Comparing rate of reaction of acetic acid and sodium bicarbonate at different temperatures Given the reaction between $\ce{CH_{3}COOH {(l)}}$ and $\ce{NaHCO_3 {(s)}}$, which happens at the same pressure but at different temperatures, is there any way to mathematically compare the rate of reactions?

I thought using the rate equations of order 2, but I use the same concentration of materials. Suppose I could have found the connection between temperature of he material and its concentration, I would be able to find my required rate-indicators.

How to find the connection between temperature and concentration, or how to describe the rate of the reaction more exactly?

Yes, there is, it has been done by others.  Typical acid-base reactions that occur in solution are diffusion controlled reactions. Such reactions do not follow second order kinetics because at virtually every encounter of the acid and base, rapid proton transfer occurs and the reaction is complete. The reaction rate is limited only by how fast the acid and base can diffuse and encounter each other. It is usually easy to identify diffusion controlled reactions because they are extremely dependent upon the rate of stirring or mixing.  They are also very sensitive to the solution viscosity. Rate constants for these reactions are typically $>\pu{10^11 sec^-1}$.  Here is a link to a full paper that provides considerable experimental detail on one way to measure the rate of diffusion controlled acid-base reactions.


How can I build the molecular orbital diagram of the complex tetraammincopper(II) (square planar structure)? I have to build the molecular orbital for the $\ce{[Cu(NH_3)_4]^{2+}}$ complex.

I know I have to use the characters table, the irreducible representations, the rule 5 and, somehow, determine the energy levels of the orbitals sigma and pi.

With this I should be able to build the MO but I don't know how to do it.

PS: I know we usually find this complex as $\ce{[Cu(NH3)4(H2O)2]^{2+}}$, but what I need is a theoretical answer to explain the $\ce{Cu-N}$ bond using MOT.

In principle, it does not matter which central metal you choose and which ligands; if the geometry is identical, the resulting MOs will be identical, too. The problem is as you state that the complex you wish to determine does not have a square-planar geometry (nor is it a tetracoordinated copper) so whatever you calculate will be off. At first, I was thinking of using a nickel(II) analogue (which would be identical save a missing electron) until I realised that $\ce{ls{-}[Ni(NH3)4]^2+}$ likely doesn’t exist and $\ce{hs{-}[Ni(NH3)6]^2+}$ is, of course, octahedral. A little bit of additional searching led me towards the square-planar complex $\ce{[Ni(\textbf{L})](ClO4)2}$, where $\mathbf{L}$ is basically a 3,10-bissubstituted 1,3,5,8,10,12-hexaazacyclotetradecane.[1] We can pretend that this perfectly fits the $D_\mathrm{4h}$ point group (it doesn’t) and that the ‘ammine ligands’ are fully identical (at least they’re symmetry-equivalent) and coordinate exactly from the coordinate axes (well, almost).

The first step is to consider the symmetry of the metal orbitals. We have five of them, and we expect them to group up into one $\mathrm{e}$-representation (two degenerate orbitals) and another three non-degenerate orbitals. A table of the irreducible represenations can be found on meta (soon™) or we can use the one from TU Braunschweig. Thankfully, it already includes the representations of the d orbitals, but for comprehensiveness you can also determine it yourself by thinking ‘does the orbital stay the same ($+1$) or invert lobes ($-1$) if I apply that symmetry operation. Here is the answer for the d orbitals:

\begin{array}{cccccccccccc}\hline
\text{Orb.} & E & 2C_4 & C_2 (z)& 2 C_2' & 2 C_2'' & i & 2 S_4 & \sigma_\mathrm{h} & 2 \sigma_\mathrm{v} & 2 \sigma_\mathrm{d} & \text{Rep.}\\ \hline
\mathrm{d}_{z^2} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & \mathrm{A_{1g}}\\
\mathrm{d}_{x^2-y^2} & 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 & 1 & -1 & \mathrm{B_{1g}} \\
\mathrm{d}_{xy} & 1 & -1 & 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 & \mathrm{B_{2g}} \\
\mathrm{d}_{xz}/\mathrm{d}_{yz} & 2 & 0 & -2 & 0 & 0 & 2 & 0 & -2 & 0 & 0 & \mathrm{E_g} \\
\mathrm{4s} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & \mathrm{A_{1g}} \\
\mathrm{4p}_z & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & 1 & 1 & \mathrm{A_{2u}} \\
\mathrm{4p}_x/\mathrm{4p}_y & 2 & 0 & -2 & 0 & 0 & -2 & 0 & 2 & 0 & 0 & \mathrm{E_u} \\ \hline
\end{array}

$\mathrm{d}_{xz}$ and $\mathrm{d}_{yz}$ must be grouped together; only as a group they will be transformed onto each other by $C_4$ or $S_4$. Always if you transform an orbital into empty space with a symmetry operation, find out which orbital was there and include that into a degenerate group.

Thus, the d orbitals transform as $\mathrm{a_{1g} + b_{1g} + b_{2g} + e_g}$. I also included the 4s ($\mathrm{a_{1g}}$) and 4p ($\mathrm{a_{2u} + e_u}$) orbitals for completeness; we’ll need them.

Next, do the same thing for the ligands. Here, we have to make use of group orbitals, i.e. group all the four ammine ligands together. Once we have the represenation of the group orbitals, we will need to reduce that to their corresponding irreducible represenations. Thankfully, we don’t have to consider whether the orbitals are pure or hybrid; we just need to consider σ-type orbitals and π-type orbitals. See the table below.

\begin{array}{cccccccccccc}\hline
\text{Orb.} & E & 2C_4 & C_2 (z)& 2 C_2' & 2 C_2'' & i & 2 S_4 & \sigma_\mathrm{h} & 2 \sigma_\mathrm{v} & 2 \sigma_\mathrm{d} & \text{Rep.}\\ \hline
4 \sigma & 4 & 0 & 0 & 2 & 0 & 0 & 0 & 4 & 2 & 0 & ? \\
8 \pi & 8 & 0 & 0 & -4 & 0 & 0 & 0 & 0 & 0 & 0 & ? \\ \hline
\end{array}

Here is where your ‘rule 5’ — or more accurately, the reduction formula — comes into play. It allows us to determine which irreducible representations a reducible one is made up of. (For smaller point groups you can often just derive it logically, but I wouldn’t do that for a group this size.) The general reduction formula is given in equation $(1)$ ($n$ is the number of times a given irreducible representation $\chi$, whose character with respect to a symmetry operation is $\chi_i$, is part of the reducible representation $\xi$. $h$ is the order of the group.) and the results for the σ and π type orbitals is given in the table below.

$$n = \frac1h \sum_i \chi_i \xi_i $$

\begin{array}{ccccccccccc}\hline
\text{Group} & \mathrm{A_{1g}} & \mathrm{A_{2g}} & \mathrm{B_{1g}} & \mathrm{B_{2g}} & \mathrm{E_g} & \mathrm{A_{1u}} & \mathrm{A_{2u}} & \mathrm{B_{1u}} & \mathrm{B_{2u}} & \mathrm{E_u} \\ \hline
4\sigma & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
8\pi & 0 & 1 & 0 & 1 & 1 & 0 & 1 & 0 & 1 & 1 \\
\hline \end{array}

By doing this, we see that the σ-type orbitals transform as $\mathrm{a_{1g} + b_{1g} + e_u}$ while the π-type orbitals transform as $\mathrm{a_{2g} + b_{2g} + e_g + a_{2u} + b_{2u} + e_u}$. (That is all assuming I didn’t miscalculate.)

The next step is to start drawing an orbital diagram. The five d-orbitals should initially be on one level corresponding to the ionisation energy of nickel(II). The virtual 4s and 4p orbitals somewhat higher. The π-type ligand group orbital should be on a level corresponding to the ionisation energy of ammonia, the σ-type ones slightly below. I don’t have values here so I’m going to ballpark the relative energies draw a merely schematical diagram for simplicity. Any orbitals that have the same irreducible representation will mix with each other. If there are only two, these will form a bonding/antibonding pair with the antibonding orbital being relatively higher than the bonding one low. If there are more than two, they will form some kind of ladder with intermediate levels preserving the overall orbital count. Scheme 1 shows what I would end up with.

Scheme 1: Approximate orbital scheme for a fully square-planar, ideal $D_\mathrm{4h}$ $\ce{[Ni(\mathbf{L})]^2+}$ $\mathrm{d^8}$ ion.

Note that there are a number of estimates in there. For example, both $\mathrm{b_{2g}}$ and $\mathrm{e_g}$ are made up solely of d-π interactions; I can’t guess a priori which one would have a higher or lesser energy. Likewise for the relative orderings of other orbitals. The only thing clear is that any interactions stemming from the σ orbitals should be stronger than those stemming from π orbitals due to σ typically displaying better overlap.

To get back to your hypothetical $\mathrm{d^9}$ system, just add the missing electron into the $\mathrm{b_{1g}}$ orbital.

Reference:

[1]: M. P. Suh, B. Y. Shim, T.-S. Yoon, Inorg. Chem. 1994, 33, 5509. DOI: 10.1021/ic00102a026.


Hardness and softness are terms used to describe which aspect of acidic behaviour? What do hardness and softness mean?

Is it something to do with the polarisability.

The terms softness and hardness are used in HSAB theory to distinguish Lewis acids and bases by charge density, polarizability, electronegativity, and, in the case of molecular compounds, HOMO and LUMO energies.

Charge density is inversely proportional to atomic/ionic radius and proportional to oxidation state (in case of a neutral atom) or charge (in case of an ion). Hence charge density increases with decreasing radius and increasing oxidation state.

Polarizability is the tendency of the electron shell of an atom/molecule to become deformed in an electric field, which can originate e.g. from the charge of an adjacent ion or the partial charge of a dipole. An electron cloud of an atom or ion is more easily deformed with increasing volume, as it is more diffuse, and its outer electrons tend to be more loosely bound. Therefore, larger atoms and molecules tend to be more polarizable, and higher polarizability is often accompanied by lower charge density or oxidation state.

In the case of molecules, when a Lewis acid reacts with a Lewis base, electrons from the highest occupied molecular orbital (HOMO) of the Lewis base are transferred to the lowest unoccupied molecular orbital (LUMO) of the Lewis acid. How well this transfer of electron density works depends on the relative energies of the acid LUMO and the base HOMO. The stability of the reaction product will increase with the energy difference between its HOMO and LUMO (source).

The HSAB theory predicts that hard Lewis bases react preferentially with hard acids, and that likewise soft bases form stronger bonds with soft acids. For example, $\ce{Li+}$ and $\ce{BF3}$ are hard acids and $\ce{F-}$ and $\ce{NH3}$ are hard bases due to their small size and high charge density/low polarizability of the ions. $\ce{BF3}$ therefore readily forms Lewis acid-base adducts with fluoride and ammonia, and $\ce{LiF}$ is less soluble in water than $\ce{LiCl}$ and $\ce{LiBr}$. $\ce{Ag+}$ is an example for a soft acid (large ion with low charge), and the halide ions become increasingly softer bases with increasing ionic radius. Silver therefore forms stronger bonds with heavier halides like $\ce{I-}$ and $\ce{Br-}$, and this also explains the decreasing solubility of the silver halides in water, which is classified as a hard solvent. In the case of soft acids and bases, the difference of electronegativity tends to be smaller, and the bonds have more covalent character (e.g. silver iodide) than those between hard acids and bases (e.g. lithium fluoride) with larger difference in electronegativity (source).

The shift of reaction equilibria can also be predicted with this concept, like in the following example (reference):

$$\ce{HgO + H2S \rightleftharpoons HgS + H2O}$$

This equilibrium will be shifted to the right side, because $\ce{Hg^2+}$ as a soft acid prefers to bond to the soft base $\ce{S^2-}$. Likewise, oxygen as a hard base forms stronger bonds with hydrogen (hard acid).


Specific kind of plastic that won't make fabrics smell? I store various fabrics (clothing, cloths, linens, etc.) in RubberMaid RoughNeck plastic bins.  It seems to work fine for a while, but given enough years, the contents take on a strong chemical smell.  These are not stored in a musty or damp area -- they are stored in my living room.  I assume, therefore, that the smell is from the plastic.  Before using the contents again, I need to wash them.  This means competing for machines in the laundry room of a highrise.  This entirely defeats the point of keeping the seldomly used items, which was to have them on stand-by for immediate use upon demand (as opposed to going out to re-buy them).

Are there specific types of plastic that I should look for in a bin to avoid this kind of smell from the plastic?

It is likely a plasticizer in polyethylene bins, rather than polyethylene itself, that has an odor. Storing the garments in clear polyethylene plastic bags (perhaps inside the bins), which have different plasticizers, should reduce the odor -- and some bags are designed to compress clothing with use of a vacuum cleaner, saving space.

Another option is to get an inexpensive wooden (cedar lined?) chest, or a cloth storage bag. Advantage: breathable. Disadvantage: does not protect from moisture.

BTW, some of the plasticizers include polybutylene, dioctyl  phthalate, glyceryl tribenzoate, polyethylene glycol, sunflower oil etc. Handbook of Plasticizers Ch. 11, G. Wypych. Perhaps rancid sunflower oil is the odor?


Why should ether O be named as OXA in a bicyclo ring when it is a substituent? I was asked to name the molecule below .

For the following molecule, since ether is a substituent here (keto is more prior group) my answer was:

3-methoxy-6-methy-7-oxy bicyclo[2.2.1]hept-5-en-2-one

.

My answer is wrong. The correct answer is

3-methoxy-6-methyl-7-oxabicyclo[2.2.1]hept-5-en-2-one

My point is, at C-3, ether is methoxy, but at 7 the same ether is OXA. Ether as substituent outside the bicyclo ring is alkoxy, but inside bicyclo compound it is OXA.
My question is, why should ether O be named as OXA in a bicyclo ring?

The suggested name 3-methoxy-6-methyl-7-oxabicyclo[2.2.1]hept-5-en-2-one is correct. Indeed, the ketone is the principal characteristic group in this case; however, this is not the reason why the other oxygen is named as ‘oxa’. Already the unsubstituted parent structure is named 7-oxabicyclo[2.2.1]hept-2-ene.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), such bicyclic heterocycles are named using skeletal replacement (‘a’) nomenclature (a system in which a heteroatom replaces skeletal atoms of a parent hydride).

  P-31.1.4.4 Bi- and polycyclic von Baeyer heterocycles named by skeletal replacement (‘a’) nomenclature. In heterocyclic compounds formed using skeletal replacement (‘a’) nomenclature, low locants are assigned to heteroatoms, in accord with the fixed numbering of the system, then to unsaturated sites.

Nondetachable prefixes, called replacement (‘a’) prefixes representing the element being introduced, are used to designate the replacing skeletal atoms with their standard bonding number, for example ‘aza’ (N), ‘oxa’ (O), or ‘thia’ (S).


Does chemistry need machine learning? In many fields of science (e.g. biology, medicine, psychology, statistics, physics), machine learning and artificial intelligence techniques are becoming more and more popular to analyze data. Is it also so in chemistry?

The short answer is yes. Machine learning, data mining, AI and other techniques are highly useful in chemistry.

I completely agree with Fred's answer that lots of machine learning, expert systems and statistical analysis in chemistry goes back a long time. This is particularly true in analytical chemistry - match a mass spec or NMR or IR against a library of known compounds.

Now I saw your tag "computational chemistry" and there are some newer applications here. Basically the question is whether you can predict properties (e.g., the heat of formation, dipole moment, etc.) that would normally require quantum mechanics.

There are a few groups attempting this, but one of the more successful attempts so far has come from Anatole von Lilienfeld.

  The researchers found that with a landscape of more than 5000 molecules, the error for predicting atomisation energies of new molecules drops below 10kcal/mol, approaching the 5kcal/mol accuracy of hybrid DFT. 'Calculating a molecule's atomisation energy using hybrid DFT would take on average one hour on a single CPU,' says von Lilienfeld. 'With machine learning, it's milliseconds.'

There are more conventional approaches to use machine learning, evolutionary algorithms and the like to perform parameterization of force fields, semiempirical quantum mechanics, etc.

Basically, if you have a lot of data, machine learning techniques can be effective ways to analyze the data and use it for other purposes.


Mono-addition of HCl to unsymmetrical dienes? I completed 3/4 of the answers, by choosing the correct reaction pathways leading to each of the 3 different products. But I don't know how to draw the reactant diene, I really thought what I initially had was correct, should I make a trans reactant or is there something else missing?

Do we agree that the first step ist the addition of a proton to the diene? This is bit different from the addition of HCl to an alkene, where you would rely on the Markovnikov rule.

Here, stabilization of the cation through resonance comes into play.

  I don't know how to draw it [...]

Draw stepwise!

In principle, a proton could add to one of the positions a to d.

How do the resulting cations look like and which of them are stabilized by resonance?

Is it now a bit more clear?


Determination of brominated aromatics by 13C and 1H NMR spectroscopy How can I identify 4-bromoazobenzene, 4-bromoazoxybenzene, and 4,4’-dibromo-$\ce{N,N}$’-diphenylhydrazine products using $^1\ce{H}$ and $^{13}\ce{C}$ NMR?

I’m really confused by the general shape of NMR peaks of aromatic protons. I want to distinguish each product from the other products, knowing that the three products are synthesized starting by the reaction of 4-bromonitrobenzene with sodium borohydride.

I don't find any experimental spectra online, but I have been suggesting NMRDB.org to students and colleagues. I don't know how accurate the predictions are, but the site is easy to use and for the compounds you're suggesting, rule-based methods should be fairly reliable.

Predict $\ce{^1H}$ NMR
Predict $\ce{^13C}$ NMR

In the 4,4’-dibromo-N,N’-diphenylhydrazine, you should have two sets of "double doublets" (i.e., four peaks in each set of equal height).

In the 4-bromoazobenzene, you break symmetry, so there should be four signals which integrate to two protons, and one additional signal integrating to one proton (i.e., all ~7.3-7.8 ppm, so a big mess).

In the $\ce{^1H}$ spectra of 4-bromoazoxybenzene, the protons closest to the azoxy -N+-O- should be deshielded a bit (maybe ~7.8-7.9 ppm?) compared to the aza analogue. Certainly it's also going to be a fairly congested spectrum.

I'll leave the $\ce{^13C}$ NMR as an exercise (or to others to answer).


Is 6-ethenyl-4-(prop-2'-enyl) octa-6-en-3-ol a valid name? I recently study organic chemistry and I read a book by Sana Jamshaid. I came across this compound and I cannot understand why it named 6-ethenyl-4-(prop-2'-enyl) octa-6-en-3-ol:

I understand that the longest chain is 8, hence octa (but if we go up on the 4th longest can be 9) but I don't understand the 4-(prop-2’-enyl). Are there any explanations to why the compound named in such way? The double bond also sits on the 7th and not the 6th.

The suggested name
“6-ethenyl-4-(prop-2'-enyl) octa-6-en-3-ol”
is not correct; the closest valid name is
6-ethenyl-4-(prop-2-en-1-yl)oct-6-en-3-ol.
However, this name does not describe the compound that is shown in the picture. Note that the parent chain actually is non-7-en-3-ol instead of oct-6-en-3-ol, and the longest side chain actually is but-3-en-1-yl instead of prop-2-en-1-yl.
Therefore, the correct name for the compound that is shown in the picture is
(7E)-4-(but-3-en-1-yl)-6-ethenylnon-7-en-3-ol.


Why butane exist as a liquid in a lighter ? This is my go at explaining after researching ...

Butane is light and normally would be a gas at standard atmospheric temperature and pressure due to Low boiling point (-1degrees)

The butane gas is in liquid state in the lighter . This is because , when it is placed in the body of the lighter , it is placed there under very high pressure as compared to the open air in standard atmospheric temperature and pressure .

Increasing the pressure of the butane gas forces the molecules closer together . This is because increase in pressure will also lead to increase in average kinetic energy meaning molecules need to 'hold' onto each other more to prevent themselves from breaking their bond. (LDF)
With that theory, that means increase in pressure leads to increase in boiling point of the molecules because more heat is needed to break the bonds and move apart .

Hence , When the butane gas are compressed at high pressures , (at a certain pressure for butane) , the alkane will undergo a phase change from gas to liquid.

Now , am I right to say that the vapour pressure is equal to the pressure of the butane gases in the small container , That's why they undergo a phase change immediately ?

Now, am I right to say that the vapour pressure is equal to the
  pressure of the butane gases in the small container, that's why they
  undergo a phase change immediately?

You seem to have answered your own question.

The pressure inside the lighter container is indeed the vapour pressure of the mixture of n-butane and iso-butane that it contains, at ambient pressure.

Here is the pressure-temperature dependence for iso-butane:

(Source)

The high vapour pressure inside means that when you open the valve, a gaseous mixture of both isomers will come streaming out and some of the liquid mixture will spontaneously evaporate to replace it.


Gold nanoparticles in a water/oil environment I am currently investigating the assembly of gold nanoparticles in a mixture of water and oil. The thing is the following: If we have gold nanoparticles (35nm diameter) dispersed in water and we put some oil on it. After that we shake it, then the gold will be kept between the layer between water and oil and now I was wondering why this happens.

So why is it not possible that the gold nanoparticles stay solely in the water or go over in the oil completely? Or it could be possible that the gold particles agglomerate, why does this not happen? Or maybe it does happen and I do not know this.

Does somebody have some background information on this topic?

It is caused by the wettability of the surface of the nanoparticles. For naked particles this will be determined by the properties of gold but, as Daniel mentioned, it is likely that there is a capping agent, in which case the wettability will be governed by the properties of the capping agent. From this point on I will just call this the solid.

Collecting at the interface

Depending on the surface energies between the solid and water (sw), water and oil (wo) and solid and oil (so) you can have 2 situations both with their own extreme case. We can understand which of the three by looking at the famous Young's equation$^1$ :
$$\tag{1} \gamma_{so}-\gamma_{sw}=\gamma_{wo} \cos \theta $$
where $\theta$ is called the contact angle and in this case I have defined it through the water phase as shown in the picture below.

If the surface energies are such that a high contact angle occurs then the solid particles will be mostly on the oil side of the interface (situation 1). But given the surface energies it is still beneficial (energetically) to have some contact with the water, hence the angle. If the surface energies are such that $\cos \theta < -1 $ it means that there is no longer a contact angle and it is preferable for the solid particles to be completely submerged in the oil.

Situation 2 is exactly the opposite, the solid 'likes' the water resulting in a low contact angle. Here again an extreme case can occur in which the equation requires $\cos \theta > 1$ which means that the particles will fully submerge in the water.

Because many liquid-liquid-solid combinations have at least some contact angle (i.e. not 0 and not 180) the particles tend to collect at the interface of the two fluids.

Agglomeration

The reasoning for whether agglomeration will occur is in fact similar. Here again you will look at the surface energies and determine whether it is energetically favorable to have reduce the solid-liquid area by making solid-solid contact or whether it is favorable to have the solid in contact with the liquid. Capping agents are typically designed to repel themselves thus making sure that agglomeration doesn't occur (because it is usually not desired)

1: note that the original essay doesn't contain the equation in mathematical form, only in wording


If a chiral molecule reacts with an achiral molecule will the product be chiral or achiral? My book says that a pair of enantiomers will react the same way with an achiral substance.

How is this possible? Does this imply that the product is achiral, even though the enantiomers are chiral? If so, why?

The key here is to look at the transition state through which the reaction proceeds.

The short answer is that the transition states are enantiomeric, i.e., also mirror images of each other, and in an achiral environment, enantiomers have the same energy. If the transition states have the same energy, then the reactions proceed at the same rate.

The long answer is, unfortunately, convoluted by the fact that a new bond may be formed in the transition state, creating a stereogenic center. In other words, you might have something like:

$$\ce{($S$)-A + B -> [($S$,$R$)-AB]^{\ddagger} + [($S$,$S$)-AB}]^{\ddagger}$$

These are diastereomeric and the rates through these two transition states is different.

But $\ce{($R$)-A}$ does the exact same thing. So the rate of reactivity is controlled via 4 transition states:

$\ce{[($S$,$R$)-AB]^{\ddagger}}$
$\ce{[($S$,$S$)-AB]^{\ddagger}}$
$\ce{[($R$,$S$)-AB]^{\ddagger}}$
$\ce{[($R$,$R$)-AB]^{\ddagger}}$

Here, 1 and 3, and 2 and 4, are enantiomeric, respectively, and the reaction will proceed through pairs of enantiomeric transition states, but since they are enantiomeric, the two sets transition states provide matched rates from the enantiomeric reactant pair.

So, we can safely conclude that indeed enantiomeric reactants react at the same rate with achiral reagents.


How am I to interpret the phrase "a solution of acid"? The question title is a bit odd, I couldn't find a plainer way of saying it.

I'm given the concentration and conductivity of a solution of a particular acid, let's just call it $\ce{AH}$. I then have to deduce that acid's acidity constant.

My first instinct is to write out the expression for the solution's conductivity... but then I realize I don't know how to do that. I have the coefficients for all the relevant ions (I'm not sure what the technical term is, but essentially I mean the lambdas in the second equation here), in particular, the ions $\ce{H_3O^+}$, $\ce{OH^-}$ and the conjugate base $\ce{A^-}$.

So if I had the concentrations of those ions in the solution, I could solve the problem! But hang on... I have no idea how to figure them out. I know the concentration of the acid in the solution. Suppose the acid has just been added to the water. Then it won't have started to disassociate yet, so the water will be exactly as it was before we added the acid: $[\ce{H_3O^+}] = [\ce{OH^-}] = 10^{-7}$. But what if originally there was a lot more acid, but it's had time to react with the water? Then the water will have been filled with an amount of $\ce{A^-}$ dependent on how long ago the acid was dissolved in it.

Do I assume the acid is at equilibrium? The question doesn't specify that.

The acidity constant is defined as the equilibrium constant for the reaction:

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

$$K_\text{a} = \dfrac{[\ce{H^+}][\ce{A^-}]}{[\ce{HA}]}$$

Note: $\ce{H3O+}$ can be considered equivalent to $\ce{H+}$

Since you have to find an equilibrium constant, (and no further info is given) you can safely assume the solution to be in equilibrium.

Plug in the various conductivities in the second equation (Kohlrausch's Law) you mentioned getting the limiting conductivity ($\lambda^o$) of the acid. This is the conductivity of the acid if it had completely dissociated into its constituent ions.

EDIT: To find the limiting conductivity of the acid using the equation, you must have the values of $\lambda^o_{\ce{H^+}}$ and $\lambda^o_{\ce{A^-}}$.

$$\lambda^o_{\ce{HA}} = \lambda^o_{\ce{H+}} + \lambda^o_{\ce{A-}}$$

The degree of dissociation of the acid $\alpha$ (The number of acid molecules which have dissociated at equilibrium per each acid molecule present originally -- it is fractional for weak acids and bases) would be the ratio of the measured conductivity of the solution ($\lambda$, which is given) to the limiting conductivity ($\lambda^o$).

$$\alpha = \dfrac{\lambda_{\ce{HA}}}{\lambda^o_{\ce{HA}}}$$

This alpha would be the same as the ratio of concentration of dissociated acid molecules at equilibrium (since each dissociated acid molecule gives one $\ce{H^+}$ ion, this is equal to the number of $\ce{H^+}$ ions at equilibrium) to the initial concentration of acid taken (which you have mentioned as given).

Therefore,
$$\dfrac{\lambda_{\ce{HA}}}{\lambda^o_{\ce{HA}}} = \dfrac{[\ce{H^+}]}{[\ce{HA}]_\text{initial}}$$

You can find $[\ce{H^+}]$ at equilibrium using this equation and then find the other concentrations using that.

Note that all this assumes that the acid is monoprotic (i.e. there is only one proton released per acid molecule). You would have to change all the equations accordingly in case more than one proton is released.


How would you Synthesise LY-293,284? How would I synthesise LY-293,284?

Figure 1: LY-293,284

I have satisfied the conditions at the page Should we allow Synthesis questions subject to the six constraints listed below? so please don't just write off this question and close it. The constraints are satisfied as followed:

Starter material(s): 5-acetylindole.

Figure 2: 5-acetylindole

I have completed a Google search to check for LY-293,284 synthesis pathways by typing "LY-293,284 synthesis" and there were no applicable results.
I would imagine the synthesis would occur via the following pathways:

Phenylation of the 5-acetylindole to phenyl-5-acetylindole.
Oxidation of the resulting phenyl-5-acetylindole with oxidising agents such as potassium nitrate?
Di-n-propylamination of the resulting 5-acetylindolylphenol with di-n-propylamine to LY-293,284.

The synthesis I'm hoping is simple.
LY-293,284 is a 5-HT1A receptor full agonist.

If you look at Eli Lilly patents and other literature, there doesn't seem to be a simple synthesis... I've put together an example of how it could be done. Although there are many ways you could go depending on starting materials, this way is based primarily on literature reactions with the exception of the last 3 steps which are Eli Lilly patents. I stopped after about 16 steps as I think that's more than enough to demonstrate how complex these designer molecule syntheses can be. I also left out reagents/reaction conditions just to show the general pathway, and to save on eye strain.

So really the simplicity (or complexity) will depend on where you are able to start on the synthetic pathway (e.g. what starting materials are available to you).


TLC: Why is a more polar eluent more effective at displacing the points? If the points are non-polar, I don't see how making the eluent more polar will make them travel faster. Anybody care to explain?

Most likely you are using "normal phase" TLC plates. The stationary phase is silica gel, which has free hydroxyl groups making it a polar surface. In TLC, there is a competition for the plate between the analytes and the eluent. As you make the eluent more polar, the eluent will interact more with the plate leaving the analytes to be carried with the mobile phase.


What does it mean by 'photons of higher intensity'? I found an Assertion and Reason question related to transition energy in Bohr's orbits.

Assertion In the H-atom, photons of higher intensity are emitted when electron falls from the 2nd to 1st orbit than when it falls from the 4th to 2nd orbit.

Reason Transition energy from the 2nd to 1st orbit is greater than for the 4th to 2nd orbit.

The answer says the Assertion is false and Reason is true.

What does it mean by photons of higher intensity? Is it talking about the frequency? But if it was talking about the frequency, then the Assertion would be true.

The assertion is incorrect. Intensity is related to the number of photons emitted and has nothing to do with Bohr orbit transitions.

The reason is correct. Rather than higher intensity, photons of higher energy are emitted when an electron falls from the 2nd to the 1st Bohr orbit as compared to when it falls from the 4th to the 2nd orbit.  This is because the energy emitted or absorbed in a Bohr transition is given by the following equation where Z is the atomic number of the atom, R is a constant and n represents the quantum number of the final and initial Bohr orbits.

$\mathrm{\Delta E=-Z^2R \left ( \dfrac{1}{n_f^2} - \dfrac{1}{n_i^2} \right )}$

since

$\mathrm{\frac{1}{1^2}-\frac{1}{2^2}=3/4}$

is greater than

$\mathrm{\frac{1}{2^2}-\frac{1}{4^2}=3/16}$

more energy is emitted in the first case than the second.


What is the major side product of the reaction between 2-bromo-2-methylpropane and hydroxide ion? I know that one of the products will be 2-methylpropan-2-ol, arising via a $\mathrm{S_N1}$ reaction. But what else is formed?

Specifically, I am looking for a side product that is gaseous.

On addition of hydroxide to t-butyl bromide there are two competing reactions, substitution and elimination. Depending on the conditions one of these reactions will predominate but both will always be present. If you are taking the $S_{N}1$ as the main reaction then the elimination ($E1$) side product will be 2-methylpropene, which is a gas at room temperatures and pressures.


Which mechanism should a reaction between the cyclopentylmethanol and the HBr should follow? I just thought that the reaction should follow $\mathrm{S_{N}2}$ mechanism this is because the alkyl halide is primary but the book says that the reaction will follow $\mathrm{S_{N}1}$ mechanism resulting in the formation of tertiary carbonation and a ring expansion product. Can anyone please explain the reason behind this ?

It's a tricky example, but the $S_N1$ process is favored because the solvent is protic and the resulting 6-member ring is more stable than a 5-member ring.

If ring expansion weren't possible, I believe your intuition is correct that the $S_N2$ mechanism would dominate. See this discussion. You're not alone!


Why is so different the temperatures of a compressed gas compared to a liquid/gas mixture of the same fluid and they are both at the same pressure? Why does $\ce{C3H6}$ gas when compressed (in a centrifugal compressor) comes out with a pressure of $\pu{16 bar}$ and $\pu{90 °C},$ and when it is a mixture of $\ce{C3H6}$ liquid/gas stored inside of a vessel reaches also the $\pu{16 bar},$ but at a lower temperature of around $\pu{38 °C}?$

I know that when it is a vessel with stored $\ce{C3H6}$ liquid, some part of the liquid will vaporize until it reaches the equilibrium with the ambient temperature (vapor pressure). I know that if the vessel is exposed to a ambient temperature of $\pu{38 °C}$ and if has enough $\ce{C3H6}$ liquid to vaporize the equilibrium will be reach at a pressure of $\pu{16 bar}.$

But what I cannot understand is why if it is only gas and if I compress the gas to $\pu{16 bar},$ why does it come out with a temperature of $\pu{90 °C}?$ Shouldn't it come out with the same temperature of $\pu{38 °C},$ like the vessel?

If anyone at least could give me the subject/effect/topic so I can study just to get an answer to this question, I would be very grateful.

Note: This isn't a homework question. I work on a LPG storage facility and the other day I started thinking about this, but I cannot understand why this happens.

Generally, compression of any gas transforms mechanical energy to thermal energy and the gas warms up, until heat dissipates.

If you have ever inflated bike tyres, you would notice the air pump gets warm ( MTB ) up to hot ( high pressure road bikes ).

When air drops down by 1500 m ( from 0.85 to 1 bar ) after passing over mountain range, it warms up by adiabatic compression by 15 deg C. Warming up to 90 Deg C to get pressure 16 bar does not seems to be much.

See Adiabatic_process

A gas, with or without its liquid phase, has 3 variables: temperature, pressure and volume. You adjust 2 of them and the gas adapt to it by adjusting the 3rd one according laws of gas behaviour.

When you keep gas/liquid mixture at the given temperature and volume, it adjusts to it by reaching its saturated vapour  pressure for given temperature  by either evaporation either condensation.

When you set a gas to have a particular volume applying a particular pressure ( by pumping it in or out ), it adjusts to it reaching a particular temperature.

For propane gas and adiabatic compression ( without heat exchange with surrounding ):

$$p_1^{1-\gamma} T_1^\gamma=p_2^{1-\gamma} T_2^\gamma $$

where $\gamma=1.13$

If we consider as initial conditions 1 bar and 20 Deg C:

$$293.15^{1.13}=16^{-0.13} \cdot  T_2^{1.13 }$$

$$ T_2^{1.13 }=   293.15^{1.13} \cdot  16^{0.13} $$

$$ t =  T_2 - 273.15 =  {( 293.15^{1.13} \cdot  16^{0.13} )}^{\frac1{1.13}} - 273.15$$

For given data it leads to about 130 deg C.

If you suddenly compress propane to 16 bar, it would get hot. If you compressed it enough and there was some oxygen, it would  self ignite. This effect  is used in diesel engines.

But if the compression of propane is gradual and gas is in contact with surrounding,  heat gets time to partially dissipate.

If the hot compressed propane is kept at constant external pressure 16 bar, it gradually contracts while cooling down.

When it's temperature reaches the temperature, at which saturated propane vapour pressure, it becomes gradually condensation.

Gradual condensation at constant pressure maintains constant temperature, as dissipative heat loses are compensated by releasing the condensation heat.


Why can't light pass through a gold foil, but alpha particles can? Rutherfords experiment showed that most of the alpha particles passed through a thin gold foil undeflected. So why can't light pass through the foil, which is supposedly thinner than the heavy alpha particles?

Light can pass through a gold foil though, it just has to be thin enough.

Pure gold is a very malleable substance and can be beaten with a hammer into foils of around 100 nm thickness. Sources suggest the gold foil used in the Geiger–Marsden experiment (known more commonly as the Rutherford gold foil experiment) was about 86 nm thick. Somewhere around this region of thickness, gold starts letting light through, as well as alpha particles. It is well known that ~20 nm thick gold (or any other metal) deposited onto a surface by thermal evaporation in a vacuum is essentially invisible and can be used as a transparent conducting surface/electrode. As the foil gets thicker, the gold starts to absorb a noticeable amount of photons, especially in the violet/blue part of the electromagnetic spectrum, while scattering away the red light. The light scattered off a semi-transparent gold film acquires a reddish/yellow tinge, while the light that still does make it through is blueish. When the film reaches 100 nm or so, it becomes practically as opaque as the bulk metal.

In truth, comparing the mean free path or collision cross-sections of photons and alpha particles through a gold sheet is somewhat more complicated than just saying alpha particles should penetrate less because they're massive, so it would be conceivable for a film of appropriate thickness to block photons yet still allow alpha particles through. A very coarse, hand-waving explanation is that photons are absorbed in the electrosphere of a gold atom, while an alpha particle is only significantly pushed away if it bangs almost straight into the nucleus, a region composing approximately $10^{-15}$ of the volume of the atom.


Why oxides with the element in higher oxidation state are more acidic We see this trend of acidic-basic property of oxides in most cases:
$\ce{Mn2O7}$ is acidic, $\ce{MnO}$ is basic, the intermediate oxides are progressively less acidic and more basic, as the oxidation state of $\ce{Mn}$ decreases.

What is the reason of this general trend in almost all cases?

I have seen many explanations of this, most of them involving the proton releasing capacity of $\ce{E-O-H}$ bond, in the hydrolysed oxide. But most oxides of transition metals do not hydrolyze in water. So, I am seeking an alternative explanation without having to involve the hydrolysis of oxide. Also, most explanations on the internet lack any reference, so I prefer a reference.

It does not appear to be always true.  According to Wikipedia, copper(I) becomes acidic, forming an anionic hydroxide complex, at about 1.5 pH unit lower than copper (II).  The higher oxidation state of copper qualifies as less acidic.


Is E1cB possible in the reaction of cyclopentadiene with acetone in presence of ethanolate? When cyclopentadiene 1 and acetone 2 are reacted with sodium ethoxide in ethanol, what product is obtained? I was told that the correct product 3 is formed by nucleophilic addition of cyclopentadienyl anion to the carbonyl group.

However, when I thought about it more, I wondered why an E1cB reaction to form the dimethyl fulvene 4 isn't possible in this case. I believe it should be possible because the aromaticity of the cyclopentadienyl carbanion allows the rate-determining deprotonation of 3 to proceed under strongly basic conditions.

On the other hand, there would be more ring strain in the more unsaturated compound 4.

How do I decide if E1cB is possible or not here?

I believe you were right to conclude E1cB would occur. Both of your considerations are exactly on point. The amount of ring strain is minimal -- if we think that it's OK to deprotonate to form a sp2-like carbon in the first place, then we've already decided that the added ring-strain is not a huge issue. Of course, the added conjugation and the entropic component of eliminating a hydroxide ion also help to drive the reaction forward.

A literature source that confirms our thinking is shown below, for your specific example no less, and it looks like there are many derivatives for which the elimination is observed. [1]

I'll go further to note that Scifinder didn't turn up any reaction results for  for your alcohol product, or any similar derivatives, under equilibrium basic conditions. This suggests that elimination is actually quite likely, and that your answer key is wrong. It is however, possible to isolate the alcohol if acid-base equilibrium between the alkoxide intermediate and solvent/conjugate-acid is minimized.[2]

References

Ragauskas, A. J.; Stothers, J. B. 13C magnetic resonance studies. 119. Tricyclo[3.3.0.0] and [3.3.1.0]octanones from substituted norbornenones via cyclopropanation and homoketonization. Can. J. Chem. 1985, 63 (11), 2961–2968. DOI: 10.1139/v85-491.
Smits, G.; Audic, B.; Wodrich, M. D.; Corminboeuf, C.; Cramer, N. A β-Carbon elimination strategy for convenient in situ access to cyclopentadienyl metal complexes. Chem. Sci. 2017, 8 (10), 7174–7179. DOI: 10.1039/C7SC02986A.


Dihedral angle of gaseous and crystalline HOOH Why does hydrogen peroxide exhibit a dihedral angle of $111.5^\circ$ in the gaseous state? And a dihedral angle of $90.2^\circ$ in the crystalline phase?

I know that in general, there is likely to be more hydrogen bonding in the crystalline phase than in the vapor phase.
But what's up with 1) the bond angles themselves in the first place and 2) the bond angle expansion in the gas phase? How can one rationalize them?
I know that there exist lone pairs on the oxygens and lone-pair/lone-pair repulsion should probably be minimized, but what it is about these conformers that minimizes energy? I'm guessing that we'd also want to minimize eclipsed and gauche interactions ... Newman projection time?
It seems that there is only one Newman projection possible for $\ce{HOOH}$; i.e. we'll have the inevitable lone-pair/lone-pair interaction, and the lone-pair/hydrogen interaction. I guess all we can do is minimize these interactions; i.e. the energy-minimized conformer is probably not the one with them all eclipsed. However, why does it seem that the energy-minimized conformer for vapor $\ce{HOOH}$ is one that puts the lone-pairs so close to each other in terms of dihedral angles?

You've already answered your own question - hydrogen bonding.
First, don't think of it as to why is the gas state different than solid-state.  Rather, think about why is the solid-state different than the gas state. In the gas state, there is very little intermolecular interaction (very, very few species do not follow the ideal gas law to many significant digits). The gas state geometry is close to what is expected for $\mathrm{sp^3}$ tetrahedral geometry and may be considered normal. The link below is a paper discussing the gas phase microwave spectroscopy of $\ce{H2O2}$ in far greater detail than I can hope to here. The description of the torsional rotation energies was not trivial.
Internal‐Rotation in Hydrogen Peroxide: The Far‐Infrared Spectrum and the Determination of the Hindering Potential by
Robert H. Hunt, Robert A. Leacock, C. Wilbur Peters, Karl T. Hecht, The Journal of Chemical Physics 1965, 42, 1931-1946, https://doi.org/10.1063/1.1696228,
mirror
When condensed to the solid-state, molecules get close enough to allow intermolecular forces (in this case hydrogen bonding to a much greater extent than van der Waals). Additionally, the energies of the crystal lattice must be considered.  In this case, th


Why isn't washing up liquid dry Washing up liquid is wet and it is added to water. Why does it not come in a half size package and you add the water yourself? In the same way washing powder is?

There's no technical reason why you couldn't make a powdered dishwashing detergent and I suspect that someone probably does sell it, just as there are liquid, tablet, and powdered automatic dishwasher detergents and laundry detergents. It does seem more convenient to use a liquid detergent for washing dishes in the sink. A liquid doesn't need time to dissolve, is easy to dispense without a scoop and if you're only washing a couple of dishes, you can simply squirt a small amount directly on a dish or use one of those brushes with a detergent reservoir.

I haven't seen any dishwashing detergent that had a full ingredients list, but the one I have says "contains anionic surfactants and enzymes". Anionic surfactant typically implies sodium dodecyl sulfate, which is itself a solid, usually found as a powder. The enzymes, while they could be freeze-dried, could retain more activity if stored in solution. Likewise, the fragrances often added to these detergents may be more suited to a liquid form.


Household Melting Point Standards I'd like to accurately calibrate a digital thermometer I'm building in the range of $50^\circ$ to $120^\circ$ Celsius.

I'm aware that one can buy various chemicals that have melting points in this range, but are there any readily available household / pharmaceutical items I could use instead?

From what I have read, it looks like the two easiest candidates would be vanillin or Sodium acetate (Thanks Curt F.), but I'm not sure how one would go about creating a pure enough calibration standard at home.

According to this cooking site, Crisco brand vegetable shortening melts between $\mathrm{47^oC-48^oC}$ (yes, I know $\mathrm{48^oC}$ is less than $\mathrm{50^oC}$, but it seems pretty close to me). The best way to use such a non-certified melting point "standard" like this is to use your thermometer side-by-side with a properly calibrated thermometer.

Vanillin may be your best bet (melting point of $\mathrm{81^oC - 83^oC}$).  Again, using a calibrated thermometer to check the actual melting point would be the ideal situation.

Although your question asks about calibrating based on melting point measurements of household items, household items just aren't usually of the purity required for accurate melting point based thermometer calibration. In the absence of a calibrated thermometer as discussed above, the best way to calibrate your thermometer is probably to use the freezing and boiling points of water (don't forget to adjust boiling point for altitude if you don't live at sea level), then check the melting points of Crisco and vanillin if you want. Good agreement with these stated melting points would be a good indicator of proper thermometer calibration, while poor agreement could either be due to improper thermometer calibration or impure material.


How do I determine which liquids are immiscible with one another? I am trying to work out which type of ferrofluid I can use to form a weak magnetic seal for a container of liquid photopolymer resin. Is there a mathematical approach to determine what substances will form a solution, and which will remain immiscible, or will I have to buy each ferrofluid and try each one?

If you want high accuracy for solubility, UNIFAC is a good way to go. The basic idea behind it is to break the molecules down into groups then generate correlations for how each of the groups interact. The weighted sum of all the group contributions gives you the total molecular interactions. From there you can get solubility.

The Dortmund Data Bank maintains a comprehensive list of all the group coefficients and offers a program to get the various properties, like solubility. Unfortunately, the free version is pretty limited. You an only access a handful of fluids.

If you want to put in a bit more work, someone has put an Excel version on the internet. It doesn't explicitly do solubility, but basically if the activity coefficients are very far from 1, you can assume they are insoluble.

Furthermore, I'm not sure what the ferrofluids will do. I would guess that the concentrations would be low enough that you could just calculate solubility using the surfactant. Otherwise, you might have to search the literature for how to handle magnetic nanoparticles with UNIFAC.


Number of geometrical isomers when a cyclic ring is involved along with axis of symmetry This is an question from my Chemistry study material and it asks to find the number of Geometrical Isomers of the following compound in the picture.

  Basically the method/formula given in my book to find number of Geometrical
  Isomers of a compound is:

Using this formula I put $n=3$.Morever since both ends are same I get $p=(3+1)/2=2$.So final answer according to me is $2^{3-1}+2^{2-1}=4+2=6$.

But the answer given is $2^3=8$.

Where did I go wrong?

Related: How to derive these general formulae for number of stereoisomers of a compound with a possible plane of symmetry?

The answer in your textbook is correct. Here are all possible isomers:

I added a symmetry axis to the right, so you can see that these compounds are mirror images and are not superimposable. The formula you used for calculations is too complicated. The way to calculate it is simple: there are four stereogenic centers so: $$2^4=16$$There is one axis of symmetry so:$$16/2=8$$Have more faith in your books:)

EDIT:
Because my answer has raised an interesting discussion, I will try to present my point of view. First of all, the cis/trans isomerism is a descriptive way to represent a stereochemistry of a molecule, but its scope is limited. It is useful for teaching organic chemistry, but not for a systematic nomenclature (especially for a ring chirality). I think that the main source of confusion here is understanding the difference between a local and global (absolute) stereochemistry. In the above picture the controversy is whether the four enantiomers in the two rightmost columns can be treated as geometrical isomers. Let's simplify the problem and consider 1,2-dimethylcyclopentane:

The situation is clear for the first two isomers (a and b) - these are geometrical isomers, no doubts about it. What about b and c? We see that in both isomers the methyl groups are in the trans orientation, so one may conclude that they are not geometrical isomers. Period. However, cis/trans notation describes only the local geometry (E/Z notation obviously cannot be used in this example). Isomers b and c have the same relative orientation of methyl groups, but in a global scope these are two different molecules, and thus two different geometrical isomers. Therefore, 1,2-dimethylcyclopentane has one cis and two trans isomers. This may be counterintuitive at first, because there are no such situations in case of double bonds, where the cis/trans notation is more frequently used, but in rings we have an additional layer of complexity. I am aware that some people may disagree with this explanation, but as far as I know there are no official (i.e. IUPAC) directives to resolve this controversy.


Does helium have a 2s orbital? Part 1: According to what I've learned so far, helium is in the first row of the periodic table, so its $n$ should always equal $1$. However, a question in my text asks about helium 2s. What is helium 2s? If helium 2s exists, why does the periodic table list helium in the first row instead of the second?
Part 2:
I believe that He 1s is more stable than He 2s. I've reasoned thus:

The numerals '1' and '2' in 'He 1s' and 'He 2s' express the size of the orbital. The greater number represents a larger orbital.
The electrons of smaller orbitals exist closer to the nucleus than the electrons of larger orbitals do.
The intensity of the attractive force between electron and protons increases the nearer the protons are to the electron
Therefore, the attractive force between the protons and the electron will be greater in He 1s than in He 2s.
The greater the attractive force between the protons and the electron, the greater the stability of the electron.
Therefore the electrons in He 1s are more stable than the electrons in He 2s.

Is that correct?

Note the following is the original question:

For each pair of orbitals determine which is more stable and explain why:
(a) He 1s and He 2s;
(b) Kr 5p and Kr 5s;
(c) He 2s and He+ 2s

Part 1: The elements on the periodic table are arranged according to their ground-state electron configurations. He 2s is an excited state of He 1s. There are even He 3s and 4s depending on whether one gives enough energy for the atom to excite to such levels. In general, higher principal quantum number (n) indicates a higher energy state (less stable)

Part 2: Your reasoning is correct.


How do I synthesise this tertiary amide from these 5 compounds? So basically I'm at a loss as to how to proceed. I started with the idea of converting the primary alcohol into an aldehyde and that is where I stoped. I understand that each substituent of the Nitrogen is somehow made from B, C and D and that the cynaonitrile is supposed to help you make the amide and the Gringnard Reagent is for adding a methyl group somewhere, maybe to C. Im completely at a loss and I'm just getting frustrated now :(

You seem to have got the uses of the compounds right, but could not bring that knowledge together to apply it here. As you said, you first use the Grignard to add a methyl group to D

Let's call this compound K. Then, we use $\ce{NaCN}$ to convert the alcohol into a nitrile, then to a carboxylic acid rather than an amide, as this would be easier to work with for us.

Let's call this compound L. Now we are ready to start creating the actual compound. Nitrate C and then reduce it to create the amine and reduce the ketone.

Then add L to the product, followed by K. This order is followed as adding the alcohol before the acid may create a tertiary amine rather than the amide we need.


Speed distribution of lighter vs heavier gases I cannot understand why the range of molecular speed is not always wider for a lighter gas as compared to a heavier gas .

If the same energy is supplied to both gases then wouldn't the molecular speed of the lighter gas be more?

To my understanding, the speed distribution is wider for lighter molecules. The Wikipedia page for the Maxwell Boltzmann distribution has two images at the top that convey this very well.

In the first plot, $a=\sqrt{k_bT/M}$ where $M$ is the molar mass and $k_b$ Boltzmann's constant. For a Maxwell-Boltzmann distribution, the variance is given by $$\sigma^2= \frac{a^2\cdot(3\pi-8)}{\pi}=\frac{k_b\cdot T\cdot(3\pi-8)}{M\cdot\pi}$$ So, this shows that the speed distribution will be wider for higher temperatures and smaller molar mass.


What compound does this chemical formula represent? I recently saw this:

What chemical does this formula represent?

The structure represents adrenaline or 'epinephrine' (en.wikipedia.org/wiki/Epinephrine). As in:

  Go out and have an adrenaline rush on your birthday!!

The image on wikipedia has a slightly rotated ring, so doesn't look identical.


Chemical expression and terminology Is this expression correct "the reaction was not accomplished " when we check the final product finding two or more spots on TLC paper ?

Because "finding two spots" was difficult to understand, my answer to this includes an idealized graphical representation of a TLC.  It consists of two lanes A, and B for two starting materials. There is a reference lane (D) of the already known, confirmed and purified product -- for example obtained by a previous batch of synthesis, or (as met once in lab classes) the TA offers to spot it, using his/her known reference.
If lane C was used to co-spot both the starting material(s) as well the reaction product, yet you do not detect (by UV-Vis, or derivatization) a spot corresponding in $R_f$-value to your reference (in lane D), than your analyte analyzed does not contain significant amounts of your intended product.

Depending on how / if at all a work-up prior to spotting the sample of the unknown was executed, this observation indeed may refer to "the chosen conditions did not yield significant amounts of the intended product".
Yet if you forgot to quench/neutralize (and subsequently extracted) your analyte sample prior to spotting, your product may be still at the ground line -- esp. for salty samples on a silica TLC -- and hence not elute at $R_f = 0.75$. Second potential pitfall I imagine: you quench your reaction with water, and extract at different pH values (separation of strongly acidic / acidic / basic compounds) -- your intended product simply is in the subsequent extract that accidentally was not spot on lane C.

The TLC was idealized that the compounds would nicely separate from each other, without any smearing/fronting/tailing either individually, nor in presence of each other (+ other components of the reaction mixture not detected by the TLC [e.g. solvent(s)]).  Simultaneously, it is assumed there is no decompostion of any of the products while eluting the TLC, that could be checked by 2D-TLC.


Why does a lowered dielectric constant cause RNA/DNA precipitation in ethanol precipitation? For ethanol precipitations, as mentioned in https://en.wikipedia.org/wiki/Ethanol_precipitation, given that the attraction of water molecules to charged ions like phosphates is based on Coulombic attraction, shouldn't the water-phosphate attraction be increased in magnitude by the same ratio as that by which the phosphate-cation in increased in magnitude? I.e., should there not be a lack of increased "preference" for phosphate-salt cation ionic bonds (and thus precipitation) over water-phosphate hydrogen bonds (and thus polynucleic acid solvation) as a result of the increased permittivity to charge-charge interactions that ethanol brings over water?
An argument based on the relative higher amounts of cations and ethanol, and lower in water, would make sense in increasing the likelihood of a cation-phosphate bond over a water-phosphate hydrogen bond, but then any mention of the dielectric constant in explaining ethanol precipitation seems entirely superfluous.

For ethanol precipitations [...] shouldn't the water-phosphate attraction be increased in magnitude by the same ratio as that by which the phosphate-cation in increased in magnitude?

The interaction between negatively charged nucleic acid and positively charged counter ions is an ion:ion interaction. The interaction between ions and water is an ion:dipole interaction.

  An argument based on the relative higher amounts of cations and ethanol, and lower in water, would make sense in increasing the likelihood of a cation-phosphate bond over a water-phosphate hydrogen bond, but then any mention of the dielectric constant in explaining ethanol precipitation seems entirely superfluous.

The presence of water between ions of opposite charge diminishes their long-range attraction (the dielectric constant is related to how well solvated ions are "shielded" by the oriented dipoles of the solvent).
Ion:dipole interactions are short range and rely on the dipole to be properly oriented. Adding ethanol to the aqueous DNA solution does not make the ion:dipole interactions of water with ions stronger. In contrast, it does make the ionic interactions stronger because ethanol is less effective in shielding charges than water.

It is important to compare dielectric constants of water and ethanol because both molecules are uncharged and polar (and capable of ion:dipole interactions), but water is the better solvent for ionic compounds for the reasons discussed above.

  [from comments] I don't understand why [we add] ethanol. Ion-dipole interactions and ion-ion interactions are both interactions based on charge and should thus both follow Coulomb's law. Ion-dipole interactions are certainly weaker, and hence shorter range, but that doesn't seem like it should exempt them from Coulomb's law. Dipole orientation is something I hadn't considered.

A dipole has a positive and a negative partial charge that add up to zero. Imagine you are the cation that interacts with a dipole. When the dipole is near you, the negative partial charge will be right next to you, and there will be a large attraction. The positive partial charge will be a bit further, so there is a smaller repulsion. Overall, it adds up to an attractive force. Now move the dipole away from you, maybe by a factor of 10. Now, the positive charge will attract much less, and be at approximately the same distance as the negative charge, so the repulsion and attraction will almost cancel out.

Take a look at pentavalentcarbon's answer to a question on intermolecular interactions. There is a nice table with the dependence of interaction strength on distance.


Synthesis of ZnO by wet chemical method and reducing particle size by ball milling I synthesized bulk $\ce{ZnO}$ which is pure white in colour, but after high energy ball milling colour changes from white to yellow, why?

Figure 1: Zinc oxide (courtesy of Wikipedia).

I had a parallel experience and noted it in my thesis, 20 years ago.

Was working on an oxide ternary phase diagram (not wet chemistry, traditional shake and bake).  There were some conducting phases (dark) and some insulating ones (white).  I had a ZnO pellet (post firing) and I ground it under acetone (just mortar and pestle).  There was a definite color change to yellow.

I had a lot of other results and never ran this issue to ground.  I did note it and speculated that it had to do with the grain interiors having some conductivity (e.g. from oxygen defects or from impurities) and thus a different color, with the grain exteriors more insulating and white (not having the defects).  This would explain the color change.  There have been papers written (more than the 20 years ago) about differences in defects at grain boundaries and interiors of ZnO.

Very much a speculation at the time and never looked into it more.  But very interesting to see you have a parallel experience.  Maybe you will do more than I did and run it to ground?

P.s.  Consider the appearance of In2O3 and Ga2O3.  Indium oxide out of the jar is yellow and has semiconducting properties, from oxygen defects or the like.  Gallium oxide is much more insulating and is bright white.  Perhaps your ZnO grain interiors are a little In2O3 ish?  After all, both materials are dopable (despite the large band gap) and are TCO materials.  Ga2O3 on the other hand is extremely insulating and undopable, with a wider band gap.


Show that this mechanism is consistent with the rate law - how to? The problem is from Principles of general chemistry, 2nd edition by Silverberg, chapter 16 problem no.81:

  "Even when a mechanism is consistent with the rate law, later work may show it to be incorrect. For example, the reaction between hydrogen and iodine has this rate law: $rate = k\ce{[H2][I2]}$. The long accepted mechanism had a single bimolecular step; that is, the overall reaction was thought to be elementary:

  $$\ce{H2_{(g)} + I2_{(g)} -> 2HI_{(g)}}$$

  In the 1960s, however, spectroscopic evidence showed the presence of free I atoms during the reaction. Kineticists have since proposed a three-step mechanism:

  $$\quad \ce{ I2_{(g)} <=> 2I_{(g)} \quad\quad[fast]}\tag {i }$$
  $$\ce{H2_{(g)} + I_{(g)} <=> H2I_{(g)} [fast]\tag{ii}}$$
  $$\ce{ H2I_{(g)} + I_{(g)} -> 2HI_{(g)} [slow]}\tag{iii}$$

  Show that this mechanism is consistent with the rate law."

I cannot understand what the problem asks. I think, this reaction is consistent with the rate law by the third reaction, $\ce{H2I + I->2HI}$, which is the rate-determining step. Is my guess correct?

You have a three step mechanism where only one step is slow and the other steps are fast.

Since the first two steps (which are, in fact equilibria) are fast, it can be assumed that each of the equilibria holds throughout the reaction. Therefore, the expression of the equilibrium constant for each step can be used:
$$\ce{K_1=\frac{[I]^2}{[I2]} \,\,\, and \,\,\, K_2=\frac{[H2I]}{[H2][I]}}$$
So, $\ce{[I]=\sqrt{\ce{K_1[I_2]}}\,\,\,\,and,\,\,[H2I]=K_2[H2][I]=K2[H2]\sqrt{\ce{K1[I2]}}}$

As only the last step is slow, the rate of formation of the product (which is the rate of the overall reaction) only depends on the last step.

$$\ce{rate=\frac{1}{2}\frac{d[HI]}{dt}=k_s[H2I][I]}$$
Here, $\ce{k_s}$ is the rate constant of the last, slow step. Substituting the expressions for $\ce{[H2I]}$ and $\ce{[I]}$ in the rate law should give you an expression of the form $\ce{rate=constant\times[H2][I2]}$, which is what the given rate law is.


Is a highly oxidized element more electropositive than a lesser oxidized counterpart? My issue

I have a significant problem with the following excerpt from a general chemistry text:

Note that ON stands for oxidation number (state). The excerpt is with regard to polyatomic anions - specifically, chloride ion and its related polyatomic anions.

  It now follows that as oxygen-atom content increases by one, ON for the atom of variable ON increases by two because each added O has an ON of -2 and O is more EN than the atom of variable ON - thus, bonding O to this atom makes the atom more electropositive.

If electropositivity is defined as the ability to stabilize positive charge, how does a highly partially positive central chlorine atom have this ability? How  can EP increase?

If electropositivity is more formally defined as the ability for an atom to give up electrons within a covalent framework (note that this is the converse of the definition of electronegativity), how exactly does a highly oxidized chlorine atom, such as the chlorine atom in the perchloric ion, have the ability to give up any more electrons? It already has no valence electrons to give up if we stick with an ionic model to assign oxidation states!

If we look at this intuitively, and agree that more electropositive elements are more likely to bear positive charges - i.e. sodium (Na) is highly electropositive and it is very easy for Na to be stripped of its one electron to become isoelectronic with a noble gas - then why would chlorine become more electropositive as it is further oxidized?

If we consider the successive ionization energies of chlorine, they become more and more massive! So how in the world would chlorine be more electropositive (willing to give up electrons within a covalent framework) as it is further oxidized?

(I do recognize that ionization energy of an electron is not a measure of electronegativity but still it is highly correlated with EN).

Professor's response

My professor, respectfully, didn't offer much a defense. He said something about how positive charge doesn't exist because it's the lack of valence electrons rather than the presence of something material. This, to me, is sidestepping the argument.

He also asked me which would be more likely to bear a positive charge - Cl or O? Obviously Cl; it has electrons way further from the nucleus than oxygen, so of course Cl would more easily be stripped of an electron than O. However, this again, I believe, is a mistake, because we cannot consider electronegativities of elements and expect them to hold within ions and molecules. Carbon, for example, isn't very electronegative at all. But no one doubts the electron-withdrawing powers of a carbonyl carbon, or better yet, the electron-stabilizing powers of a ... CARBOCATION.

Further, last semester, I distinctly remember correcting him with regard to this electronegativity business; he said that a more negative ion would be more electronegative. I had to correct that statement; a more negative ion would be less able to stabilize valence electron density and less able to attract valence electrons if bound within a covalent framework, so the correct term would be instead more electropositive.

Does anyone here agree with me? Sometimes I feel that it's not the argument that matters as much as your tenacity and ability to present the argument. I lacked both tenacity and presentation skills when presenting the above argument.

I came close to telling him that his entire take of electronegativity was fundamentally flawed when I told him that "I don't associate a positive charge with electropositivity and a negative charge with electronegativity" but that was met with nothing to the contrary (which would have been welcome!) - because no, positive charge has nothing to do with electropositivity; in fact, it suggests the opposite of electropositivity!

There are two definitions of electropositive:

  1:  charged with positive electricity

  2:  having a tendency to release electrons

The author is using the first definition, while you are thinking of definition 2, which is the one that usually appears in chemistry books.

Instead of "makes the atom more electropositive" it could say "makes the atom more electrically-positive".

For example according to the calculation data from Paul J. Schupf Computational Chemistry Lab

Starting from a -1 charge for Cl-

Adding one oxygen atom (ClO-) increases the charge on the Cl atom to -0.298.

Adding two oxygen atoms (OClO-) increases the charge on the Cl atom to +0.212.

Adding three oxygen atoms (chlorate) increases the charge on the Cl atom to +0.510

Adding four oxygen atoms (perchlorate) increases the charge on the Cl atom to +0.907.

Another example of a different author using the word "electropositivity" in the same way, but who makes the meaning explicitly clear is Interaction of indole derivatives with biologically important aromatic compounds. Part 22. Importance of simultaneous co-operation of hydrogen-bond pairing and stacking interactions for recognition of guanine base by a peptide: X-ray crystal analysis of 7-methylguanosine-5′-phosphate–tryptophanylglutamic acid complex

  MNDO calculations showed that the protonation of the
  guanine N(7) atom increases the electropositivity of the 8-H
  atom [ -0.014 e for guanine and 0.079 e for N(7)-protonated
  guanine],

(bracketed text in original document)

Specifically regarding chlorine, the term "electropositive chlorine" has been used to refer to Cl+, for example in the Canadian Journal of Research

Another example to look at is from the thesis Synthesis of functionalised silanes for use in the asymmetric allylation reaction:

  In silicon based C-C bond forming reactions the important factor key to their
  success is the ability of the coordination number of the silicon atom to be
  varied. As the number of bonds to silicon increases the electropositivity and
  consequently Lewis acidity on the atom increases.[reference 1]

  Upon expansion (Scheme 3; 11 to 13), and ensuing reduction in the s-character
  orbital composition at the silicon centre, the electron density decreases.

  Therefore the electropositivity and Lewis acidity at the silicon centre is
  increased. These tetravalent and pentavalent Lewis acidic silianes have been
  exploited in several Lewis acid-catalysed transformations. Once the shell has
  expanded to incorporate six subsituents it is unlikely for any further extension of
  the valence shell to occur.[reference 2]

  Silicon becomes more positively polarised with the addition of each ligand. As
  this occurs there is a shift in electron density. While the electron density is
  increasing at the ligand, it is decreasing at the silicon centre. The magnitude of
  this polarization is also dependent on the electronegativity of these ligands.

For a very old example, showing the original meaning of "electropositive" to be the first definition, see William Henry's  1823 The Elements of Experimental Chemistry, Volume 1 where on page 229 there is a section "Electro-positive Bodies" which defines electropositive bodies as bodies attracted to a negative electrode and whose inherent electrical state is therefore positive.

Also, there is an historically-interesting article "Electropositivity" and "electronegativity" J. Chem. Educ., 1938, 15 (6), p 291 discussing the inconsistent usage of the term electropositive and advocating for saying "electron donor" instead (if meaning #2 is intended) to avoid confusion.

There is also a 1949 textbook General Chemistry which uses both meanings in the same sentence, using the older meaning to define the newer meaning:

  The electropositive character (tendency to go to a more electropositive condition) of elements, in general, decreases across the periods

Overall, it is confusing to use the word "electropositive" contrary to the definition that usually appears in recent chemistry books, but not really wrong.


What products did this aldol condensation create? I followed a procedure in my organic chemistry lab, but I don't understand what product the procedure created. Can someone tell me what chemicals reacts with which to create the product?

First $\ce{NaOH}$, water, and ethanol were mixed together and cooled. Then acetone was added, followed by benzaldehyde (the reaction mixture then turned yellow/orange). The reaction was stirred for a while. The product was washed with water and ethanol before being recrystallized from ethyl acetate. After crystal formation the product was reduced then cooled.

If I understand correctly the acetone and benzaldehyde reacted together and the ethyl acetate was only used as a solvent?

Straightforward aldol between benzaldehyde and acetone. Enolate of acetone adds to benzaldehyde (which cannot enolise) then eliminates water to give the enone. Product recrystallised from EtOAc.


what is the issue in evaluating two electron integral? What sort of issue is there in two electron integral in quantum chemistry ? Is it a problem of convergence when we do numerically? Can someone suggest any review article or reference articles which has the numerical algorithm to evaluate and to understand it fully ?

The main problem of 2-electron integrals is their amount. They are given by

\begin{equation}
  I_{ijkl} = \langle ij|r^{-1}_{12}|kl\rangle
\end{equation}

where $i$, $j$, $k$ and $l$ run over all $N$ atomic orbitals. So we need to calculate $\mathcal{O}(N^4)$ integrals, which makes this a bottleneck (in terms of memory and computing time).

For Gaussian Type Orbitals we can evaluate the integrals analytically, which is much faster than numerical approaches. But this does not help with the $N^4$ scaling.

Evaluation for Gaussian Type Orbitals is explained in Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory by Szabo and Ostlund for s orbitals ($l=0$). More general evaluation schemes for higher angular momentum are explained in detail in Molecular Electronic-Structure Theory by Helgaker, Jorgenson and Olsen.


Understanding gas stoichiometry for the reaction of xenon and fluorine A $\pu{20.0 L}$ nickel container was charged with $\pu{0.500 atm}$ of xenon gas and $\pu{1.50 atm}$ of fluorine gas at $\pu{400 ^{\circ}C}$. The xenon and fluorine react to form xenon tetrafluoride. What mass of xenon tetrafluoride can be produced assuming $100\%$ yield?

Attempt at solution: From the description, the reaction can be written as
$$\ce{Xe + 2F2 -> XeF4}.$$
I first calculated the amount of substance of $\ce{Xe}$ using the ideal gas law:
$$n(\ce{Xe}) = \frac{pV}{RT} = \frac{\pu{0.5 atm} \cdot \pu{20 L}}{0.08206 \cdot \pu{673 K}} = \pu{0.18 mol}$$
Doing the same for fluorine gives:
$$n(\ce{F2}) = \frac{\pu{1.5 atm} \cdot \pu{20 L}}{0.08206 \cdot \pu{673 K}} = \pu{0.54 mol}$$
Then, for every mole of $\ce{Xe}$ we need two moles of $\ce{F2}$. Since we have more than $2 \cdot \pu{0.18 mol} = \pu{0.36 mol}$ of fluorine, xenon is the limiting reactant. The molar mass of $\ce{XeF4}$ is $207.1\,\mathrm{g/mol}$.
So now I would just do:
\begin{align}
n(\ce{XeF4}) M(\ce{XeF4}) &= m(\ce{XeF4})\\
\pu{0.18 mol}  \times \pu{207.1 g mol-1}  &= \pu{37.278 g}
\end{align}
However, the answer at the back of my textbook says it should be $\pu{37.5 g}$. So did I make a mistake somewhere or is this just a roundoff-error?

Yes, this is a rounding error.
Where you calculate:
$$n = \frac{pV}{RT} = \frac{\pu{0.5 atm} \cdot \pu{20 L}}{0.08206 \cdot \pu{673 K}} = \pu{0.18 mol}$$
The actual unrounded answer is $\pu{0.181072885 mol}$ - a little bit higher than the rounded value.
So, in your final equation:
$$\pu{0.181072885 mol} \times \pu{207.1 g//mol} = \pu{37.5 g}$$
I find using the full value of an intermediate step usually eliminates these types of errors in the final calculation.


Determination of carbon content of steel by liquid chemistry It seems like the typical laboratory methods for determining the carbon content of steel are to cuppellate it in a stream of oxygen and then measure the volume of the evolved $\ce{CO}$ and $\ce{CO2}$, which involves many high-temperature complexities.

Why can't the carbon content of steel be measured by dissolving it in acid and reacting out the carbon at low temperature?

I think cuppellate is a foreign word meaning "burn" essentially.

The basic technique would be to have the iron sample in a small ceramic boat which is put into a hot stream of $\ce{O2}$ so the iron and carbon both oxidize. The iron oxide stays in the boat, and the $\ce{CO2}$ is sweep away and trapped/reacted somehow. We talking small samples 10 grams perhaps.

If you try to dissolve the steel in an acid solution then the $\ce{CO2}$ gas is formed in solution. It is really hard to get a small amount of $\ce{CO2}$ out of solution given that you are starting with a small amount to begin with. Also if you try to scale up such a reaction it gets to be a problem. A $2~\mathrm{cm^3}$ block of iron takes a long time to dissolve in acid.

So choosing the analytical method depends on a lot of factors. Time per sample, sensitivity, and overall cost being the main ones.


How to detect equivalence points? How do we detect equivalence point from a titration of an acid with a basic titrant from thу titration curve ($\mathrm{pH}$ vs. Volume of titrant added)?

Can we predict an equivalence point just by seeing the shape?
Some people say that

  when $$\frac{\mathrm{d(pH)}}{\mathrm{d}V} = 0,$$ equivalence point has been reached.

Is it correct? I doubt if it is correct. It is also written that there is a distinct "break" in the curve at equivalence point. What is this break?

You do not detect equivalence points. What you detect is the endpoint. This is a very critical distinction.

You have some feature you are explicitly measuring, be it pH or color or opacity.

Once this feature changes you have reached the endpoint. The endpoint is not the equivalence point because you will never have exactly the right amount. Frequently, you overshoot a bit.

Also, the reason that the endpoint is detectable is because the transition is very sharp and sudden. That is because near the equivalence point, the pH (and whatever pH derived feature you have) changes very rapidly. The derivative is definitely not zero. In fact, the derivative you've provided is never zero but monotonically increasing or decreasing depending on whether you're tritrating with acid or base. It has to. If you add more base, the pH is only going to increase in value.


How do I (further) lower the pH of a pyrrolidine solution? Pyrrolidine ($2~\mathrm{mol}$) is slowly added to $164~\mathrm{ml}$ conc. $\ce{HCl}$ in $200~\mathrm{g}$ ice water to adjust the $\ce{pH}$ to $3-4$. When it is mixed, the $\ce{pH}$ is $14$, how do I lower the $\ce{pH}$ to $3-4$? I cannot add more $\ce{HCl}$ because it only calls for $164~\mathrm{ml}$.

A pH of ~ 4 would correspond to a solution of pyrrolidinium chloride.  Use less pyrrolidine or more hydrochloric acid to arrive at that point.

The commercial concentrated hydrochloric acid is 37 % (w/w), this would be a concentration of ~ 12 M, and 164 ml of that is slightly less than 2 mol HCl.  It's very strange that you are seeing a basic pH.


Balancing the equation for the preparation of dichromate from chromite ore $\ce{FeCr_2O4 + Na2CO3 + O2 -> Na2CrO4 + Fe2O3 + CO2}$

I have tried balancing it by writing the reduction and oxidation half equations viz. :

$\ce{2Fe^{+II}Cr_2^{+III}O4 -> 4Na2Cr^{+VI}O4 + Fe2^{+III}O3 + 14e^-}$

And:

$\ce{3O2^{0} + 12 e^- -> 2Fe2O3^{-II} } $

Then I multiplied the first equation by 6 and second by 7 and added them and though the electrons got cancelled, it turned out to be really hard to make any progress.

Could someone please provide the technique for balancing such an equation?

As of someone mentioned here, balancing most chemical equations is a a case of trial and error. Your equation is easy to balance by this way than go to extremes.

1) A relatively complicated formula within this equation is $\ce{FeCr2O4}$ on the reactants' side (LHS) of the equation. Since $\ce{Fe}$ atom in it is converted to the resultant product, $\ce{Fe2O3}$ and no other $\ce{Fe}$ bearing products on products' side (RHS), you should start with putting coefficient $2$ in front of $\ce{FeCr2O4}$ to balance $\ce{Fe}$ in LHS.

2) It would make four chromium atoms in LHS. Thus, you need to balance that by putting coefficient $4$ in front of $\ce{Na2CrO4}$ on RHS because it is the only $\ce{Cr}$ bearing molecule among products.

3) That action would make eight sodium atoms in RHS. Thus, you need to balance that extra $\ce{Na}$ by putting a coefficient $4$ in front of $\ce{Na2CO3}$ on LHS because it is the only $\ce{Na}$ bearing molecule among reactants.

3) Now, you would find four carbon atoms in RHS. To balance that extra $\ce{C}$ you may need to put a coefficient $4$ in front of $\ce{CO2}$ in LHS because it is the only $\ce{C}$ bearing molecule among products.

4) Now you have balance all atoms except for oxygen atoms. Thus, count the number of oxygen atoms in RHS and LHS, separately. If you count correctly there are 27 $\ce{O}$ atoms in RHS and only 20 of them in LHS. To balance that extra 7 $\ce{O}$ atoms, you should put a coefficient $\frac{7}{2}$ in front of $\ce{O2}$ in LHS. This would complete the balancing your equation:
$$\ce{2FeCr2O4 + 4Na2CO3 + 7/2 O2 -> 4Na2CrO4 + Fe2O3 + 4CO2}$$

In addition, some people do not like to keep fractions as coefficients in their balance equations. So, you may multify whole equation by 2 to get integers as a coefficients:
$$\ce{4FeCr2O4 + 8Na2CO3 + 7O2 -> 8Na2CrO4 + 2Fe2O3 + 8CO2}$$


Do I understand why heat capacity is called 'heat *capacity*'? The notion of heat capacity seemed a little odd to me. I thought that the word 'capacity' could be throwing me off. Do I understand the notion correctly?

I take the word 'capacity' to express, roughly, 'the maximum amount of something that a thing can take'. My first intuition is that the maximum temperature of a system equals whatever temperature corresponds to every molecule in that system moving at the limit of the speed of light. Accordingly, if 'heat' denotes a certain species of energy transfer, then the heat capacity (i.e. the maximum amount of energy that a system can receive) should be equal to the number calculated by, for each molecule in the system, taking the difference between the speed of light and the molecule's starting speed, summing those differences, and then dividing by the number of molecules in the system. Accordingly, if everything else is equal (i.e. pressure and volume), then the heat capacity of a system is a function of the starting temperature of the system and its volume. At first blush, that conception of heat capacity seems compatible with the equation
$C_x = \frac {Q}{\Delta T_x}$.

Expanding on Nicolau Saker Neto's comments above, the heat capacity dictates the (reversible) heat flux required to increase the system's temperature by one unit.

Definitions of the heat capacity ($C_V$ and $C_p$)

Another way to look at it is that the heat capacity dictates the response in internal energy ($U$)/enthalpy ($H$) of the system to temperature, as described below. Note that the heat capacity at constant volume ($C_V$) is not the same as the heat capacity at constant pressure ($C_p$) - and that both may be defined in several ways, the most common being:

$$ C_V = \left(\frac{\partial U}{\partial T}\right)_{V,\mathbf{n}} = \left(\frac{\partial Q}{\partial T}\right)_{V,\mathbf{n}} $$

$$ C_p = \left(\frac{\partial H}{\partial T}\right)_{p,\mathbf{n}} = \left(\frac{\partial Q}{\partial T}\right)_{p,\mathbf{n}} $$

Contributions to the heat capacity

In should be added that while translation - the kinetic energy of molecules in motion - are an important contribution to the heat capacity of a gas, both vibrations and rotations also contribute for all but the simplest molecules. For high temperatures (approx. $T \geq 3000\ \mathrm{K}) $, the excitation of outer electrons will also contribute to the heat capacity. For even higher temperatures, the substance in question will behave as plasma.

In short

There are many complicating factors to the precise understanding of the origin of contributions to the heat capacity - both at the high end and low end of the temperature range. In all cases, however, the heat capacity will give you the answer to the question:

  "How much energy do I have to add to this system at the given
  conditions in order to increase the temperature with one unit?"

Further reading on Wikipedia is recommended for anyone interested.


Which are the major truths of paradigms on which the whole body of chemistry is built? For example, as evolutionary theory is fo biology or the existence of numbers for math, the existence of an independent reality for physics, etc.

Well the most important driving force I see in chemistry is for atoms and molecules to be most stable when the atoms have a noble gas electronic configeration. Many reactions will occur to try to reduce the energy of the atoms and molecules by gaining / losing or rearranging electrons to bring as many atoms as close as possible to a noble gas configeration.

Another important driving force is to reduce or avoid the mechanical strain in a molecule (Steric effects)

There are effects in pi systems such as aromaticness (4n + 2 pi electrons in rings)

There are entropy effects, by increasing the entropy of a system then an endothermic reaction can be made to occur. For example the dissolution of ammonium nitrate in water occurs with ease but this is an endothermic process.


Why would diclofenac have a higher skin permeability when formulated as diclofenac epolamine rather than diclofenac sodium? There is a suggestion on the literature that the formation of ion pairs creates a permeation enhancement.
Usually only non-ionised compounds penetrate the skin.
Is it possible that the conjugate species does so?
Some sources claim the epolamine salt has enhanced solubility, others say it is about the same.
Appreciate any help

Voltaren is a 1 -2% paste of the ammonium salt of diclofenac (epolamine is hydroxyethyl pyrrolidine) and is used a topical pain reliever. You can smell the ammonia.

It could be that the increased pH of the paste softens skin and makes it more permeable. High pH will peel your skin right off!


Significance of phase of atomic orbitals I am learning about orbitals and bonding and antibonding MOs. So far, I know that when you combine s orbitals, you form both bonding and antibonding MOs- bonding when the wavefunctions are added in phase and antibonding when added out of phase. I have had several questions along the way:

What is the significance, if any, of the phase of the wavefunction for electrons in atoms when the atom is not interacting? Suppose you had a free hydrogen atom in space. Would it make a difference if you switched the phase of its 1s orbital? What about if you had a multi-electron atom? Would there be a consequence of switching the phase of the 1s orbital but not of the other orbitals? My guess is that, if this is at all possible, then yes this would make a difference due to the penetration of higher orbitals- i.e. the 2s orbital has some penetration into the 1s orbital. Thinking about it in this way, I think atoms are most stable when all the s orbitals are in the same phase and the p orbital phases are aligned. Otherwise you would get cancellation of the wavefucntion in regions closer to the nucleus so the electrons would be found further out, and therefore would have a higher potential energy. There would also be more localisation of electrons further from the nucleus so they would have a higher kinetic energy too. I think it therefore would be significant when we change the phase of orbitals relative to other orbitals (whether in the same atom or in interacting atoms), but if you changed the phase of all wavefunctions in the universe right now then it wouldn't matter.

My second question:

How do we form the bonding and antibonding MOs? Is it becaue each orbital is actually in a superposition of being 'phase 0' and 'phase pi' which is why you get both bonding and antibonding MOs? I'm just not sure if superposition of states is at all important here, or if I have misunderstood this.

Finally, when I was considering p orbitals, I noticed that th phase of the two lobes in a single p orbital is opposite.

So here we have two opposite phases existing together. Is there a significance of this? Or is it perhaps again only important in the interaction of p orbitals?

You have quite a few questions indeed!

  What is the significance, if any, of the phase of the wavefunction for electrons in atoms when the atom is not interacting? Suppose you had a free hydrogen atom in space. Would it make a difference if you switched the phase of its 1s orbital?

No, see this question: Physical intuition behind negative values for wave function? and the answers within.

The phase is not related to any of the physical properties of the atomic orbital. For example the probability density of the electron in the orbital is given by $|\psi|^2$. If we write $\psi_\mathrm{ps} = \mathrm{e}^{\mathrm ix}\psi(r,\theta,\phi)$ ("ps" for phase shifted) it is easy to see that the probability density is independent of the phase chosen $x$:

$$|\psi_\mathrm{ps}|^2 = \psi_\mathrm{ps}^*\psi_\mathrm{ps} = \mathrm{e}^{-\mathrm ix}\psi^*(r,\theta,\phi)\mathrm{e}^{\mathrm ix}\psi(r,\theta,\phi) = |\psi^2|$$

so the phase shift has not done anything. Likewise, the energies etc. are independent of any constant factor that is placed in front of the wavefunction. This is easy to show using

$$E = \frac{\langle \psi|\hat{H}|\psi\rangle}{\langle\psi|\psi\rangle}.$$

  What about if you had a multi-electron atom? Would there be a consequence of switching the phase of the 1s orbital but not of the other orbitals? My guess is that, if this is at all possible, then yes this would make a difference due to the penetration of higher orbitals- i.e. the 2s orbital has some penetration into the 1s orbital. Thinking about it in this way, I think atoms are most stable when all the s orbitals are in the same phase and the p orbital phases are aligned. Otherwise you would get cancellation of the wavefunction.

In hydrogen the 1s and 2s orbitals are orthogonal to each other (in general all AOs are orthogonal to all other AOs). This is because the 2s orbital has an inner radial node and two regions which are phased differently (conventionally, positive + negative; however you can draw it negative + positive and the overlap would still be zero).

  How do we form the bonding and antibonding MOs? Is it becaue each orbital is actually in a superposition of being 'phase 0' and 'phase pi' which is why you get both bonding and antibonding MOs?

The absolute phase of the orbital has nothing to do with it, but rather the relative phase. See the linked question above.

  So here we have two opposite phases existing together. Is there a significance of this? Or is it perhaps again only important in the interaction of p orbitals?

I'd be inclined to say that there's no real significance, except that that's how the maths works out. More fundamentally the presence of a node (in this case an angular node) means that there must be regions of different phases. The same applies to the 2s orbital which has a radial node instead of an angular node. This is analogous to the solutions to the particle in a box: the ground state solution has no nodes and is therefore positive throughout, but the first excited state has one node and therefore has one positive and one negative region.


Mechanism of Grignard addition to carbonyls So from the general reaction we can see that the carbonyl is reduced while the Grignard is oxidized to an alkene.

So for this hypothetical reaction (does this really occur?) we are asked to provide a plausible mechanism. This is what I got, but the prof. wants something better - why transfer negative charge from oxygen to an $\text{sp}^3$ carbon?

Click on the image to enlarge, if you wish.

Here are his hints:

  Remember, prior to Grignard reaction the $\ce{Mg}$ center swaps out an ether
  solvent molecule for a reactant oxygen.  This interaction will lead to
  the reaction but in this case, due to steric interference a hydride is
  delivered instead as the Grignard is oxidized to an alkene.  In this
  way the Grignard reagent is almost reacting in the way $\ce{NaBH4}$ would in
  the presence of this ketone. Try to get all of this action working
  together in a single, concerted step.

So ... I get the first part just fine. The $\ce{Mg}$ center has an octet; it's solvated by two ether molecules in solution usually. Okay and in this case the carbonyl oxygen helps solvate the $\ce{Mg}$ center. And then hydride transfer occurs ... but how exactly do we get the hydride to leave without creating a carboanion? And is it really that important that everything be concerted?

This is my second attempt ... if this is all concerted in real life then I must have hit the orbital alignment lottery! All the stars must have been aligned ... !

The only thing I see about my second attempt is that in the end the $\ce{Mg}$ center doesn't have a complete octet - perhaps I should have shown another ether molecule popping in to help $\ce{Mg}$ out with its octet fantasies.

This reaction can be viewed as an example of a pericyclic (concerted) Ene reaction.  The ene reaction is actually fairly common in organic chemistry.   The following diagram details the mechanism as a concerted reaction with a cyclic 6 membered transition state involving 6 electrons (an "aromatic" transition state).

A note on the Ene reaction...

The Ene reaction is really just a variation on the Diels-Alder theme.  The Diels-Alder reaction is a concerted [4 pi + 2 pi] reaction, whereas the Ene reaction is a [(2 pi + 2 sigma) + 2 pi] concerted reaction.