Fragmentations and rearrangements in n-(4-alkyl phenyl) alkanoic acids in mass spectrometry I have this example exam question for which I don't have the answer. I have been working on it the whole morning but I can't figure it out. The question suggests three molecules, only one of which corresponds to a given fragmentation table and mass spectrum following EI-based MS.

The molecules are 4-(p-hexyl phenyl) butyric acid:

and two variants where the phenyl is shifted one to the left (pentylphenyl pentanoic acid) or one to the right (heptylphenyl propanoic acid).

The table and EI-MS spectrum are unfortunately not of ideal quality, but they are as follows:

The 177 and 188 were given to both be directly stemming from the molecular ion, and with one of them having an even and the other an odd mass, this suggests a simple cleavage and a cleavage following rearrangement.

With pentylphenyl pentanoic acid I got close, but the fragment following rearrangement is 192 Da instead of the 188 Da seen in the spectrum.

UPDATE - Let me add two other examples of what I have tried so far. Maybe there is something wrong with my approach or someone can see the obvious change that would get to an observed peak at 188 m/z from here. The fragmentation on the left (below), is where there actually are 188 and 60 fragments, but the 60 fragment carries the charge (this peak at 60 m/z can be seen in the spectrum). On the right, I tried to get to a similar fragmentation while starting by putting the charge on the phenyl.

In your update you nearly got to the point: the McLafferty rearrangement from the carboxylic acid can lead to two products, either the ionized acetic acid (ML product) or the ionized alkene (ML'). In this case, ionization energy of the enol tautomer of acetic acid is probably much larger than that of alkenyl substituted benzenes (for instance the closest I could find in structure is o-Allyl-Toluene for which the IE is 7.78 eV). Thus from application of the Stevenson-Audier rule, the charge should remain on the moiety with the lowest ionization energy, leading to the formation of the m/z 188 moiety (ionized p-pentyl-allylbenzene).

This hypothesis is further reinforced as it also explains the formation of m/z 117 from a oss of 71 ($C_5H_{11}^\bullet$) from fragmentation of the m/z 188 radical cation, through a pathway similar to that leading to m/z 177 from the molecular ion.

(This is likely oversimplified mechanism interpretation: I strongly suspect that some other rearrangements or cyclisation likely occur in the process. I did not run a complete database search for other EI spectra of similar molecules, but for instance 10-phenyldecanoic acid leads to only a very minor loss of 60. Thus I would suggest that the m/z 188 probably undergoes cyclization and / or hydrogen migration processes leading other products.)


Typical reaction yield for combustion of hydrogen I am writing an essay on the chemistry of rocket fuel, and I was wondering what, if any, would be a commonly accepted value for the actual yield of the reaction between hydrogen, $\ce{H2},$ and oxygen, $\ce{O2}.$ I would also love it if you could cite some sources so I can find more information on this exothermic reaction:

$$\ce{H2 + O2 -> H2O + heat}$$

The equilibrium constant for the reaction $\ce{2H2 + O2 <=> 2H2O}$  is colossal, equal to $\mathrm{2.4 \times 10^{47}}$ at $\mathrm{500\ K}$ (source). This would ensure virtually 100% yield for the combustion of a stoichiometric mixture of oxygen and hydrogen in equilibrium conditions. However, a rocket engine is a reaction vessel that is far from equilibrium, and therefore oxygen could remain unreacted. This is in fact a problem, as it could damage the engine at its operating temperature. This, among other reasons, is why oxygen/hydrogen mixtures for rocket engines are actually very hydrogen-rich, containing up to twice the stoichiometric amount of hydrogen required.

So in practice, taking everything into account, rocket engines running on oxyhydrogen fuel are designed to have a  practically 100% combustion yield based on the limiting reagent, oxygen, while possessing a substantial excess of hydrogen fuel. There is a strong engineering pressure to maximise energy output by ensuring the oxygen is completely consumed; unreacted oxygen is a waste of lift capacity, storage space, materials, cost, etc.


What exactly is the use of photographic films in cameras? How are the pictures generated? By definition:

  Photographic film is a strip or sheet of transparent plastic film
  base coated on one side with a gelatin emulsion containing microscopically small light-sensitive silver halide crystals. The
  sizes and other characteristics of the crystals determine the
  sensitivity, contrast and resolution of the film.

But how exactly is a photographic film utilized in a camera?

I know that it has a sort of a silver coating. But how does it work? I read the Wikipedia article but found it a bit difficult to understand.

A photographic film consists of a light sensitive emulsion of fine particles (grains) of silver salts in gelatine
spread on a clear celluloid strip or a glass plate.

The grain size is very important to photographers, as this
affects the quality of the pictures produced.

AgBr is mainly used as the light sensitive material.

Some AgI is used in "fast" emulsions. The film is placed in a camera. When the photograph is exposed, light from the subject enters the camera
and is focussed by the lens to give a sharp image on the film.

The light starts a photochemical reaction by
exciting a halide ion, which loses an electron.

The electron moves in a conduction band to the surface of the
grain, where it reduces a $\ce{Ag+}$
ion to metallic silver.

  $\ce{2AgBr + h\nu \rightarrow 2Ag + Br2}$

In modern photography only a short exposure of perhaps 1/100th of a second is used.

In this short time, only a few atoms of silver (perhaps 10-50) are produced in each grain exposed to light.

Parts of the film which
have been exposed to the bright parts of the subject contain a lot of grains with some silver.

Parts exposed to paler parts of the subject contains a few grains with some silver, whilst parts not exposed contain none.

Thus the film contains a latent image of the subject.

However, the number of silver atoms produced is so small that the image is not visible to the eye.

Next the film is placed a developer solution. This is a mild reducing agent, usually containing quinol.
Its purpose is to reduce more silver halide to $\ce{Ag}$ metal.$\ce{Ag}$ is deposited mainly where there are already some $\ce{Ag}$ atoms. Thus the developing process intensifies the latent image on the film so it becomes visible. The
correct conditions for processing must be used to obtain an image of the required blackness.

If the film was brought out into daylight at this stage, the unexposed parts of the emulsion would turn black
and thus destroy the picture.

To prevent this happening any unchanged silver halides are removed by placing
the film in a fixer solution.

A solution of sodium thiosulphate is used as fixer. It forms a soluble complex with silver halides.

  $\ce{AgBr + 2Na2S2O3 \rightarrow Na3 [Ag(S2O3)2] + NaBr}$

After fixing, the film can safely be brought out into daylight. Parts blackened by silver represent the light parts of the original picture. This is therefore a negative.

To obtain an image with light and dark the right way round, a print must be made.
Light is passed through the negative onto a piece of paper coated with $\ce{AgBr}$ emulsion. This is then developed and fixed in the same way
as before.

EDIT: Wikipedia writes:

  In black-and-white photographic film there is usually one layer of
  silver halide crystals. When the exposed silver halide grains are
  developed, the silver halide crystals are converted to metallic
  silver, which blocks light and appears as the black part of the film
  negative. Color film has at least three sensitive layers,
  incorporating different combinations of sensitizing dyes. Typically
  the blue-sensitive layer is on top, followed by a yellow filter layer
  to stop any remaining blue light from affecting the layers below. Next
  come a green-and-blue sensitive layer, and a red-and-blue sensitive
  layer, which record the green and red images respectively. During
  development, the exposed silver halide crystals are converted to
  metallic silver, just as with black-and-white film. But in a color
  film, the by-products of the development reaction simultaneously
  combine with chemicals known as color couplers that are included
  either in the film itself or in the developer solution to form colored
  dyes. Because the by-products are created in direct proportion to the
  amount of exposure and development, the dye clouds formed are also in
  proportion to the exposure and development. Following development, the
  silver is converted back to silver halide crystals in the bleach step.
  It is removed from the film during the process of fixing the image on
  the film with a solution of ammonium thiosulfate or sodium thiosulfate
  (hypo or fixer).Fixing leaves behind only the formed color dyes, which
  combine to make up the colored visible image. Later color films, like
  Kodacolor II, have as many as 12 emulsion layers,with upwards of 20
  different chemicals in each layer.


In the photoelectric effect, why we use momentum to calculate kinetic energy? If we know the wavelength of an ejected electron, why do we use the momentum equation ($E_\mathrm k=\frac{p^2}{2m}$, $\lambda = h/p$). Why can't we use $\frac{1}{2}mv^2$ instead, using $c/\lambda = v$?

Does it have something to do with particle-like or wave-like behavior?

Relativity
Since electrons are travelling at near the speed of light, we need to use relativity.
In relativity, the correct equation relating energy to mass and momentum is the energy-momentum relation:

$E^2 = (m_0 c^2)^2 + (p c)^2$
where:

$E$ denotes the total energy of the particle.
$m_0$ denotes the (rest) mass of the particle.
$c$ denotes the speed of light.
$p$ denotes the momentum of the particle.

The momentum of a particle can be found from its (rest) mass and its velocity:

$\displaystyle p = \gamma m_0 v$
where:
$\displaystyle \gamma = \frac {1} {\sqrt{1 - \left({ \frac v c }\right)^2}} = \left({1 - \left({ \frac v c }\right)^2}\right)^{-0.5}$

Note that $p = m v$ is only true when $0 \lt v \ll c$, when $\gamma \approx 1$.
Kinetic energy
When the particle is not moving, $p = 0$, so $E = m_0 c^2$ (mass-energy equivalence) is denoted the rest  energy.
Substituting the definition of momentum gives:

$\displaystyle E^2 = m_0^2 c^4 + \gamma^2 m_0^2 v^2 c^2 = m_0^2 c^4 \left({\frac {c^2} {c^2 - v^2}}\right) = m_0^2 c^4 \gamma^2$

Now, the kinetic energy is defined as the total energy minus the rest energy:

$E_{\mathrm k} = E - m_0 c^2$
$\displaystyle E_{\mathrm k} = m_0 c^2 \left({\gamma - 1}\right)$

Classical approximation: $\displaystyle E_\mathrm k = \frac 12 m_0 v^2$
Now, the classical formula $\displaystyle E_\mathrm k = \frac 12 m_0 v^2$ is an approximation of $E_\mathrm k$ when $0 \lt v \ll c$:

$E_{\mathrm k}$
$\displaystyle = m_0 c^2 \left({\gamma - 1}\right)$
$\displaystyle = m_0 c^2 \left({\left({1 - \left({ \frac v c }\right)^2}\right)^{-0.5} - 1}\right)$
$\displaystyle \approx m_0 c^2 \left({1 - (-0.5) \left({ \frac v c }\right)^2 - 1}\right)$
$\displaystyle = \frac 1 2 m_0 v^2$

However, as I have mentioned, it only works when $v$ is very small, which is not the case for electrons.
Another approximation: $\displaystyle E_{\mathrm k} = \frac {p^2} {2 m_0}$

$E_{\mathrm k}$
$\displaystyle = \sqrt{m_0^2 c^4 + p^2 c^2} - m_0 c^2$
$\displaystyle = m_0 c^2 \sqrt{1 + \frac {p^2} {m_0^2 c^2}} - m_0 c^2$
$\displaystyle \approx m_0 c^2 \left({1 + \frac {p^2} {2 m_0^2 c^2}}\right) - m_0 c^2$
$\displaystyle = \frac {p^2} {2 m}$

Comparison
Now, for reasons unknown, the second approximation is actually better than the first:

(For convenience reasons, $c$ is took to be $1$ in the graph. It can be seen that the second approximation (blue) goes to infinity with the real kinetic energy (black) as $v \to c$, while the first approximation (red) does not.)
Perhaps this is the reason why the second approximation is used instead of the first.
Note that it is still an approximation.

Another reason is that, $\displaystyle \nu = \frac c \lambda$, but that is $\nu$ (Greek letter nu) for frequency not the $v$ (Latin letter vee) for velocity. The former has a unit of $\mathrm{s^{-1}}$ while the latter has a unit of $\mathrm{m~s^{-1}}$.


Is calcium chloride an acidic or basic salt? What I reason:
Calcium chloride is the salt of hydrochloric acid and calcium hydroxide. Calcium hydroxide is usually not considered a strong base, and I believe this is because of it's low solubility. $\ce{HCl}$ is a strong acid, and so the salt should be slightly acidic. Wikipedia states the $\mathrm{p}K_\mathrm{a}$ of $\ce{CaCl2}$ is between 8-9, which is in fact slightly acidic, confirming my theory.

What I've experienced:
I've prepared a number of aqueous calcium chloride solutions from distilled water, all of which are purple in the presence of universal indicator and tested with a calibrated $\mathrm{pH}$  probe to be basic. I've tried multiple sources of calcium chloride and each is basic.

TLDR:
Theoretically and literature values state that $\ce{CaCl2}$ is acidic, but my empirical evidence shows that it is basic.

Calcium hydroxide has solubility about 1.9 g/L. This is enough to create pH above 11, i.e. a strongly basic solution.

$\ce{CaCl2}$ solutions should be very slightly acidic if they were made from pure $\ce{CaCl2}$. This might not be the case. In industry calcium chloride is produces by reaction of calcium hydroxide with ammonium chloride, so industrial-grade calcium chloride is likely to be contaminated with calcium hydroxide.

Another potential reason for contamination is hot drying. Calcium chloride is quite hygroscopic, so it needs to be dried before use. When heated to hight enough temperature, moist calcium chloride hydrolyzes https://link.springer.com/article/10.1007/BF02654424


Can nitric acid be used to create an acidic medium for redox reactions? While reading about redox reactions, every time I found that $\ce{H2SO4}$ is used for creating an acidic medium.
My question is - can I use nitric acid to create an acidic medium?

Yes, you can.

However, nitric acid is an oxidizer, so it can interfere in some cases. If a strong reducer (like magnesium) is involved, nitric acid may act as an oxidizer even in diluted solutions. Obiously, it acts as an oxidizer in concentrated solutions, producing rather nasty nitrogen oxides, and if used without prior knowledge, it may produce unstable explosives.

For this reason, one generally use hydrochloric acid or sulfuric acid. Mind you, in case oxidation is not a problem, nitric acid may be used with no problems. Another reason to use sulfuric acid is that it is dirt cheap. It is the cheapest alternative.

Still, in concentrated solutions sulfuric acid does show oxidating properties, so in some cases sulfuric acid is not suitable either. Unfortunately, there is no common and reasonable alternative in the case, except significanlty weaker phosphoric acid.


Preparing a solution of hydroxylamine hydrochloride in HCl, struggling I recently came across a paper with a methodology description i cant seem to get the grasp of.

I need to make a particular solution for Fe extraction. The protocol's description is from this paper.

  Take a 100 μL aliquot of X (a soil sample previously mixed with 0.5M HCl) and mix it with 2 mL of a solution of 0.25M hydroxylamine hydrochloride in 0.25M HCl and incubate at 60 °C.

What I don't understand is how to make the solution when the volume of the two compounds is not specified? Does it imply that the 2mL of solution is made up of equal parts (1mL) of 0.25M hydroxylamine HCl and 0.25M HCl? Any help is highly appreciated. Thanks a lot in advance!

Does it imply that the 2mL of solution is made up of equal parts (1mL) of 0.25M hydroxylamine HCl and 0.25M HCl?

No, if you mixed 1 mL of 0.25M HCl and 1 mL of hydroxylamine hydrochloride, you'd have 2 mL of a solution that was 0.125M in HCl and 0.125M in hydroxylamine hydrochloride.

I think the purpose of the hydrochloric acid is to ensure that the pH stays acid.  One way you could make this solution is by:

Decide on a volume of solution you want to use, say $X$ mL.
Prepare at least $X$ mL of 0.25M hydrochloric acid by dilution from a more concentrated stock solution.
Weigh out the amount of $\ce{NH2OH.HCl}$ that you need.  It has a formula weight of 69.49 g/mol, so for $X$ mL of a 0.25 M solution you will need:

$$0.25 \frac{\mathrm{mol}}{\mathrm{L}} \times \frac{69.49\mathrm{g}}{\mathrm{mol}} \times \frac{X\;\mathrm{mL}}{1} \times \frac{\mathrm{L}}{\mathrm{1000mL}}=\frac{69.49\;X}{4000}\mathrm{g\;of\;\ce{NH2OH.HCl}}$$

Dissolve the weighed out powder with your 0.25M $\ce{HCl}$ solution in a volumetric flask, first fully dissolving the powder and then slowly adding more HCl until the target volume is reached.

The resulting solution will be 0.25M in both hydrochloric acid and in hydroyxlamine hydrochloride.


What is non-expansion work? I learnt that Gibbs free energy is the maximum amount of non-expansion work.
But the phrase non expansion work confuses me. Work is defined as pressure times change in volume. If there is no expansion, how can work be done?

Pressure-Volume work is a type of mechanical work .
There are also non-mechanical forms of work where in pressure and volume terms are not involved .
The most famous example is electrical work . The work done to move electrical charges ( like motion of electrons in a conductor) comes under this category .


Acid-base reaction vs. salt dissolving reaction I was watching a video in Khan academy about acids and bases. In the video, they asked the $\mathrm{pH}$ of a solution that is $\pu{0.15M}$ of $\ce{NH3}$ ($K_\mathrm{b}=1.8\times10^{-5}$) and $\pu{0.35M}$ of $\ce{NH4NO3}$. So, it was obvious to me that the first reaction is going to be that of a weak base because the $K_\mathrm{b}$ is given. But what wasn't clear to me that the second reaction is going to be that of a salt dissolving reaction that is dissolved completely as is claimed in the video.

Since this problem is presented in a beginners learning video I guess there is a way to determine that $\ce{NH4NO3}$ is a salt that is dissolving completely in water and not an acid or base according to the information given in the video.

tl;dr

  What wasn't clear to me that the second reaction is going to be that of a salt dissolving reaction that is dissolved completely as is claimed in the video.

Remember the rule: "All ionic salts and strong acids/bases are 100% dissociated in aqueous solution"

  I guess there is a way to determine that $\ce{NH4NO3}$ is a salt that is dissolving completely in water and not an acid or base

Realize that all such types of species of the form $\ce{A+B-}$ are ionic salts. If you try draw out the Lewis structure of $\ce{NH4NO3}$, you'll quickly realize it to be very different from that of, say, water or acetic acid.

In ammonium nitrate, you have two separate ions - ammonium cation and nitrate ion - held together by strong electrostatic forces of attraction (often called an "ionic bond"). Water, having a high dielectric constant, is able to break these electrostatic forces easily and dissolve the salt in 100% concentration.

Hence, we say that "All ionic salts and strong acids/bases are 100% dissociated in aqueous solution" i.e. when dissolved in water, they'll dissociate completely.  It seems you already knew the latter part, while the part for ionic salts is new for you.

As I said before, note that ionic salts are different from weak acids or weak bases like $\ce{CH3COOH}$. The $\ce{O-H}$ bond here is covalent and breaking it is an altogether different job as compared to breaking ionic bonds. Here, the reaction (or "equilibrium") has to favor the dissociation of acetic acid. This is ensured by the high stability of the conjugate base. If you've taken GOC classes, you would quickly realize that the conjugate base $\ce{CH3COO-}$ is much weaker than say $\ce{HSO3-}$ or $\ce{NO3-}$ (guess which acids the latter two are? ;) )

Again, realize that the rule is actually not completely correct. Later down the equilibrium road, you'll come across the equilibrium constant $K_\mathrm{sp}$ that defines exactly how much soluble certain salts are in aqueous solutions. Certainly, many ionic salts would be in that list, and that too with very low solubilities instead! Again, the way these people simplify calculations at an elementary level is to implicitly assume that "All ionic salts whose $K_\mathrm{sp}$ is provided are only partially soluble, while those whose $K_\mathrm{sp}$ is not provided are completely soluble (i.e. 100% dissociated)"


Smallest nearest neighbor distance among elements Of all the elements in periodic table, which one has smallest nearest neighbor distance? I tried searching on net, but could not find any reference. My earlier guess was one which goes into FCC with smallest unit cell dimension should be one. But I am not sure.

In an FCC, if unit cell dimension is a, then nearest neighbor distance will be a/sqrt(2) because nearest atom to an atom on the edge of the cube will be face atom. But this is for FCC. For SC of size a, nearest atom will be at a distance of a. So, for different elements, there will be different nearest neighbor distance based on its unit cell geometry and unit cell length. I want to know which combination of these result in shortest nearest neighbor distance.

Both FCC and HCP packings have nearest neighbours as close as possible (in nature at least, the simple cube is even closer but doesn't exist). Their nearest neighbour distances in terms of $a/R$ are $2\sqrt{2}$. Here $a$ is the length of a side of the unit cell and $R$ is the radius of the atom the cell consists of.

When you are looking for the smallest nearest neighbour distance this means that you are looking for the smallest $a$ in an FCC or HPC packing. You can translate this into: you are looking for the atom which forms an FCC or HPC packing and has the smallest value for $2\sqrt{2}R$ so for $R$.

Now, I don't know all the options but for example $\ce{Cu, Au, Ag}$ form FCC structures and $\ce{Zn}$ forms an HCP structure (see e.g. here). You can look up the atomic radii and find that Zinc has the smallest: $122$ pm and will thus have a nearest neighbour distance in the lattice of $345$ pm.


η1 or sigma complex involving cyclopentadienyl anion When $\ce{\eta^1-C5H5-}$ acts as a ligand, does the donor carbon become $sp^3$ hybridized, and would this be a sigma donor through sigma orbitals? Or, is it $sp^2$ where the lone pair is donated through a pi orbital, making it a sigma donor through a pi orbital?

The second option seems much more likely as the ligand changes the $\ce{C}$ attached to metal randomly with time through rearrangements.

Also, does "sigma complex" and "pi complex" only differentiate the hapticity? as $\ce{\eta^1-C5H5-}$ is a sigma complex and $\eta^3$, $\eta^5$ are pi complex.

Even though depictions of $\eta^1$-coordination might suggest a $\mathrm{sp^3}$-carbon donating, there is little evidence, that it actually is. Taking the example you have presented in your answer from Casey et. al.[1] I have run a geometry optimisation to obtain molecular orbitals at the DF-BP86/def2-SVP level of theory. The following geometry parameters around the coordinating carbon already suggest only little pyramidalisation:
\begin{array}{lrr}\hline
\text{Parameter} & \text{Experimental [1]} & \text{DF-BP86/def2-SVP}\\\hline
\mathbf{d}(\ce{Re-C^1}) & 236.0 & 240.2\\
\mathbf{d}(\ce{Re-H})   & 288.7 & 282.6\\
\mathbf{d}(\ce{Re-C^2}) & 317.8 & 321.3\\
\mathbf{d}(\ce{Re-C^5}) & 315.6 & 315.6\\
\angle(\ce{Re-C^1-H})   & 114.3 & 100.7\\
\angle(\ce{Re-C^1-C^2}) & 110.8 & 109.5\\
\angle(\ce{Re-C^1-C^5}) & 108.5 & 106.4\\
\angle(\triangle[\ce{C^1C^2C^5}]\ce{-H}) &
                        122.8   & 139.1\\
\text{RMSD} & 0.0 & 0.1\\\hline
\end{array}

We know that crystal structures are not really that accurate with protons, hence I'd put a little more trust into the calculation at this point. The paper does not give a value for any protons.
Here is a picture of the calculated structure.

When we look at the canonical MO we will see that all (occupied) π-obitals of the Cp-ligand are still there. Obviously they are distorted, but clearly recognisable.

The second orbital is the HOMO-1, which is the main contribution for the coordination, and the last is the HOMO (click for large). The only s-contribution to the HOMO-1 is from the carbon of the $trans\text{-}\ce{CO}$-ligand (and some hydrogen).
In output numbers, missing from 1 are minor contributions (<0.05):

 Alpha occ 81 OE=-0.276 is C39-p=0.19 C42-p=0.12 C41-p=0.12 C43-p=0.11 C40-p=0.11
 Alpha occ 87 OE=-0.185 is C39-p=0.26 C42-p=0.23 C41-p=0.22 Re24-d=0.08 C27-s=0.05
 Alpha occ 88 OE=-0.153 is C43-p=0.32 C40-p=0.32 C41-p=0.13 C42-p=0.12

When we localise the orbitals with the Natural Bond Orbital (NBO) analysis, we find a lone-pair at $\ce{c^1}$ (blue/orange) which donates into the anti-bonding $\ce{Re-^{$trans$}CO}$ orbital (yellow/red):

The composition of the lone pair orbital is

     (Occupancy)   Bond orbital / Coefficients / Hybrids
 36. (1.27157) LP ( 1) C 39            s(  8.77%)p10.40( 91.23%)d 0.00(  0.00%)

The low occupancy is normal for donor-acceptor-complexes with some covalent character.

TL;DR Most $\eta^1$ complexes are in first approximation π-coordination complexes like the definition suggests[2]. This may change significantly when moieties are altered.

Further thoughts:
Wikipedia makes one important statement (it rarely does, but here it is true)3:

  Molecules with polyhapto ligands are often fluxional, also known as stereochemically non-rigid.

You can expect that in $\eta^1$-ligands the metal can easily migrate from one carbon to another, as the MO show. I would expect some kind of pseudo-rotation.

Ref.

Charles P. Casey, Joseph M. O'Connor, William D. Jones, Kenneth J. Haller. Organometallics 1983, 2 (4), 535–538. (link via doi, pdf via researchgate.net) Crystallographic data via CCDC number: 1118829.
η (eta or hapto) in inorganic nomenclature
Wikipedia: Hapticity and fluxionality


Heat effect on diluting concentrated NaOH solution I am an engineer in water treatment and have a chemistry question. In water treatment we dose $32\%$ sodium hydroxide solution for $\mathrm{pH}$ adjustment and there is a concern that our dosin lines must be $100\%$ dry before introducing the $\ce{NaOH}$ solution. This is the problem as they have been pressure tested with water and there is water residue in them.
Will $32\%$ sodium hydroxide solution react violently with cold water? It is my understanding that the reaction only occurs when solid sodium hydroxide (lye) is dissolved in water and since my $\ce{NaOH}$ is already in solution then this reaction will not take place. Is this correct?

The solution gets warmer by diluting with extra water, but no violent reaction. I suggest to do a test, if the temperature raise is concerning you, e.g. because of the tubing temperature resistance.
What  can also be done is applying solution with graduating concentration, so warming would be spreaded in sevceral steps. But as hydroxide is already quite diluted, I expect temperature change rather few than many degrees.
Another concern is initial dosage would be affected by water, decreasing concentration of the applied volume.


Ventilating combustion byproducts in water heaters I'm designing a system which will heat up water by burning natural gas which is commonly available. I would like to have a slow flame because water will need to reach $\mathrm{70^\circ C}$ and stay there. Having in mind that even weak winds can extinguish the flame as well as the system will be put together with other objects and systems in a tight space I decided it would be a good idea to build a chamber around the flame nozzle thats directly attached to the water container. I will build a small opening in the chamber to which I will attach a tube which will feed atmospheric air into the chamber through some kind of a filter (probably a piece of cloth) to keep it clean. I read that the byproducts of burning natural gas are $\ce{CO2}$ and water vapor which raises 2 questions:

Do I have to worry about humidity increasing in the chamber and eventually water forming on the inside, in extreme conditions water in the system can be as low as $\mathrm{-20^\circ C}$ (I call it water but it will be actually antifreeze)
Do I have to worry about the chamber filling up with $\ce{CO2}$ and eventually putting the flame out which will then be unable to be lit again. Is there anything I can do such that the byproduct will exit the chamber naturally without me having to do any work for it.

Edit:

So far from the comments it was made clear that it would be better to use a condensing boiler design efficiency wise, however due to construction complications that I'd like to avoid I'll stick with only the hot half of the design in question where the water will pass through a tubing inside the chamber where it will be heated, instead of applying heat to the container directly. There will be an exhaust that will utilize natural convection to expel the exhaust gasses in addition to the inlet.

Rather than design something that offers danger of overheating, carbon monoxide poisoning, gas and over-pressure explosion, etc., why not use an off-the-shelf "instant" tankless water heater, available from sources such as Home Depot, Amazon or Lowes?

Find one certified for your area. Making your own is a tankless task!


What is a benzene ring minus a hydrogen atom called? I thought it was Phenyl but the teacher said it wasnt and that it started with "A" I cant seem to figure it out

You could be referring to "aryl".  "Phenyl" is specific to $\ce{C_6H_5}$, "aryl" is generic to monovalent benzenoid  aromatic ring groups.


Different possible Lewis structures of peroxymonosulfate anion I was trying to draw the Lewis structure of $\ce{HSO5-}$ (peroxymonosulfate), so I came up the structure on the left that satisfies the octet rule. Then, I saw that the correct structure is the one on the right. Why is the structure on the right favored over the one on the left? They both have the same formal charges. So shouldn't the one that satisfies the octet rule be favored over the one with an extended octet?

$\hspace{15 mm}$

So shouldn't the one that satisfies the octet rule be favored over the one with an extended octet?

That's only true when you are comparing resonance structures. Resonance structures only involve movement of electrons and not atoms.

It's not true for different Lewis structures, which are the ones you drew. There are lots of compounds which are perfectly stable but do not (in principle) satisfy the octet rule.

I would have mentioned the extreme instability of your multiple $\ce{O-O}$ single bonds in your alternative proposed structure, but the other answer beat me to it.


Determination of spinel structure by CFSE $\ce{CuFe2O4}$ has inverse spinel structure, but $\ce{ZnFe2O4}$ has normal spinel structure. Explain.

I know that normal spinels have a structure like $\ce{(A^{II})^{tet}(B^{III})^{oct}_2O_4}$-like structure, while inverse spinels have a $\ce{(B^{III})^{tet}(A^{II}B^{III})^{oct}O4}$-like structure.
Also, I know that if $\ce{B^{III}}$ ion has greater CFSE gain in octahedral site than that of $\ce{A^{II}}$ ion, a normal spinel structure is expected. If $\ce{A^{II}}$ ion has greater CFSE gain in octahedral geometry than the trivalent $\ce{B^{III}}$ ion, then we get an inverse spinel structure.
For the given question, $\ce{Fe}$ has a $\mathrm{3d^6}$ configuration having a structure $\mathrm{t_{2g}^4 e_{g}^2},$ giving a CFSE
$$2(0.6 \Delta_\mathrm{o}) + 4(-0.4 \Delta_\mathrm{o}) = -0.4 \Delta_\mathrm{o},$$
while $\ce{Zn}$ has a $\mathrm{3d^{10}}$ structure having 0 CFSE. So, CFSE of $\ce{Zn}$ is greater than CFSE of $\ce{Fe}.$ So, according to the rule stated before, it gives me an inverse spinel structure.
Similarly for $\ce{Cu},$ it should have a $\mathrm{3d^9}$ $(\mathrm{t_{2g}^6 e_g^3})$ structure which gives a CFSE of
$$3(0.6 \Delta_\mathrm{o}) + 6(-0.4 \Delta_\mathrm{o}) = -0.6 \Delta_\mathrm{o} < -0.4 \Delta_\mathrm{o},$$
which gives a normal spinel structure.
What is wrong with this calculation so that I'm getting opposite structures to those predicted in the question?

I'm not really getting your numbers here.
In your situation the iron is Fe(III), so a $d^5$ system. For $d^5$ it does not matter if it's in an octahedron or a tetrahedron since the energies will just cancel out. So in both situations it will be determined by the other ion. For Zn(II) it also doesn't really matter. So I guess here the regular spinel dominates as Zn(II) just prefers more tetrahedral coordinations in general, especially in solid states while Fe(III) is happy with an octahedron.
For the Cu(II) however you've got a $d^9$ system. So the octahedron will give you something like $(6*2/5)-(3*3/5) = 3/5$ (ignoring things like spin pairing energ here).
But in the tetrahedral case you also have to consider that the ligand field splitting for a tetrahedron is only $4/9$. So it will be something like $4/9*[(4*3/5)-(5*2/5)] = 8/45$ and therefore lower.
So the Cu(II) will prefer the octahedral position here and hence form an inverse spinel. Feel free to correct me here I was just drawing the diagrams as I read the question, perhaps I did a mistake somewhere.


Heat Transfer during Phase Change When ice is melting, we know that the temperature stays constant at 0 ℃. If we were to place ice at 0 ℃ in a sufficient amount of water at ambient temperature, would the temperature of the water itself also stay constant during the time the ice is melting, or is it only the temperature of the ice that stays constant? Does the actual cooling of the water occur after or during phase change?

No, as the ice is melting, the water is cooling down. Ice melts in water because the water molecules in solution collide with those in the ice, giving them energy. In exchange, the water molecules in solution slow down a little during the collision. While the Ice does not change temperature as it is melting, the energy from the water is going into breaking the crystal structure of the ice, turning it into water. This process consumes energy, and the energy to fuel this process comes from the water in solution. Another factor to consider is conservation of energy - if energy is being put into breaking the ice, where else could that energy come from but from the water?


Why is Nitrogen 15 found almost exclusively in marine environments? Prelude

The following story was told to me earlier this summer

Every year during October, the Pacific rain creates an ideal spawning condition for the wild Pacific salmon in the coastal forests of British Columbia. Of course, when we think of salmon swimming upstream in the Pacific Northwest, some images might come to mind.

Black bears and grizzly bears carry hundreds of kilos of fish out of the water and into the forest to consume. Naturally in the areas proximal to the rivers, fish carcasses are littered throughout, with the bears having ate what they wanted, leaving behind the rest to decompose.

Scientists several years ago took to examining the nitrogen uptake and found that trees in nearby the river received a significant amount of their nutrients from the fish. Through their findings they showed a case that reverses the land to sea flow of nutrients. The study has been used to argue for the dedication of resources to rebuild and maintain wild pacific salmon runs.

Source: David Suzuki Foundation

Chemistry and the Question

The study above uses the abundance of $\ce {^{15}N}$ found naturally in the ocean, and transported via salmon, to support their findings. $\ce {^{15}N}$ has a 0.365% natural abundance, and according to the story, it is almost exclusively found in marine environments. Which leads us to my question: why? Why is this isotope largely found in the ocean and not on the land?

Let's start that the sentence found in the source you provided:

  the nitrogen isotope 15, which is found almost exclusively in marine environments

is misleading and simply incorrect.

Now to some basics. When we discuss isotopes in environmental sciences, we never use the absolute values of an isotope content. Instead we use isotope ratios:

$$\text{Ratio (R)} = \frac{\text{heavy isotope abundance}}{\text{light isotope abundance}}$$

However, the ratios themself are not very useful, so the delta ($\delta$) notation is used:

$$\delta[^0\!\!/\!_{00}] = \left(\frac{R_{\mathrm{sample}}}{R_{\mathrm{standard}}}-1\right)\times1000$$

All standards have the delta value equal $0$ by convention. There is an international standard material for each isotope, for example for $\ce{^13C/^12C}$ ratio it is VPDB (Vienna PeeDee Belemnite), and for $\ce{^15N/^14N}$ it is atmospheric nitrogen. The latter standard was chosen because the content of the $\ce{^15N}$ isotope in atmospheric nitrogen is uniform throughout the world and it does not vary significantly over time. Many chemical, biochemical and physical processes cause the so-called isotopic fractionation to occur - it results in the increased (enrichment) or decreased (depletion) isotope ratios.

OK, we know the terminology, so now to the point. In general, the metabolic processes of living organisms cause enrichment of nitrogen. The extent of the enrichment depends on the trophic level (position in the food chain) and dietary sources of nitrogen. The highest enrichment is found in terrestial herbivores, while marine organisms usually show lower $\delta\ce{^14N}$ values. Given those facts the cited sentence makes no sense, and is probably a result of a misinterpretation of some scientific paper.

There is a lot of literature on the subject if you are interested. I can give you one good reference as a starting point:

M.A. Vanderklift, S. Ponsard. Sources of variation in consumer-diet
$\delta\ce{^15N}$ enrichment: a meta-analysis. Oecologia (2003) 136:169–182 DOI 10.1007/s00442-003-1270-z.


How to know if a reaction forms H3O+ ions or OH- ions? How would you know when an acid or base is paired with $\ce{H2O}$ that it will form an $\ce{OH-}$ ion or a $\ce{H3O+}$ ion? I just started the acid and base equilibrium unit, and I'm just confused on what conditions $\ce{OH-}$ or $\ce{H3O+}$ ions form.

H3O+ is just the combination of a a $\ce{H+}$ ion, which we know to be released from the dissociation of an acid in an aqueous solution, with a water molecule. It happens because the $\ce{H+}$ is so, so positive (and therefore so reactive) that the water molecules, with its lone pairs (which are locally negatively charged), are willing to form a dative covalent bond with the $\ce{H+}$ ion.
So like Ivan said, whenever you see an acid in an aqueous solution you can consider $\ce{H+ \equiv H3O+}$. The key there though, is that it must be in aqueous solution so that it can dissociate (and the $\ce{H3O+}$ form in the first place). That is the condition that is needed to form it. This is the same for $\ce{OH-}$, which requires an aqueous solution to form and thus release the ions in say NaOH, forming $\ce{Na+}$ and $\ce{OH-}$.


Number of stereoisomers of a polyene Find the number of stereoisomers of the following compound:

I think the answer is 16:

The double bond in the ring cannot show geometrical isomerism.
The first double bond on the top substituent can show cis-trans isomerism and the one on the end will not show cis-trans isomerism because of two same groups on one carbon.
The substituent on the bottom that is: $\ce{-CH=C=C=CHMe}$ can also show cis-trans isomerism.
Then, there are two chiral centres and the number of optical isomers is $2^2=4$. The total number of stereoisomers is 16.

But the answer given is 4. Am I correct or not?

A picture is worth a thousand words so I've redrawn your compound and have numbered the areas to be considered.

Double bond "1" is not capable of displaying cis-trans
stereoisomers since there are two identical substituents (hydrogens)
on one end of the double bond.
Double bond "2" is capable of displaying cis and trans stereoisomers since there are two different substituents on each end of the double bond.
Double bond "3" can only exist as a cis isomer due to the geometric constraints imposed by the 6-membered ring.
Double bond "4" is a 1,2,3-triene (allene would be a 1,2-diene); such trienes  display the same stereochemical cis-trans possibilities as a double bond.
The substituents on the two cyclohexene carbons contained in circle "5" also have cis-trans arrangements; these are further illustrated in the cyclohexene drawings pictured below.
Finally, each of the carbons contained in circle "5" are chiral (they each have 4 different substituents) and can exist as R and S stereoisomers.

Considering only the circles that can have stereoisomers (2, 4 and 5) we can have the following spatial arrangements:

2-cis, 4-cis, 5-trans, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.
2-trans, 4-cis, 5-trans, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.
2-cis, 4-trans, 5-trans, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.
2-trans, 4-trans, 5-trans, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.
2-cis, 4-cis, 5-cis, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.
2-trans, 4-cis, 5-cis, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.
2-cis, 4-trans, 5-cis, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.
2-trans, 4-trans, 5-cis, and this geometric isomer can further exist as an enantiomeric pair; so 2 stereoisomers in all.

Therefore, there are a total of 16 stereoisomers for your molecule, 8 diastereomers, each of which can exist as an enantiomeric pair.

Thanks to Loong for pointing out that the cis and trans geometry in circle "5" limits the R and S possibilities at the two chiral carbon atoms in circle "5".  That is, these carbons in (for example) the cis isomer cannot be R,S R,R S,R and S,S, - in the cis isomer they can only be either R,S and S,R or R,R and S,S.


etymology of phosphatidates and phosphatidic acid A phosphatidic acid is an ester of a glycerol, two fatty acids and a phosphoric acid.
Who coined this term, and is the word's "phat" part related to english "fat"?

Apparently, "phosphatidate" is from "phosphate" + "-ide" + "-ate".

The word "phosphatidic acid" is from "phosphate" + "-ide" + "-ic acid".

Phosphate ($\ce{PO4^3-}$) is the anion of phosphoric acid ($\ce{H3PO4}$).

The suffix "-ide" is used in biochemistry to just about anything. It just denotes an element of a group of related compounds.

Now, "-ate" and "-ic acid" denotes the acidic nature of the compounds.

And no, it is not related to the English word "fat".


Stability order of octahydrobiphenylenes I am facing trouble in the following question

  What is the stability order of the following compounds?

In my view cis is less stable than trans. So if prior groups are on the same side of the double bond then it must have less stability. Also conjugated double bond in short non aromatic cyclic system is highly unstable. So 1,2,3,4,5,6,7,8-octahydrobiphenylene (X) should be the least stable. However I could not guess the correct order.

While solving other related questions, I came up with a idea that

  More is the number of $\alpha$ hydrogen present(hydrogen attached to carbon which is adjacent to a carbon containing double bond ) more stable is the compound.

In the question $Z$ has maximum 10 $\alpha$ hydrogen so it is the most stable.$X$ and $Y$ both have 8 $\alpha$ hydrogen.But $Y$ is more stable since conjugated double bond in short non aromatic cyclic system is highly disfavoured.

Hence the correct stability order is $Z>Y>X$

Side note- I am still not satisfied with my second argument.Any other explanations or thought will be highly appreciated.


Rigorous reason behind internal energy change being zero while mixing A container is divided in two parts: one part contains oxygen gas $(n_1$ moles, at temperature $T_1)$ and the other part contains helium gas $(n_2$ moles, at temperature $T_2).$ The partition separating them is removed. Find the final temperature of the mixture $(T).$

While solving this (multiple choice exam) question, since I was constrained by time, I started to look for hints within the question. I realized the question has explicitly mentioned the gases, i.e. it gave us information about their degrees of freedom (since oxygen is diatomic and helium is monoatomic). So my instinct was that this had to do something with $C_V.$ At the time, I hypothesized $\Delta{U}_\mathrm{sys} = 0.$ Thus:
$$n_1\frac{5R}{2}(T - T_1) + n_2\frac{3R}{2}(T - T_2) = 0$$
$$T = \frac{5n_1T_1 + 3n_2T_2}{5n_1 + 3n_2}$$
Which was indeed one of the options, and in fact, the correct one.
However, is there any rigorous method/reasoning that tells us why exactly is it that $\Delta{U}_\mathrm{sys} = 0?$ I can't think of a satisfactory explanation. I tried to analyze it using the first law of thermodynamics, but that clearly seems to be the wrong approach.

The process in question is presumably adiabatic, since we are told that the temperature changes and are not given any information about the temperature of the surroundings (adiabaticity can also sometimes be assumed if gases are mixed rapidly). Secondly, the final volume of the system (the entire container) is the same as the initial volume (and in addition we can assume that each gas expands without performing any work, i.e. the expansions can be assumed to be free). That means that neither heat nor work exchanged with the system during the process, which means that in accordance with the first law $\Delta U = 0$.


When I dissolve sugar in my cup of tea/coffee, does it become a liquid? When I dissolve sugar in my cup of tea/coffee, does the sugar go from being a solid to being a liquid?

Related: https://physics.stackexchange.com/questions/5562/what-keeps-the-sugar-suspended-in-the-tea

When the sugar dissolves, it is broken up into individual particles; each molecule is surrounded by water molecules. Essentially, you are trying to decide if a single molecule is a solid or liquid when dissolve.

The best fit is a liquid since the molecules are free to move independently and will take on the shape of the container.


Comparing Acidic Strength of H3PO4 and H3AsO4 The $\mathrm{p}K_\mathrm{a}$ values for $\ce{H3AsO4}$ and $\ce{H3PO4}$ are $2.30$ and $2.12$ which indicate that $\ce{H3PO4}$ is a stronger acid compared to $\ce{H3AsO4}$. I looked up their electronegativity values to find that they are almost the same for both arsenic and phosphorus. I feel $\ce{H3AsO4}$ should be the stronger acid because arsenic atom is larger in size and the charge is spread out in much more volume which should make the conjugate base more stable. What am I missing out?

The stability of an acid is judged by the strength of it's conjugate base. If the conjugate base is stable, it implies that it is a weak base and hence the corresponding acid is stronger.
Now consider $\ce{H3PO4}$. The structure has one $\ce{P=O}$ bond and $\ce{3 P-OH}$ bonds.

The conjugate base would be one Hydrogen short from an $\ce{OH}$ group. Similarly with $\ce{H3AsO4}$. However, the conjugation (resonance) will be better in the case of $\ce{P=O}$ bond, since it is an orbital overlap of oxygen's $\mathrm{2p}$ and phosphorus' $\mathrm{3p}$ as against arsenic's $\mathrm{4p}$. $\mathrm{2p}$ and $\mathrm{3p}$ are of comparable size and account for a better overlap and better resonance, distribution of electronic charge and hence a more stable conjugate base.
Consequently, $\ce{H3PO4}$ is better than $\ce{H3AsO4}$ as an acid.


Are sp, sp2, and sp3 hybridisation only relevant to the carbon atom? When we are talking about $\mathrm{sp, sp^2}$ and $\mathrm{sp^3}$ hybridisation, is it only relevant to the carbon atom only?
For the following molecules:
Acetone (propanone), acetic acid (ethanoic acid), tert-butanol (2-methyl-2-propanol), benzene, cyclohexane, 2-butyne.
Which of these contain at least one $\mathrm{sp}$ atom?
I am confused because for acetic acid and tert-butanol, the oxygen atom in the two molecules are bonded to a carbon and a hydrogen - that means 2 bonds.
Would this be and $\mathrm{sp}$ hybridisation? But doesn't $\ce{O}$ forms $\mathrm{sp^3}$ hybridisation as an alcohol function?

Not only $\ce{C}$,but also other elements (e.g., $\ce{B}$, $\ce{N}$, $\ce{O}$, etc.), which have partially filled $\ce {p}$ orbital can undergo $\ce {sp^3}$, $\ce {sp^2}$, or $\ce {sp}$ hybridization.

To understand how it works for molecules other than carbon, which do not show four bonds, let's take a water molecule as an example. Oxygen atom has $\ce {6 e-}$s in its outer most energy level (n = 2), which has electron configuration: $\ce {2s^2 2p^4}$. When water molecule is formed, it was quantum mechanically argued that all these 4 orbital ($\ce{2s, 2p_x, 2p_y, and 2p_z}$) undergo hybridization to form four equally energized $\ce {sp^3}$ orbitals (the name, $\ce {sp^3}$ is given to the newly formed hybridized orbitals because they were made up by $\ce {1\times s + 3\times p}$). Since they are equally energized, they would occupy the space in such a manner that they have equal distances and equal angles to each other. The only way it can arrange to achieve that is in tetrahedral shape. That four orbital can have maximum of eight electrons. When you put six electrons, which are belong to oxygen in its four $\ce {sp^3}$ hybridized orbitals, you end up with two fully filled and two half-filled orbitals, because you are supposed to avoid pairing them whenever possible to avoid pairing energy so that you keep the atom in its lowest possible energy state. Now, two hydrogen atoms can occupy these two half-filled orbitals with one  $\ce {s}$ electron of each, making 2$\sigma$ bonds. We called it $\sigma$ bonds because there are two nuclei involved to make the bond and it was head-to-head overlapping (precisely, $\ce {sp^3-s}$ bond). What about other two $\ce {sp^3}$ orbitals, which are already filled with electrons from oxygen? We called they are lone pairs.

Now $\ce {O}$ of $\ce {H2O}$ molecule is $\ce {sp^3}$ hybridized. Therefore, the theoretical shape of $\ce {H2O}$ molecule is tetrahedral. Yet, actual shape is angular because if you get a snap shot of $\ce {H2O}$ molecule, you can't see the electrons. Only you see is three nuclei, one from oxygen and other two from hydrogen, arranged in angular manner.

Similar to water molecule, oxygen atom in tert-butanol (2-methyl-2-propanol), in addtion to all four carbon atoms, is $\ce {sp^3}$ hybridized. The $\ce {O-H}$ bond is $\ce {sp^3-s}$ bond while the $\ce {O-C}$ bond is a $\ce {sp^3-sp^3}$ bond.

Now, you can look how it work for $\ce {sp^2}$ and $\ce {sp}$ hybridization: Only three $\ce {sp^2}$ hybridized orbitals can be made since only three orbitals used to make them. Only way three can arrange in space with equal energy is with trigonal planar geometry (say they are in $xy$-plane). Thus, the $\ce {p}$ orbital not used for the hybridization occupy as $z$-axis to that $xy$-plane. For example, oxygen and middle carbon it attached to in acetone are $\ce {sp^2}$ hybridized so that the $\ce {O-C}$ bond is a $\ce {sp^2-sp^2}$ bond. Similarly, one $\ce {O-C}$ bond in acetic acid is $\ce {sp^2-sp^2}$ bond while the other $\ce {O-C}$ bond is $\ce {sp^3-sp^2}$ bond (both oxygen attached to $\ce {sp^2}$ hybridized carbon).


What's the purpose of a Grignard reagent initiation? Preface: I've read this question about water preventing an initiation from occurring.

I understand why it's important to slowly add the alkyl halide to the dry ether and magnesium for exotherm control. But why is the first addition called an initiation and what's its purpose?

Is it just to confirm that the conditions are ready to continue slowly adding the rest of the reagent? I know it's a free-radical reaction, maybe that's where the term "initiation" comes from. But I've seen the reaction described as "self-sustaining" after the initiation, but it's not a chain reaction. Perhaps that's referring to a predictable reaction proceeding forward instead of stalling then blowing up your glassware.

The first step in the Grignard reaction involves transfer of an electron from the surface of the metal to the alkyl halide to form a radical anion.  The radical anion then disproportionates into a free radical and halide anion.
\begin{aligned}
\ce{R−X + Mg &-> R−X^{•−} + Mg^{•+}}\\
\ce{R−X^{•−} &-> R^{•} + X^{−}}\\
\ce{R^{•} + Mg^{•+} &-> RMg^{+}}\\
\ce{RMg^{+} + X^{−} &-> RMgX}\\
\end{aligned}

When you take the magnesium out of the jar, the surface has a thin oxidation layer on it that prevents the initial electron transfer from occurring.  We used to take one piece of magnesium and scratch it good with a knife to expose a clean, fresh magnesium surface.  We would place this piece of magnesium in the flask (dry ether and the rest of the magnesium already added) and drip in a few drops of the alkyl halide.  Without fail, within a minute or two, mild bubbling would be noticed (local ether boiling), which became vigorous as the initiation phase of the reaction "cleaned" the surface of other pieces of oxidized magnesium.  After the vigorous bubbling started to subside we would drip in the remaining alkyl halide plus ether solution at a rate to maintain ether reflux.  So the reaction does involve a free radical but the initiation phase really has more to do with removing the oxidation layer from all of the magnesium and allowing that initial electron transfer step to occur.


Is there an iron–iron bond in diiron nonacarbonyl? Diiron nonacarbonyl, $\ce{[Fe2(CO)9]}$, is often depicted with an Fe–Fe bond as shown at the left-hand side:

The Fe—Fe bond is usually invoked in order to (1) explain the observed diamagnetism and (2) satisfy the 18-electron rule. However, I'm also aware that the presence of such a bond has been challenged numerous times in some textbooks and also by theoretical studies. Being untrained in computational chemistry, though, means I don't fully understand some of these papers.

So: is there a bond? If there isn't (which seems to be the case), can the bonding in this complex still be reconciled with the 18-electron rule?

I'd especially appreciate it if answers include some explanation of what is going on in the literature.

tl;dr: There probably isn’t an Fe–Fe bond. We should probably stop drawing an Fe-Fe bond. Textbooks and lecturers should really stop teaching students that there is an Fe-Fe bond. But all that said, there isn’t one single satisfactory explanation for why there isn’t one…

Despite a lot of studies (and despite $\ce{[Fe2(CO)9]}$ being one of the earliest examples of a molecule with symmetrically bridging carbonyls) , the nature of bonding in $\ce{[Fe2(CO)9]}$ is still a little hard to get a clear picture of.
Whilst at a research level, most academics now accept that there is no significant bonding interaction between the two metal centres, this is not echoed at the undergraduate level where the Fe–Fe bonding model is still often taught.
Most textbooks either avoid the issue entirely (never providing a structure), or show the Fe–Fe bond (with some footnote that it might not be correct), and even those that do acknowledge the lack of the metal-metal bond (Elschenbroich being a good example) don’t really discuss how this works in practice, or how one would go about electron counting (a source of many a nasty tutorial problem).
Historical context
The simple answer is that there was never really any strong evidence for the presence of the Fe–Fe bond. As is often the case, it was adopted by several (well known) chemists at the time, and subsequent publications (and especially textbooks) propagated the idea, largely due to its convenience with respect to electron counting.
The earliest data that really looked at the nature of the Fe–Fe bond in $\ce{[Fe2(CO)9]}$  can be attributed to a crystal structure measured by Powell (2.46 Å)[1](later corrected by Cotton (2.53 Å)).[2]
Although these values didn’t necessarily prove the existence of an Fe–Fe bond (you can’t see a bond with X-ray, only the atoms, from which you can infer bonding), its presence was necessary to fulfil the 18-electron rule, and the values obtained were well in-keeping with other M–M bond lengths known.
The issue with the above argument (as convenient as it is) is that it’s impossible to tell whether the 2.53 Å represents a true bonding interaction or whether it’s simply a consequence of the bridging Fe–C(O)–Fe bonds pulling the metals close together. The iron itself is relatively contracted, so unlike in some other bridged systems, the steric problem of bringing the two metals together is relatively minimal.
The Beginning of Change
Although the Fe–Fe bond was fairly widely adopted, not all were in agreement.
Roald Hoffmann (of Woodward–Hoffmann fame) published several papers which appeared to contradict the presence of an Fe–Fe bond, showing that the interaction was in-fact “antibonding and repulsive”.[3]
Though not fully quantitative (simple Hückel calculations), Hoffmann was able to construct qualitative molecular orbital diagrams using molecular symmetry and isolobal/electronic fragments (from well behaved molecules) to piece together the likely bonding in $\ce{[Fe2(CO)9]}$. To date, his findings are largely still valid (though exact energies have now been calculated with far more accuracy).
Quantitative Computational Studies[#]
Modern computational studies agree that overall, the Fe–Fe interaction is net anti-bonding (though some bonding interactions have been found) with no genuine covalent Fe–Fe bond, as was initially proposed.
In Liddle’s Molecular Metal-Metal Bonds, McGrady provides a qualitative  molecular orbital diagram for the complex. The σ and σ* orbitals, as well as the δπ and δπ* orbitals (sandwiched between the former pair), are fully occupied and provide no net bonding character. However, the population of the Fe–Fe antibonding πδ* orbitals leads to a formal bond order of –2.[4]

This diagram correlates well to the results in one of the earliest computational studies on this system comparing various methods (DFT, MP2, SCF) by Reinhold.[5]
In the study, various properties of the molecules are calculated (M–M bond length, M–C bond lengths, bond angles etc.) and this data is compared to known data from crystal structures/other experimental work. They determine that only DFT gives accurate results (casting some doubt on earlier computational work using less rigorous methods).[##]
The molecular orbital diagram constructed (in agreement with the one from Ref [4] above) crucially shows a negative overlap of the Fe–Fe orbitals is observed, suggesting a direct repulsion. To quote directly from the paper:

in the iron systems, with more contracted metal orbitals, the direct metal–metal repulsion is relatively weak while the metal–bridge bonds are sufficiently strong

Other computational studies have looked to calculate properties of the complex, such as vibrational frequencies of any proposed Fe–Fe bond, or the NMR shieldings that would be present on the ligands with and without the presence of the Fe–Fe bond. This work is referenced in Comprehensive Organometallic Synthesis II (Vol. 6), but doesn’t seem to have lead to very useful results.
Conclusion
The take home message from this is that there really is unlikely to be an Fe–Fe bond in the way it is commonly taught to undergraduates, and certainly not for the purposes of electron counting.
A nice possible explanation is given by McGrady,[4] which provides some explanation for why the 18-electron rule isn’t important in this case (and at a level that an undergraduate could understand):

There are, in fact, many parallels between 34-electron $\ce{[Fe2(CO)9]}$ and square-planar 16-electron species, not least the tendency of both to react with nucleophiles. The breakdown of the 18-electron rule in square-planar complexes occurs because the vacant pz orbital fails to find an appropriate symmetry match among the linear combinations of ligand-based orbitals, and so is not used, either to form a bond or to accommodate metal-based lone pairs of electrons. [...]
In $\ce{[Fe2(CO)9]}$, there are two vacant $\mathrm{a_2''}$-symmetric metal-based combinations (primarily linear combinations of 4s and 4pz) but only one occupied ligand-based combination of the same symmetry. Thus, one of the two metal-based orbitals remains “unused” in the same sense that the pz orbital remains unused in square-planar species. This simple symmetry-based analysis suggests that there is no need to invoke an Fe–Fe bond in $\ce{[Fe2(CO)9]}$ simply because the metal centers do not conform to the 18-electron rule.

References & Notes
[1]: Powell, H.M.; Ewens, R.V.G J. Chem. Soc 1939, 286
[2]: Cotton, F.A.; Troup, J.M. J. Chem. Soc. Dalton Trans. 1974, 800
[3]: Lauher, J.W.; Elian, M.; Summerville, R.H.; Hoffman, R. J. Am. Chem. Soc., 1976, 98, 3219
[#]: Full disclosure, I’ve read most of the papers that will follow at some point whilst trying to answer this in the past, but (big but) I’m not a computational chemist. I’m hoping someone else may be able to shed light on how valid these computational results actually are.
[4]: McGrady, J.E. in Molecular Metal–Metal Bonds; Liddle, S., Eds.; Wiley-VCH: Weinheim, 2015
[5]: Hunstock, E.; Mealli, C.; Jose-Calhorda, M.; Reinhold, J. Inorg. Chem. 1999, 38, 5053. Note that this isn’t a complete geometry optimisation, they start with some known parameters. New J. Chem. 1994, 18, 465 is a paper supposedly attempting the complete optimisation from scratch, but I can’t find it anywhere right now to access it
[##]: Even now there are a lot of issues with accurately calculating properties of organometallic complexes… it takes a vast vast amount of computational time due to the need for basis sets far expanded compared to those developed for simple organic systems


How is laughing gas produced in the Nef reaction? I can't think of a mechanism for this reaction to give laughing gas.

Attempt:

I used the base to take out the acidic hydrogen at alpha position with respect to nitro group, but that makes the intermediate formed even more unstable. What should be the proper mechanism?

The Nef reaction does not produce nitrous oxide directly. It produces nitroxyl as a reactive intermediate, which rapidly decomposes via a bimolecular reaction to nitrous oxide and water$^{[1]}$. You are correct that the base removes the alpha hydrogen; the nitro group is more than capable of accepting this negative charge. The crux of the Nef reaction is this:

The mechanism for the formation of nitrous oxide is as follows:

$[1]$ Wikipedia: Nitroxyl, Reactions


How to predict the decomposition products of for example barium carbonate? I am a high school student studying basic chemistry and I am confused about the process of writing decomposition reactions.

The question I have encountered asks me to decompose $\ce{BaCO3}$.

Online, I have found that this decomposes to $\ce{BaO + CO2}$.

My question is, essentially, is there a method to determine which elements go where. I'm having trouble finding answers.

Chemistry, despite being several hundred year old science, it is still very much empirical. In the beginning you will have to learn and accept some facts. Later you will start to see patterns (hence the periodic table). If you learn that the carbonate of barium decomposes into carbon dioxide and barium oxide, it will be safe to say that Mg, Ca, Ba, Sr, will do the same. Here you have learnt four more chemical reactions. If you recall the periodic functions in mathematics, which repeat themselves periodically, in the same way, elements repeat their properties in the periodic table.

When you strongly heat inorganic salts, usually you get smaller products (e.g., carbon dioxide and barium oxide) which are energetically more stable with respect to the starting material.


How do we know which compounds have higher solubility? Is there any way we can look at the chemical formula of various compounds, do some calculations or something and figure out which compounds will have higher solubility as compared to the others.

It depends on the solution and the solute. Usually, the rule is "like" dissolves "like". A polar solution will dissolve polar compounds and a non-polar solution will dissolve non-polar compounds.

https://www.youtube.com/watch?v=PVL24HAesnc

As far as I know, to find the solubility you have to determine it experimentally. This can be done by "feel" first. If you have a compound that doesn't have many polar groups on it, chances are it can be dissolved by a non-polar solution. If you know what the structure looks like try to match it with a solvent that looks similar to it.


How many electrons are around the platinum in an ethenetriplatinate complex? In $\ce{[Pt3(C2H4)]-}$, how many electrons are there around platinum?

While solving this problem I thought three $\ce{Pt}$ formed a triangle with ethylene above. But I was confused about how the $\ce{C2H4}$ gave electron pairs to $\ce{Pt}$. In addition, the answer gives $16$, but I only see the initial $10$ electrons around a $\ce{Pt}$ and $2$ given by another two $\ce{Pt}$.
What is the exact structure of this coordination compound, and how can it be deduced?

Looking up your proposed sum formula, I arrived at a quantum mechanical paper that used said complex (triangular $\ce{Pt3}$ moiety, whereonto an ethylene molecule coordinates onto the longer edge of the triangle) as a modelling study for the interaction of ethylene with platinum metal surfaces.[1]

As far as I can tell,[2] the authors in now way account for a surface of platinum; their three platinum atoms in vacuum are supposed to model an entire platinum surface. I cannot say whether that type of approach is typically used or whether it is considered good by quantum chemists.

Searching further, I also found a 1981 paper describing $\ce{[Pt3(C2H4)(cod)2(cot)2]}$, a neutral complex that includes two additional cycloocta-1,5-diene ($\ce{cod}$) and two additional cycloocta-1,3,5,7-tetrene ($\ce{cot}$) ligands.[3] In this complex, the bonding situation and electron count can be well answered, but that wasn’t your question.[4]

Furthermore, I find a complex $\ce{[Pt3(C2H4)]-}$ to be highly questionable: Platinum is in the 10th group, ethylene is a singlet molecule so adding a single additional electron will create a radical anion. Another reason why your question and answer cannot be serious, since in no way can a radical end up at $16$ electrons. I would even go a step further and say that it is likely an unstable complex that would decompose depending on which species are close to it. Note that the authors of the first paper calculate an uncharged complex $\ce{[Pt3(C2H4)]}$.

Notes and references

A. Cruz, V. Bertin, M. Castro, Int. J. Quantum Chem. 2000, 80, 298. DOI: 10.1002/1097-461X(2000)80:3<298::AID-QUA3>3.0.CO;2-2.
I am not a quantum chemist, and my last calculation efforts date back to a research project during my master’s studies. Those acquainted with methods will have to comment on how good the paper is.
N. M. Boag, J. A. K. Howard, J. L. Spencer, F. G. A. Stone, J. Chem. Soc., Dalton Trans. 1981, 1051. DOI: 10.1039/DT9810001051.
Two platinum atoms with 18 electrons in a square-planar environment, one with 16 electrons in a trigonal environment. The former coordinated by $\ce{cod}$ and $\ce{cot}$ (each contributing two double bonds), the latter coordinated by two $\ce{cot}$ molecules (each contributing one double bond) and an ethylene. The three platinum centres are too far apart to be considered bonded, according to the authors.
The bonding of the two 18-electron platinums to their respective $\ce{cot}$ unit is described as $\ce{cot^2-}$ and $\ce{Pt^2+}$, resulting in oxidation states of $\mathrm{+II}$ and $\mathrm{\pm 0}$ for platinum.


What is generally meant by distinct line spectrum? In a single hydrogen atom the electron is excited to 6th orbit. The book says maximum 5 distinct spectral lines are possible when the electron comes to the ground state.

Looks like they have only considered those lines when the electron jumps from shell 6 to ground, shell 5 to ground, shell 4 to ground etc. But in my opinion we should also consider the lines produced while transfer of electrons from 6 shell to 5 shell etc. As the elctron can jump from shell 5 to shell 1 directly only when it has jumped from shell 6 to 5 earlier. That would lead to more number of spectral lines than just 5.

So, am i thinking something wrong here (the word "distinct"  seems to be the reason)?

Your intuition is right: all the transitions are possible.

But the language is a little vague and might just be a simplification.

The emission spectrum of hydrogen atoms contains a lot of lines which are often grouped into series (which are named Lyman, Ballmer and Paschen, for example). What groups the series is the lower energy level in the transition responsible for the line: the Lyman series are transitions to the lowest energy level; Ballmer is transitions to the second lowest level...). A good summary of all this is given here.

Your book may have been assuming you only counted the lines corresponding to the final drop to the ground state (the Lyman series) ignoring the intermediate drops to other energy levels. If not, it is clearly wrong as all the transitions are possible.


Le Chatelier's principle: Are there any exceptions? The way Le Chatelier's principle is presented in most introductory chemistry books (high-school) is as though it's an indisputable law of the physical world (in the sense that we're never shown an exception, not that its universality is explicitly stated).

This is supposed to differ from the more frequently encountered  half-assed  "laws" like Ohm's Law or Hooke's Law (which are really just generalizations drawn from a few, everyday cases... not real "laws" as such).

Now I'm curious. Is there really no exception to Le Chatelier's principle (for a closed, isolated system)?

Wording this differently: Is there any situation (closed, isolated system) where conditions conducive to Le Chatelier's principle are in place, yet the expected Le Chatelier's "response" (the tendency to oppose change) is not observed upon effecting the said change (in parameters such as concentration, temperature, volume, etc.)?

If there are any exceptions that one comes across in everyday life, I'd prefer to hear those (but since that's extremely unlikely, I suppose just about any exception will do).

Are there any theoretical grounds for exceptions to Le Chatelier's principle?

If Le Chatelier's principle cannot be violated, then why is it so?

What sparked this question, is the following statement from a (school-issued) book of mine:

  The dissolution of NaOH in water is exothermic. According to Le Chatelier's principle decreasing the temperature of the solution should favor solubility, however, in reality solubility of NaOH increases with temperature. This is therefore an exception.

However, if it weren't for the sharp users of Chemistry.SE I would've accepted my book's claims (since it corroborates seemingly well with common experience). O:)

I did find this article on the ACS page, but it's paywalled >_<

There are many exceptions to Le Chatelier's Principle (LCP) so it can certainly be violated. Recently, I have read many articles pertaining to this violation so I believe I should be able to provide an informed answer. Most of the violations of the principle can be seen in gaseous systems. I would not call them "violations", rather, just cases where LCP does not work so well

One of the most classic cases for such an exception, as presented by Cheung (2004), is the addition of nitrogen gas, at constant temperature and pressure, to an equilibrium mixture containing $\ce {N2}$, $\ce{H2}$ and $\ce {NH3}$. The mechanical approach invoking LCP would quickly conclude that there would be an increase in the amount of the gaseous product due to an increase in the amount of one of the reactant gases, favouring the forward reaction. However, this approach fails to consider the instantaneous increase in volume of the system, which would cause the partial pressure of $\ce {H2 (g)}$ to decrease. Now, if we consider the principle again, taking into account both changes, we would realise that the principle would tell us that the equilibrium would shift in both directions. Thus, we see that there is a possibility that LCP cannot give a definite prediction if the number of moles of gaseous products in the balanced equation is not equal to the number of moles of gaseous reactants.

Another example cited by Cheung (2004) is the dimerisation of $\ce {NO2}$. Consider an equilibrium mixture of $\ce {NO2}$ and $\ce {N2O4}$ in a sealed container. When we raise the temperature, we may expect that the position of equilibrium to shift to the left since the backward reaction to form $\ce {NO2}$ is endothermic. However, there is also an increase in the total pressure of the system when the mixture is heated. Again, by LCP, the pressure increase would likely shift the position of equilibrium to the right to the side with less moles of gaseous molecules. This is another case where there are indeterminate predictions given by LCP.

References

Cheung, D. (2004). The Scientific Inadequacies of Le Chatelier's Principle. Hong Kong Science Teachers' Journal, 22(1), 35-43. Retrieved July 27, 2018, from https://www3.fed.cuhk.edu.hk/chemistry/files/LCP.pdf.

Cheung, D. (2009). The Adverse Effects of Le Chatelier's Principle on Teacher Understanding of Chemical Equilibrium. Journal of Chemical Education, 86, 514-518. doi:10.1021/ed086p514

Quilez-Pardo, J., & Solaz-Portoles, J. J. (1995). Students' and teachers' misapplication of le chatelier's principle: Implications for the teaching of chemical equilibrium. Journal of Research in Science Teaching, 32(9), 939-957. doi:10.1002/tea.3660320906


Heterocyclic Compounds - Glucose Why cyclic glucose is not a heterocyclic compounds if it consist Oxygen as a Heteroatom ?

Waylander is right, many sugars are heterocyclic compounds. While sugars such as glucose can exist in an open chain form, many of them are normally found in the form of six or five membered rings which consist of carbons and an oxygen.

Thus glucose fits the very definition of a heterocylic compound. It is important to keep in mind that any cylic (ring) compound where more than one element type is found in the atoms which make up the ring is a heterocylic compound.

For example S4N4, pyridine, ethylene oxide, glucose, 9-BBN and tetrahydrofuran are all heterocycles. They have very different chemistry to each other but they are all heterocylic.


How do I solve for Ksp The problem gives that the solubility of Silver dichromate is $8.3\times10^{-3} g/100mL$. I need to find $K_{sp}$ which is supposed to be $2.8\times10^{-11}$.

$K_{sp}$ is $\ce{[Ag+]^2[Cr2O7^{2-}]}$

I changed the $8.3\times 10^{-3}$ from g/mL to mol/L and got $1.9\times 10^{-8}$.

Silver dichromate breaks into 2 Ag and one dichromate.

So that would be $(2 \times 1.9\times10^{-8})^2 \times ( 1.9\times10^{-8})$

That's what I did but it's not giving me the right answer. Where am I going wrong here?

The problem gives that the solubility of Silver dichromate is $\mathrm{8.3\cdot10^{-3} g/100\,mL}$.
[…]
I changed the $8.3\cdot10^{-3}$ from g/mL to mol/L and got $1.9\cdot10^{-8}$.
[…] Were am I going wrong here?

Was the solubility given in $\mathrm{g/100\,mL}$?
$$\mathrm{8.3\cdot10^{-3} g/100\,mL = 8.3\cdot10^{-2}\,g/L = \frac{8.3}{431.72}\,mol/L = 1.92\cdot10^{-4}\,mol/L}$$


How much precipitation occurs for a given ionic product? Let us say we have a salt AB in water. Suppose the ionic product exceeds the solubility product at a particular temperature. Now, I know that the ions will precipitate out of the solution. My doubt is, how much of the ions will precipitate? By saying "precipitation starts", do we mean that all the ions will precipitate, or will some ions remain in the solution?

Solubility product is the product of concentration of ions that can be soluble / in aqueous ionic state under given set of conditions. When ionic product is more means, the remaining is not soluble. Only that amount precipitates out. Still there are ions in water, product concentration of which is exactly equal to solubility product.

For basic understanding of equilibrium constant, you can refer: http://www.adichemistry.com/physical/equilibrium/introduction/chemical-equilibrium.html


How to find the temperature relationship between the isotherms in a compressibility factor (Z) vs pressure graph? The following graph denotes the variation of the compressibility factor (Z) with pressure at different temperatures for a real gas. Simply each of the curves represents an isotherm.

Now, suppose we are not given the temperatures and we are asked to find the temperature relationship between the isotherms, how to prove the topmost curve corresponds to the higher temperature and the lowermost curve corresponds to the lowest temperature? Is there any mathematical proof for this or is this just an experimental result?

The compression factor $Z$ is defined as the ratio of the real molar volume with respect to the theoretical ideal

$$Z = \frac{V_m}{V_m^{ideal}}$$

where if $Z$ = 1, the gas behaves ideally.

If we want to get some sort of understanding for how this function behaves when you change the Pressures and Temperatures of your system, we need to substitute into this equation our variables of interest. To take the simplest approach, lets just use the ideal gas law as a substitute for $V_m$ (from a quantitative sense this doesn't make sense to do but qualitatively we can still derive some relationships therefrom)
$$Z = \frac{RTP^{-1}}{V_m^{ideal}}$$
What this shows us is that $Z \propto T$. Thus, an increase in $T$ demonstrates an increase in $Z$, which is the qualitative understanding we're looking for herein.

To note, using a more realistic model for $V_m$ gets complicated; for example if you wanted to use the Van der Waals equation
$$P = \frac{RT}{V_m-b} - \frac{a}{V_m^2}$$
instead of the ideal gas equation, the $V_m$ term doesn't reduce too nicely because of that $\frac{a}{V_m^2}$ term. Gets even worse for the Virial equations.


Reaction of moist egg shell powder and acid? So I performed an experiment to investigate the effect of fluoride ions on the rate of reaction between egg shells (calcium carbonate) and acid.

When doing this I had a dry sample of crushed egg shells and a sample previously soaked in distilled water which had dried for 3 days but was still slightly moist in a 'gloopy' texture.

When I reacted the moist egg shells with hydrochloric acid they made a frothy mixture of bubbles where the carbon dioxide was trapped in bubbles while the dry egg shells fizzed and the carbon dioxide was not trapped in bubbles.

What could be a reason for the differences in the way the carbon dioxide was trapped?

(membranes were removed from both samples)

I don't think anything chemically has occured. It's likely that the membrane of the dry eggs became rigid and brittle and thus fractured into pieces when the bubbles formed where as the moist membranes would be more deformable allowing them to trap $\ce{CO2}$ bubbles.


what is the difference between a redox reaction and an ionic bond? I am new into chemistry...
I'd like to know what the difference is between an ionic bond: which involves a bond with an electron transfert, and a redox reaction; which seems to be exactly the same to me.

Thanks

An ionic bond is formed by electrostatic attraction of electrons between one atom and the other whereas in redox reactions the electrons are transferred from one atom to the other. Hope this helps.


Can bridged and bridgehead positions be chlorinated using free radical mechanism? When the above compound is treated with $\ce{Cl2}$ in the presence of sunlight ($h\nu$), what is the number of monochloro structural isomers formed?

I expected only 1 product becuase I don't believe that monochlorination can take place at the bridgehead or the bridge positions. And rest of the 4 positions, by symmetry, are equivalent. This is because chlorination takes place through free radical mechanism in which a $\ce{sp^2}$ hybridised carbon free radical is formed.
But the answer given in the test's answer key is $3$. I think they have considered the products of bridgehead and bridged positions too but I don't understand how that is possible.

The "bridged position" is just another ring, when viewed differently.

Suppose you look at the molecule a little differently (rotate it around a bit), then you could have a five-membered main ring, with a two membered "bridge", as opposed to what we have now (6 membered main ring with 1 membered bridge).

It's definitely possible to have a free radical halogen attack on that position, and with two stereoisomers at the positions which you have identified, it becomes total 3 monochlorinated derivatives.

(in case you're wondering how to check if the two stereoisomers are not identical, you could try to superimpose them, or go from one form to the other without breaking any bonds. In that case, you'll find that they're indeed different. You could even check for a plane/center of symmetry)

--EDIT--

My answer is wrong, as @Abcd and @AkshatJoshi pointed out, because I counted stereoisomers along with the required structural isomers.

I read about attacks at Bridgehead positions and there's NO place that mentions the existence of Bridgehead free radicals. Thus, there will only be 2 structural isomers, but 3 isomers overall.

So, I believe the answer ought to be 2.


How to temporarily hide hydrogen atoms in Molden? I am visualizing some molecules with Molden. Is there a way to hide hydrogens, without deleting the hydrogen entries from the structure file?

As long as you are not using an ancient version of Molden, this should be no problem. (The oldest one I found in my software archive was Moleden5.1 from January 2014 and it already contains this feature.)

In the Molden control window there is a button with an italic H. This will open a menu with "Excl. All Hydrogens", "Excl. H except H-bond", "(Re)calculate H-bond", "H-bond parameters", and "Opt. H positions". I'll leave it to you to play around with it, below you find a small animation of the first two modes for the base pair GC.

And because it is a little bit too tiny in the gif, here is a crop of the Molden control window.


How to determine and derive the protonation state of a functional group based on its pKa? In our biochemistry class we were introduced to a rule of thumb, which goes something like this:

"If the pH of a solution is one or more units below the $\mathrm{p}K_\mathrm{a}$ of a group, then said group is protonated and if it is one or more unit above, then the group is de-protonated"

However, we weren't provided with a mathematical derivation of this fact.
I attempted one by myself using:

$$\ce{HA + H2O <=> A- + H3O+}$$
thus, $$\ce{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{\ce{A-}}{\ce{HA}}$$
or, $$\ce{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{1-\alpha}{\alpha}$$

and then define $\mathrm{p}K_\mathrm{a}$ as the point where ${\alpha=0.5}$ i.e proton occupancy is 50%

$$\alpha(\ce{pH}) = \frac{1}{1+\exp\left\{-\ln10(\mathrm{p}K_\mathrm{a}-\ce{pH})\right\}}$$

where $\alpha$:= degree of proton occupancy.

By looking at this function, I can sort of get that if pH goes up, $\alpha$ goes down i.e case of protonation, and if pH goes down  $\alpha$ goes up i.e case of deprotonation.

This seems a little bit "back of the envelope" to me. Is this sufficient, or am I missing something.

You're on the right track.  The Henderson-Hasselbalch equation is where this rule of thumb comes from; the key thing to remember is that, in biochemistry, your $\mathrm{pH}$ is fixed by the environment (there's always something acting as a buffer).  This is in contrast with typical H.H. problems in general chemistry where the amount of molecule and its $K_\mathrm{a}$ is changing the $\mathrm{pH}$ of the solution.

Anyhow, given that $\mathrm{pH}$ and $\mathrm{p}K_\mathrm{a}$ are fixed values, we're free to simply compare the ratio of $[\ce{A-}]$ and $[\ce{HA}]$:

\begin{align}
\mathrm{pH} &= \mathrm{p}K_\mathrm{a} + \log\frac{[\ce{A-}]}{[\ce{HA}]}\\
\log\frac{[\ce{A-}]}{[\ce{HA}]} &= \mathrm{pH} - \mathrm{p}K_\mathrm{a}\\
\frac{[\ce{A-}]}{[\ce{HA}]} &= 10^{\mathrm{pH} - \mathrm{p}K_\mathrm{a}}\\
\end{align}

If $\mathrm{p}K_\mathrm{a} = \mathrm{pH} + 1$ then we have $\frac{[\ce{A-}]}{[\ce{HA}]} = 10^{-1}$.  In other words, we have 10 times as much $\ce{HA}$ as we have $\ce{A-}$ (i.e., we have $\approx91\%\ \ce{HA}$) as we should expect for something that is mildly basic.

If $\mathrm{p}K_\mathrm{a} = \mathrm{pH} - 1$ then we have $\frac{[\ce{A-}]}{[\ce{HA}]} = 10^{1}$.  In other words, we have 10 times as much $\ce{A-}$ as we have $\ce{HA}$ (i.e., we have $\approx91\%\ \ce{A-}$) as we should expect for something that is mildly acidic.

Clearly, if the $\mathrm{pH}$ differs from the $\mathrm{p}K_\mathrm{a}$ by only 1 $\mathrm{pH}$ unit, we don't really have a completely protonated or deprotonated molecules, but it's certainly mostly protonated or deprotonated.  As the $\mathrm{pH}$ and $\mathrm{p}K_\mathrm{a}$ get further apart, then we continue to change this ratio by orders of magnitude.

Let's look at a concrete example.  Physiological $\mathrm{pH}$ is 7.4 and the side chain of aspartic acid has a carboxylic acid group with a $\mathrm{p}K_\mathrm{a} = 3.9$, so we predict

$$\frac{[\ce{RCOO^-}]}{[\ce{RCOOH}]} = 10^{7.4 - 3.9} = 10^{3.5} = 3200$$

In other words, we still have some protonated sidechain, but we have >3000 times more deprotonated sidechain than protonated sidechain (did you notice that? the difference was >3 so we have >1000 difference in concentration ratio).  For all practical purposes, the aspartic acid side chain is "completely deprotonated" at physiological $\mathrm{pH}$.


How to convert ethanol to benzoic acid? I have tried this method:

Can anyone suggest me a shorter route?

  Criteria: Benzene has to be produced, it is not available.

Edit : I don't understand why the question has been put on hold. They provide the reason that there may be possibly to many good answers or the question is very broad. As far as my knowledge goes, I haven't received any correct answer that addresses what I have asked.

I would request the authority to reconsider on their decision.

Here is how I would do it.

Part I: Blue

First, ethanol 1 is oxidised to acetaldehyde 2, which undergoes nucleophilic addition with $\ce{KCN}$ and $\ce{HCN}$ to form the hydroxynitrile 3, and finally dehydrated and hydrolysed at the same time with sulfuric acid to give acrylic acid 4.

Part II: Red

First, ethanol 1 is reacted to form the Grignard reagent 5, then combines with the aldehyde 2 to form the alcohol 6. It is dehydrated to form butene 7, dihydroxylated to form the diol 8, and dehydrated again to get the diene 9.

Part III: Black

In the final part, 4 and 9 participate in a Diels-Alder reaction with the Lewis acid $\ce{TiCl4}$ to give 10. Selenium dioxide gives the diol 11 and subsequent dehydration gives us benzoic acid, 12.

Although it may be longer, it does not involve things such as high reaction temperatures and therefore may be more feasible in the context of a laboratory.

EDIT: Here is another attempt:

Ethanol 1 is oxidised to acetaldehyde 2, and subsequently with excess formaldehyde and acidic conditions forms pentaerythritol 3, and is oxidised again to form the quadruple aldehyde 4. Strong heating causes 3 to turn into the diol 5. The Grignard reagent 6 is formed and oxidation and reduction of the alkene forms the reactant 7.

Now, both 4 and 7 react to form the ring product 8, and with strong heating thus produces benzoic acid 9.


Is there a common name to refer to the groups 13 and 14? The group 15, 16, and 17 are called the pnictogens, chalcogens, and halogens respectively. Is there a name for the groups 13 and 14 as well?

It seems that the carbon group is known as either the tetrels or the crystallogens; while the boron group is known as triels or icosagens. These names are very rarely used.

Source: Wikipedia articles on Carbon family (also, crystallogen) and Boron family. Although other (seemingly) respectable sites use them.


Mechanism for synthesising azetidine Problem
   A synthesis of azetidine is shown below. Provide a mechanism for all the steps. Why does the formation of azetidine in the last step require high temperature?

My working is thus:

I am stuck on the last step and my version seems wrong. Can someone please guide me in the right direction?

There is not much to say other than: your version is more or less correct. It is debateable whether the ester hydrolysis happens before or after the cleavage of azetidine (I would suspect before due to the lower overall charge). Remember that you always have to consider aquaeous work-up at the end of a reaction, which will give you the three protons you need to regenerate acrylic acid, ethanol and azetidine.

You also have another question written on your paper after the halogenation of the terminal alcohol, namely:

  What is the point of this step?

It’s not really a point. It’s just that the reaction conditions contain $\ce{HCl}$ which will protonate any strong enough base around and the secondary amine is a strong enough base. It’s more of a consequence of the reaction conditions than an actual pointful step.


Balancing the redox disproportionation of chlorine by half-reactions I am having some difficulty balancing the following reaction using the half-reaction method:

$$\ce{Cl2(g) -> Cl-(aq) + ClO3-(aq)}$$

My attempt at the problem:

By determining oxidation numbers it is possible to see what is being oxidised and what is being reduced:

$$\ce{\overset{0}{Cl}_2(g) -> \overset{-1}{Cl}^-(aq) + \overset{+5}{Cl}\overset{-2}{O}_3^-(aq)}$$

From this I determined the following two half-reactions:

Reduction Half: $\ce{Cl2(g) + 2 e- -> 2 Cl-(aq)}$

Oxidation Half: $\ce{Cl2(g) + 3 H2O (l) -> 2 ClO3-(aq) + 6 H+ + 4 e-}$

Net ionic equation: $\ce{6 Cl2(g) + 6 H2O(l) -> 8 Cl-(aq) + 4 ClO3-(aq) + 12 H+}$

The answer to this problem is however given as

  $$\ce{2 Cl2(g) + 6 H2O(l) -> 2 ClO3-(aq) + 12 H+ + 10 Cl-(aq)}$$

I have tried this problem several times over and cannot work out how this answer was reached.

Update:

Silly mistake, oxidation reaction should be: $\ce{Cl2 + 6H2O -> 2 ClO3- + 12 H+ + 10 e-}$ hence solving the mystery of why my answer is wrong.

You have an error half-way through your oxidation half-reaction which probably multiplies itself. You correctly determined the oxidation states so it should be simple to spot it. But for reference, here they are:

$$\begin{align}&\overset{\pm 0}{\ce{Cl2}}/\overset{\mathrm{+V}}{\ce{Cl}}\ce{O3-}\\[0.9em]
\ce{Cl2 \phantom{\ce{{} + 6 H2O }} &-> 2 ClO3- + 10 e-}\tag{Ox1}\\
\ce{Cl2 \phantom{\ce{{} + 6 H2O }} &-> 2 ClO3- + 10 e- + 12 H+}\tag{Ox2}\\
\ce{Cl2 + 6 H2O &-> 2 ClO3- + 10 e- + 12 H+}\tag{Ox3}
\end{align}$$

Thenceforth it should be clear to see that you need five reductions per oxidation, thus the redox equation is:

$$\begin{align}\ce{6 Cl2 + 6 H2O &-> 2 ClO3 + 10 Cl- + 12 H+}\tag{Redox1}\\
\ce{3 Cl2 + 3 H2O &-> ClO3 + 5 Cl- + 6 H+}\tag{Redox2}\end{align}$$

(The second version is the first but divided by two; you only have even coefficients.)


Molarity vs Normality in labeling sulfuric acid Molarity (M) is defined as moles of solute per liter of solution while Normality (N) is defined as moles of reacting units per liter of solution.  Most texts use sulfuric acid as an example to demonstrate the difference: a 1 M solution of sulfuric acid is 2 N because each mole of sulfuric acid gives two moles of $\ce{H+}$ (the "reacting unit"). Why then is concentrated sulfuric acid labeled as 18 N when it should be 18 M and 36 N (according to my text's example above)?

Who labels concentrated sulfuric acid as 18N?  36N sounds right to me and Sigma-Aldrich:

  Concentration: The commercial reagent contains 93-98% H2SO4, the balance being water. Its effective concentration is 18 M (36N).

https://www.sigmaaldrich.com/content/dam/sigma-aldrich/docs/Sigma-Aldrich/Product_Information_Sheet/258105pis.pdf

IUPAC discourages use of normality because it is ambiguous.  The number of equivalents can be ambiguous depending upon the purpose (equivalents of protons, equivalents of electrons, etc.).

See the following IUPAC document: http://old.iupac.org/publications/analytical_compendium/Cha06sec3.pdf


Precipitation of a lead(II) sulfate Lead (II) sulfate precipitate can be dissolved if:
a) We add an acid
b) We add EDTA
c) None of the above is true.

I think EDTA is a ligand, and if you want to dissolve the sulfate, then Pb sticks with EDTA to form a complex. And so more Pb will be formed to balance the reaction.

The stability constant of $\ce{Pb−EDTA}$ complex is $\ce{10^{18}}$, which is huge. The solubility product of $\ce{PbSO_4}$ is $10^{−8}$
, which cannot be compared to the effect of $\ce{EDTA}$. So the precipitate $\ce{PbSO4}$ must be soluble in $\ce{EDTA}$ solutions. These numerical data have been taken from J. G. Stark, Chemistry Data Book, John Murray, London, $1994$


Materials that resist adhesion with molten glass even when hot I'm looking for ways to manipulate molten glass with (e.g. bending a heated glass tube).

This article mentions that oxides tend to adhere, hence molten glass to heated stainless steel.

What materials could I use to manipulate hot glass? Perhaps I could coat the glass or instrument doing the manipulation with some sort of powder or liquid?

The glass is soda lime glass which has a transition temperature of 550 degrees C, so the temperature it needs to cope with will be a fair bit above that.

Thanks

Professional glass blowers use graphite paddles to "push" the glass around.  But to bend a heated glass tube you need to know that glass is a very poor conductor of heat.  Thus it is possible to grasp the heated tube six or so inches from the heated spot and just bend the glass.  This works well for small bore tubing.  Just be careful putting the bent tube down.  Putting it on a high heat conductivity surface will result in the glass breaking.  In the old days asbestos pads were used, but that's a big no-no now.


Why can hydrogen ions be added to balance a reaction? This link explains the process behind balancing redox equations.

  Now you have to add things to the half-equation in order to make it
  balance completely.

  All you are allowed to add are:

  electrons
  water
  hydrogen ions (unless the reaction is being done under alkaline conditions - in which case, you can add hydroxide ions instead)

Adding electrons makes sense, because this whole process models electron transfer. Any added electrons will be accounted for elsewhere in the reaction.

Adding water makes sense as well, because it's an entire molecule -- it is essentially part of the reaction at this point.

I don't understand why $\ce{H^+}$ ions can be added simply for balancing purposes.

In the second example on the same site, the final reaction is this:

$$\ce{2MnO4- + 6H+ +5H2O2 -> 2Mn^{2+} + 8H2O + 5O2}$$

Where does the $\ce{6H^+}$ come from, in the real world? It seems like it just appears out of nowhere to make the math work out.

What is actually happening, on a physical basis, to make this type of balancing check out?

You should really only add $\ce{H+}$ to balance if a reaction is explicitly said to take place in acidic solution. For example, in your link it says.

  The reaction is done with potassium manganate(VII) solution and hydrogen peroxide solution acidified with dilute sulfuric acid.

So the sulfuric acid allows you to add these $\ce{H+}$. Alternatively, if it is in basic solution, $\ce{OH-}$ atoms can be added instead.


Which C-C bond would break first, the one in ethane, or 2,2-dimethylpropane? I have drawn the figures and determined intuitively that the larger molecule's $\ce{C-C}$ would break first. I made this assumption that ethane has the stronger $\ce{C-C}$ bond and I am correct, however I do not understand the underlying reasons why.

In both molecules we have single bonds, but in ethane we have more space for the hydrogen atoms to fill. This was my initial reasoning, that the larger molecule is less compact and will have longer $\ce{C-C}$ bonds. Am I correct in my reasoning, or does it have something to do with the radical stabilities? My book is saying "radical stability increases along the series from primary to secondary to tertiary; consequently, the energy required to create them decreases."

In ethane we would have primary carbons because they bond to only one other carbon. In 2,2-dimethylpropane the central carbon is bonded to four other carbons, making it quaternary. Is this the underlying reason, and if so then what is the reason for this trend?

It does not have to do with steric effects as you suggested. Rather, the reason behind this trend is the inductive effect. Figuratively speaking, carbon side-chains attached to an central atom "push" electron density towards this atom, while hydrogen atoms don't. If you break a C-C bond and produce a carbon radical, it will be stabilized if electron density is donated towards the site of the unpaired electron, which is the case in the tert-butyl radical formed from dimethylpropane, but not in the methyl radical from ethane.