Why does formaldehyde exist primarily as the gem-diol in aqueous solution? From Wikipedia,

Methanediol is the product of the hydration of formaldehyde $\ce{H2C=O}$, and predominates in water solution: the equilibrium constant being about $10^3$ and in a 5% by weight solution of formaldehyde in water, 80% is in the methanediol form.

As far as I know, carbonyl compounds usually don't exist in their gem-diol form, unless there's a stabilizing effect (like H-bonding, release of angle strain, inductive effect etc.), which we observe in ninhydrin, chloral, hexafluoroacetone, etc. (relevant image) But such effects don't seem to be the case for formaldehyde.
My question: What is the stabilizing factor in the case of hydration of formaldehyde?

In acetone the carbonyl form is more stable than the gem-diol, whereas with formaldehyde the situation is reversed and the gem-diol form is preferred. Overall, the equilibrium shifts by 6 powers of 10 between these two compounds.

It takes $\pu{\approx1.4 kcal/mol}$ to shift an equilibrium by 1 power of 10. Therefore in the acetone-formaldehyde comparison there is a stabilization change of $\pu{\approx8.4 kcal/mol}$.

It is difficult to see how changing the substituents in the gem-diol could have such a large effect, after all the gem-diol carbon is a saturated  $\mathrm{sp^3}$ carbon, no p-orbitals are avaliable for resonance interaction. Considering steric effects as proposed in the other answer, methyl groups and hydroxyl groups are not particularly large, nor is the gem-diol carbon especially congested. Hence it is difficult to see how steric factors could provide the necessary $\pu{8 kcal/mol}$ change in stabilization. For reference, Cis-di-t-butylethylene only has $\pu{9.3 kcal/mol}$ of strain energy.(1)

On the other hand we might expect the substituents on the carbonyl carbon (an unsaturated $\mathrm{sp^2}$ carbon with a p-orbital available for interaction) to have a significant effect on stability. Just as alkyl substituents stabilize a carbon-carbon double bond through both an electron-releasing inductive effect (2) and a hyperconjugative effect, we would expect the same effects to operate with a carbon-oxygen double bond.  In fact, we might anticipate that the alkyl stabilization effect will be even greater with the carbonyl double bond since the carbonyl carbon is more electron deficient than the carbon in a carbon-carbon double bond.

If we compare the heats of combustion for the isomers acetone ($\pu{-427.9 kcal/mol}$) and propanal ($\pu{-434.1 kcal/mol}$), we see that attaching a second alkyl group to the carbonyl carbon stabilizes the system by $\pu{6.2 kcal/mol}$.(3) Of course, adding the "first" alkyl group will also provide significant stabilization. Hence, it is clear that adding two alkyl groups to formaldehyde will provide enough electronic stabilization (inductive and hyperconjugative) to shift the equilibrium by $10^6$ from the enol to the keto form.

see here
the alkyl group has an $\mathrm{sp^3}$ carbon while the double bond carbon is $\mathrm{sp^2}$ hybridized; since the latter is more electronegative, electron density will flow inductively (through sigma bonds) from the alkyl group to the $\mathrm{sp^2}$ carbon.
data from K. Peter C. Vollhardt, Neil E. Schore, "Organic Chemistry, Fourth Edition: Structure and Function" p. 125; see here


Why do NMR spectrums have the higher chem shift values closer to x-axis? Im not understanding what the intuition is behind why the graph is represented so that the higher chem shift value is closer to x-axis (shouldnt it be the other way around?)

I understand that downfield = deshielded and will have higher ppm value which is closer to x-axis, but i dont understand why this is reversed to how an x/y graph is usually represented.? Thanks for any help

This convention stems from historical reasons only. In the early days of NMR, there were a number of different ways to report NMR data; values for 'shielding constants' of nuclei was of most interest, rather than what we commonly refer to as chemical shifts. The two most common scales were the tau and delta scales . The tau scale had the chemical shifts running in the opposite direction, with increasing shielding constants running from left to right.

A third scale, the sigma scale was very commonly used, especially in heteronuclear analyses, and again had the shielding constant increasing to the right. This scale was favoured often by theorists, as it was usually a direct measurement of the shielding constant.

A mixture of tau and delta values was regularly seen in the literature up until IUPAC made their initial recommendations, in 1970. This was to adopt the delta scale, and report chemical shift relative to TMS at delta 0ppm, as values of chemical shift, which are dimensionless and, hence, field independent.


What is the dimension of Avogadro's constant? What is the dimension of Avogadro's constant ($N_\mathrm{A}$).
On Wikipedia it says it is dimensionless, but in Nigel Wheatley's article (pdf) On the dimensionality of the Avogadro constant and the definition of the mole it says it is $\mathsf{N}^{-1}$.
\begin{align}
\text{Number of particles} &= N_\mathrm{A}\times \text{Amount of substance}\\
[\text{Amount of substance}] &= \mathsf{N}\\
[\text{Number of particles}] &= \mathsf{1}
\end{align}
If $[N_\mathrm{A}] = \mathsf{N}^{-1}$ then the above equation follows, otherwise it doesn't. Is this the right value?

The dimension/unit of the Avogadro constant $N_\mathrm{A}$ is actually $1/\mathsf{N}$ or $\pu{mol-1}$. It is shown that way in your first reference as well, i.e.
$$ N_\mathrm{A} = \pu{6.02214076E23 mol-1}. $$
What is "dimensionless" is the Avogadro number or the number of atoms / molecules in a single $\pu{mol}$ of that substance, sometimes written as $N$ or $N_0$.
The numerical value is the same, but the concept is different. Basically the relationship between the two is:
$N_0 = (\pu{1 mol})\times N_\mathrm{A} = \pu{6.02214076E23}$
See also the answer discussed in the comment by Ed V, where the new definition of the constant is explained: Effects of Changing Avogadro's Constant.


Suitable safe cleaning solution for aliphatic urethane I am looking for a suitable cleaning solution for a sprayed on coating that is an acrylic aliphatic urethane. My end goal would ideally be something that can neutralize rust, whiten and brighten the surface, and also include a powerful enough solvent that can loosen stuck on dirt/soap scum but not harm the urethane.

The urethane is approximately 70% solids (don't know if that matters or not). I'm not a chemist so layman's terms would help greatly.

Oxalic acid is good at removing rust (iron oxides) and generally not agressive against polymers like polyurethanes.
Chemical resistance of polymers is however a complicated area, because polymers comes in a wide spectrum of formulations, additives etc. Some indications can be found on different manufacturers websides


Does any known substance ignite on cooling? As the title says, I'm interested in knowing if there is any substance — or combination of substances — that ignites (or even increases its chance of spontaneous ignition) when cooled.

I've never heard of such a thing, nor can I find it in Atkins' Physical Chemistry, but I might easily have overlooked it, since I don't know what to call it. Googling gives me information on substances that lower the freezing point or ignition point of explosives/fuels.

It seems that this should obviously be forbidden on thermodynamic grounds, but I can't quite rule out a phase change allowing such behaviour, for instance.

Actually... yes! Iron(II) oxide is thermodynamically unstable below $848~\mathrm K$. As it cools down to room temperature (it has to do it slowly) it disproportionates to iron(II,III) oxide and iron:

$$
\ce{4FeO -> Fe + Fe3O4}.
$$

The iron is in a form of a fine powder, which is pyrophoric (it may catch a fire when exposed to air). You can see it in action here.


Ingredients for cleaning optical elements I have recently purchased a cleaning solution for precision optics which has the following ingredients (amounts were not specified):

Ethyl Alcohol
Methylethylketon
Aqua
Diethyl Ether
Isopropyl

Typically reagent grade isopropyl alcohol alone is used for cleaning optical elements. So, does the mixture above provide any advantages over using pure isopropyl? (Possible advantages may include a faster dry time, better removal of oils/smudges, and less toxicity)

Any explanations would be greatly appreciated.

It might possibly be intended to dissolve an adherent organic film, such as that left by evaporation of plasticizers from foam packaging.

However, it also might well partially dissolve paints and cements used in the lens assembly, leaving yet harder-to-remove dirt on the lens. Unless you have a specific need for those solvents, use distilled water and/or isopropanol, according to B&H.


Do lithium batteries 'explode'/'burn' because of the electric energy stored within? I've been watching some videos where people show lithium ion batteries burning/exploding. What usually happens is someone punctures the battery, and a jet of gas comes out the puncture hole, followed by the gas catching fire and the entire thing going up in flames.

I often hear that the energy for this reaction comes from the chemical energy stored in the battery. Thing is; i'm pretty sure that there wouldn't be enough chemical energy in the battery to cause such an intense reaction. I'm thinking that it's probably the lithium inside the battery reacting with the air in an exothermic manner.

So what exactly causes the reaction that we see? Does it have anything to do with the electric energy stored in the battery or is it just a lithium/oxygen reaction?

Edit: It was suggested I actually do the math. So here goes; typical energy density of a lithium ion battery would be around 200 watt-hours per kg of battery mass. Therefore a 1kg battery would have about 720 kJ of chemical energy. Proceeding from here we would need to know at the very least a.) the mass of the substance we see being 'exploded', b.) the specific heat capacity of said substance.

For example; if I assume that from a 1kg battery we'd have 300 grams of material being heated up and expelled from the battery, and that said material has a specific heat capacity of 2 kJ/(kg K) , and that said material starts at 293 kelvin, then I'd know that it would take .6 kJ of energy to heat said material up by 1 degree Kelvin. 720 kJ would be enough to heat it up to 1493 Kelvin, assuming 100% of the energy went into heating up just that 300 grams of material and not the other 700 grams. With a specific heat capacity of 3kj/(kg*kelvin) we'd get a temperature of 1013 Kelvin.

I suppose it'd really come down to what the substance we see shooting out of the battery is, and how much is shooting out of the battery, and what it's heat capacity is (and whether it's approximately constant across the temperature range of interest).

"Electrochemical" shows the correspondence between chemical and electrical energy. The energy in a cell can be released slowly over time, or rapidly when short-circuited internally or externally, or the ingredients meet under the wrong conditions. No additional chemicals, such as oxygen from the air, are needed.

You state, "there wouldn't be enough electrochemical energy in the battery to cause such an intense reaction." Calculate how much energy is stored, instead of guessing:

The common Li-ion 18650 cell stores about 2.5 ampere-hours at 4 volts (rounded). That's 10 watt-hours, and multiply by 3,600 sec/hr to get 36,000 watt-seconds or joules (J). That is a lot of energy compared to some projectile weapons. For example, the largest cartridge shown in this reference, for the .50 caliber Browning Machine Gun, packs 18,000 J, half that of the inexpensive 18650 cell.

So the question is, does the energy release come from thermal effects due to ohmic heating, or from the direct combination of chemicals. Most likely, once the cell is physically damaged, whether by overheating or mechanical stress, direct chemical combination takes place. Though it can be exacerbated by water or atmospheric oxygen, there is more than enough energy in even small Li-ion cells to cause spectacular fires or explosions.


Compressibility Factor and Generalized Compressibility Chart From my understanding, the compressibility factor is defined by
$$Z=\frac{pv}{RT}=\frac{p\bar{v}}{\bar{R}T}$$
It can also be defined by the ratio of the real molar volume of a gas to the ideal molar volume of a gas at the same temperature and pressure. Essentially it corrects for the deviation of a real gas from an ideal gas.
On a generalized compressibility chart, the compressibility $Z$ is plotted as a function $f=f(p_R,T_R)$ of the reduced pressure and temperature. I don't understand why exactly; it would be nice if someone could explain that a little more.
Another thing I'm confused about is the psuedoreduced specific volume, given by
$$v'_{R}=\frac{\bar{v}}{\bar{R}T_c/p_c}$$
Why don't we use the reduced specific volume as opposed to the psuedoreduced specific volume?
In my thermodynamics class, I was presented with the following question:
Determine the temperature, in °C, of air at 30 bar and a specific volume of 0.013 $m^3/kg$. Use compressibility chart.
I was confused about how to figure out this problem since I can't find the reduced temperature without the actual temperature of the air.

You start out with an initial guess for the temperature.  I suggest starting out using the value that you would calculate using the ideal gas law.  You then,

Calculate the reduced temperature and pressure
Get z at this reduced temperature and pressure
Use these to calculate a value for the specific volume.
If the value for the calculated specific volume does not match 0.013, adjust your guess for the temperature and go back to step 1.

Continue doing this until the calculated specific volume matches the prescribed value of 0.013


What would the Electronic Configuration of Ru²⁺ be? I'm doing homework for Inorganic Chemistry, and I am a little confused.

The Question is:

  What is the electronic configuration of $\ce{Ru^{2+}}$?

If I were to remove from the highest energy level it would be $\ce{[Kr]~5s^1~4d^5}$
leaving $6$ unpaired electrons.

If I were to take into account electron pairing it would be $\ce{[Kr]~4d^6}$ with $4$ unpaired electrons.

What would be the correct approach?

You've already figured out that the electron configuration of $\ce{Ru}$ is $\ce{[Kr]\,4d^7\,5s^1}$.

All you need to know know is the order of energies for the $\mathrm{4d}$ and the $\mathrm{5s}$ orbitals. You might want to have a look at the Aufbau principle (and possible pitfalls/exceptions) and another thread on this site.


How to visualize or think about spin waves (magnons)? According to Wikipedia: "A magnon is a quasiparticle, a collective excitation of the electrons' spin structure in a crystal lattice."

I have little pictures in my mind for other quasiparticles. For example, I can picture a phonon as (quantized) vibrations in a crystal lattice, or I can picture excitons as an electron jumping between band gaps (or HOMO-LUMO, etc.), or plasmons as quantized oscillations/sloshing of an electron gas.

I have no mental image for what a spin wave looks like. Can someone help me understand this? What would a collective spin excitation look like? Do the spins just precess faster or something of the sort? Also, how would we measure a spin excitation or spin wave?

In short this is the way I imagine them, as slight directional changes in spins propagating through the system.

The Heisenberg Hamiltonian for the exchange energy associated with magnetic coupling is a pairwise sum over an exchange integral $J_{ij}$ for two sites $i$ and $j$, with spin moments $\hat S_i$ and $\hat S_j$.
\begin{equation}
H_{ex}=-\sum _{i<j}J_{ij}\hat S_i\cdot \hat S_j
\end{equation}
Imagine a periodic array of lattice points each with a spin of two possible orientations $\pm\frac 12$, for a ferromagnetic ground state $(J_{ij}>0)$ with all the spins aligned, denoted as;
\begin{equation} |\uparrow \uparrow \uparrow \dots \uparrow\uparrow\rangle
\end{equation}
A low energy excitation would intuitively be a single flip of a spin somewhere in the lattice e.g.,
\begin{equation}
|\uparrow \downarrow\uparrow\dots\uparrow\uparrow\rangle
\end{equation}
However, this is not an eigenstate of the spin Hamiltonian. Instead low lying excitations have a complex wave character that is akin to a fractional displacement of a spin with respect to its neighbour. Such infinitesimal variation I personally imagine as a single reversed spin distributed over all the spins in the lattice.

Therefore you could view a spin wave as a superposition where all the spins contain a bit of this single inverted spin in the first excited state. It is easy to see that this is the case since we are using ladder operators to change our spins from up to down. Acting as a superposition over the lattice (spin wave) reduces the disturbance between any given adjacent pair, whereas the naive expected eigenstate would include a really big difference in exchange energies at a localised region.

As you may be thinking, how can I have an infinitesimally displaced spin when there are only two spin orientations? Well that is a bit of a misnomer as this oversimplified model has a discrete symmetry of 2. In reality there is a smooth symmetry associated with these spins so that the spin wave is actually a more realistic picture than the 2-fold orientation (Ising model).

Hope that helps :)


Find the pH of a sodium chloride solution using the extended Debye-Huckel equation Question
  Use activities to calculate the pH of each of the following solutions, being sure to use $\mathrm{\alpha}$ values and the extended Debye-Huckel equation.

  The first one the concentration is $0.020 \mathrm{M}$ $\ce{NaCl}$
  with the answer being : $\mathrm{pH}$ = $6.99$

  The second one the concentration is $0.10 \mathrm{M}$ $\ce{NaCl}$
  with the answer being : $\mathrm{pH}$ = $6.98$

The thing is i have no idea how to find the activity coefficient of $[\ce{H}^+]$

I know ionic strength $u$ is calculated as
$\mathrm{\frac{1}{2}\left(\sum C_iZ_i^2\right)}$

but how does $[\ce{H+}]$ work into the charge balance?

I also know that the activity coefficient is calculated as

$$\log\gamma=\frac{-0.51(z^2)(\sqrt{u})}{1+(\alpha(\frac{\sqrt{u}}{305})}$$

The best I have is:
$$\ce{Na+ + Cl- + H+ + OH-}$$

and that raises the question of how do I know what $[\ce{H+}]$ is and $[\ce{OH-}]$ for calculating the ionic strength? Am I approaching this wrong? Where do I start?

Let's first look at ionic strength:

Ionic strength is a measure of the total ion concentration in solution, however, species with higher charge can also have elctrostatic interactions with other species.

$$\mu = \frac{1}{2} \sum_i C_iZ_i^2 $$

What is the strength of 0.02 M NaCl solution?

$$\mu = \frac{1}{2}([\ce{Na+}]z^2 + [\ce{Cl-}]z^2) = \frac{1}{2}([0.02](1)^2 + [0.02](-1)^2 ) = 0.02 M$$

With the ionic strength, we can now calculate the activity coefficients ($\gamma_i$) of ions in solution. The activity is merely the product of the concentration of each species and the activity coefficient.
$$\textrm{activity} = a_i = \gamma_i[i]$$

Let's look at a typical equilibrium expression:

$$\ce{A + B <=> C + D} $$

$$K_{eq} = \frac{[C]\gamma_c[D]\gamma_d}{[A]\gamma_a[B]\gamma_b}$$

The Extended Debye-Huckel equation is used to find the effect of ionic strength on the activity coefficent. It is given as above:

$$\log_{\gamma \ce{H+}}=\frac{-.51(z_i^2)(\sqrt{\mu})}{1+(\alpha(\frac{\sqrt{\mu}}{305})}$$

where $\alpha$ is the size of the hydrated ion in pm.

The Extended Debye-Huckel equation is valid from $\mu = 0 \ \text{to} \ 0.1 \ \text{M}.$

To find pH, you must first find $\mu$ of the hydrogen ion. You can then use the Extended Debye-Huckel equation to find the $log_{\gamma \ce{H+}}$. Take the antilog.

$$a_{\ce{H+}} = \gamma_{\ce{H+}}[\ce{H+}] $$

Take the negative log and you will get pH.

$$ pH = -\log( a_{\ce{H+}} ) = -\log( \gamma_{\ce{H+}}[\ce{H+}] ) $$

In context with the problem:

$$\log_{\gamma_{\ce{H+}}} = \frac{-.51(1^2)(\sqrt{0.02})}{1+(900 (\frac{\sqrt{0.02}}{305})}$$

$$\log_{\gamma_{\ce{H+}}} = -0.0508886 $$

$$\gamma_{\ce{H+}} = 1.1243166 $$

$$\textrm{pH} = -\log a_{\ce{H+}} = -\log (\gamma_{\ce{H+}} \cdot 1.0 \cdot 10^{-7}) = 6.95 $$

Your error is when you take the antilog of the $\log_{\gamma \ce{H+}}$. You raised it to the $10^{\log_{\gamma \ce{H+}}}$ which is $0.89429197$. It is actually supposed to be $10^{-(-0.0508886)}$ which comes out to be $1.124316$.

For the 0.10 M of $\ce{NaCl}$ I calculated the pH to be $6.92$.


How does mass spectroscopy of nicotine work? Take this nicotine molecule:

And this mass spectrum:

I'm not sure how to explain all the peaks.

This might explain the peak at 42:

But can this peek als be explained by something else in the molecule?

Like the decomposition of the ring with the double bonds:

This might explain the peak at 84:

And I'm clueless on the peak at 133

The peak at 133 corresponds to a loss of 29 mass units (nicotine is 162).  N-CH3 has a mass of 29


What chemical available in 18th century would explosively react with phosphorus? For fictional use, I'd like to "construct" a clockwork device featuring phosphorus which could reasonably be expected to:

A) glow consistently.

B) explode upon some trigger (e.g., influx of air, mixture with another chemical, a spark), with enough explosive force or incendiary strength to kill someone standing very close.

Constraints:

1) this takes place in the early 18th century, so phosphorus has been discovered, but most of the modern types of phosphorus used in weaponry have not.

2) the quantity is no larger than a couple of liters.

Progress:

I've read the relevant chapters in Emsley's 13th Element, and then moved on to the primary sources for the time – mostly records of demonstrations held for the Royal Society and Boyle's Aerial Noctiluca.
It's hard to tell exactly what sort of phosphorus any of these early chemists were working with, but it looks like I can achieve goal (A) by just letting in a bit of air now and then (based on Boyle's description / Wikipedia on oxidation).

Goal (B) is more difficult. Is just oxygen enough?

Slare added sulfuric acid ("oil of vitriol") and achieved something, but it doesn't seem to have been an explosion exactly:

  being well shiken together did first heat, and then throw up such fiery Balls, which like so many stars will adhere to the sides of the Glass and continue to burn for some time

I think if you just add oxygen you'll just end up with a lump of burning phosphorus. I would suggest having a timed explosive charge at the centre of a phosphorus payload to essentially achieve something like a white phosphorus bomb.

Since white phosphorus is pyrophoric, you only need to disperse it for it to ignite. To do this, you could either use an automated striking mechanism to ignite a fuse to something like black powder, or my Improvised Munitions Handbook suggests a combination sulfuric acid/potassium chlorate or perchlorate (note: not chloride) + sugar device, with the mechanism keeping the two components separate, would be suitable.

The main problem with your design specifications as I see them is having an excuse to have the external surface of phosphorus exposed to the air: white phosphorus will only glow from the air-exposed surface, so if you put it behind glass, for example, it won't glow. A fine wire mesh as your external covering might do the job, but I'm not sure exactly what effect that has on reliability or effectiveness (and may cause the device to spontaneously and accidentally catch fire, especially if not packed properly).

Near explosion + being covered in burning white phosphorus = much sadness, probable death. See: White Phosphorus - Effect on People.

(Usual disclaimer: I am not a munitions designer.)


The Chemistry behind creating Polylactic Acid (PLA) I've seen a few videos of persons creating plastics from starch by adding an acid, glycerol, and water, however no explanation of the chemistry.
Does anyone know what $\ce{(C6H10O5)_{n} + C3H8O3 + C2H4O2 + H2O -> (C3H4O2)_{n} + (what)}$?
or does it?

I only have a basic understanding of organic chemistry, so sorry if the equations don't make sense.

Polylactic acid polymers are creating a lot of interest these days because they're biodegradable. On the downside, corn starch is often one of the starting materials...and corn is also used as a food source, so there's some controversy over the wisdom of this approach.

I think I've found a site that takes the approach you're basing your question on...but there are a couple of issues with the equation you've written. So, I'll answer this question in two parts. Plastic from starch is what you saw in the video; PLA from starch is what you asked about.

Plastic from starch. The recipe for the plastics-from-starch uses starch, glycerol, acetic acid (the $\ce{C2H4O2}$ in your equation) and water as a solvent. (They do not add lactic acid, the $\ce{C3H8O3}$ in your equation, but lactic acid can be produced by hydrolysis of the starch.) With the starch-acetic acid-glycerol recipe, you do not actually make a PLA polymer, but a starch polymer that is somewhat like it instead. (Ref)

So, starch is a polymer that looks like this:

In water, it tends to ball up into globular clusters due to hydrogen bonding. By adding acetic acid and heating the starch in water, you "denature" the starch and turn it into "disordered chains" that stretch out. When you let these disordered chains dry, they become entangled and form a flexible film, the plastic. In this recipe, the glycerol is included as a  plasticizer, so it is unchanged chemically in the reaction. The glycerol also helps to keep the starch stretched out by stabilizing it with hydrogen bonding. Plasticizers (as the name would indicate) are additives that increase the plasticity of a plastic and make it more durable or flexible. Changing the amount of glycerol in the recipe can have a big effect on the strength of the plastic film you make. This is the process you're seeing in the videos, I think.

Making PLA from starch. This is going to be a bit of a challenge to explain depending on your understanding of organic chemistry, but hopefully the pictures will help. When you make PLA commercially, you digest starch, usually with a bacteria, to turn it into lactic acid (see picture below for the structure of lactic acid). This lactic acid must be purified to make a high quality product. If you just combine a bunch of lactic acid molecules to make PLA, one water molecule would be lost each time they combine...sort of like this. (This is an acid and alcohol combining, not two acids, but the idea is the same.) The water in the polymerization reaction leads to a poor quality plastic.

So, in commercial synthetic schemes, two purified lactic acid molecules are reacted to form lactide...they come together in a controlled condensation reaction, with the loss of water. (Ref for picture.)

Now you can use the lactide, i.e. "lactic acids with the water removed", to do the polymerization to PLA. You add an initiator to open the lactic acid by breaking the bond between the $\ce{O-C}$ in  $\ce{O-C=O}$ and this gives you 2 lactic acid molecules hooked together...which react with another lactide molecule to add two more lactic acid molecules to the growing chain, and so on. And that's how the PLA forms.

So the reaction sequence is strach-> lactic acid-> lactide + $\ce{H2O}$ -> PLA

Probably more than you wanted to know!


Calculating the concentration of acetic acid in vinegar from pH DavePhD was kindly helping me with this problem but the site indicated the discussion had gone on too long. Calculating vinegar strength I don't have enough units to start a chat so I am starting a new thread.

The formula above is for determining the concentration or percentage of acetic acid in a vinegar sample when the ph is known. The farmer will get the ph with a ph meter, do the math and calculate the percentage acidity of her vinegar. The Ka, according to the formula DavePhD gave, is .000174 because the pKa for vinegar is 4.76. The molar mass is 60.05196 g/mol. So assuming the ph is 3, could someone write out the "long math" for this? I need someone to walk me through this, step by step to solve for c.

The last time I took algebra was about 50 years ago so, while I remember some things, it is probably better to assume I don't know much :-). Let me know if you need other numbers. Dave says the "M" is just a place holder

First off $$\ce{HC_2H_3O_2}\leftrightarrow \ce{H^+} + {\ce{C_2H_3O_2}}^-\\and\\ K_a = \frac{[\ce{H^+}][{\ce{C_2H_3O_2}}^-]}{[\ce{HC_2H_3O_2}]}$$ So let's do some simplifying. The amount of acetic acid left is the orignial molarity minus the amount that's dissociated. The amount of Hydrogen ion and conjugate base left is equal to the amount of that's been dissociated from the original amount of acid. So let's let HA equal the molarity of the acid and x equal to the amount dissociated. Soooooo $$K_a = \frac{[x][x]}{[HA-x]} = \frac{x^2}{HA-x}\\\text{Now let's solve}\\\rightarrow K_a(HA-x) = x^2\\\rightarrow HA -x= \frac{x^2}{K_a}\\\text{ switching the x interms of ph}\\ HA -10^{-ph} = \frac{(10^{-ph})^2}{K_a}\\\text{ plugging in}\\HA- 10^{-3} = \frac{(10^{-3})^2}{.0000174} = .0575$$ This is the molarity of the acid at equilibrium. Add 10^{-ph} for the concentration of the acid before dissociation.


Does the outcome of aromatic nitrations depend on the concentrations of nitric acid and sulfuric acid? I came across a few equation and observed that when concentration of the $\ce{HNO3}$ and $\ce{H2SO4}$ with benzene ($\ce{C6H6}$) or its derivative varied the product changed. I did not perform any experiment and all of this I ask here is from the stuff I read in various books.please tell me about this nitrating solution and give me all the cases if possible.

Nitration occurs as electrophilic aromatic substitution and activity of substrate varies depending on exact structure of ring system and substitutes in the ring. In general electron acceptors (nitro-, acyl-, nitrile- and sulfogroups) deactivates the reaction, while electron donors (hydroxy-, aryl-, alkenyl- and to lesser extent alkylgroups) activates the reaction. As nitration occurs step by step, it will advance to dinitroarene only if mononitroarene is not deactivated too much by introduced nitrogroup, and same for other nitroarenes. However, different nitration agents have different nitration activity with 100% nitric acid in oleum being one of the strongest, usual nitration mixture (70% nitric acid + 98 % sulfuric acid) being strong, 70% nitric acid being moderate, diluted nitric acid being gentle and sodium nitrite in acidic solution being one of the gentlest. The picture is further complicated because strong nitration agents are also strong oxidizers, capable of breaking apart delicate molecules like phenols. Given that, nitration is often performed in several steps, with first 'gentle' nitration to introduce the first nitrogroup without breaking the molecule and then further nitration with stronger agents (see, for example, synthesis of styphnic acid).

Exhaustive review of the topic is beyond what a simple post can cover. For further insight into the reaction I recommend to consult a decent university-grade organic chemistry textbook, like 'March's Advanced Organic Chemistry'.


Is it possible to construct a 3D molecular model using x-ray powder diffraction data? I've got x-ray powder diffraction (XRPD) data on bendamustine HCl from a patent and I want to know how to construct a 3D model using this data with software.

What do you hope to achieve?  It is possible to obtain accurate interatomic distances from powder data (here is an example, the structures of several alkali metal cyclopentadienides), but the method is non-routine, and the data-parameter ratio is low, which means that you cannot refine many parameters and must put much information in through appropiate restraints.  You still need accurate peak shape data.

If you just want the general shape of the molecule to illustrate Wikipedia, you are better off with obgen from the OpenBabel project.  It usually gives good results.


What is a highest oxide of element? When I go through oxides of period 3 elements in periodic table I came across a term "highest oxides" and when I searched I didn't found any appreciative answer for my question.

What are highest oxides and why they are called so?

Highest oxide of an element is the oxide compound of the element which has highest possible value of oxidation state of the element to which the oxygen is bonded. Particularly, the highest oxide of period 3 is chlorine heptoxide, $\ce{Cl2O7}$ in which chlorine's oxidation state is +VII.

In general, the highest oxide of transition metal which has oxidation value of + VIII is ruthenium tetroxide, $\ce{RuO4}$ and osmium tetroxide, $\ce{OsO4}$.


Methods to increase the speed of dissolving potassium sulfate in water apart from heating, grinding or shaking I bought potassium sulfate to be used as a fertilizer. The fertilizer comes in the form of small pebbles 1-2mm in diameter. Wikipedia states that the solubility is around 111g/liter at 20°C.

I tested several concentrations

100g / liter

50g  / liter

25g  / liter

at room temperature (22° C). After 2 hours there are still solids remaining in the solution. The water here is hard (EC= 0.6ms / cm) which hinders the dissolution. What can I do to improve the solubility of this fertilizer in water apart from heating, using reverse osmosis, grinding the pebbles or vigorous shaking of the solution (which worked) but is very tedious.

Since you're really asking about speeding dissolution, not increasing solubility, use a mixer; either:

From above, e.g., a drill attachment for mixing paint, or
From below, e.g., a magnetic mixer/stirrer.

The links are just as examples; you could devise your own mixer.


How can I easily identify if a substance is an acid or base based on its chemical formula? I know of the Arrhenius definition, Bronsted Lowry definition and the Lewis definition for acids and bases but I don't know how to apply them on the spot. Also, some chemical formulas (such as acetic acid) throw me off since it has an OH at the end which causes me to think that it's a base when it really isn't.

Any tips/ways to identify if something is an acid or base simply based on its chemical formula?

I was going to mark this as too broad, but it's not too broad, because the answer is quite simple: in general, you cannot determine the acid or base properties of a substance from its formula.

As a concrete example, consider the formula $\ce{C4H7NO3}$.

Here are two possible isomers for that formula:

The left compound is an acid. The right compound is a base.


Average energy concept Problem

When the number of electrons in bonding molecular orbitals is equal to number of electrons in anti-bonding molecular orbitals bond cannot be formed between two atoms because
A) Bonding effect by the electrons in bonding M.O.s is cancelled by the antibonding effect of electrons in antibonding M.O.s
B) Antibonding effect by the electrons in antibonding orbitals is more than the bonding effect by the electrons in bonding M.O.s
C) Average energy of bonding and anti-bonding M.O.s is more than the average energy of atomic orbitals from which they are formed
D) Average energy of bonding and antibonding M.O.s is equal to the average energy of atomic orbitals from which they are formed

Answer

 B and C

My thoughts
I get the reason for option B as anti-bonding orbitals have higher energy comparing to the bonding orbitals, so naturally their destabilizing effect must be greater than the stabilizing effect offered by the bonding orbitals.
But I really can't understand the average energy concept of orbitals as expressed in the option C . could you help clarify what exactly is the energy concept regarding the molecular orbital and atomic orbital ? and how they can be compared?
Correct me wherever I'm wrong.

Choices B and C are two different ways of saying the same thing. Say $E_{MO}(A)$ and $E_{MO}(AB)$ are the energies of electrons in bonding and anti-bonding orbitals, respectively, relative to the electrons in the original atomic orbitals from which the molecular orbitals are formed, or, to make matters simple, set the energy of electrons in the original atomic orbitals to zero, that is, set as a reference $E_{AO}(1)$ and $E_{AO}(2)=0$ (assuming for simplicity that there are two MOs formed from two AOs).
To say bonding occurs is equivalent to saying that the energy of electrons in the separate atomic orbitals is higher than the energy of the electrons in the bonding molecular orbital. Using that reference system, positive energies (>0) disrupt bonding, and negative lead to bonding. Then saying that the effect of the antibonding electrons is greater than that of the bonding ones is equivalent to the statement
$$0\lt E_{MO}(B) + E_{MO}(AB)$$
or equivalently
$$0\lt  \frac{E_{MO}(B) + E_{MO}(AB)}{2} = E_{MO,\text{avg}} $$
but this is just what choice C says, since the average energy of the atomic orbitals is just zero, that is
$$  \frac{E_{MO}(B) + E_{MO}(AB)}{2} = E_{MO,\text{avg}} \gt 0 = E_{AO,\text{avg}}  $$


Ceramics in "bullet proof" vests? Kevlar is an aramid fiber that is used in ballistics protection vests. From the structure (structure shown from Wikipedia article), you can see why it would work to dissipate the energy from a bullet since there is an extensive network of interactions between the linear chains. The wearer ends up bruised from the force, but unpunctured and avoids serious injury from handguns bullets.

More modern ballistics protection devices use ceramic plates (along with Kevlar) to give greater stab resistance and protection from higher velocity rifle rounds. My question is, what ceramics do they use? Do they have a "flexible" ceramic that will deform to absorb the energy and not break? Or do the plates work by breaking to absorb the energy and, thus, have to be replaced after each event?

As far as I can tell there are two material types used for this - pure ceramics or ceramic mixtures that are designed to break and be replaced, and composite metal + ceramic materials that are designed to be able to handle a couple of rounds without losing too much effectiveness. There are also techniques involving purely metal layers in overlapping structures with fibers of materials like Kevlar or ultra-high-molecular-weight polyethene (UHMWPE) (of which Spectra is a brand).

Shopping around, it seems the best materials are boron and silicon carbides (extremely hard materials, though not quite as hard as diamond), which can be formed into plates with multiple levels of structure by starting with hexagonal plates and forming larger plates from those. In addition to personal armour, the same materials and techniques are used for military vehicle (including aircraft) armour. In the ceramic/metal composite materials, you get ceramic plates formed up within a metal sheet, where a ceramic element will shatter to take most of the impact, while the rest will be absorbed by the metal matrix. The metal matrix then remains intact (but may be deformed) so that the armour is still useful but weakened around the area of impact.

Good places to start from are the Wikipedia pages for Chobham armour (the common name for the tank armour that includes composite metal/ceramic matrix materials), and Small Arms Protective Insert.


What does all chromatography have in common? We're learning about chromatography in class, and I'm confused about all the different types (e.g. those that separate liquid-liquid solutions vs. those that separate liquid-solid solutions vs. those that separate solid-solid solutions, etc.).
Could you provide a summary of the similarities between all forms of chromatography? (just an overview, do not require specifics)

Here are some similarities:

There is always a mobile (e. g. Gas - GC, Liquid - HPLC; GPC) and a stationary phase (liquid; gel - GC, solid - LCs, GPC).
Compounds in the sample interact different with both phases and are therefore held back stronger or lesser. This results in the accumulation of compounds that interact similar at some point in the system $-$ with some chromatographys, especially affinity chromatography, one might even get pure substances.
The mobile phase carrys the sample and is commonly less polar than the stationary phase, except for RPCs.
The structure of a chromatograph is basically like this:

transport system: reproductive transport of the mobile phase, adjustable (e. g. piston pump - LCs, compressed gas cylinder - GC)
sample application: import of the sample into the flowing mobile phase (e. g. injector, valve)
column: stationary phase and its mounting (e. g. capillary column in the column oven)
detector: transformation of the chemical or physical properties, which change with substance concentration, into electrical signals (e. g. UV/VIS-detector, thermal conductivity detector)
plotting unit: visualisation of the electrical signals (e. g. integrator, computer)


How to Experimentally Quantify the Response of a Receptor to a Ligand? In medicinal chemistry, there are multiple reasons why one would want to experimentally determine the effect of a particular molecule on a receptor e.g. measuring how 'strongly' that molecule brings about a chemical response (an agonist) or how strongly it prevents a response (an antagonist).

If we were looking at an insulin receptor antagonist (for whatever reason), then it would be easy to quantify the strength of the antagonist by a simple blood test of the test subject (human or animal), to see how much insulin was present.

What happens when we're faced with a scenario inherently difficult to quantify? How do we measure the action of a drug then?

Say we've found a natural neurotransmitter receptor agonist and want to compare this to a synthetic drug of a similar nature. I'd struggle to see how we'd experimentally be able to analyse the amount of neurotransmitter present in the synapses in each of the cases to compare.

Do we then look at less direct, perhaps qualitative methods of analysis? Such as 'state of agitation' or 'how many times the subject's arm twitched in a minute'?

What you're asking about is a wide topic.

You have the basic ideas of agonist and antagonist, but there are also molecules called inverse agonists. Every receptor has some base level of activity, even without an agonist to activate them, or in the presence of an antagonist. Inverse agonists push this activity even lower. There are even "partial agonists" which are molecules that activate receptors, but not as well as the endogenous ligand, so act like competitive inhibitors because they occupy some of the receptors the endogenous ligand would have. In addition, the binding site is important. The endogenous ligand binds to a location called the orthosteric binding site, but other molecules may bind to other locations on the receptor. These are allosteric binding sites. Allosteric binders may activate or inhibit a receptor, but they may also be modulators. Modulators don't directly activate a receptor, but allow a stronger response when another ligand binds at the orthosteric site. Here's a schematic to help show how different types of ligand affect receptor activity.

Anyway, ligands have two important properties, affinity and efficacy. Affinity is how strongly a ligand binds to a receptor, and efficacy is how the ligand affects the receptor's activity.

Before we can dig too deep into agonists or antagonists, we should think about receptor theory. Receptors, like all proteins, have complex 3 dimensional structure, or conformation. This conformation is affected by solvent, pH, temperature, and other factors, but always has some fluctuation. Classical receptor theory used a lock and key model, where an inactive receptor bound a ligand and became active. However, this did not explain the base activity and activity of inverse agonists, so an equilibrium based model was developed. Here, the population of receptors has an ensemble of conformation states, some of these are active. When an agonist binds, it drives the equilibrium of the population toward the active state by stabilizing the active conformation. This diagram shows a simplified example of how drugs affect the equilibrium. Ri is inactive receptor, Ra is active. D is drug. Ra leads to some activity, while Ra-D leads to strong activity.

Affinity is relatively straightforward to measure becuase the same idea can be applied to almost any receptor-ligand system. Usually, a radiolabled version of the drug is added to a known amount of receptors in a binding assay. The goal is to determine a dissociation constant, Kd. Kd is used to quantify how well a ligand binds to a receptor, and smaller numbers are tighter binders.

Look at this reaction: D + R <--> DR with forward rate constant k1, reverse rate constant k2. ( how to get the right arrows here? )

At equilibrium, k1[D][R] = k2[DR]
and $K_d =\frac{[D][R]}{[DR]}$ = $\frac{k_2}{k_1}$

So incubate the radioligand and receptor until some DR complex is formed, then seperate the bound and unbound ligand so you can measure D and DR. The seperation method is important because if it's too slow the DR complex will revert to D and R, creating inaccurate data. But this is commonly done using dialysis, centrifugation, and vacuum filtration. Dialysis works well when dissociation rate is slow. You put receptor in one side and drug in both sides. It's slow, but it's theoritically most accurate becuase you can measure the true steady state by taking an equal volume of solution from each side of the membrane without affecting equilibrium. But you also need a large supply of receptor and drug.
Centrifugation works when you can immobilize the receptor on a large particle, usually you do this by using whole cells. You add an excess of of drug, then spin it down and take off the supernatent. Timing is still an issue and free drug can get caught in the pellet.
Vacuum filtration is usually considered the best because it can seperate the components very quickly, by pulling the solution through a filter and trapping the receptor on the surface. However this requires specialized and expensive equipment.
Once you measure bound and free drug at several concentrations, you can do a double reciprocal plot

Measuring efficacy is different because you need to choose an activity to measure. Idealy it will be objective and easy to quantify. If the receptor is an enzyme, you can measure its enzyme activity. If it activates a gene transcription factor, you can use cells modified to express green fluorescent protien or luciferase when that transcription factor is activated. If it's an ion channel you can measure a cell's membrane potential. If you cannot measure anything else, you can use behavioral assays. There are several assays used to measure mouse activity and try to get at certain aspects of their brain and are often used to study depression and other mental illnesses. These are subjective. Efficacy is quantified by EC50, the concentration at which response is half of maximum response. Smaller EC50s are more efficacious.

This field is much much larger than my answer. For more you should read this article : http://www.ncbi.nlm.nih.gov/pubmed/15063082


Chemical reactions with a room-scale cooling effect Are there chemical reactions that could cool down an average sized room by a noticeable amount (say 5 °C)?

I would like to investigate if it is possible to have a situation where I can mix 2 reagents at room temperature and pressure and in open air then they should react and become colder than room temperature without evaporation of some type, with an eye to making a noticeable drop in the room temperature.

One way to do it is, as user41631 says, with an endothermic reaction.

Another is to use phase-change materials, from which you remove the thermal insulation, to allow the phase-change to take place. Now, all materials change their phases, at some combinations of temperature and pressure. The phrase "phase-change material" is used not to classify materials, but to identify the paricular use of a material - that is, the material is being used because one of its phase changes happens at a temperature and pressure that makes it useful for a specific process.

Such materials are being looked at for the temperature regulation of buildings.

Let's take a material with a melting point of, say, 18°C, and a high latent heat of fusion. Freeze it, insulate it, and take it into a room that's at 22°C. While the material melts, it will absorb heat from the air, cooling the room.

This principle has been applied in building material design, to provide smoothing of temperature variations. See, for example, Schossig et al's "Micro-encapsulated phase-change materials integrated into construction materials", [DOI], from which the following graph is taken, showing the smoothing of temperatures over a week in August , when a phase-change material (PCM) is used (T_Wall_PCM) - this is compared to an identical room without phase-change materials (T_Wall_REF) - as you can see, the PCM reduces the amount of time that the room spends above 26°C, providing more thermal comfort for people in the room.


reaction of diketone with ammonium carbonate at 100-115 degree celsius Please explain what reaction is this and what are the products of reagent $\ce{(NH4)2CO3}$
at 100-115 C with ketone.

1,4-diketones are likely to give pyrroles in a Paal-Knorr reaction.


Alternative mechanism for carbocation rearrangements Is there no alternative mechanism for carbocation rearrangements rather than hydride shifts? I've been killing myself over this thing for over two days, and I do believe that hydride/alkyl shifts are not only unfavourable, but unnecessary as well.

What follows might be a little weird to comprehend at the first try, but please, stay with me here...

Let us deal with one of the simplest situations we can think of, say, 1-butanol in acidic medium.

The H+ ion will interact with the lone pair in oxygen and H2O goes out. Till here we are all in the same boat. Now comes the interesting part.

We're all familiar with the classical theory, so I shall not waste time discussing that mechanism (kindly refer to the enclosed series -1 if in doubt).

I shall only point out what seems, to me, a major flaw.

When the primary carbocation is formed, it causes a number of things to happen:

(1) The electronegativity of the carbon holding the positive charge increases drastically.

(2) This increase in electronegativity causes a dissatisfaction in the adjacent carbon, increasing, in turn, the second carbon's electronegativity significantly.

As it is, carbon has a higher electronegativity than hydrogen. now, you've gone and increased it even further. Now, when the carbon(2)-hydrogen bond is cleaved, what would one expect? One would expect that the electron pair is retained by the carbon and hydrogen goes out as H+. The classical theory suggests that the opposite of this is true. How, then is that kind of bond cleavage feasible?

I say that it is not. It is my strong conviction that nature, in fact, doesn't work that way. Let us look at an alternative mechanism to this kind of a reaction.

(I request all of you to kindly assist me in finding the flaws in what I shall put forth now, for if I can think of it, so can every other person on the planet, and it is impossible that some organic chemist didn't think of it before me. Since it has not been accepted, there HAS to be a flaw. What I think has been expressed in the enclosed Series-2.)

According to the alternate theory, the carbon(2)-hydrogen bond is cleaved as one would expect it to. A comparatively stable intermediate, namely 1-butene, is formed.

There is a comparatively higher electron density in the region of the pi bond. Hence, in absence of any other option, the proton stays in the vicinity of this pi bond, due to electrostatic force of attraction.

This system, when given enough energy(via collisions with other molecules), cleaves the pi bond and the electron pair, now retained by one of the carbons   (most of the time with carbon(1) since a primary carbanion is much more stable than a secondary carbanion), is given to H+, thus giving you your desired secondary carbocation.

After the formation of the secondary carbocation, the formation of 1-butene/ 2-butene is but a trivial question, which happens in the same way as what the classical theory suggests.

So... Is there any other theory as well? Is what I propose feasible? We can never know the absolute truth, of course, but how close are we to it?

PostScript: I am very much aware that what I state is not something new. It must have crossed many other minds before mine.

Scheme 1: Orbital model of a Wagner-Meerwein rearrangement.

You may think you are proposing something new, but you aren’t.

While these Wagner-Meerwein rearrangements are typically referred to as ‘hydride shifts’, that name should not be taken too literally (as in your first scheme). Instead, it is a concerted orbital interaction. There is never any hydrogen cation or hydride anion formed, hydrogen remains in the vicinity. The entire process is concerted and thus proceeds suprafacially.

Writing the central structure as a double bond with proton in the vicinity, as you did, or as ‘two cations coordinating a hydride anion’ is really only a formality. It is not an intermediate in any way, it is a transition state.

Being a transition state, by the way, means that it is a directed process. If the hydrogen went half way, it is going to proceed on (law of inertia, if you wish) and not go back. However, given a sufficiently stabilised cationic system, it may reverse at a later moment in time.


Does Reverse Osmosis remove elemental ions from water? From what I understand, RO is able to remove (or at least considerably reduce) the salts dissolved in water.

But thinking about it just doesn't make sense. RO works by filtering the solution using a membrane with pores so fine that only water can go through them. But elemental ions are WAY smaller than molecules of water.

So how does RO prevents salts from going to the other side?

The short answer is pressure. What you've described above is basically just osmosis, and without applied pressure to overcome the osmotic pressure of the system, reverse osmosis doesn't happen.

This Wikipedia article gives a good description of the concept, and the difference between filtration, osmosis and reverse osmosis:

  Osmosis is a natural process. When two solutions with different
  concentrations of a solute are separated by a semipermeable membrane,
  the solvent has a tendency to move from low to high solute
  concentrations for chemical potential equilibration.

  Formally, reverse osmosis is the process of forcing a solvent from a
  region of high solute concentration through a semipermeable membrane
  to a region of low solute concentration by applying a pressure in
  excess of the osmotic pressure. The largest and most important
  application of reverse osmosis is the separation of pure water from
  seawater and brackish waters; seawater or brackish water is
  pressurized against one surface of the membrane, causing transport of
  salt-depleted water across the membrane and emergence of potable
  drinking water from the low-pressure side.

  The membranes used for reverse osmosis have a dense layer in the
  polymer matrix—either the skin of an asymmetric membrane or an
  interfacially polymerized layer within a thin-film-composite
  membrane—where the separation occurs. In most cases, the membrane is
  designed to allow only water to pass through this dense layer, while
  preventing the passage of solutes (such as salt ions). This process
  requires that a high pressure be exerted on the high concentration
  side of the membrane, usually 2–17 bar (30–250 psi) for fresh and
  brackish water, and 40–82 bar (600–1200 psi) for seawater, which has
  around 27 bar (390 psi)[6] natural osmotic pressure that must be
  overcome.


How to interpret spectroscopic data to obtain the structural formula? I am currently struggling with an assignment question regarding structural determination using both $\ce{^1H}$ and $\ce{^{13}C}$ NMR data.

The first thing I have done is calculate my double bond equivalences which I believe is 3. I believe the IR data corresponds to a $\ce{C=O}$ bond which is supported by the number of DBEs. As it states this is a cyclic compound I believe that the there is a ring containing a double bond and a carbon double bonded to an oxygen which accounts for the 3 DBEs.

Putting together this information which I believe to be correct with the spectroscopic data above is where I am really struggling.

I am quite confident in saying that the 6.8 triplet is due to an aromatic carbon with two hydrogens on the adjacent carbons so I have a fragment there that I believe is correct but I cannot get anything more than that.

Your calculations of the double bond equivalents seems correct, and there is also another hint towards a $\ce{C=O}$ double bond: the corresponding $\ce{^{13}C}$ chemical shift.

Also, while it says you have $\ce{^1H}$ decoupled $\ce{^{13}C}$ data, this is clearly wrong, because they’re showing you multiplets. The $\ce{^{13}C}$ data is $\ce{^1H}$ coupled data, so you can use the information there to deduce the number of carbons with which number of hydrogens.

A third hint given to you is the amoun of signals: You see signals that add up to six carbons and eight hydrogens, meaning there is no symmetry in your molecule.

The 6.8 hydrogen could, by its chemical shift, be aromatic, but you do not have enough DBE’s for an aromatic ring (you would need at least 4). Do you know other types of protons that give chemical shifts in similar ranges, and are consistent with both your DBE count and the structure element(s) you have already found out?

Further ideas to think of:

You have certain integrals on your hydrogen signals, especially one that stands out (not the $6.8\,\mathrm{ppm}$). What could this integral mean, and how does it affect the rest of the structure?
Make a list of the carbons you have and the amount of hydrogens attached to them.
How can you turn all that into a cyclic compound?
Once you have, you can check if the multiplets of the hydrogen signals can be fitted to that structure. Hint: You seem to have no H–H couplings other than across three bonds. (So no long-range couplings.) And no H–C couplings other than one bond length.

   Right, in this first spoiler tag I’m just gonna state the different signals. Check the second one, if you really want to know the solution.
 The carbon at 209.6 (singlet) is most likely not an aldehyde but a ketone. If it had a proton connected, it would need to be a doublet.
 The singlet at 1.8 is a methyl group, belonging to the carbon at 12.1.
 The triplet at 6.8 needs to be next to a double bond (strong low-field shift). The corresponding carbon needs to be 157.9 (doublet).
 Since it’s a triplet, it couples to two hydrogens. There is no other hydrogen on a double bond, but there is a triplet and a quartet left. One of these two must couple to the proton integral one at 6.8. Luckily they are both integral 2. A little further reasoning tells us, it must be the quartet, because the other triplet (2.4) has to couple to two protons, which have to couple back, therefore no other possibility.
 There is another carbon at a double bond (142.1) and it has no hydrogens attached to it (singlet). It needs to have the methyl group on it.

   The structure I deduced is 2-methylcyclopent-2-en-1-one.


Cyclopentyl to cyclohexyl carbocation rearrangement Here is my problem. The mechanism seems pretty straightforward, but I am having trouble converting the cyclopentane to a cyclohexane. During a synthesis problem, does the starting material add the extra C onto its chain during carbocation rearrangement or a methyl/hydride shift?

As for the problem at hand, I believe the first step is for the pi bond to attack the $\ce{H+}$, and that will leave me with a carbocation at the more substituted carbon. From there, I know $\ce{HOCH3}$ will attack, but somehow, the molecule needs to rearrange its format in order for the $\ce{HOCH3}$ to come in. What needs to happen here?

Your first step is correct. The reactive exomethylene group attacks the proton leaving a secondary carbocation (remember Markovnikovs rule). Now the five-membered ring can rearrange into a more stable six-membered ring by releasing ring strain. This is a fast intramolecular step. Finally methanol attacks the rearranged carbocation releasing a proton which is used catalytically in the next reaction.

Remember though that you can often not be completely sure about a mechanism before it is verified in practice (i.e. in the lab through an experiment). Sometimes true mechanism can be quite different from what appears to be most reasonable on paper!


ORCA: How to plot an adiabatic potential in dihydrogen H2 molecule? I stared to study ORCA, and I try to obtain classical results for dihydrogen as for example. I need to get a starting point to understand what needs to be done.
So, How to plot an adiabatic potential in dihydrogen H2 molecule with ORCA calculation?
! RHF OPT def2-QZVPP
%geom Scan
B 0 1 = 1.0, 3.0, 12
end
end

* xyz 0 1
   H        0.000000       0.00000        0.00000
   H        0.800000       0.00000        0.00000
*

By adiabatic, I presume you mean the Born-Oppenheimer approximation (which is usually used). If you need a plot of potential against H-H distance, then you need to do a relaxed potential energy surface (PES) scan.
! UHF OPT def2-QZVPP
%geom Scan
B 0 1 = 0.3, 1.3, 30
end
end

* xyz 0 1
   H       -4.61685        1.79381        0.00000
   H       -4.14986        1.26166        0.00000
*

The Scan directive in %geom section will perform a relaxed geometry scan (although in this case you have only two atoms so there is no difference between relaxed and unrelaxed). The B 0 1 = 0.3, 1.3, 30 line will scan the bond (B) between atom 0 and atom 1 (in Orca, atom counting starts from 0). The scanning will start from the bond distance 0.31Å and will end at 1.3Å and will go through 30 steps (i.e. the PES will have 30 points in total).
Another thing is that you need to use the unrestricted hartree-fock (UHF) because when the H atom splits the two electrons go into different orbitals, so RHF will give the wrong behaviour at higher bond distances (it will force double occupancy in one orbital, resulting in $\ce{H+}$ and $\ce{H-}$ fragments). UHF gives the right behaviour but may not reproduce the correct energy. Multi-configurational methods might be required.
The input file that I have shown will show you the energy behaviour of the ground state (in your graph, $\mathrm{U_S}$). Plotting the first excited state (antibonding) i.e. the $\mathrm{U_A}$ graph would be more difficult. You would probably need to do some sort of orbital rotation to converge to the excited triplet state, and then use that as the guess for the geometry scan (or maybe do TDDFT). I don't know much about that.
After running the calculation, the energy values at each bond length will be printed out at the end of the .out file, and also as a .dat file. You can import the .dat file into Excel or other programs to get the actual visual plot.
Edit: If I am not mistaken, the $\mathrm{U_A}$ graph indicates the first excited triplet state, so converging to that UHF solution is easier, all you have to do is to set the multiplicity to 3 (*xyz 0 3). Using UHF/def2-SVP, the scan looks like this—

As you can see, the two curves are crossing whereas they shouldn't really cross. I am not entirely sure why this happens, but I suspect that using multiconfigurational methods will solve this. Also note that the energy plotted is the absolute SCF energy, so the graph's don't go to zero at high bond distances.


What causes the DNA fragments to stop moving in gel electrophoresis? I'm currently studying VCE BioChemistry, and we're studying the separation of DNA strings of different lengths via gel electrophoresis.

(This involves having 'clumps' of DNA at one end of a gel medium and applying an electric current, pushing the DNA strands across the gel. Different sized strands move at different speeds through the gel)

At some point, the strands stop moving such that there is a distribution of strands across the gel. What causes the strands to stop?
Surely they'd continue until the end of the gel.

If suppose the electric current could be stopped, thus causing the strands to stop moving, but how would one know when to stop the current?

There are electrophoretic methods where the molecules stop automatically like Isoelectric Focusing, but in the common agarose gel electrophoresis of DNA you mention this is not the case.

The DNA is negatively charged, it will move along the electric field you set up until you stop applying any current. It will happily move out of your gel if you let it run for too long.

To see the progress of your gel some dyes are usually added to the sample like bromophenol blue or xylene cyanol. They don't interact with the DNA, they just provide you a reference on how far your gel has run. You still need to have a rough idea how your samples run compared to those dyes if you want to use the full length of your gel.


Cerium electronic configuration change from 0 to +2 oxidation state Cerium in $0$ oxidation state has electronic configuration

$$[\ce{Xe}]\mathrm{(4f)^1(5d)^1(6s)^2}$$

But when it gets oxidised to $+2$ state, it becomes

$$[\ce{Xe}]\mathrm{(4f)^2(5d)^0(6s)^0}$$

This is a change as in other elements in the lanthanides series get their $\mathrm{6s}$ electrons abstracted and the other shells remain at liberty.
Could you please explain this phenomenon to me? And also correct me if anything unsaid above was incorrect.

The lanthanides are most stable in their 3+ oxidation state. For example, lanthanum is perfectly happy changing from

$$\mathrm{La}^0 = [\ce{Xe}]\mathrm{(5d)^1(6s)^2}$$ to

$$\mathrm{La}^{3+} = [\ce{Xe}]\mathrm{(5d)^0(6s)^0}$$

Then, as you go along the lanthanide series, you're adding more 4f electrons. For example:

$$\mathrm{Ce}^{3+} = [\ce{Xe}]\mathrm{(4f)^1(5d)^0(6s)^0}$$
$$\mathrm{Pr}^{3+} = [\ce{Xe}]\mathrm{(4f)^2(5d)^0(6s)^0}$$
$$\mathrm{Nd}^{3+} = [\ce{Xe}]\mathrm{(4f)^3(5d)^0(6s)^0}$$

etc.

This electronic configuration is extremely stable for all lanthanides, and there are very few exceptions to it.

Ce(II) is extremely unstable, and if there's one way to stabilise it, that would be adoption of the Pr(III) electronic structure. This way Ce has no 5d or 6s electrons, which makes lanthanides happy (apologies for the non-scientific way of framing this).


Boiling point of vodka What is the boiling point of vodka?

I have found the boiling point of ethanol, ~173 (degrees) F.

However, I am unable to find the boiling point of vodka. I found some information on it, but I do not trust the source.

Vodka is a solution of water and ethanol.

Water boils at 100 °C
Ethanol boils at 78.37 °C

A mixture of both liquids will start boiling close to the boiling point of ethanol (otherwise distillation wouldn't work).

The lowest boiling temperature is a mixture of:

95.63 mass-% ethanol
4.37 mass-% water

which boils at 78.2 °C.

Read more about this effect at: Azeotrope There is also a picture of a phase diagram there.


How to reduce propyne selectively for the trans product In my textbook it is given that when alkynes react with sodium in liquid ammonia, hydrogenation takes place and a trans isomer is formed.

It is also given (in another part of the chapter) that when terminal alkynes react with sodium in liquid ammonia, they form a salt with sodium.
I assume the first mentioned reaction doesn't take place in the case of terminal alkynes.

Then, how do we, for example, add deuterium to propyne so as to form trans 1,2-dideuteropropene?

As you point out, a dissolving metal reduction using $ \ce{Na}$ and $ \ce{NH3}$ in the presence of a proton source such as $ \ce{tBuOH}$ is able to reduce a disubstituted alkyne to the corresponding trans alkene.

The mechanism is understood to go via the following stepwise mechanism:

Both protons on the product  come from the $ \ce{tBuOH}$, rather than the $ \ce{NH3}$ as suggested in the comments, so in theory, using $ \ce{tBuOD}$ would give you the desired product.

One issue with this however, is the ease of exchange between $ \ce{H}$ and $ \ce{D}$– any residual water present would be able to convert $ \ce{tBuOD}$ to $ \ce{tBuOH}$, which would then give you the protonated rather than deuterated product.  Experimentally, you'd probably end up getting a mixture of products with varying degrees of deuteration (mono and di).


Which combinations of orbitals lead to pi bonds according to Valence Bond Theory? How many of the following combination of the orbitals will lead to formation of $\pi$-bonds with $z$ axis being the internuclear axis:
$$p_x+p_x,\,p_z+p_z,\,p_y+p_y,\,d_{zx}+p_x,\,d_{zy}+p_y,\,s+p_y,\,d_{yz}+d_{zy},\,d_{zx}+d_{zx},\,d_{z^2}+s$$

My attempt/Understanding:
According to  Valence Bond Theory, any orbital having any relations with $z$-axis or $z$ related planes should form $\sigma$-bonds. This is because $z$ is internuclear axis and these orbitals will have direct overlapping, forming $\sigma$-bonds.
Rest orbitals will form $\pi$-bonds by lateral overlapping, so I counted them as answer $(4)$:
$$p_x+p_x,\,p_y+p_y,\,s+p_y,\,d_{xy}+d_{xy}$$
But answer given is $(6)$, and it is not mentioned which are orbitals are counted. So, please tell which orbitals are to be counted for, and tell any flaws in my understandings, if any.

Here are the requirements for a $\pi$ bond:

Each contributing orbital has to have exactly one nodal plane containing the bond axis.  There can be other nodal planes that might cut through the axis, but don't contain it so those planes don't count.  Two nodal planes containing the axis, as with a $d_{xy}$ orbital, also fail because that's too many.

The nodal planes identified in (1) that contain the bond axis must line up, so they end up as a single nodal plane going through the bond.  Draw, let us say, the $p_x-p_x$ pair, which is easy to see it's a $\pi$ bond, to see what I mean.

If you sketch the orbital combinations for each of the nine given cases you see immediately that three fail test (1) (the $s$ orbital is no-go for a nodal plane, and $p_z$ has its nodal plane oriented wrong), but the others pass both tests (1) and (2) and they are the six combinations identified by the book.
The hidden lesson here is that $d$ orbitals can form $\pi$ bonds too.  You will see them do so when you study transition-metal complexes.


Are there any STP liquids not containing large amounts of hydrogen and oxygen that are safe for human contact? I'm working on a science fiction short story and was wondering if there were any liquids that would be safe to bathe in while traveling on a space craft with artificial (spin-based) gravity.  In such a context hydrogen and oxygen are valuable resources not to be wasted.  While bath water could be recycled, it is not available for other uses while someone is sitting in it, so I was wondering if there are other liquids at Standard Temperature Pressure that would be safe to bathe in (potentially ingesting small amounts).

I thought of using carbon bucky-ball fullerenes in a fluid but the viscosity looks to be much too high (essentially tar-like).  I know there are waterless chemicals that can be used for cleaning (e.g. DryBath), the purpose of this is more to have a hot-tub to relax in than to disinfect and clean.  If the addition of some hydrogen and oxygen (either as H2O or in other compounds) can improve the viscosity or is necessary to form the liquid, that is acceptable as long as less than 10% of the liquid is comprised of hydrogen and oxygen.  The purpose is to save water and oxygen for air while still being able to soak in a hot-tub.

Unfortunately most oils contain a substantial amount of hydrogen and some oxygen, so they don't qualify unless the amount of carbon offers substantial savings in volume.  I'd prefer something without any hydrogen or oxygen if such a chemical exists.

This is my first post here and I'm unfamiliar with the tags, so feel free to edit or add any you feel are appropriate.

Perfluorocarbons would be a possibility, $\ce{(-CF2-)_{n}}$, they contain only carbon and fluorine and are relatively low cost and widely available in large quantities.  They are inert (e.g. non-reactive) and depending upon chain length and structure, you could pick one that can be warmed up nicely to a bath temperature.  They are also used industrially in degreasing operations, so they might have some cleaning ability as well.

Some perflurocarbons are actually being investigated as human blood substitutes - they can transport and release oxygen to tissues.  Perfluorodecalin is a prime example. As shown below, it is basically a naphthalene ring in which all of the hydrogens and double bonds have been replaced with fluorine.

Ingesting small amounts would not be a problem. In fact, these same materials have been used in liquid breathing experiments.

There are a number of other scientific aspects of these perfluorocarbons that could play into science fiction.  For example, the small size of the molecules (compared to say, haemoglobin) allows it to travel to smaller places that haemoglobin cannot access due to its larger size. Perhaps the perfluorcarbon could also carry small drugs to relax, rejuvenate, repair body parts, or enhance sensory experiences during the bath. Since these perfluorocarbons act as oxygen transport agents they could also be worked into the story as a bath medium that also scavenges "loose" oxygen.


How to Predict the Formation of Racemic Mixtures or Single Products from a Diels Alder Reaction? So far, I understand the mechanism of the Diels Alder reaction and am able to correctly predict the product of some slightly complex examples.

However, it seems that in some examples, a racemic mixture is formed, whereas in others that's not the case.

I have basic understanding of the Endo Rule (endo product favored over exo product), but have trouble with the stereo chemistry involved.

I also know that the symmetry of the diene and dienophile has to do with it, but can't seem to find a consistent pattern.

Any help would be appreciated.

It is very simple: if you do not have any enantiopure compound present at the beginning of the reaction, the end product will either be inherently symmetric (contain a plane of centre of symmetry) or racemic. This will be because there are two symmetry-equivalent paths of attacks that will lead to the two enantiomeric products (if enantiomers are even generated) and neither of the two paths is inherently more probable. It is a great exercise to try getting both enantiomers for a stereouncontrolled reaction!

If you do have any enantiomeric excess in your reactants, your products will also have some enantiomeric excess. If the chiral centre is far from the reactive site, the ee will be low; if it is very close or part of the reactive site the ee will typically be much higher. This is because the transition states will now be diastereomeric rather than enantiomeric, meaning they differ in energy and one will be more likely.

While it is still a great exercise to draw the two different transition states leading to different products it can be much harder to distinguish the more likely major product; especially if it is a chiral catalyst giving enantioselectivity.


Do multielectron atoms have nodes? For hydrogen, other than in the 1s state, the electron wavefunctions have radial and/or angular nodes where the electron probability density is zero.

In helium or further atoms with more than one electron, considering correlation of electrons, what happens to these nodes?  Are there still radial and angular nodes where electron probability density is zero?

For example, if helium is in a 2s2s state, is there anywhere that an electron cannot be?

My answer to this question is the following. Nodes can happen for highly excited states, whose average distances are typically much larger than the Bohr radius. Consider Na, Cs etc, which have one outermost shell electron. When the outermost shell electron is highly excited, the case is not much different from that of a hydrogen atom (i.e. the indistinguishability of electrons may be ignored when electrons are well separated.) Thus there can be nodes in these highly excited states.

A more non-trivial question is, whether the total electron density can have nodes for the ground state. I tend to think that it's not possible. My following naive, sketchy explanation may be seriously wrong. For a second-order eigenvalue problem, we have the sturm theorem (?; in doubt), which guarantees that the roots of successive states occur in an interlacing manner. That means, where state $i$ has a node at $r_i$, state $i+1$ cannot have a node at $r_i$. Thus the density, which is equal to the sum of square of wave function of all the occupied states, cannot be zero at one point. We have a similar second-order eigenvalue problem for each angular momentum $l$, and so the argument goes.

The above too sketchy, so let's formalise it a bit based on the one-particle approximation. The total ground-state electron density is proportional to $ \sum_{\{n,l, m, s\}; \textrm{occupied}}|\psi_{nlm}(r)|^2 |Y_{lm}(\theta, \phi)|^2$. Here $s$ is for the spin, $n$ labels the energy state (1 for ground, 2 for first excited, etc), and the sum is over all occupied states. Note that in order for the electronic density to be zero at $r_0$, each term has to vanish at $r_0$. But remember that for $l=0$, $m=0$, the $n=1$ radial wave function $\psi_{100}(r)$ can have no node, and $Y_{00}$ is a constant, which means the term $\psi_{100}(r)Y_{00}$ can vanish nowhere in space. This state has to be occupied in the ground state.


Reporting reaction energies calculated using electronic energies rather than enthalpies I understand that electronic energy includes a combination of

electron-electron repulsion energy,
nuclear-nuclear repulsion energy, and
electron-nuclear attraction energy

at 0 K, and the energies are relative to a zero with all particles infinitely far apart.

So is this when it's appropriate to use electronic energies rather than enthalpies when determining reaction energies at 0K and when particles are infinitely far apart?

In the case of atoms or molecules in a vacuum, at $0~\mathrm{K}$, and infinitely apart, the "reaction energy" of

$$
\ce{A + B -> A \bond{....}\bond{....}\bond{....} B},
$$

where $\ce{A \bond{....}\bond{....}\bond{....} B}$ is effectively at infinite separation, is

\begin{align*}
E_{\text{rxn}} &= \sum_{i}^{\text{products}} E_{i} - \sum_{j}^{\text{reactants}} E_{j} \\
&= E_{\ce{A \bond{....}\bond{....}\bond{....} B}} - (E_{\ce{A}} + E_{\ce{B}}) \\
&= E_{\ce{A}} + E_{\ce{B}} - (E_{\ce{A}} + E_{\ce{B}}) \\
&= 0.
\end{align*}

The reasoning is that energies are extensive, and there is no change in the total energy of the system. The total energy of infinitely separated particles is their sum, as there is no attractive or repulsive interaction between them. They're effectively completely isolated. In that sense, there is no reaction, which is I put reaction energy in scare quotes.

In practice, when you have a finite interaction such as a chemical reaction, the total interaction energy should include electronic, vibrational, rotational, and translational components, of which the latter 3 will grow as temperature increases. Most likely the thermal vibrational contribution is the only significant one.

One missing term that is not dependent on temperature and should be added is the zero-point (vibrational) energy,
$$
E_{\text{ZPVE}} = \frac{1}{2} \hbar \sum_{c}^{\text{normal}\\\text{modes}} \omega_{c},
$$
which is the non-zero amount of non-electronic energy that molecules have even at zero kelvin and is purely quantum mechanical in nature. The ZPVE or ZPE can alter energy barriers non-linearly, since vibrations are not necessarily localized, and is the most significant non-electronic contribution to gas-phase reaction energies.


Does calcium hydroxide react with citric acid in orange juice? I noticed that calcium-fortified orange juice on the shelf at my local grocery store contains calcium in the form of calcium hydroxide.

And I also noticed a couple other ingredients that ordinary orange juice doesn't list explicitly:  citric acid and malic acid.  Are these meant to neutralize the calcium hydroxide to form calcium citrate (commonly listed on calcium supplement pill bottles)?

I'm just curious. I want to know what final chemical(s) will end up in my stomach if I swallow this concoction.

Don't know if this is the brand but here is such a label

Therefore calcium hydroxide is an actual additive. In this case advertised along with vitamin D (calcium fortification).

As for what gets to your stomach, there are reasons why ingredients, not compounds produced via cooking are listed. Cooking unlike chemical synthesis isn't supposed to create specific compounds, but food in which all compounds are mixed and every possible equilibrium happens at once.

In the juice there's just more calcium cations and pH may be slightly affected, if it was significantly changed you'd find it by your taste.


Reason for protonation of amino group in amino acids at near neutral pH What happens to the amino group when amino acid is added to a near neutral solution? How does it come about gaining an extra proton?
I understand what happen to the carboxyl group, since it is a strong acid it will dissociates quickly to donate a proton.

Prerequisite knowledge

A proton is deprotonated when the $\mathrm{pH}$ exceeds the $\mathrm pK_\mathrm a$ of the proton.

Analysis
There are 3 different forms that an amino acid can exist in. For the sake of simplicity of analysis, I am taking the simplest amino acid - glycine.
$$\underset{\mathrm{glycine}}{\ce{HOOC-CH2-NH2}}$$
Continuing to state the conditions of analysis, let's take a highly acidic medium to begin with and observe the changes that take place on increasing $\ce{pH}$.
Case 1: pH of solution < 2.34
In such a scenario, the $\mathrm{pH}$ is lower than the $\mathrm p K_\mathrm a$ of the acid group and lower than the $\mathrm p K_\mathrm a$ of the amino group. This means that both the amino group and acid groups are protonated. The amino acid is positively charged at this $\mathrm {pH}$.
$$\underset{\mathrm{glycine\,(pH \,<\,2.34)}}{\ce{HOOC-CH2-NH3+}}$$
Now, we increase the $\ce{pH}$ to a value greater than $2.34$ but less than 9.6  Specialty of these numbers? We'll see at the end.
Case 2: pH of solution greater than 2.34 but less than 9.6
In such a scenario, the $\mathrm{pH}$ is higher than the $\mathrm p K_\mathrm a$ of the acid group but lower than the $\mathrm p K_\mathrm a$ of the amino group. This means that the amino group remains protonated and the acid group is now deprotonated.
$$\underset{\mathrm{glycine\,(2.34\,<\,pH \,<\,9.6)}}{\ce{^-OOC-CH2-NH3+}}$$
Case 3: pH of solution greater than 9.6
In this last case, the $\mathrm{pH}$ is higher than both the $\mathrm pK_\mathrm a$s of the amino group and the acid group. This means that both groups are now deprotonated. The amino acid is now negatively charged.
$$\underset{\mathrm{glycine\,(pH \,>\,9.6)}}{\ce{^-OOC-CH2-NH2}}$$

Observations and Conclusion
As you noticed, there are some specific numbers that are so important they are mentioned. They are the $\mathrm p K_\mathrm a$ values of the acid group ($\mathrm pK_{\mathrm a_1}$) and the amino group ($\mathrm pK_\mathrm {a_2}$). There is one more important number and that is the exact point of net average neutrality. It is known as the isoelectric point.
$$\mathrm {pI} = \frac{\mathrm pK_\mathrm {a_1} + \mathrm pK_\mathrm {a_2}}{2}$$


Is lithium L-threonate a potential brain medicine? Lithium ions have known neuroprotective qualities. That would explain its therapeutic benefit in some mental disorders. It is also well known that lithium carbonate must be taken in toxic doses to get even small amounts into the brain. This makes lithium carbonate a very non-optimal medicine. On the other hand, lithium orotate promises to reach the brain much more efficiently and therefore require smaller doses. Unfortunately it contains orotic acid, which is a known carcinogen.

Magnesium L-threonate is a form of magnesium that easily penetrates the blood-brain barrier and therefore confers therapeutic effects on the brain. Is it then conceivable that lithium L-threonate is both safe and effective as a medicine for brain disorders?

Lithium has a diagonal relathionship with magnesium, so in effect, it makes sense to compare it to magnesium compounds. It may also substitute for sodium ions in transporters. I doubt the counter ions is relevant, since most Li coumpounds are quantitatively dissociated. Stable concentrations of lithium are only achieved after several week of treatment. Lithium carbonate, citrate and sulfate are used to my knowledge. Plasma concentrations are monitored so as to avoid reaching toxic levels (more than 1 milli molar). Some physicians also test for the concentration inside red blood cells to get an idea of the concentration in the brain, but this has not been confirmed to be relevant. What is important is not the amount of it, but the clinical effect : is the patient responding to treatment ? does he/she support it ?

Lithium has not only neuroprotective effects, but also mood-stabilizing effects. It is mostly used in case of bipolar disorder (mood swings), but also as an antidepressant. The mechanism of action is still elusive. It may enhance serotonin pathway, and, by competing with magnesium ions, inhibit some magnesium-dependent enzymes (typically enzymes handling MgATP). One of these enzymes is called glycogen-synthase-kinase (GSK).


How do I solve this problem on mass balance from blast furnace iron making? Question

An ironmaking blast furnace produces hot metal of the composition: $\pu{93.6\%}$ $\ce{Fe},$ $\pu{2.1\%}$ $\ce{Si},$ $\pu{3.6\%}$ $\ce{C},$ $\pu{0.7\%}$ $\ce{Mn}.$ The composition of the iron ore is $\pu{78\%}$ $\ce{Fe2O3},$ $\pu{9\%}$ $\ce{SiO2},$ $\pu{5\%}$ $\ce{Al2O3},$ $\pu{1\%}$ $\ce{MnO},$ $\pu{7\%}$ moisture. Coke rate is $\pu{850 kg/thm}.$ Composition of the coke is $\pu{80\%}$ $\ce{C},$ $\pu{10\%}$ $\ce{SiO2},$ $\pu{5\%}$ $\ce{CaO}.$ Enough limestone is charged to make $\pu{45\%}$ $\ce{CaO}$ in slag and composition of limestone is $\pu{100\%}$ $\ce{CaCO3}.$
Calculate (i) Weight of ore used per tonne of hot metal production; (ii) weight of slag produced per tonne of H.M.; (iii) weight of limestone used to produce given slag composition.

Doubt
In such problems all I need to balance is iron, silica, $\ce{MnO}$ etc. However, in case of $\ce{MnO}$ the solution in my book reads $\ce{MnO}$ from ore equals $\ce{Mn}$ in hot metal plus $\ce{MnO}$ in slag.
My question is for balancing $\ce{MnO}$: why do I need to drag in $\ce{Mn}$ content? The $\ce{Mn}$ in iron ore should be equal to $\ce{Mn}$ in slag. In fact, I feel that any content apart from $\ce{MnO}$ should not be dragged into picture. The rest of the contents like alumina, iron etc. quite comply to this argument. I would require a vivid justification to this.

"The Mn in iron ore should be equal to Mn in slag." vs
"the solution in my book reads MnO from ore equals Mn in hot metal plus MnO in slag."

It seems that the calculation desired involves breaking down the MnO in ore to Mn + O, and putting some of the Mn into the metal; the rest of the Mn goes into slag, as MnO. The hot metal would contain only metallic Mn, not MnO. It shouldn't be a huge adjustment in the answers.


Photoelectric effect and kinetic energy According to my textbook, electrons which are emitted due to the photoelectric effect have kinetic energy given by:
$$KE_{electron}=h\nu-h\nu_0$$
Where $\nu_0$ is the activation frequency of the metal.

My question is: when electrons are emitted in this way, is all of the photon's energy really converted to kinetic energy in the electron? Is there no loss of energy in this process?

If there is energy lost (which I suspect must be true), is it a sufficiently small amount so that we can say it is negligible for most purposes?

$h\,\nu$ is photon's energy; $h\,\nu _0$ is the potential energy of the electron. All energy excess is converted in kinetic energy of the eletron - that's the left side of the equation.


How could I easily quantify the amount of EGCG in a sample of green tea? I am currently conducting an experiment to study the effects of pH on the rate of thermal decomposition of Epigallocatechin Gallate (EGCG) in green tea.

One of the key steps in my procedure is quantifying this EGCG content in the green tea. What are some of the methods which I could use to determine and quantify this EGCG content?

To analyze specifically this component, you will have to use HPLC.
In case you can use a more generic approach, you can analyze the total phenolic content with the Folin-Ciocalteu method.

The Folin-Ciocalteu method is based on a reaction with polyphenols that gives a blue color.
This color can then be measured in a UV-Vis spectrophotometer and the concentration can be established.

This method has been used to quantify polyphenols in many plants, as for example here.


Which occurs first: nitration or oxidation of aniline by concentrated nitric acid? Which of the cases is possible or both or none?

Concentrated $\ce{HNO3}$ is ‘only’ around $70~\%$. To add a nitro functionality by electrophilic aromatic substitution, typically the $\ce{NO2+}$ cation is required which is formed only by strong dehydrating agents such as sulfuric acid:

$$\ce{O2N-OH <<=>[\ce{H2SO4}][-\ce{HSO4-}] O2N-OH2+ ->[][-\ce{H2O}] O=\overset{+}{N}=O}\tag{1}$$

Concentrated nitric acid is not strong enough to dehydrate iself. Thus, nitration will typically not be observed.


Common ion effect and acids If I have $0.01~\mathrm{M}$ $\ce{HBr}$ in $0.09~\mathrm{M}$ $\ce{KBr}$, why would the presence of $\ce{Br-}$ ions from the $\ce{KBr}$ not reduce the dissociation of $\ce{HBr}$ based on Le Chatelier's principle? Or would it?

The $\ce{Br-}$ would make less $\ce{HBr}$ dissociate, But $\ce{HBr}$ is such a strong acid ($\mathrm{pKa}$ is $\mathrm{-9}$) that still only a trivial amount would be molecular.


How do we calculate a character table from its basis functions This question is a actually a homework question, I'm given the basis functions of all the irreducible representations of $C_{2v}$ group, then I'm asked to complete the character table of $C_{2v}$ point group.

For example, for $A_{1}$ irreducible representation, I'm given the basis functions {$z$, $x^2$, $y^2$, $z^2$, $z^3$, $z(x^2-y^2)$ } which I can choose from one of them.

My questions are:

Why we need to complete the character table based on those basis functions? Simply by using the two orthogonality theorems of characters we can readily fill the character table.
How can I use the basis functions given to complete the character table? Should I represent those symmetry operations by using the basis (x, y, z), then transform the basis to those given to calculate the characters? I think it is too tedious.

I think now I answer my question.

Although in principle, I can calculate the full character table totally from the orthogonality theorems of characters. Namely, I need to use the two relations:

$$\sum_{k} N_k\chi^{(\Gamma_i)}(C_{k})\chi^{*(\Gamma_j)}(C_k) =\delta_{ij}$$

$$\sum_{\Gamma_j}N_k\chi^{(\Gamma_j)}(C_{k})\chi^{*(\Gamma_j)}(C_{k'})= \delta_{kk'}$$

But in practice, it is not always easy to use these two relations to calculate the characters. Even for $D_3$ group which has only two three irreducible representations, we need proper initial guess if we don't want to solve the set of equations.

On the other hand, if we know the basis functions in the first place, then we can effectively get the characters.

For example, if we want to calculate the characters of $A_1$ of $C_{2v}$ group. First, I need to choose a basis function. Let's say I choose $z$ as my basis function (I have listed the basis functions of $A_1$ in my question), since the irreducible representation is of one dimension, I simple need to see how the operations will change the sign of $z$.

There are four classes for $C_{2v}$ group, and the representatives are $E$, $C_2$, $\sigma_v(xz)$ and $\sigma_{v}^{'}(yz)$

$E$
It is evident that $E$ won't change the sign of $z$, so the character is 1.
$C_2$
The principle axis is $z$, so the character is also 1.
$\sigma_v(xz)$
It is a reflection across $xz$ plane, but it doesn't change the sign of $z$, so the character is still 1.
$\sigma_v^{'}(yz)$
It is a reflection across $yz$ plane, the character is also 1.

So, based on the above discussion of using basis function to calculate characters, we see that basis functions can be readily used for calculating characters. Although the example is really trivial (we can write down the characters of $A_1$ even without thinking), but the example indicates that if we need to calculate the characters of $A_2$, we just don't need to solve the set of equations.


First point of difference rule 1,6-dimethylcyclohex-1-ene According to the first point of difference rule, second one should be correct right?

First of all, the ‘lowest sum rule’ mentioned in the picture in the question does not exist in IUPAC nomenclature.

As already explained in a related answer, the most important simplified criteria for the numbering in such cases are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

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

  P-14.4 NUMBERING

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

  (…)

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

  (…)

  (e) saturation/unsaturation:

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

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

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

  (g) lowest locants for the substituent cited first as a prefix in the name;

  (…)

The compound that is given in the question doesn’t have any principal characteristic group that could be expressed as a suffix. Therefore, Rule (c) is not relevant in this case.

Thus, a low locant is assgined first to the double bond according to Rule (e). Therefore, the name of the parent structure without any further substituents is cyclohex-1-ene. (Without further substituents, however, the locant ‘1’ would be omitted.)

Finally, low locants are assigned to the remaining substituents as a set according to Rule (f).

  P-14.3.5 Lowest set of locants

  The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set ‘2,3,5,8’ is lower than ‘3,4,6,8’ and ‘2,4,5,7’.

Thus, the correct name is 1,6-dimethylcyclohex-1-ene rather than 2,3-dimethylcyclohex-1-ene since the locant set ‘1,6’ is lower than ‘2,3’.


Naming a coordination compound I was asked to determine the coordination compound which has the name:

$$\text{triamminetrinitritocobalt(I)}$$

I thought it must be: $$\ce{[Co(NH3)3(ONO)3]^2-}$$

The $-2$ charge can be deduced by adding up the oxidation numbers.

However, in this case, shouldn't the name be written as cobaltate instead of cobalt? Where is my mistake and what did I miss here?

Yes, you are correct: an -ate prefix is used when the complex has a net negative charge and the name should be triamminetrinitritocobaltate(I) ion

Reference:

  If the overall coordination complex is an anion, the ending "-ate" is attached to the metal center. Some metals also change to their Latin names in this situation. The rest of the metals simply have -ate added to the end (cobaltate, nickelate, zincate, osmate, cadmate, platinate, mercurate, etc. Note that the -ate tends to replace -um or -ium, if present).

Chemistry Libretexts - Nomenclature of Coordination compounds


Is lanthanum a lanthanoid? Is lanthanum classified as lanthanoids? If yes, then can somebody explain in detail why La has 6s2 5d1 as its electron configuration and Ce has 6s2 4f1 5d1?

Yes, Lanthanum is classified as a lanthanide although it does not have any electrons in its f orbital.

La does not obey the general rule of lanthanides ([Xe]4f^n 6s^2), but is included in this group due to its characteristics (particularly trivalency) being in accordance with the rest of the group. Such as its atomic radius & ionisation energies.

Ce is another exception to the rule (as are Ga & Lu). It ought to be [Xe] 4f^2 6s^2, but the effective nuclear charge, which is still relatively small, is not great enough for f-orbital contraction to be sufficient for 2e in the f orbital. Thus the energy required to fill the f orbitals with 2e is still greater than the energy required to fill the d orbitals.
That being said, it is equally as difficult to fill the 5d orbitals with both electrons, as the f orbital contraction has lowered the energy required enough such that 1 electron can occupy the f orbitals and the other can occupy the d orbitals.

I understand if this isn't the exact answer you are looking for as many areas of periodicity have discrepancies, but reasons can essentially be derived back to a balance between the stability of the species & the energetics arising from the occupation of the electron in available orbitals.