Are alkylidene cycloalkanes planar molecules? Above is an Alkylidene Cycloalkane. Its structure closely resembles that of allenes.

According to my knowledge, allenes are non-planar molecules due to the two $\pi$-bonds being formed on almost perpendicular planes.

So due to the similarity in structures and optical activity, are Alkylidene Cycloalkanes also non-planar?

The alkylidene cycloalkane contains only two C($sp^2$) hybridised atoms, ajacent to each other, as in the instance of a "normal" double bond.  Hence I expect that in the first sphere / in $\alpha$ position to this double bond, all four substitutents adjacent to this double bond may be found almost in the same plane including the axis along the $\ce{C=C}$ bond, too.  The difference to a "normal" case of a single / isolated double bond more frequently seen in textbooks is that two of the substitutents of one side of the double bond are tethered together.  On the level of the double bond, they remain planar.

In the allene example, there are three consecutive C($sp^2$) hybridised atoms, with two adjacent $\ce{C=C}$ bonds. This differs from the instance above, and here the allene protons of one side and their carbon atom they are bond to describe -- as you correctly referred to -- a plane  almost perpendicular to the plane defined by the other two allene protons and carbon atom of the other side of this molecule.  At the level of the double bonds, they are non-planar.

Regarding the methyl group across the cyclohexane, I expect this (again) is preferentially oriented equatorial, rather than in axial alignment towards the main plane of the cyclohexane ring.


Do buffers only work for species of the opposite variety? (I.e. weak acids buffer strong bases and vice versa) I have looked at several sources including Khan Academy, ChemCollective and the answer to a similar question:
Can someone please explain buffers to me?

However, it is not clear whether buffers can only protect against the opposite species. I.e. a weak acid can only protect against the pH change caused by strong bases, or if it can also protect against the pH change caused by strong acids.

This is my thought:

If we have $\ce {H_2CO_3_{(aq)} + H_2O_{(l)}<=>H_CO^{-}_3_{(aq)} +  H_3O^+_{(aq)}}$

Then we add a strong acid $\ce{HCl + H_2O-> H_3O^+ + Cl^-}$

Then we have $x$ amount of protons in the solution and for every one of the weak acid's conjugate base, there is a proton. So, to me it appears that there is no way for the weak acid to act as a buffer for a strong acid. Is this correct or am I missing something?

Think of it in terms of the equilibrium constant:

$$K_{c}\; =\; \frac{[\ce{H3O+}][\ce{HCO3-}]}{[\ce{H2CO3}]}$$
Let's assume for a moment that there are 2 mols of $\ce{H3O+}$, 2 mols of $\ce{HCO3-}$, and 1 mol of $\ce{H2CO3}$. In our pretend scenario, $K_c$ is 4 (however in reality the $K_c$ is much lower and is about $2.5 \cdot 10^{-4}$). When you add an acid, the $\ce{H3O+}$ concentration rises. Lets assume there are now 9 mols of $\ce{H3O+}$. So now the reaction quotient is

$$Q\; =\; \frac{9\cdot 2}{1} = 18$$

Because $Q > K$ the reaction has to shift left in order to compensate and go back to the equilibrium state. If 1 mol of $\ce{H3O+}$ recombines with 1 mol of $\ce{HCO3-}$ to give back 1 mols of $\ce{H2CO3}$ now you have

$$Q\; =\; \frac{8\cdot 1}{2} = 4$$

and because $Q = K$ the reaction is now at equilibrium.

So while the total number of mols of $\ce{H3O+}$ did increase, the effect of it was dampened by the buffer (as it only went up to 8 mols instead of 9 mols).


Why does water evaporate at room temperature? When water temperature reaches $100\ ^\circ \mathrm{C}$, the molecules get so excited that the hydrogen atoms lose the bonds to the oxygen atom and therefore the water starts to become gas. I get that, but at room temperature ($23\ ^\circ \mathrm{C}$), is there no excitation in the atoms or is there?

First, I think I should make it clear that when water boils, the bonds in the water molecule linking the hydrogen and oxygen atom are not broken. During boiling, the intermolecular bonds in water are the ones that get broken, that is the bonds that link the water molecules together.
At room temperature, there is evaporation (I wouldn't call it excitation). This is because there are a few molecules of water which can manage to muster enough energy to escape from the large body of molecules and escape into air.
This can be explained through a graph depicting the distribution of speed among water molecules worked out by Maxwell and Boltzmann.

As you can probably see, there are a lot of water molecules with lower kinetic energy than with higher kinetic energy. Those that have the higher kinetic energy are the ones that are able to break through the water surface to become vapour.
Even at low temperatures, there are some water molecules are have enough energy to escape and that's why evaporation in water can occur at any temperature (yes, even if the water is in ice).
When the temperature increases, there are more molecules with higher kinetic energy and thus, more water can evaporate.


Why is Rf called Retardation Factor? I understand that $$R_\mathrm{f} = \frac{\text{distance traveled by center of analyte spot } (b)}{\text{ distance travelled by solvent front } (a)}$$ What I do not understand is why this is called retardation factor; an analyte that travels further in the mobile phase seems to me to be less retarded by the stationary phase, while one that doesn't travel as far seems more retarded. If we're using this terminology why doesn't $R_\mathrm{f} = 1- \frac ba$?

The clearest answer is that an analyte is retained by both the mobile and the stationary phase and the retention factor is a measure of the ratio between retention in the mobile phase and in the stationary phase.
So you could think of it as a higher Rf value indicates a higher retention in the mobile phase.

(Also, in answer to some of the comments, the retardation factor is the inverse of the retention factor)


When fugacity and z-compressability equations instead of Gibbs free energy equations? In very high pressure environments such as mud circulating systems, conditions for polymers contain fugacity/z-compressability criteria instead of Gibbs free energy, more in the answer. So

When fugacity and z-compressability equations instead of Gibbs free energy equations?

and helper questions to the latter

  when are fugacity and z-compressability equations used?
  When are Gibbs free energy equations used?
  By which criteria can you justify your decision to choose between Gibbs free energy criterion and fugacity/compressability criterion?

Elastic solids (like gelled drilling muds and elastomers) can not be treated in the same way as liquids and gases, using the z factor and fugacity (unless the loading on the solid is purely isotropic).  This is because, at equilibrium, a solid is capable of storing elastic energy (involving anisotropic loading and strains) after it is deformed.  As a result, the equilibrium state of an elastic solid cannot be described in terms of only two intensive variables (like it can for a liquid or gas).  In the case of an isotropic elastic solid, you need to specify either (a) temperature and three principal stresses or (b) temperature and 3 principal strains).  So, rather than 2 intensive variables, you need to specify 4 intensive variables.  All the thermodynamic functions, U, H, S, A, and G are functions of these 4 intensive variables.


Why does water break salt really A kid asked me this question just the other day, and I said that the positive part of $\ce{H2O}$ attracts the negative $\ce{Cl}$ ion and the negative part of the $\ce{H2O}$ attracts the positive Na ion.

Then he asked me several questions which stumped me:

Why doesn't $\ce{NaCl}$ dissolve water instead (break water into $\ce{H}$ and $\ce{O}$)?
If $\ce{H2O}$ is electrically neural, why would it attract any ions?
Why doesn't $\ce{NaCl}$  stick together, how can one know if one molecule will attract the ions hard enough that the ions breaks away?
Why is Na positive and $\ce{Cl}$ negative?

Hopefully someone can help me answer the above questions as I have not done chemistry in years!

1.) The covalent OH bond strength in water is slightly stronger than the NaCl ionic bond strength.

2.) Water might be neutral but each atom exhibits a partial charge.  Oxygen sucks up most of the electron density away from the hydrogen atoms leaving O with a partial negative charge and hydrogen with a partial positive charge.  This gives rise to the famous hydrogen bonding behavior of water.

3.) You would have to analyze the energies of each of the interacting components to know.

4.) Cl is more electronegative than Na due to the larger number of protons in the nucleus.  This means that Cl will suck electron density away from Na as opposed to the other way around.


Use household ammonia damage aluminum? I'm sorry if this has already been asked but I could not find it on a search here though I also could not determine how to search for the presence of two words ('ammonia' and 'aluminum')

I've read mixed results about whether using household ammonia on aluminum is damaging or not. In my case I want to use it to clean baking trays by putting some (~ 1 tablespoon to 1/4 cup) ammonia on the pan and cover it with a plastic bag and set it aside for hours or a day or so to allow it to soften the baked on grime. I also want to and am using it for a variety of other household and tinkering projects. I read on one site that it is a very bad idea to use ammonia on an aluminum bike frame because it will severally weaken it.

Does household ammonia damage aluminum?

Yes, aqueous ammonia is likely damaging in time to metal Aluminum in the presence of air and any salts (from outside dust,...) that may further serve as an electrolyte.

Ammonium salts, themselves are good in transporting solvated electrons, and here you are, unintentionally, likely creating a metal-air battery scenario.

Possible half-cell reactions:

$\ce{ Al -> Al(3+) + 3 e- }$

$\ce{ O2 + 2 H2O + 4 e- -> 4 OH- }$

Resulting in a net cell reaction of:

$\ce{ Al(3+) + 3 OH- -> Al(OH)3 }$

The aqueous ammonia may also assist here in the removal of the Aluminum hydroxide by forming a "dissolved/peptised/supersaturated hydrated alumina" (per a paper 'The Precipitation of Aluminium Hydrous Oxide and Its Solubility in Ammonia' by Prideaux and Henness).

In essence, Aluminum is now a more rapidly corroding anode.

Nice experiment, albeit, unintended.


Meaning of m2 in the symbol for an isotope of an element What does the notation $^{197\mathrm{m}2}\ce{Pb}$ mean? Specifically, the '$\mathrm{m}$2' part.

I've found this and it appears to have something to do with charge distribution. The original notation was found here.

As per this Wikipedia page on isotopes:

  The letter $m$ is sometimes appended after the mass number to indicate a nuclear isomer, a metastable or energetically-excited nuclear state (as opposed to the lowest-energy ground state), for example $\ce{^{180\mathrm{m}}_73Ta}$ (tantalum-180m).

Note that the lower subscript to the left of the chemical symbol is the atomic number of the element (in the example above, 73 is the atomic number for tantalum) and is omitted in your example.


Is mixing ammonia and bleach spray in the same parcel of air considerably dangerous to health? I've read a few articles advising not to mix bleach and ammonia in the same room because it makes chloramine gas. But my boyfriend and I are having an argument - he thinks that since those articles don't specifically say that mixing them in the same room (in our case, our small, poorly-ventilated bathroom) is bad, the only situation in which we shouldn't do this is spraying them directly on the same surface.

I don't see any reason why the chemical reactions producing chloramine gas wouldn't occur if they're mixing in the same room, from our bleach cleaner and glass cleaner spray bottles, though they would be less concentrated. However, he would only trust an answer from a chemistry expert. Please advise!

This information is only for academic and educational purpose only. Please never ever try to mix ammonia and bleach.

Mixing bleach and ammonia is extremely dangerous, since many toxic vapors (nitrogen-chlorine species) will be produced. The primary toxic chemical formed by the reaction is chloramine vapor and hydrazine. (very, very toxic).

The following is the list of chemical species produced by the reaction:

$\ce{NH3}$ = ammonia
$\ce{HCl}$ = hydrochloric acid
$\ce{NaOCl}$ = sodium hypochlorite
$\ce{Cl2}$ = chlorine gas
$\ce{NH2Cl}$ = chloramine
$\ce{N2H4}$ = hydrazine
$\ce{NaCl}$ = sodium chloride or salt (non-toxic)
$\ce{H2O}$ = water (non-toxic)

The bleach decomposes to form hydrochloric acid:

$$\ce{NaOCl → NaOH + HOCl}$$

$$\ce{HOCl → HCl + [O]}$$

And then the ammonia and chlorine gas react to form chloramine, which is released as a vapor:

$$\ce{NaOCl + 2HCl → Cl2 + NaCl + H2O}$$

$$\ce{2NH3 + Cl2 → 2NH2Cl}$$

If ammonia is present in excess, toxic and potentially explosive liquid hydrazine may be formed.

$$\ce{2NH3 + NaOCl → N2H4 + NaCl + H2O}$$

References:

http://chemistry.about.com/od/toxicchemicals/a/Mixing-Bleach-And-Ammonia.htm
http://home.onehowto.com/article/what-happens-when-you-mix-bleach-and-ammonia-7512.html


Why are isomers difficult to separate? I recently learned that attempts to compare the spectra of two isomers of $\ce{C_3H_3^+}$ were frustrated by the difficulty of separating the two species.

What makes these isomers difficult to separate?

Molecules which are made up of exactly the same elements but differ in atomic arrangement are called isomers.

Since the two isomers have the same mass and charge, any seperation techniques which rely on a difference in mass and/or density would be very difficult.

Mass spectrometry is an analytical technique in which components of different mass are separated and detected, typically by ionizing (charging), accelerating through an electric field to a given velocity and finally, 'bending' through a magnetic field. The final position of the is detected as a function of its charge-to-mass ratio.

Any two isomers of $\small\ce{(C3H3)^+}$ have the same charge-to-mass ratio and therefore would be very difficult to separate and detect with mass spectrometry.


Correlation between Coulomb's law and VSEPR theory I've always been told in chemistry lessons that lone pairs repel more due to a greater charge density than bonding pairs. And that makes sense when steric effects are involved when bond orbitals overlap.

But this conflicts with what I have learnt in physics i.e. Coulomb's law, which states that the only thing affecting the force between two charged bodies is the magnitude of the charge and the distance between them. Therefore it seems weird to me that a lone pair should "repel" more because of greater charge density, despite having the same charge magnitude as a bonding pair.

What am I neglecting or if I'm right, does that mean for large atoms (such as iodine, where electron pairs are further apart) the lone pairs and bonding pairs repel equally as per Coulomb's law?

In bonded pairs of electrons, the repulsion of the negative charges is somewhat reduced by the positive charge of the bonded atom's nuclei. Since lone pairs don't have to deal with this positive charge, naturally their repulsion is stronger.


Can the product be called "vinegar"? For a product to be called "vinegar", it must contain at least $5.0$% acetic acid by mass. A 10.0 g sample of a "raspberry vinegar" is titrated with $0.125 \mathrm{M}$ barium hydroxide and required $37.50 \mathrm{mL}$ for complete neutralization. Can the product be called "vinegar"?

My professor said that the product is vinegar, but I did not understand how he solved it. Could you please tell me what to do? What formulas should I use to solve this kind of problem?

0.125M of $\ce {Ba(OH)2}$ means 0.125 moles of $\ce {Ba(OH)2}$ per volume unit (so far I've only seen liters been used as volume unit) .
Assuming that all the $\ce {Ba(OH)2}$ you used was consume on the reaction with the acetic acid, and knowing the ratio of $\ce {Ba(OH)2}$ and acetic acid on the reaction is 1/2 you`d have this reaction:

$$\ce {Ba(OH)2 + 2CH3CO-OH -> Ba(CH3COO)2 + 2H2O}$$

All you need to do is calculate the amount of acetic acid consumed. Knowing the amount of acetic acid consumed based on the amount of moles of $\ce {Ba(OH)2}$ that are present on 37.5mL. Then you need to get the molar mass of acetic acid  and calculate the amount of mass of acetic acid you had. Then divide the total mass of vinegar by the mass of acetic acid  you found, multiply it by 100 and you'll have the mass percentage of acetic acid.


How do flame tests give off color if they only emit the colors when the electrons return to ground state? Since we are constantly holding the element in the fire and the elements only give off the color when the electrons emit the energy that was supplied and return to ground state, why do we only see the colors when in the fire and don't see the colors when we take the supplied energy ( i.e., the fire) away?

The flame area is a violent place due to high temperature favoring ionization. The color appears due to the dynamic process of the formation of short-lived species formed briefly in the flame. As we are burning hydrocarbons, they get split into unstable radicals easily loosing unpaired electrons. In the area of the flame enriched with electrons metal ions from the probe such as $\ce{Na+}$ are reduced to the metal atoms:

$$\ce{Na+ + e- -> Na^0}$$

which are emitting yellow light in a form of observable doublet of spectral lines ($λ_1 = \pu{589.0 nm}$, $λ_2 = \pu{589.5 nm}$) arising from the electronic transitions $\ce{3s^1 -> 3p^1}$. Note that it is the metal atom, not the ion, that is responsible for the emission of light of the given color.

So, as long as we keep the test probe (metal salt) in the flame, we can observe spectral lines of the given element as both conditions are fulfilled (energy supply (heat) and reducing atmosphere (free electrons from the fuel)). By taking the sample out of the flame, electron transitions are no longer possible as there is neither metal atoms capable of any, nor enough energy to promote the electron to the higher level.


Effect of effective nuclear charge increase on the stabilization of 2s and 2p orbitals Are 2s or 2p orbitals more stabilized when going from left to right in the period? There are good arguments for both possible answers:

1) 2p orbitals are more stabilized because they penetrate less in the space of the 1s orbital, so the increase of the effective nuclear charge affects them more. When plotting the 1st ionization energy with the increase of the atomic number, it can be seen that the ionization energy increases with the increase of the atomic number, but the slope is bigger for O-F-Ne (3.97eV) than for B-C-N (3.12eV). 3s and 3p orbitals aren't stabilized so much (because they both penetrate into the space of the inner orbitals), therefore it is much easier to get a high oxidation state of sulfur than of oxygen.

2) 2s orbitals are more stabilized because the electron spends more time closer to the nucleus, hence the orbital is more affected by the increase of the effective nuclear charge. Proof: the MO diagrams for B, C and N differ from the ones for O, F, and Ne, because s-p mixing in B, C and N happens as the 2s and 2p orbitals are similar in energy. The energy difference increases towards the right (because 2s orbital is more stabilized), so 2s and 2p can't mix efficiently in O,F and Ne. If the 2p orbital was more stabilized, then the difference between 2s and 2p would be smaller through the period.

I really need some clarification here.

2s orbitals are stabilized more than 2p orbitals by the effective nuclear charge because of better penetration.

Everything you wrote above is correct except for,

  2p orbitals are more stabilized because they penetrate less in the
  space of the 1s orbital

It is correct that they penetrate less, but they are not stabilized more.  Because they penetrate less, they are stabilized less by the effective nuclear charge.  Electron penetration correlates with electron stabilization.

Edit: Response to OP's first comment

  explain then why 2nd period non-metals can't easily form higher
  oxidation states while 3p elements can?

First off, second period non-metals can form higher oxidation states, for example $\ce{NO3^{-}}$ where the nitrogen is in the +5 oxidation state.  But I understand what you're getting at, and the answer comes back to the difference in effective nuclear charge that 2p and 3p electrons feel.  3p electrons are further from the nucleus and screened by all of the 2s and 2p electrons.  3p electrons are therefore held more loosely than 2p electrons.  This leads to lower electronegativities,

image source

lower ionization potentials

image source

and higher oxidation states being more common for third period non-metals compared to second period non-metals.

Edit: Response to OP's second comment

  3p electrons penetrate well into the space near the nucleus (link)
  Therefore it
  should be easier for 2p elements to form higher oxidation states than
  3p elements, but it's not, how?

Your link (Rose-Hulman) to the 2p-3p radial electron distribution is good, it shows that the 3p electrons are "on average" further from the nucleus than 2p electrons - which is the key point.  Here is a 3-dimensional representation of the same thing.  Look how much further away the 3p electrons are from the nucleus than the 2p electrons.  Given that the 3p electrons are 1) better screened (more interior electrons) and 2) much further away from the nucleus than the 2p electrons, it will much easier to remove more 3p electrons than 2p electrons and achieve higher oxidation states.  This is consistent with the electronegativity and ionization potential arguments presented in my first edit.

Edit: Response to OP's third comment

  the effective nuclear charge has bigger increase for 2p than 2s
  (except N-O) according to this chart. How so?

The key here is that the s orbital is spatially symmetric, it has a spherical pattern.  An electron in an s orbital can screen outer electrons from the nucleus equally in all directions.  On the other hand, a p orbital is directional, it is not spherically symmetric.  An electron in the 2px orbital will not screen outer electrons from the nucleus equally in all directions.

As we proceed from boron to carbon, that first 2s electron is effective in screening the second 2s electron because of its spherical orbit (2s effective nuclear charge increase=0.641).  However, the first 2p electron is not as effective in screening the nucleus because of its non-spherical shape.  So when we add the second 2p electron, it "feels" the nucleus more strongly than the second 2s electron did so the change in effective nuclear charge is greater for the second p electron than it was for the second 2s electron (2p effective nuclear charge increase=0.715).

You noticed the "anomaly" for the 2p electrons as we go from nitrogen to oxygen.  This is because with 3 2p electrons, all of the nitrogen 2p orbitals (px, py, pz) have an electron and produce a spherical electron distribution pattern.  The nucleus is better-screened by the 3 2p electrons in nitrogen, so when we add one more p electron to yield oxygen, the effective nuclear charge increase is noticeably smaller (N -> O 2p effective nuclear charge increase=0.619).


Using RDKit to count "RR, the number of rigid single or fused ring systems in the molecule" I am a computer scientist, not a chemist, working with RDKit. I need to compute something described to me as "RR, the number of rigid single or fused ring systems in the molecule."
I know I can access the rings of a molecule with AtomRings and BondRings. However, I don't know how to select only the "rigid" ones. Perhaps all ring systems are "rigid" and the word is just there for emphasis and redundancy? If not, how can I tell?

My group has thought about this as we pursue better methods for finding conformers which would ideally include flexing rings.
I'm not aware of a current descriptor for "rigid" rings because it's one of those concepts that chemists discuss but is more of a heuristic for "I can move some of the atoms." (Which would imply you need some sort of molecular dynamics to determine - solvable but not fast to compute.)
Let's try something like this:

To get the number of fused ring systems, you'd go through the SSSR (smallest set of smallest rings) and see if a ring bond is also a member of another ring.  If so, that's a fused ring system. You'd then need to do graph traversal to find all ring bonds in the fused ring system (i.e., in cholesterol there's only one fused ring system).

To get "rigid single rings" you'd need to go around the ring atoms and see if there is at least one "flexible" ring atom. The best heuristic I've derived for a potentially flexible ring atom seems to be "both neighboring atoms only have single bonds and are not connected to each other."

This last heuristic would ignore 3-membered rings (e.g., neighboring atoms are connected), benzene (i.e., atoms have double/aromatic bonds), .. seems to get most things right.
IMHO, even fused ring systems can flex, which is why my group has centered on item #2 - finding ring atoms that can move. But if you're implementing someone else's heuristic from a paper, you'll need to carefully do the ring traversal in #1 to accurately count the fused ring systems.


Basicities of nitrogen atoms in purine Arrange the sites in purine in order of basicity.

I feel that 9 is the least basic as nitrogen's lone pair is delocalised. However, I am unable to rank 1, 3, and 7 in a particular order. Is there a particular theory to decide basicity of sites?

I don't think that question is trivial and except for what you have already deduced, I would not have an idea how to solve that without employing computational chemistry.

The problems already start that I suspect that the proton from position 9 readily exchanges to position 7 in any kind of polar solvent (and in absence why not with itself).

Well from that point on forward, I'd say you need at least a lot of acid to protonate 9, as the π system is aromatic and will most likely not react first.

After that you could probably deduce from the approximate bond angles which of the nitrogen would have the largest s- and p- character, then rank them from highest to lowest p contribution. The one with the most will probably be the HOMO and therefore react first. To be honest though, this is just guessing.
I have absolutely no clue what the lesson with this exercise it, except for: "They're pretty much the same."

Let's have a closer look anyway. I have computed the neutral molecule at the DF-B97D3/def2-TZVPP level of theory, and used NBO6.0 for the partial charges:

As expected, there is not much of a difference; I'd even go as far and say they are the same. Position 9 is a little bit more negative, because it has a hydrogen to draw from.

Next up, let's have a look at the highest molecular orbital, as we would assume to protonate there:

Again, there is only little to no difference. Or in numbers (contributions > 3%):

Alpha occ 31 OE=-0.217 is
N1-p=0.3635 N1-s=0.0382
N3-p=0.2551
N7-p=0.0667
C5-p=0.0563
C4-p=0.0473
C2-p=0.0452
C6-p=0.0382

From this we would conclude that since N1 has the largest contribution, the proton is most likely to go there.

Therefore I also calculated all the protonated species, and their relative energies are:
\begin{array}{lr}\hline
\text{Position} & \Delta G(\pu{298.15 K}, \pu{1 atm})/(\pu{kJ/mol})\\\hline
1  &     0.0  \\
3  &    41.2  \\
7  &    26.1  \\
9  &   196.8  \\\hline
\end{array}

From this we conclude that the most likely position to be protonated is N1, followed closely by N7, and also N3. Off the charts is, as expected, N9 as it breaks the aromaticity.

TL;DR: Order of protonation $1 > 7 > 3 > 9$, calculated at DF-B97D3/def2-TZVPP.


Nomenclature of alkoxy and halo groups CH3-O-CH2CH2Cl

How will this compound be named? Both, methoxy and chloro are given equal preference, since they are prefix functional groups. The answer given in the book is 2-chloro-1-methoxyethane, but why is it not 1-chloro-2-methoxyethane, as the C of chlorine comes before M in the alphabetical order?

Your name, 1-chloro-2-methoxyethane, is correct.

Here is a link to the relevant page of the 1993 edition of the IUPAC Blue Book - Nomenclature of Organic Chemistry, which is published online by ACD Labs with permission of IUPAC.

Here are some example ethers and sulfides from that page, and we see a similar case to yours:

There is no difference in preference or priority between halo and alkoxy substituents (or any of the other prefix only substituents). Thus, in a case where there are two such groups and assignment of locants generates the same combination either way, you choose to assign the lower locant number to the alphabetically first.


Mustard gas hydrolysis (at a very high rate) Explain mustard gas hydrolysis at a very high rate:

  $\ce{Cl-CH2-CH2-S-CH2-CH2-Cl}$

I tried searching online but did not find this reaction anywhere. Can someone explain me the steps/mechanism of this reaction? Even some hints might help.

My attempt:

According to Wikipedia

  Solvolysis is a special type of nucleophilic substitution ($\mathrm{S_N1}$) or
  elimination where the nucleophile is a solvent molecule. For certain
  nucleophiles, there are specific terms for the type of solvolysis
  reaction. For water, the term is hydrolysis; for alcohols, it is
  alcoholysis; for ammonia, it is ammonolysis; for glycols, it is
  glycolysis; for amines, it is aminolysis.

It clearly states that hydrolysis/solvolysis can only be $\mathrm{S_N1}$ or elimination. However I'm wondering how can $\mathrm{S_N1}$ occur on a primary carbon! Morever water doesnt seem to be a strong enough base for elimination. So what will be the preferred reaction mechanism?

Luckily I got the mechanism via Google Image Search.

Internal attack by sulphur atom takes place during the reaction. This is called anchimeric assistance, neighbouring group effect or $\mathrm{S_N{NGP}}$. Wikipedia describes it well.


Relevance of strain energy, total energy and stability of molecule? Would anybody please highlight the difference between total strain energy of a molecule, total energy of a molecule and what we can infer about the stability of the molecule using the total strain energy of the molecule as well as the scope of using such inferences made.

Would anybody please highlight the difference between total strain energy of a molecule, total energy of a molecule

Strain energy is usually defined as the sum of all angle and steric strains that a certain molecule possesses, compared to an unstrained analogue molecule.

For example, here is a table that compares various cyclic alkanes to cyclohexane:

wikipedia

In this way, stain energy can be thought of as a positive contribution to potential energy that destabilizes a molecule above the stability that its ideal bonding would suggest.

Total energy of a molecule could be defined in lots of ways, often it is done as a $\Delta H_f$, which could be positive or negative, and compares the potential energy of a molecule (i.e. how strong the bonds are, stronger means lower potential energy) to the elements from which it is formed. However, this will also take into account any intermolecular forces present and these are not inherent to a single molecule, but to a collection of molecules in standard states.

We could also define the energy of a molecule by thinking of the sum of the bond enthalpies, which define how strong the bonds are compared to unbonded atoms (specifically by breaking bonds homolytically, forming radical species, in the gas phase). The larger the value, the stronger the bonds are and the lower the potential energy of the molecule.

In computational chemistry, the total energy of a molecule can theoretically be calculated using quantum mechanical means. This is done using energy minimization software. There is a good post here about the theory in this process. This value will take into account the bond enthalpy/strength and strain within a molecule.

  and what we can infer about the stability of the molecule using the total strain energy of the molecule as well as the scope of using such inferences made.

In general, the greater the strain energy and the higher the total energy, the less stable/more reactive the molecule.


Formal charge of oxygen in protonated alcohol When ethanol is attacked by positive hydrogen ion, how can the oxygen atom acquire formal positive charge?
I tried calculation it as : $\frac{6-4}{2-1}$ which gives answer 3, but the answer should be 1. Can someone explain this?

I guess you need the answer using Lewis dot structure, so I used the same.


Nuclear Fusion and Fission When a heavy nucleus  undergoes nuclear fission , it splits to two smaller, more stable nuclei and produces heat, similarly, when two light nuclei fuse, to make a heavier, more stable nucleus and produces energy.

The produced energy, is due to the mass defect between the reactants and the products, which is transformed to energy, according the mass-energy equivalence.

  But if the products of each process(fission, and fusion) are more stable, that means that they have  more binding energy, thus, shouldn't the mass defect be transformed to nucleus binding energy, rather than emitted as heat or another sort of energy ?

Note that binding energy of nucleons does not mean contained energy, but the energy needed to break it.

Therefore kernels with lower binding energy per nucleon contain more nuclear energy per nucleon and have higher mass per nucleon.

It means forming kernels with higher binding energy per nucleon releases energy and vice versa.

It is the same as for chemical bonds, just some 6-7 orders higher.

E.g. nitroglycerine has lower binding energy per atom than reaction products formed by its explosion.


Can anyone identify this glassware? Hoping for a little help identifying this lab glass , I don't know anything about lab glass , this was in a storage unit I'm purchased at auction,,, there is what I think is called a condense & 2 round beakers or flasks,, one is 50ml & the other is 200ml ,,, all the male & female connector are marked SJ GSA CO 28/15 ...PLEASE is there anything you can tell me about what there are called & what they would be used for.

The glassware isn't particularly valuable. Finding someone who wants it will be the problem. If you were going to buy it new, I'm thinking maybe $150.

It is indeed a distillation setup. The large flask is the pot in which you'd put the stuff that you wanted to boil. The small flask would collect the stuff that you boil off. The nipples on the condenser are for rubber hoses to connect to a water supply so that the condenser jacket can be water cooled. Water goes in on the bottom and out on the top. In order to use this you're missing some parts.

You need a couple of ringstands and clamps to hold the whole thing together.
Two thermometers would go into the top of the condenser. One to monitor the pot temp, and the other to monitor the condensate temp.
You need some metal clamps to hold the flasks to the condenser.
You need something to heat the pot, a Bunsen burner would do.


How to separate paracetamol and caffeine? How do you separate paracetamol and caffeine using high-school laboratory equipment?

I was thinking about using the fact that paracetamol has a lower melting point (169 °C) compared to caffeine (around 235 °C): I would heat them up till paracetamol melts while caffeine is still solid. What I can't figure out is how you filter the caffeine from the liquid paracetamol (filter paper would burn at such a high temperature).

I would like to know how you do the filtration at high temperature; or, if you have a better method, please let me know!

The answer that my teachers had in mind was column chromatography. Essentially we use the van der Waals force of silicon gel to do an elution of the caffeine / paracetamol solution dissolved in water.

More details can be found here:
http://www.chemguide.co.uk/analysis/chromatography/column.html


How can I determine/quantify the presence of boron nitride on a metal surface? We use micron-grade hexagonal boron nitride (hBN) to coat the surface of bullets that are jacketed in gilding metal.  This coating is mechanically embedded/plated onto the surface in order to increase the lubricity of the bullets as they are propelled down the rifled bore of a gun.

We can capture fired bullets in a water tank.  There is some question as to whether any of the hBN survives the temperature, pressure, and friction involved in shooting the bullet through a firearm.

How can we determine whether the hBN coating survives the firing process?  Or better yet, can we characterize the quantity that remains on the bullet, or any part of it, after firing and recovery?

Destructive methods of analysis are fine.

I'd suggest looking into the following:
scanning electron microscopy (should be able to see the film and how intact it is on the surface); Inductively Coupled Plasma tests (destructive).  I believe after your sample is prepared and presented ICP can find B and N emission lines


Finding out the concentration of oxygen in water I read in a biology course that you can determine the concentration of oxygen in water by adding some $\ce{NaOH}$ and then a small amount of $\ce{MgCl2}$. They say that a dark coloured precipitate is formed.
I can't see a way of writing this reaction. Can anyone help me?

This is the old Winkler titration, published as

Die Bestimmung des im Wasser gelösten Sauerstoffes by Ludwig Wilhelm Winkler in Ber. Dtsch. Chem. Ges., 1888, 21, 2843-2854 (SRC)

How does it work?

Add $\ce{MnSO4}$ to your water sample, then $\ce{NaOH}$ and $\ce{KI}$.

Under alkaline conditions, oxygen dissolved in your water sample oxidizes $\ce{Mn(II)}$ to $\ce{Mn(IV)}$.
\[\ce{2Mn^{2+} + O2 + 4 OH- -> MnO(OH)2 v}\]
This is the brownish precipitate mentioned in the question.

Now, the sample is acidified (Winkler used hydrochloric acid) and the solution becomes yellowish due to the iodine formed. Under these conditions, $\ce{Mn(IV)}$ oxidizes iodide:

\[\ce{Mn(IV) + 2 I- -> Mn^{2+} + I2}\]

The amount of iodine formed, which repesents the aount of oxygen in the water sample, can be determined by titration with thiosulfate solution.


The particle in a two-dimensional box model and graphene I have been having trouble with understanding and solving the following problem.

For two-dimensional conjugated systems, we may use the particle in a two-dimensional box model. In this case the total energy can be written as follows:
$$E_{n_1,n_2} = \frac{h^2}{8m} \left( \frac{n^2_1}{L^2_1}+\frac{n^2_2}{L^2_2} \right)$$
where $L_1$ and $L_2$ are the lengths and $n_1$ and $n_2$ are the quantum numbers of the first and second dimensions, respectively, $m$ is the mass of the particle and $h$ is the Planck constant.
Graphene is a sheet of carbon atoms in the form of a two-dimensional hexagonal lattice in which one atom forms each vertex.

The distance between two adjacent carbons in the hexagonal 6-carbon unit is approximately $1.4$ Å. For a square shaped graphene sheet with $L_1$ = $L_2$ = $11$ Å:
1) Calculate the number of electrons in a ($11$ Å $×$ $11$ Å) sheet of graphene. For this problem you may ignore edge electrons.
2) Calculate the energy of the HOMO.
3) Calculate the energy of the LUMO.

For the first question, I simply divided the area of the graphene sheet by the area of a hexagon shaped graphene unit. This resulted in approximately 24 units, which means  there is a total of 48 carbon atoms, since each unit contributes 2 carbon atoms. Therefore there would be a total of 288 electrons in the sheet (from multiplying 48 by 6, since each carbon atom has 6 electrons).
In case this is right, I don't know how to go about calculating the HOMO and LUMO energies. I'm having a bit trouble understanding the quantum numbers in the formula given above: I don't know what the values of the quantum numbers should be when calculating the HOMO or LUMO energy. I do understand how the quantum numbers combinations and energy are related, but I don't know, which specific combinations are used when calculating the HOMO and LUMO energies.

The key to answering this is to remember the following principles:

1) electrons are fermions and therefore two electrons cannot share all quantum numbers (degeneracy of all QM numbers is forbidden)

2) electrons have spin quantum number, which means an energy state corresponding to a given set of $n_1$ and $n_2$ quantum numbers can be doubly occupied (doubly degenerate)

3) the Aufbau principle, which means lowest energy states are occupied first

Once you know the total number of electrons, you have to assign each a pair of quantum numbers $n_1$ and $n_2$ (ignoring $n_s$, the spin quantum #). The first two principles listed above mean that to each $(n_1,n_2)$ state you can assign two and only two electrons (with opposite spin quantum #). The third principle means that, since the energy increases with $n_i$, and the sides of the box are identical (making states that permute the $n_i$ degenerate), you should assign the $n_i$ in increasing order of energy as follows: the lowest E state has $n_1=1,n_2=1$, or (1,1); followed by (2,1) or (1,2) - these are degenerate in energy; then (2,2); then (3,1) or (1,3) -again degenerate; then (2,3) and so on.

In tabular form the energies increase as follows:

$$\begin{array}{c|c|c|c}\hline i & n_1 & n_2 & n_s  \\ \hline 1  & 1 & 1 & \alpha \\   & 1 & 1 & \beta \\ 2  & 2 & 1 & \alpha \\   & 2 & 1 & \beta  \\   & 1 & 2 & \alpha \\   & 1 & 2 & \beta  \\ 3  & 2 & 2 & \alpha  \\   & 2 & 2 & \beta\\ 4  & 1 & 3 & \alpha \\   ...&...&...& \\ \hline\end{array} $$

If you counted 288 electrons you need 144 states that are doubly degenerate (permutations of $n_i$, differing only in $n_s$) to populate the box. The HOMO is the occupied state with highest energy.

Using a script to crack the problem (rather than searching for a mathematical formula that encodes the pattern) produced the following list of states (each can be populated by two electrons, one with $n_s=\alpha$, the other $n_s=\beta$). The HOMO is triply degenerate (ignoring spin) and the LUMO doubly degenerate.

$$\begin{array}{c|c|c|c}\hline i & n_1^2 +  n_2^2 & n_1 & n_2  \\ \hline 1 & 2 & 1 & 1 \\
     2 & 5 & 1 & 2 \\
     3 & 5 & 2 & 1 \\
     4 & 8 & 2 & 2 \\
     5 & 10 & 1 & 3 \\
     6 & 10 & 3 & 1 \\
     7 & 13 & 2 & 3 \\
     8 & 13 & 3 & 2 \\
     9 & 17 & 1 & 4 \\
     10 & 17 & 4 & 1 \\
     11 & 18 & 3 & 3 \\
     12 & 20 & 2 & 4 \\
     13 & 20 & 4 & 2 \\
     14 & 25 & 3 & 4 \\
     15 & 25 & 4 & 3 \\
     16 & 26 & 1 & 5 \\
     17 & 26 & 5 & 1 \\
     18 & 29 & 2 & 5 \\
     19 & 29 & 5 & 2 \\
     20 & 32 & 4 & 4 \\
     21 & 34 & 3 & 5 \\
     22 & 34 & 5 & 3 \\
     23 & 37 & 1 & 6 \\
     24 & 37 & 6 & 1 \\
     25 & 40 & 2 & 6 \\
     26 & 40 & 6 & 2 \\
     27 & 41 & 4 & 5 \\
     28 & 41 & 5 & 4 \\
     29 & 45 & 3 & 6 \\
     30 & 45 & 6 & 3 \\
     31 & 50 & 1 & 7 \\
     32 & 50 & 5 & 5 \\
     33 & 50 & 7 & 1 \\
     34 & 52 & 4 & 6 \\
     35 & 52 & 6 & 4 \\
     36 & 53 & 2 & 7 \\
     37 & 53 & 7 & 2 \\
     38 & 58 & 3 & 7 \\
     39 & 58 & 7 & 3 \\
     40 & 61 & 5 & 6 \\
     41 & 61 & 6 & 5 \\
     42 & 65 & 1 & 8 \\
     43 & 65 & 4 & 7 \\
     44 & 65 & 7 & 4 \\
     45 & 65 & 8 & 1 \\
     46 & 68 & 2 & 8 \\
     47 & 68 & 8 & 2 \\
     48 & 72 & 6 & 6 \\
     49 & 73 & 3 & 8 \\
     50 & 73 & 8 & 3 \\
     51 & 74 & 5 & 7 \\
     52 & 74 & 7 & 5 \\
     53 & 80 & 4 & 8 \\
     54 & 80 & 8 & 4 \\
     55 & 82 & 1 & 9 \\
     56 & 82 & 9 & 1 \\
     57 & 85 & 2 & 9 \\
     58 & 85 & 6 & 7 \\
     59 & 85 & 7 & 6 \\
     60 & 85 & 9 & 2 \\
     61 & 89 & 5 & 8 \\
     62 & 89 & 8 & 5 \\
     63 & 90 & 3 & 9 \\
     64 & 90 & 9 & 3 \\
     65 & 97 & 4 & 9 \\
     66 & 97 & 9 & 4 \\
     67 & 98 & 7 & 7 \\
     68 & 100 & 6 & 8 \\
     69 & 100 & 8 & 6 \\
     70 & 101 & 1 & 10 \\
     71 & 101 & 10 & 1 \\
     72 & 104 & 2 & 10 \\
     73 & 104 & 10 & 2 \\
     74 & 106 & 5 & 9 \\
     75 & 106 & 9 & 5 \\
     76 & 109 & 3 & 10 \\
     77 & 109 & 10 & 3 \\
     78 & 113 & 7 & 8 \\
     79 & 113 & 8 & 7 \\
     80 & 116 & 4 & 10 \\
     81 & 116 & 10 & 4 \\
     82 & 117 & 6 & 9 \\
     83 & 117 & 9 & 6 \\
     84 & 122 & 1 & 11 \\
     85 & 122 & 11 & 1 \\
     86 & 125 & 2 & 11 \\
     87 & 125 & 5 & 10 \\
     88 & 125 & 10 & 5 \\
     89 & 125 & 11 & 2 \\
     90 & 128 & 8 & 8 \\
     91 & 130 & 3 & 11 \\
     92 & 130 & 7 & 9 \\
     93 & 130 & 9 & 7 \\
     94 & 130 & 11 & 3 \\
     95 & 136 & 6 & 10 \\
     96 & 136 & 10 & 6 \\
     97 & 137 & 4 & 11 \\
     98 & 137 & 11 & 4 \\
     99 & 145 & 1 & 12 \\
     100 & 145 & 8 & 9 \\
     101 & 145 & 9 & 8 \\
     102 & 145 & 12 & 1 \\
     103 & 146 & 5 & 11 \\
     104 & 146 & 11 & 5 \\
     105 & 148 & 2 & 12 \\
     106 & 148 & 12 & 2 \\
     107 & 149 & 7 & 10 \\
     108 & 149 & 10 & 7 \\
     109 & 153 & 3 & 12 \\
     110 & 153 & 12 & 3 \\
     111 & 157 & 6 & 11 \\
     112 & 157 & 11 & 6 \\
     113 & 160 & 4 & 12 \\
     114 & 160 & 12 & 4 \\
     115 & 162 & 9 & 9 \\
     116 & 164 & 8 & 10 \\
     117 & 164 & 10 & 8 \\
     118 & 169 & 5 & 12 \\
     119 & 169 & 12 & 5 \\
     120 & 170 & 1 & 13 \\
     121 & 170 & 7 & 11 \\
     122 & 170 & 11 & 7 \\
     123 & 170 & 13 & 1 \\
     124 & 173 & 2 & 13 \\
     125 & 173 & 13 & 2 \\
     126 & 178 & 3 & 13 \\
     127 & 178 & 13 & 3 \\
     128 & 180 & 6 & 12 \\
     129 & 180 & 12 & 6 \\
     130 & 181 & 9 & 10 \\
     131 & 181 & 10 & 9 \\
     132 & 185 & 4 & 13 \\
     133 & 185 & 8 & 11 \\
     134 & 185 & 11 & 8 \\
     135 & 185 & 13 & 4 \\
     136 & 193 & 7 & 12 \\
     137 & 193 & 12 & 7 \\
     138 & 194 & 5 & 13 \\
     139 & 194 & 13 & 5 \\
     140 & 197 & 1 & 14 \\
     141 & 197 & 14 & 1 \\
     \text{HOMO}\,142 & 200 & 2 & 14 \\
     \text{HOMO}\,143 & 200 & 10 & 10 \\
    \text{HOMO}\, 144 & 200 & 14 & 2  \\
     \text{LUMO}\, 145 & 202 & 9 & 11 \\
    \text{LUMO}\, 146 & 202 & 11 & 9 \\
     147 & 205 & 3 & 14 \\
     148 & 205 & 6 & 13 \\
     149 & 205 & 13 & 6 \\
     150 & 205 & 14 & 3 \\
     151 & 208 & 8 & 12 \\
     152 & 208 & 12 & 8 \\
     153 & 212 & 4 & 14 \\
     154 & 212 & 14 & 4 \\
     155 & 218 & 7 & 13 \\
     156 & 218 & 13 & 7 \\
     157 & 221 & 5 & 14 \\
     158 & 221 & 10 & 11 \\
     159 & 221 & 11 & 10 \\
     160 & 221 & 14 & 5 \\ \hline\end{array} $$


Reaction Mechanisms: Free Radical Halogenation In Free Radical halogenation using chlorine and sunlight, why does the chlorine-free radical get attached to the allylic carbon?
Will it be correct to say that it follows the anti-Markovnikov rule?
But if so, there is no $\ce{H2O2}$ present...
The substrate can be considered as but-2-ene.

H. C. Brown has suggested that the halogen atom does add but, at high temperatures, is expelled before the second step of free radical addition can occur$^1$.

Addition of halogen atom gives this radical, which falls apart to regenerate the starting material if the temperature is high or if it does not soon encounter a halogen molecule to complete the addition. Hence, low concentration of halogen can be used with high temperature to favour substitution over addition$^1$.
The allylic radical, on the other hand, once formed, has little option but to wait for a halogen molecule, however low the halogen concentration$^1$.

For the formation of minor product you may look into this question.
To follow the Markovnikov's or anti-Markovnikov's rule, there should be difference of  electronegativity between the two atoms. Therefore, I do not understand what you are trying to ask.
Reference

Organic Chemistry- Morrison and Boyd (6E); Page 389-390


Resonance involving carbanion Suppose a carbanion is involved in resonance and thus its electrons conjugate with unhybridised p orbital
of the nearby sp2 hybridised carbon and a pi bond is formed. Since carbanions are sp3 hybridised, how would it form a pi bond or will the hybridisation of the carbon change?

Yes, when bonds are formed, the hybridization can change in a carbon atom.
I think it’s important to remember something. Hybridization is not a real phenomena that occurs in reality. It is a model we use to simplify the complex bonding that occurs between atoms.
So, yes hybridization of the carbon will change, as it is not concrete and can change as bonds are made.


Why is Maleic Acid more soluble in water than its trans isomer Fumaric Acid? My teacher taught me that compounds having intramolecular H-Bonding are relatively less soluble in water as they will not involve in H-Bonding with water appreciably while those able to do intermolecular H-Bonding will appreciably form H-Bonding with water, and thus the latter ones will be relatively more soluble in water.
Applying this logic will make maleic acid less soluble than fumaric, but that's not the case. Wikipedia shows 478.8g/L for maleic, while just 4.9g/L for fumaric. Can someone help me with where I am going wrong?

Determining the crystal lattice energy would be helpful in comparing the starting points for a solubility investigation. Perhaps an approximation to the crystal stability is provided by the melting points: for fumaric acid, the mp is 287$^o$C; for maleic acid, 135$^o$C. 135$^o$C is a sort of normal mp for organic compounds. 287$^o$ seems quite high, suggesting a high lattice energy, and therefore, lower solubility for this isomer.

Even if molecules have the possibility of intramolecular bonding, as indicated by molecular models or drawings on paper, the existence of intermolecular bonding is not precluded.
Sometimes explanations of an observation are used as predictions which then turn out to be incorrect. The explanations may still be useful, but as explanations of actual data, rather than as reliable predictions.


What's the reason for the instability of ethylene dione? Ethylene dione, with a chemical formula $\ce{C2O2}$ and an IUPAC name of ethene-1,2-dione, is an oxocarbon.

The Wikipedia page says:

  The existence of $\ce{OCCO}$ was first suggested in 1913. However, despite its deceptively "simple" structure, for over a century the compound had eluded all attempts to synthesize and observe it. Such elusive nature had earned $\ce{OCCO}$ the reputation of a hypothetical compound and a mysterious, "exceedingly coy molecule".

I cannot understand what the rest of the article is telling me. So, why is ethylene dione unstable?

Expanding what Tan Yong Boon said, there is a full scientific article detailing why ethylene dione is unstable Chem. Eur. J., 1998, 4, 2550-2557.

Long story short: Two CO molecules are far more stable than OCCO. The ground state of ethylene dione is a "diradical". This is, OCCO has two unpaired electrons instead of forming the intuitive chain of double bonds (O=C=C=O). However, if you add one or two additional electrons, the resulting anion becomes stable and adopts the shape of glyoxal (but without hydrogen atoms).


Does metal oxidize (faster) in dirt? I had an interesting event occur recently: I stuck a screwdriver in a potted plant, for it is the only way that I am able to unlock a certain door. It was left in there for about a month before I needed it again, and, when I took it out, it had what I thought to be rust. I am doubting my original assumption.

I know for a fact that it is not dirt itself, as I vigorously attempted to clean it with suds (though I'm not entirely sure whether or not that was the best approach...).

Question

Does iron a) oxidize in dirt and b) if a), does it oxidize faster in dirt?

Iron can corrode faster in soil than in air, particularly if partly immersed in the soil. Moisture and electrolyte are retained in the soil, making it a good anode. Also, microbes in the soil can accelerate corrosion.


How does Polytetrafluoroethylene (Teflon) differ from Polyvinylidene fluoride (PVDF) as a protein binding membrane material? PTFE and PVDF (durapore) are both used in protein binding filter membranes (Millipore specifically). Chemically speaking the two polymers differ quite significantly due to the additional fluorides and molecular weight. Their protein-binding properties also differ. I'm curious about how the chemical differences affect their properties as a membrane material.
From what I understand, the durability between the two is quite different. The PTFE membranes are used for the filtration of organic solvents where as the PVDF membranes are generally for low protein binding. Furthermore the PTFE membranes typically require a polyethylene support.
I don't believe that the degree of polymerization or the architecture; if we were dealing with a copolymer I expect differently. Based on my molecular intuition, I would imagine that this is completely based on the chemistry of repeat unit.
My question concerns a few things. What makes the PTFE a better solvent resistant polymer? The other is how does the reduced number of fluroides result in reduced protein binding?

Teflon/PTFE

PVDF

Essentially you are asking why about the different surface properties of PTFE vs. PVDF.  Generally PTFE is more 'non stick' than PVDF, which is more non-stick than polyethene (PE).  PTFE is much more chemically resistant to corrosion than PVDF or PE.  You are correct to look at the chemical structure.  The more Fluorine atoms on the backbone, the more proteins will not be able to chemically latch onto the surface.  Measuring things like contact angle will confirm this.  I don't know how much the pore structure of the membranes affect things, but from a chemical standpoint, there are clear differences.


Enthalpy of a reaction changes with temperature, does this mean bond strengths change with temperature? During chemical reactions, the bonds between atoms break or form to either absorb or release energy. The result is a change to the potential energy of the system. The heat absorbed or released from a system under constant pressure is known as enthalpy.

Mathematically, we can think of the enthalpy of reaction as the difference between the potential energy from the product bonds and the potential energy of the reactant bonds.

But we know that enthalpy of a reaction depends on the temperature according to Kirchoff's equations. So does this mean that bond energies also change with temperature.

Also can someone explain to me at the molecular level why does enthalpy of the reaction change with temperature?

And since the enthalpy of a reaction depend on the constant pressure at which reaction is carried out, does this mean bond strengths also depend on pressure?

No: bond strengths as such do not change with temperature. The potential energy surface holding the atoms together, i.e. all the bonds, is constant for a given molecule.

However, all molecules have vibrational and rotational energy levels and molecules also have bulk kinetic energy due to thermal motion in a gas or liquid. As the temperature increases the bulk thermal energy increases, i.e. molecules move faster, but also more vibrational and rotational energy levels become excited, which means that the vibrational and rotational quantum numbers increase. This means that less extra energy is needed to break a bond if the temperature is already high, compared to that needed at low temperature, simply because the bond is already has a lot of internal (vibrational/rotational) energy.

Not all molecules have exactly the same energy but instead there is an exponential distribution of energies (the Boltzmann distribution) where the higher the energy is the fewer molecules are there that have this energy.

The subject of  Statistical Mechanics describes these molecular effects and also forms the foundation for Classical Thermodynamics, i.e. what we usually call 'Thermodynamics', and which was developed before atomic/molecular theory.


Activation Energy in Acid/Base reactions For simple acid/base reactions - i.e. those involving inorganic compounds - does activation energy ever play in reactions?

Or can we ignore kinetics entirely and just consider thermodynamic factors such as conjugate stability?

I'd be very interested if there were some book or article examining inorganic molecules' acid-base behavior from a kinetic standpoint.

Proton transfer, especially in aqueous media, has an activation barrier that is easily surmounted at room temperature or higher.  Such reactions are said to be diffusion controlled (the reaction happens at the rate of proton diffusion through the media) or spontaneous.  In such cases thermodynamics will control the process.   That said, I'm sure there are cases where the base the proton is transferring to, is, for example, surrounded by bulky ligands substantially increasing the activation energy for the process.


Is there a list of known rate laws for reactions? I've been scouring the web for a website/pdf that lists/tabulates reactions and their corresponding rate laws. Something like:
\begin{align}
\ce{A + B &-> C} & \text{rate}&=1.234[\ce{A}]^2[\ce{B}] &&\text{at } 25~^\circ\mathrm{C}\\
\ce{D + E &-> F} & \text{rate}&=5.678[\ce{D}][\ce{E}]^2 &&\text{at } 25~^\circ\mathrm{C}\\
\ce{G + H &-> I} & \text{rate}&=9.001[\ce{G}][\ce{H}] &&\text{at } 25~^\circ\mathrm{C}\\
&\cdots&&\cdots&&\cdots\end{align}

Is there such a list of known rate laws for reactions?

Integrated Rate Laws in Chemical Kinetics are very well known (http://www.csus.edu/indiv/m/mackj/chem142/kinetics.pdf).

Here is a good article with the list of the constants: Table of Recommended Rate Constants for Chemical Reactions Occurring in Combustion by Francis Westley


Standard reduction potential of MnO4-/MnO2 couple This is the data given:

$$\ce{MnO4- /Mn^{2+}} = 1.5~\mathrm{V}$$

$$\ce{MnO2 /Mn^{2+}}= 1.23~\mathrm{V}$$

I know that :
$$\ce{MnO4- +5e- + 8H+ ->Mn^{2+}}$$
$$\ce{MnO2 + 4H+ +2e- -> Mn^{2+}}$$

After that here's what's done which I absolutely don't get. After using $\Delta G^\circ = -nFE^\circ$, won't we have to multiply these energy values by the coefficients of the balanced equations before we can get $E_{\mathrm{cell}}$?

In particular, shouldn't the $\Delta G^\circ$ of the first equation be multiplied by two and that of the second by three?

The important thing to note with electrochemistry questions like these is that, if you add two equations together to get a third, their $\Delta G^\circ$ can be added, but their $E^\circ$ cannot.

Your half-equations are incomplete, which might not be the worst thing, but it certainly doesn't help your case. Using $\Delta G^\circ = -nFE^\circ$, we have the two equations

$$\begin{align}
\ce{MnO4- + 8H+ + 5e-} &\ce{-> Mn^2+ + 4H2O} & \Delta G_1^\circ &= -728.46~\mathrm{kJ~mol^{-1}} \tag{1} \\
\ce{MnO2 + 4H+ + 2e-} &\ce{-> Mn^2+ + 2H2O} & \Delta G_2^\circ &= -237.35~\mathrm{kJ~mol^{-1}} \tag{2}
\end{align}$$

The trick is to find the appropriate combination of equations $(1)$ and $(2)$ that give the equation $(3)$, which you are interested in:

$$\ce{MnO4- + 4H+ + 3e- -> MnO2 + 2H2O} \qquad \qquad \Delta G_3^\circ = \,\,? \qquad \qquad \tag{3}$$

If you just look at the manganese-containing species, it's actually not too hard to see that $(1) - (2) = (3)$. That's why I said it's probably not the worst thing that you omitted $\ce{H2O}$ in your half-equations - since it doesn't really make a difference in terms of finding the answer - but it's just not good practice to write something that's incorrect. Anyway, that means that:

$$\begin{align}
\Delta G_3^\circ &= \Delta G_1^\circ - \Delta G_2^\circ \\
&= -728.46~\mathrm{kJ~mol^{-1}} + 237.35~\mathrm{kJ~mol^{-1}} \\
&= 491.11~\mathrm{kJ~mol^{-1}} \\
E_3^\circ &= \frac{-\Delta G_3^\circ}{3F} \\
&= +1.697~\mathrm{V}
\end{align}$$

If you obtain a slightly different answer, it may be due to rounding inconsistencies, or perhaps because I used more accurate values for $F$ and the initially given reduction potentials. In any case, you should get an answer around $+1.70~\mathrm{V}$.


Equations for reactions dealing with coordination compounds Write a balanced chemical equation for the reaction depicted in this figure. Express your answer as a chemical equation. Identify all the phases in your answer.

This is what I have tried so far:

NH4NCS(aq) + [Fe(H2O)6]3+(aq) -> [Fe(H2O)5NCS]2+(aq) + H+(aq) + NH4OH(aq)

NH4NCS(aq) + [Fe(H2O)6]3+(aq) -> [Fe(H2O)5NCS]2+(aq) + H2O(l) + NH4+(aq)

I found the answer which I'm assuming was just getting rid of an unnecessary spectator ion:

SCN-(aq) + [Fe(H2O)6]3+(aq) -> [Fe(H2O)5(SCN)]2+(aq) + H2O(l)


Is this definition of mole correct? Chemical engineers define one mole as the amount of a substance which possess as many entities as $12\ \mathrm g$ of $\ce{^{12}C}$.

The number of atoms in $12\ \mathrm g$ of $\ce{^{12}C}$ is $6.022 \times 10^{23}$ which is a constant by its definition.

Now come to the relevant definition which is given in my textbook that defines one mole of a substance as the atomic mass, molecular mass or formula mass in grams.

Is this definition correct?

The mole is a base unit as specified in the Système international d’unités (SI) by the bureau international des poids et mesures. Its decisive definition is that published in French:

  La mole est la quantité de matière d’un système contenant autant d’entités élémentaires qu’il y a d’atomes dans 0,012 kilogramme de carbone 12 ; son symbole est « mol ».
  Lorsqu’on emploie la mole, les entités élémentaires doivent être spécifiées et peuvent être des atomes, des molécules, des ions, des électrons, d’autres particules ou des groupements spécifiés de telles particules.
  La mole est une unité de base du Système international d’unités.

The proposal was brought forth by the International Union of Pure and Applied Physics (IUPAP), the International Union of Pure and Applied Chemistry (IUPAC) and the International Organisation for Standardization (ISO). As with all SI texts, the decisive French version has a semi-official English translation:

  The mole is the amount of substance of a system which contains as many
  elementary entities as there are atoms in $0.012$ kilogram of carbon 12; its
  symbol is “mol”.
  When the mole is used, the elementary entities must be specified and may be
  atoms, molecules, ions, electrons, other particles, or specified groups of
  such particles.

(The third point is not translated.)

This definition is more or less identical with the one in your first paragraph.

In practice and well within experimental error, this means that your later definition will hold true for any substance. I.e. take $1~\mathrm{mol}$ of an entity and the combined mass of that mole will be the same numerical value in grams as a single entity has in atomic mass units ($\mathrm{u}$). It is not the correct definition, and for any entity that is not carbon-12 the masses will differ slightly (but well within margin of your macroscopic experimental error) but it is good enough for most contexts.

A redefinition of the SI units is being discussed and will likely be adopted at the 26th General Conference of Weights and Measures in autumn 2018. This would redefine the mole in a way that the Avogadro constant is defined to be numerically exactly $6.02214 \cdot 10^{23}~\mathrm{mol^{-1}}$ (with a few further digits appended to the end of the number that yet need agreement). This would mean that the new definition of the mole would be along the lines of:

  The mole, mol, is the unit of amount of substance of a specified elementary entity, which may be an atom, molecule, ion, electron, any other particle or a specified group of such particles; its magnitude is set by fixing the numerical value of the Avogadro constant to be equal to exactly $6.02214X \cdot 10^{23}$ when it is expressed in the unit $\mathrm{mol^{-1}}$.

Currently, the Avogadro constant must be measured experimentally giving a value of $6.022140857(74)~\mathrm{mol^{-1}}$; the digits in brackets express the numerical uncertainty.

This will mean that $1~\mathrm{mol}$ of carbon-12 atoms will no longer have the mass of exactly $12~\mathrm{g}$ (but again, it will be well within experimental error for everybody not practising theoretical physics).

To answer the follow-up question you asked in the comments: The coefficients in chemical equations such as

$$\ce{Zn + 2 HCl -> ZnCl2 + H2}$$

are always and exclusively to be understood as ratio coefficients. Thus, instead of thinking one atom or one mole of zinc, think amount $n$ of zinc and amount $2n$ of $\ce{HCl}$.


Why copper sulphate is a catalyst in this reaction Recently, we learnt about catalyst is a substance can increase the rate of reaction without any without itself undergoing any permanent chemical change.The experiment we did was the reaction between zinc and hydrochloric acid as the equation below

$$\ce{Zn + 2HCl -> ZnCl_2 + H_2}$$

However the thing I don't understand is why copper sulphate is a catalyst when the copper ion is reduced to copper atom as the equation below

$$\ce{Cu^2+ + Zn -> Cu + Zn^2+}$$

I mean how could it call as a catalyst when it is reduced from ion to atom

The reaction rate increases because the copper is deposited on particles of zinc, creating a galvanic couple causing galvanic (aka electrolytic) corrosion. [It could be argued that this is not catalysis, because the copper sulfate is entering into a reaction.]


Selective transfer of least/most stable anion from organometallic reagents When heterocuprate reagents participate in conjugate addition, the least stable anion is transferred from copper to the electrophile:

On the other hand, palladium-catalysed cross-coupling reactions show a different trend, e.g. in the Stille coupling, the most stable anion is transferred:

How is this dichotomy explained?

It’s not really a dichotomy, it’s just two very different reaction mechanisms.

For the cuprate addition, which also applies to general Grignard-type additions, in principle the reaction is reversible (figure 1).

Figure 1: Reversibility of Michael additions.

However, the reverse direction is only really possible if the anion released is somewhat stable. In your example, thiophenolate may attack but may also be removed again in the course of the reaction. Once an alkyl chain has attacked, there will be no reverse reaction since the eliminiation of $\ce{R-}$ is too unfavoured.

For palladium-catalysed cross-coupling reactions, we need to consider the typical palladium catalysis cycle (figure 2).

Figure 2: typical catalytic cycle of palladium-catalysed ($\ce{M} = \ce{Pd}$) cross couplings. In a Stille coupling as in the question, $\ce{Y} = \ce{SnR3}$. Image taken from Wikipedia where a full list of authors is available.

The important step here is the transmetallation step: One of the ligands of metal $\ce{Y}$ ($\ce{Sn}$) must be transferred to the palladium ($\ce{M}$) centre. This will be the ligand that dissociates most easily — the best stabilised anion. So if the tin centre has three n-butyl ligands and one phenyl one, phenyl will be preferentially transferred since it dissociates more easily.

This is also part of the reason why $\ce{C}_\mathrm{sp^3}\ce{-C}_\mathrm{sp^3}$ cross coupling are so rare.


How to synthesize cyclopentane from cyclohexane? Does this route work to synthesize cyclopentane from cyclohexane?

Free-radical halogenation.
Elimination with strong base to form cyclohexene.
Ozonolysis to make di-aldehyde.
Oxidation to make di-carboxylic acid.
Esterification with ethanol, acid, and heat.
Alpha-proton elimination with ethoxide ion.
Intramolecular attack of other carbonyl carbon; loss of ethoxide leaving group.
Acid, high heat, water to return to carboxylic acid. Expulsion of ethanol.
Heat for decarboxylication. Loss of carbon dioxide.
Not illustrated: removal of carbonyl using Wolff-Kishner and Clemmensen.

I think a few modifications I could make are:

Formation of methyl ester using diazomethane. This would not involve heat and acid. But it would involve something that is potentially explosive.
Alternately, make an acid chloride and then go down to the ester if I wanted to avoid the heat involved with Fischer esterification.

Why not just use the Aldol condensation on the dialdehyde produced from ozonolysis and save some steps?


Is sodium acetate a neutral or a basic salt? I read somewhere that salts which do not contain a replaceable hydrogen or hydroxyl group are classified as normal salts. But I also know that strong base-weak acid salts are basic in nature.

Normal salts do not have to be neutral, and vice versa. There are two different features of a salt: (1) containing a replaceable hydrogen, and (2) making an acidic solution in water. They are not necessarily related. For example:

$\ce{NaHSO4}$ has both.
$\ce{NaHCO3}$ has the first and not the second (in fact, being a salt of a strong base and a weak acid, it has to be basic in nature, as you correctly recall).
On the other hand, $\ce{CuSO4}$ is partially hydrolyzed in water solution and hence has the second feature, but not the first.
To make the picture complete, $\ce{NaCl}$ has neither.

Likewise, the opposite features of (1) having a hydroxy group and (2) making a basic solution in water are even less related. For example:

$\ce{Cu2(OH)2CO3}$ has the first feature, but not the second (for it is not even soluble in water).
$\ce{CH3COONa}$ is partially hydrolyzed in water solution and hence has the second feature, but not the first.
$\ce{MgCl2}$ and $\ce{Mg(OH)2}$ form several basic salt phases with both features (J. Chem. Eng. Data 2017, 62 (4), 1384–1396).  They are only sparingly soluble but still form measurably alkaline solutions, especially when warmed.
$\ce{MgSO4}$ has neither.


Is there any software package for quantum chemistry that includes CAMB3LYP? I want to do calculations on systems with photoinduced electron transfer, and I've read that the Coloumb Attenuating Method is a modification of the functionals that makes calculations including long-range electron transfer more accurate. Unfortunately I can't seem to find CAM in Gaussian. Is there some other software I should look at?

CAM-B3LYP is present in Gaussian, Q-Chem, GAMESS, NWChem, ORCA, DALTON, DIRAC, and perhaps other major software packages, either as cam-b3lyp or camb3lyp, however keywords are entered.

Notably it isn't available in TURBOMOLE as of version 7.1.


Why does an increase in reduced mass lead to a decrease in frequency? In terms of IR spectrum, why does an increase in reduced mass lead to a decrease in frequency?

Consider a simple molecule $\ce{A-B}$ composed of atoms $\ce{A}$ and $\ce{B}$ and let the bond connecting them be described as a spring.  Then the stretching and contracting of the $\ce{A-B}$ bond (spring) can be considered as a simple harmonic oscillation and is described by Hooke's law.
${\nu = \frac{1}{2 \pi} \sqrt{\frac{k}{m}}}$
where ${\nu}$ is the frequency of the stretch,  ${k}$ is the force constant describing the specific spring and ${m}$ is the mass of the system.  In our $\ce{A-B}$ example, the mass of the system is described by the reduced mass ${\mu}$, where
${\mu = \frac{m_A m_B}{m_A + m_B}}$
So our equation becomes
${\nu = \frac{1}{2 \pi} \sqrt{\frac{k}{\mu}}}$
From this equation we can see that as the reduced mass increase's the stretching frequency will become smaller.


In what way trans alkenes are more symmetric than cis alkenes? While providing the reason for the higher melting point of trans molecules, everyone says that trans molecules are more symmetric than cis molecules. An example can be found over here, where the OP states that:

trans-2-Butene has more symmetry than its cis isomer which results in better packing and hence higher melting point.

But to my (uneducated) eyes there doesn't seem any more symmetry (if any) in trans form than cis form of but-2-ene. I mean cis form is vertically and horizontally symmetric, whereas trans form is not.
Why do we call trans form more symmetric than cis form?

Trans alkenes have a $C_\mathrm{2h} $ symmetry, identified as such because it has a $C_2$ rotational axis (you need to rotate 180° to have an identical molecule) and a mirror plane perpendicular to that axis $\sigma_\mathrm h$. You need to follow the symmetry flowchart to assign symmetry groups.
A $C_\mathrm{2h}$ group is characterised by having:

$E$: the identity operation
$C_2$: a twofold rotational symmetry axis
$i$: a center of inversion
$σ_\mathrm h$: a horizontal mirror plane

Cis alkanes have a $C_\mathrm{2v}$ symmetry; therefore they have:

$E$: the identity operation
$C_2$ :  a twofold rotational symmetry axis
$2σ_\mathrm v$: two mirror planes that include the rotation axis

As you see a $C_\mathrm{2h}$ group has more symmetry elements and therefore sometimes people say "more symmetry" but this is not a technical way of describing it. To say which is more symmetric you need to consider the possible symmetry operations you can do.
Easier answer:
A symmetric object is an object that looks the same if you rotate, reflect etc. The more operations you can do keeping the object identical, the higher the degree of symmetry. For example a sphere is "more symmetric" than a cube because you have much more symmetry operations you can do (you can have infinite rotation axis and infinite mirror planes etc.). In the case of alkenes the main difference between the trans and cis ones is the so-called inversion point (blue drawing in first figure) that only the trans alkene has and therefore is "more symmetric".


Surprising values of intermolecular forces of ethane, oxygen and xenon I know that gases with stronger intermolecular forces should have a larger value of the van der Waals constant $a$. So I looked up a table on van der Waals constants and tried to reason about the relative values of that constant. Though I was able to explain most of the values, there was one thing I found absurd. Here:

$$
\begin{array}{cc}
\hline
\text{Substance} & a~(\pu{atm {dm}^6 mol^{-2}}) \\
\hline
\text{Oxygen} & 1.364 \\
\text{Xenon} & 4.137 \\
\text{Ammonia} & 4.225 \\
\text{Ethane} & 5.507 \\
\hline
\end{array}
$$

My intution says that dioxygen, which has a slightly higher molar mass (32 vs. 30) than ethane, but significantly less molecular size or volume, should have a only slightly lower value of $a$ than that of ethane, but here we find a significant difference! Another example is xenon; its molar mass (131) is much larger than ethane (30), but its value of $a$ is smaller. Does it have to do with the dominance of molecular size over mass? Or atomicity or atomic dipoles? I suspect that it's probably the large and branched structure of ethane that's causing this.

Why does the molecular size/volume seem to play such a dominating role in the value of the constant $a$? I thought it was primarily influenced by molecular mass.

Van der Waals forces are weak and depend on polarizability. There is a correspondence between the boiling points of oxygen (−183 °C), xenon (−108 °C), and ethane (−88.5 °C) and their constants (1.364, 4.137, 5.507). These gases are polarizable to varying extents, but not polar with charge separation. The size of the molecule and its mass will affect polarizability - perhaps a way to look at this is to say electron density in the outer regions of the molecule will determine how easily dipole-dipole interactions can evolve.

But ammonia stands out with a boiling point (−33 °C) which is too high to be explained by van der Waals forces alone. Of all the gases listed, ammonia is the only one with significant hydrogen bonding capability. Perhaps if $\ce{H2O}$ were included, it would be an even greater outlier.


Why does HgCl2 react with HCO3^- but not with CO3^2-? I have started learning Qualitative Salt analysis recently. When I was reading about tests to distinguish $\ce{CO3^2-}$ and $\ce{HCO3-}$, I came across a line in a book which states

  $\ce{HgCl2}$ forms a reddish-brown precipitate of $\ce{3HgO.HgCO3}$ with $\ce{HCO3-}$ but no reaction is reported in case of $\ce{CO3^2-}$.

I am not able to understand why there will be no reaction of $\ce{HgCl2}$ with $\ce{CO3^2-}$.

It would be better if I can get a sound logic for this so that I can remember it better.

Anirudh, which book is teaching this stuff? Sorry, I can only find this discussion in 1890s and the latest in my favorite Mellor's Treatise from 1930s.

Coming to 2020:

After doing a simple acid test for carbonates, a nice safe & sound method to distinguish carbonate vs. bicarbonate salt is

(a) to react it with a neutral barium chloride solution. If the solution is turbid, carbonate is present, otherwise bicarbonate. Of course the methods fail with a mixture of the two.

(b) Sodium Carbonate-Phenolphthalein Test: phenolphthalein is turned pink by soluble carbonates and but remains colourless by soluble bicarbonates

Your textbook is not completely correct when it states that bicarbonate does not react with $\ce{HgCl2}$. For your reference from A treatise on chemistry, Volume 2
by Henry Enfield Roscoe, Carl Schorlemmer, 1913, pg 699


Explaining boiling using internal energy Internal energy is summation of potential and kinetic energies of a substance

Using this interpretation of internal energy, how can I explain boiling?

Kinetic remains constant.
Potential should decrease because of an increase in average spacing.
But internal energy should increase.

Can somebody explain what flaw I am making?

You are wrong when you say the potential energy of particles decrease when they move  apart.

The forces between the particles are attractive not repulsive.

Recall the definition of potential energy. It is the work done by external force on a charged particle under influence of electric field.

Therefore, the potential energy for particles, which attract each other, increases when they move apart.

So when to particles get close their potential energy decreases, hence their stability increases. This also happens to be one of the reasons why they come closer.


Rotovapping Aqueous Solutions - Need a cold trap? I am working on a large scale natural product Rotovapping project. There are two botanical extracts to evolve, combined in processing. The solvents are 50% EtOH and 50% glycerin. There is a large volume of material to dry that will require several weeks of run time. I am using a diaphragm pump that has specs for chemical resistance; I am not too concerned for any issue with EtOH. My concern is that the efficiency of the pump may be impacted by water vapors if I push the parameters to expedite the process. I know that cold traps are encouraged for high volatiles...but would it be helpful to add one to condense any water vapor also, for a run this large? The pump does not have any trap system in place currently. Not sure if it is worth the trouble to create an inline trap as or if it will enhance the speed.

You will always want a cold trap not only to protect the pump but also to make sure it's running at peak efficiency at all times. Especially when you push the vacuum and heating to try to expedite the process, you never know when the whole thing might suddenly bubble up. A clean cold trap will also allow you to recover the material that has escaped the main chamber, whereas without one you'd probably be faced with loss of compound and having to repair the pump.


Why is an ionic bond a chemical and not a physical bond? Ionic bonds seem to be intermolecular but are classified as chemical bonds.

"Ionic bonding is a type of chemical bond that involves the electrostatic attraction between oppositely charged ions." -  Wikipedia, definition of an ionic bond
"The physical force of attraction which holds atoms and molecules in a matter is called a physical bond. van der Waals' forces and coulombic forces are physical forces." - Chemistry Stack Exchange, definition of an intermolecular force
"A chemical bond is an attraction between atoms that allows the formation of chemical substances that contain two or more atoms. The bond is caused by the electrostatic force of attraction between opposite charges, either between electrons and nuclei or as the result of a dipole attraction." - Wikipedia, definition of a chemical bond

Before an ionic bond is formed, oxidation and reduction must occur. For example, let's use sodium chloride. A chlorine atom will steal the outer electron from a sodium atom. The chlorine is gaining an electron, i.e. being reduced. The sodium is losing an electron, i.e. being oxidized. I agree that this is a purely chemical process.
Now that the atoms are oppositely-charged ions, they can attract and form an ionic bond:
$$\ce{Na+ + Cl- -> NaCl}$$
My question is this: How is this bond defined as a chemical bond?
Let's take water; $\ce{H2O}$. The oxygen forms two polar-covalent bonds with two hydrogen atoms. Each hydrogen shares two electrons with the oxygen. The electrons are being shared, so this bond an intermolecular force. The atoms stay together because they both depend on the electrons they share.
Now let's take more than one water molecule. These molecules will form hydrogen bonds. The hydrogens on the water molecules do not often have an electron around them because the oxygen has a greater attraction with them. This leaves the lonely proton of the hydrogen to make that end of the water positive while the oxygen is negative with its electrons. The negative end of the water (oxygen) will become attracted to the positive end of another water molecule (hydrogen), forming our anticipated intermolecular force. Does this sound familiar?
I hope it does. According to my knowledge, an ionic bond is that, but instead of a dipole-dipole interaction, it's a monopole-monopole interaction. If you look at the bond itself and not the reaction forming it, it acts just like an intermolecular force.

Short version:

We don't call bonds "physical", there are chemical bonds and other types of interactions between particles. The chemical bonds are classified this way because they make up molecules, salts, polymers and such, which are the materials chemists are interested in studying, and not particularly because of their sub-atomic/electrostatic characteristcs.

Your question relates to two frequent sources of confusion:

How do we define a chemical bond? What makes a bond between two atoms ionic or covalent?
Where does chemistry (hence chemical interactions) ends, and consequentially where does physics (and hence physical interactions) begins?

Both questions have been pretty much answered here on Chemistry SE and also on Physics SE. I'll give some links to related questions and also a shrot summary.

1. How do we define a chemical bond? What makes a bond between two atoms ionic or covalent?

Your question didn't go much into covalent bonds, but anyone trying to understand anything past the surfaces of chemical bonds should know that the definition of a bond as covalent or ionic is simply an easier way for us to classify things, because when the bonds are categorized in groups we can say that one group behaves in a way, and the other behaves in a different way. The truth is, all bonds have some ionic and some covalent character to them.

Good answers here on Chemistry SE about that can be found here:

Can 100% covalent bonds exist?
What happens if the electronegativity difference is exactly 2.1?

2. Where does chemistry (hence chemical interactions) ends, and consequentially where does physics (and hence physical interactions) begins?

For a Physics pont of view, see: What distinguishes between physics and chemistry?

For a Chemistry point of view, see: Differences between chemical physics and physical chemistry?

Basically, it's clear that both sciences have a lot of overlapping, and perhaps the main difference is the approach taken and how the results will be used. While chemists usually focuses on understanding the structure and interactions between substances and materials, in order to be able to improve them or develop new ones, physicists are mostly interested in understanding the exact laws of nature and representing them with mathematical equations that can be used to explain and predict phenomena. This definitions do not seem so different, and that's not supposed to be wrong: We do have the Physical Chemistry and Chemical Physics fields.

Ok, but Why is an ionic bond a chemical and not a physical bond?

Two important things were mentioned by @Mithoron in the comments. The term "Physical bond" is not frequently used by chemists, (or anyone I suppose), pretty much because all interactions in whole universe are physical. This single quote by @Mithoron is related to the two points I highlighted above. Both "types" bonds "follow" the same laws, and there is no hard line between them.

It's also worth noting that the definition you gave for Physical Bond was taken from an answer without a single upvote, which is an indicator that it isn't what the Chemistry SE community thinks (despite the fact that Google liked it very much and uses it in it's short-answer for the query). Another indicator that "Physical Bond" isn't really a type of bond is that there aren't definitions of it out there (Yahoo answers doesn't count), not even a Wikipedia article.

So why do we call them chemical bonds?

That's pretty much because the bonds we call chemical are those which make up molecules, salts and other materials which are mainly studied by chemistry. The definition of chemical bond varies a lot (as explained here), but they are the bonds which create what chemists study.

Other types of bonds which aren't called chemical are the intermolecular interactions, which are obviously of interest to chemistry, but also the Strong Force, which is the force that binds protons and neutrons (nucleons) together to form the nucleus of an atom. (I hope it's okay to call Strong Force a bond, I'm not sure to be honest).


Test for diethylether or tertiary alcohol? What is a simple chemical test to tell the difference between a tertiary alcohol (e.g. 2-methylpropan-2-ol) and an ether (specifically diethylether/ethoxyethane)?

Neither of them can be oxidised by reagents such as potassium dichromate, Tollens reagent or fehlings reagent.

I came up with a test in which a less than stoichiometric amount of ethanoic acid is added and left to react, then some sodium carbonate is added and if there is no effervescence that means an ester has formed hence it must have been the alcohol, otherwise it was probably the ether. However I am unsure about the extent to which this reaction completes and it's simplicity.

As mentioned in the comments, using Lucas' Reagent is the best option in this case because you're dealing with a tertiary alcohol vs an ether, but let's consider your proposal as well.

In principle, it should work because an ether will not react with ethanoic acid whereas an alcohol would. However, the formation of an ester simply by the addition of ethanoic acid to alcohol is not likely to complete (you generally need a dehydrating acid like $\ce{H2SO4}$ to drive the reaction). In other words, when you put in the bicarbonate you will most likely see effervescence.

If you're thinking of looking for effervescence, using thionyl chloride $\ce{SOCl2}$ might be an option since it will react with the alcohol to give a sulfur dioxide gas and HCl:

$$\ce{R-OH + SOCl2 -> R-Cl + SO2(g) + HCl}$$

You have to make sure the solvent is dry though.

In the specific case of t-Butanol vs diethyl ether, though, you could simply heat the solution slightly to about 40 degrees Celsius. If it boils, then you've got diethyl either. If it doesn't, it's t-Butanol.


How can geometric isomerism arise due to steric factors? I can see how a double bond and a ring structure can lead to geometric isomerism but can't see the same for steric factors. Could you help? Thank you!

Let's take a biphenyl compound:

Due to steric factors, the rings cannot be planar. Instead, they have an approximate right angle between them:

(Image source)

Note that for the right angle to "flip" over to the other side, the $\ce{NO2}$ and $\ce{CH3}$ would have to pass. These groups are bulky enough to prevent that from happening1 unless sufficient energy is provided.

Now, try to look from it along the axis. You'll notice that for this particular molecule, the nitro group farther from your eyes is a clockwise step from the closer nitro group. This happens regardless of which side of the axis (left/right) you look at it from.

On the other hand, if we twisted the molecule so that the angle is again a right angle on the other side, there is now an anticlockwise step from the closer nitro to the farther nitro.

These two configurations of the compound are stereoisomers.

Since the steric barrier can be overcome with sufficient energy, such compounds usually racemize over time (though when the groups themselves  are as bulky as a phenyl, this doesn't happen at any appreciable timescale, IIRC).

1. I actually had some trouble using ChemDoodle to draw the molecule without any overlapping bits because of that. Doesn't exactly prove the bulkiness, but one gets the idea.


Equilibrium Constants For Phenol Tautomers I encountered the following equilibrium constant values for the keto-enol interconversions of phenol in Organic Chemistry (Solomons, Fryhle, Snyder, 3rd Edition):

I wonder what could explain the differences (though I'm not sure whether they're significant) between the equilibrium constants of the two keto-enol interconversions wherein in one case the double bond of carbon with the oxygen and the hydrogens are oriented in ortho position and in para position in the other.

Edit: Can this difference concern the medium the reaction is catalyzed in or directive influence of groups?
Do explanations for equilibrium constant differences exist for other tautomer interconversions or are such queries quite speculative?

I believe that's because in the first step, phenol regains its aromaticity which makes the $K_{eq}$ substantially ($1000$ times) larger than that in the second step where it loses its aromaticity.


Are all ambidentate ligands monodentate? Several websites define an ambidentate ligand as being monodentate. What about a ligand with say 3 atoms that can act as donor atoms, but due to the sterics of the ligand, only any two atoms can coordinate bond to an acceptor at any one time?

When a ligand has multiple donor sites, but uses only say $x$ of them at a particular time, then it's called ambidentate ligand. However, if it uses different number of sites in different compounds (say $x$ in one and $y$ in the other where $x\ne y$), then flexidentate is a more appropriate term as said in the other answer. But this doesn't mean all ambidentate ligands are monodentate.

For example, dithiooxalate is bidentate and ambidentate:

This example is from the book Concise Inorganic Chemistry by J.D. Lee (Adapted by Sudarsan Guha).


Reactivity of meta- and para-bromo-nitrobenzene towards nucleophilic substitution Among 1-bromo-3-nitrobenzene and 1-bromo-4-nitrobenzene, which would be more reactive towards nucleophilic substitution?

From what I can infer, the answer should have been 1-bromo-3-nitrobenzene. Nitro is a −M/−R group, so it would create regions of high positive charge in the ortho and para positions with respect to the nitro group.

Now in 1-bromo-4-nitrobenzene, Br is in the para position with respect to the nitro group. Hence the positive charge would be created only in the ortho position, as the para position is blocked. On the other hand, in 1-bromo-3-nitrobenzene, there would be regions of positive charge in the both the ortho and para positions wrt to the nitro group.

But the answer given in my book is 1-bromo-4-nitrobenzene. Am I missing something here?

p- Bromonitrobenzene is certainly more reactive towards Nucleophilic Substitution than m-Bromonitrobenzene. The reasons are as follows,  Firstly, due to $-R$ effect of $\ce{NO2}$ group, the positive charge is created on the carbon attached to bromine in para position, which helps to increase thre electrophilicity of that carbon and that makes easier for the electrophiles to attack the benzene ring. If $\ce{NO_2}$ group would have been in  meta position, the positive charge would have been created on the adjacent carbon, so, the electrophilicity of the bromine attached carbon is not increased so much which in turn will not help it to undergo Nucleophilic attack easily. The image below demonstrates this effect on para position.

Secondly, the Nucleophilic substitution in these systems undergo by addition-elimination mechanism. So, when the nucleophile attacks and forms the intermidiate complex, that complex is much more stabilised due to charge delocalisation in the $\ce{NO2 }$ group also if it is in para position.

But if $\ce{NO2}$ is in meta position, the charge can only delocalise in the ring, but it can not delocalise in the $\ce{NO2}$ group (you can verify by looking at the  image also), so it is not so stabilised as the para isomer intermidiate.  These effects are mainly dominating which makes the p-Bromonitrobenzene more reactive than m-Bromonitrobenzene in Nucleophilic substitution.


Why is the melting point of PCl3 less than that of PCl5? From just the IB chemistry bonding chapter information, you would normally deduce that $\ce{PCl3}$ and $\ce{PCl5}$ are both covalent molecules.

However, as $\ce{PCl3}$ has a dipole moment, it has dipole-dipole intermolecular attractions, implying that  $\ce{PCl3}$ has a higher melting point than $\ce{PCl5}$, which only has London dispersion forces.

So why is it that $\ce{PCl5}$ has a higher melting point than $\ce{PCl3}$ (I saw online that  $\ce{PCl5}$ is an ionic solid)?

Because $\ce{PCl5}$ does something which is not immediately obvious from its molecular formula: it autoionizes and becomes an ionic solid $\ce{PCl4+PCl6-}$. As such, it has much stronger interactions than $\ce{PCl3}$ with its mere dipole-dipole attractions, hence the higher melting point.

If not for that fact, you deduction should have worked just fine.


In a reaction between Calcium Ammonium Nitrate or Urea with water, does Nitrogen Content and Purity affect how cold the solution gets? Recently, I bought an ice pack that contained Calcium Ammonium Nitrate, and realized that it really didn't get cold. I came to the conclusion that perhaps Urea is more effective, yet it is still puzzling to me that in a perfectly secure ice pack, it got to refrigerator water temperature.

I was thinking, perhaps purity of compound and nitrogen content (46% in Urea vs ~23% in CAN) could make a difference.

The endothermic process is just the dissolution of the calcium ammonium nitrate (CAN) or urea in water.
The most likely explanation for the lack of cooling when you used your pack is that over time, by getting partially squished at some point, or from defective manufacturing, some water leaked out so that the CAN or urea had already been partially/fully dissolved by the time you tried to use it.
These reagents are pretty stable and it is very unlikely that the nitrogen had changed form or that the system had been contaminated.


Comparision of donation power There is an R an organic group connected to either Oxygen or Nitrogen. R is such a group which has almost zero electronic effects (Resonance effect and inductive effect). I want to ask if N or O are to donate a lone pair then which one will have higher donating capacity?

According to my opinion  it should be Oxygen as it is more electronegative than Nitrogen but I'm still not confident and want some proof considering other factors also and neither do I have an answer.

...According to my opinion it should be Oxygen as it is more electronegative than Nitrogen...

It's correct that oxygen is more electronegative than nitrogen. However, this makes an oxygen group a poorer electron donating group than the corresponding nitrogen group. Electronegativity means that the atom is less willing to give up the electrons. Nitrogen is more willing to give up (donate) its electrons because it is less electronegative.


Relative levels of the two most prominant glutamate species at pH 4.7 Calculate the relative levels (i.e. ratios of concentrations) of the two most prominant glutamate species at pH 4.7.

I know that the Henderson–Hasselbalch equation is to be used which is
$$\mathrm{pH=p}K_\mathrm a + \log_{10}\mathrm{\frac{[A^-]}{[HA]}}$$

For one glutamate species, this would change to
$$\mathrm{4.7=2.19 + \log_{10}\frac{[A^-]}{[HA]}}$$

and for the other species it would be
$$\mathrm{4.7=9.67 + \log_{10}\frac{[A^-]}{[HA]}}$$

I know there are some other variables I need to figure out to solve the equation, but I'm not sure how to go about finding them with the information given.

The two species are related. They are conjugated acid and base to each other. Assuming one species $\ce{A^2-}$ with concentration $x \,\pu{mol/L}$, the second species $\ce{HA-}$ with concentration $y\,\pu{mol/L}$. There is a third species $\ce{H2A}$ with concentration $z \pu{mol/L}$.
$$\frac{10^{-4.7}\times x}{y}=10^{-9.67} \tag{1}$$
$$\frac{10^{-4.7}\times y}{z}=10^{-2.19} \ \ \tag{2}$$
Based on above two equations, the two most prominant species are $\ce{HA-}$ and $\ce{H2A}$ and ratio should be $y/z$. I utilized definition of $K_\mathrm a$ instead of $\mathrm pK_\mathrm a$ to get the above two equations. $10^{-4.7}$ is simply $[\ce{H+}]$ at $\mathrm{pH} = 4.7$
Therefore the final answer is:
$$\frac{y}{z} = 10^{2.51}$$


Best Guide to Chemical Thermodynamics I understand thermodynamics in a physicsy way - at the level of Callen's Thermodynamics. However, my chemical thermodynamics is quite rusty: concepts like fugacity and activity coefficients are only vaguely familiar from undergrad courses.

I'm looking for a well written, rigorous and succinct guide to advanced chemical thermodynamics - at the level where I can read papers on ENRTL equations of state, or understand Pitzer models fully. If it is geared towards chemical engineers (like myself) that would be a bonus.

Basically, I consider Callen's book a classic, and wonder if there's anything that good within chemical thermodynamics.

Does such a book exist? Thanks

I would suggest:

1) Introduction to Chemical Engineering Thermodynamics by Smith, Van Ness, and Abbott;

2) Chemical, Biochemical, and Engineering Thermodynamics by Sandler; and

3) This is more suitable for advanced reading or for a graduate student: Thermodynamics and Its Applications by Tester and Modell.