in many types of processed food contribute to shortening a products shelf life?  Don't bacteria love things like sugar so if a product has a lot of fructose-glucose added to it, it will need even more preservative chemicals to compensate for any increase of sugar?

Two factors may be at play. (1) While bacteria and molds do indeed grow on sugar, a very high concentration of sugar inhibits growth. So fruit preserves, syrup, honey, molasses etc. do not need to be refrigerated and generally do not require other preservatives to be added. Soft drinks contain less sugar and are more susceptible to spoilage, hence the use of preservatives. I would definitely refrigerate a soft drink after opening it, if not consumed immediately. Acidity also helps inhibit growth (e.g. sweet pickles).

(2) Another factor is that bacteria and molds require water for growth, so foods such as dried fruits, that contain a moderate level of sugar but only a small amount of water, may have a long shelf life.


What is the principle of diffuse reflectance spectroscopy behind band gap measurement? As the title suggest can you explain how DRS can be used for finding band gap of a material. Below is an DRS output result:

From the figure we can see that at about 465 nm the reflectance drops steeply with respect to wavelength and attain a plateau at around 420 nm. Using above data we can assume that the band gap is around 2.76 eV (450 nm).

Can you explain this figure in  physical sense (i.e. with respect to the material)?
Why do we take the energy of a ~450 nm photon as the band gap? What is the rationale behind it?

Reflectance spectroscopy is very closely related to UV/Vis spectroscopy, in that both of these techniques use visible light to excite valence electrons to empty orbitals. The difference in these techniques is that in UV/Vis spectroscopy one measures the relative change of transmittance of light as it passes through a solution, whereas in diffuse reflectance, one measures the relative change in the amount of reflected light off of a surface.

A solution that is completely clear and colorless has essentially 100% transmission of all visible wavelengths of light, which means that it does not contain any dissolved components that have (allowed) electronic transitions over that energy range. By the same line of reasoning, a white powder effectively reflects 100% of all visible wavelengths of light that interacts with it.

However, if the material has electronic energy levels that are separated by an energy in the visible region, then it may absorb some of light energy to move electrons from the filled energy level (valence band) into this empty level (conduction band). This causes an relative decrease in the amount of light at that particular energy, relative to a reference source. In other words, the % transmission/reflectance will decrease.

Your data suggests that the separation between the filled valence band and the empty conduction band is around 2.76 eV. Because when light of that energy or higher interacts with the material, it is absorbed resulting in an excitation of the electrons from the valence bands into the conduction band. This lowers the relative reflectance of the material (i.e. lower % transmittance).

I don't this that entirely answered all your questions, but I hope it helps a little.


Acid-Base Reaction in MacGyver Episode "Countdown" This question is about the old MacGyver show, Season 1, Episode 14, "Countdown." Could someone please explain how/why MacGyver neutralizes the acid in the bomb? He claims it is an electrolyte switch, and proceeds to neutralize it with oven cleaner and milk before draining it out.

wouldn't the milk act as a buffer?
wouldn't the neutralized solution also contain electrolytes?

Thanks!

The answer to your first question is yes, to the second maybe.

You are right that milk can act as a buffer, as explained for instance by this page from the Food Science Department of the University of Guelph:

  This leads to the concept of buffer capacity, which is an important principle in cheese making. The effect of protein removal on the TA of whey, is related to the ability of protein to 'buffer' the milk against changes in pH. That same buffer property is the reason it helps to take acidic medication, like aspirin, with milk.
  [...]
  The two most important buffer components of milk are caseins (buffer maximum near pH 4.6) and phosphate (buffer maxima near pH 7.0). The buffer maximum near pH 5.0 is extremely important to cheese manufacture because the optimum pH for most cheese is in the range of 5.0 - 5.2. As the pH of cheese is reduced towards pH 5.0 by lactic acid fermentation, the buffer capacity is increasing (i.e., each incremental decrease in pH requires more lactic acid). The effect is to give the cheese maker considerable room for variation in the rate and amount of acid production. Without milk's built in buffers it would be impossible to produce cheese in the optimum pH range.

On the other hand denaturation of the milk proteins might act to form a gel that could perhaps help stabilize the trigger fluid, but milk gelation appears to display a complex pH dependence and a mechanism of stabilization is unclear.

The electrolyte concentration might be diluted by milk, since its ionic strength is surprisingly moderate, see Ref (1) for instance:

  Taking into account these different associations, the diffusible fraction of milk at pH 6.6–6.7 appears to be supersaturated in calcium phosphate (Tab. V), and to have an ionic strength of about 80 mM.

Much of the salt is bound to protein which forms part of the buffering system. So much depends on the composition of the sinister trigger fluid.

The substantial concentration of divalent calcium (~1 g/L or ~50 mM) might be useful, but how will presumably also remain a well-kept secret of the writers privy to the composition of the very unstable trigger electrolyte fluid.

Reference

Frédéric Gaucheron. The minerals of milk. Reproduction Nutrition Development, EDP Sciences,2005, 45 (4), pp.473-483.