What is the process of connecting two molecules under removal of one molecule of water? Certain molecules can be joined by the removal of water in a process called:

Hydration.

Hydrolysis.

Neutralization.

Condensation.

I know that the process of hydration is defined as, "A compound produced by combining a substance chemically with water."
The process of hydrolysis is defined as, "The chemical breakdown of a compound due to reaction with water."
Neutralization is a chemical reaction in which an acid and a base react quantitatively with each other.
Condensation is the conversion of a vapor or gas to a liquid.
The answer is most likely (1), but I'd like some feedback and explanation of these definitions!

Since it is a removal of water, it cannot be hydration which is defined as the addition of water.

Similarly, hydrolysis includes the addition of water to something so that it degrades in some way.

Neutralisation does indeed generate water, but not always one molecule per reaction equivalent (can be two or three) and you do not connect molecules but ions.

Condensation not only has its physical definition you stated but it also means (in a chemical sense) connecting two molecules under formal exclusion of one molecule of water.

Note that in many to most processes, the water exclusion is entirely formal and what actually is excluded is something like $\ce{H2PO4-}$.


What makes a glowing splinter glow and a burning splinter burn This question bugs me for years since I started learning the oxygen, carbon dioxide and hydrogen tests. I know you can make a glowing splinter burn when doing the oxygen test due to enough oxygen. But why won't it burn in ordinary air?

So, my question is what factor differentiate a glowing splinter and a burning splinter

A splinter can undergo multiple types of thermal reactions depending on both the maximum temperature and amount of oxygen present.  The answer to the question seems to be in the difference between total and partial inceneration.

In total inceneration, which is very likely in the total oxygen atmosphere at high temperatures, all of the organic carbon is converted to carbon dioxide. Carbon dioxide is colorless so the emission of electrons, i.e. the glowing, is observable.

In partial inceneration, which occurs when oxygen is present but deficient, the organic carbon is converted into both carbon dioxide and a variety of volatile compounds observable as smoke. Smoke forms at lower temperatures than thermal emissions. Smoke can also obscure thermal emissions.


Difference between any arene and an aromatic compound? I can't seem to find what the difference between these two is. In my text book, it says that 'arenes are aromatic hydrocarbons containing one or more benzene rings' which to me suggests that arenes are a subgroup of aromatic hydrocarbons and aromatic hydrocarbons need not contain one or more benzene rings. However later on in my book it says that 'any compound with the benzene ring is classified as an aromatic compound', which to me is the exact same definition as that given for an arene. Is there a difference between arenes and aromatic compounds?

Thank you in advance :)

Arene — Compound which contains a benzene ring

Aromatic Compounds — Compounds having aroma (wait, the modern definition is different, I know)

Every arene is an aromatic compound but every aromatic compound need not be an arene.

Mathematically,

$$\text{arene} \subset \text{aromatic compounds}$$

therefore anything which contains a benzene ring has to be aromatic

Some examples of arenes:

Getting into aromatic compounds, the modern criteria for aromaticity are:

The molecule is cyclic (a ring of atoms)
The molecule is planar (all atoms in the molecule lie in the same plane)
The molecule is fully conjugated (p orbitals at every atom in the ring)
The molecule has $4n+2$ π electrons ($n=0$ or any positive integer)*

Examples (non-benzenoids):

Now, benzenoids (compounds which contain benzene) do match with this. Therefore, benzene derivatives are aromatic compounds.


SN1 or SN2 reaction? Here, this is a 1° allylic halide. If it goes by SN1, the intermediate will be resonance stabilised.

Also, since it is 1° Halide, it can also go via SN2. Which one will be preferred?

This reaction will happen by both SN1 and SN2 mechanism, apart from this SN1' and Sn2' (allylic mechanism ) will also occur. Thus from SN1 and Sn2 one will get primary cyanide and from SN1' and SN2' one will get secondary cyanide .( read organic chemistry by jerry march for the detailed explanation).
I am attaching a screenshot of clayden which explains this beautifully.


Converting from Grams to Molecules I was doing some problems from the textbook on converting between units using dimensional analysis and I came across this problem.

A vat of Hydrogen Peroxide ($\ce{H2O2}$) contains
455 grams of oxygen atoms. How many
molecules of ($\ce{H2O2}$) are in the vat?

Sorry if I didn`t format the question correctly, I am a still a newbie.
But how would I set this up? I would not like the answer to this question but I would like some input on how would I get how many molecules of hydrogen peroxide from grams of oxygen atoms? Any help would be much appreciated. Thanks in advance!

A vat of Hydrogen Peroxide ($\ce{H2O2}$) contains 455 grams of oxygen atoms.
  How many molecules of ($\ce{H2O2}$) are in the vat?

How many moles of oxygen atoms are in 455 gms of oxygen atoms? Let's call the answer $x$.
How many moles of $\ce{O2}$ are in $x$?  Let's call the answer $y$.
How many moles of $\ce{H2O2}$ could you make from $y$ moles of $\ce{O2}$?  Let's call the answer $z$.
1 mole of any substance contains Avogadro's number of molecules.  How many molecules are in $z$ moles?


Reaction of magnesium with the surrounding air In my lecture notes for inorganic chemistry, it is stated that magnesium has high affinity for oxygen but is rather stable at room temperature in air.

Can anyone explain why this is the case?

I believe that magnesium exposed to air quickly ends up with a coating of magnesium oxide on its surface.  This oxide coating is quite uniform and does not allow any further oxygen to reach the surface.

Several metals, but NOT iron, have this property of a protective oxide coating.  In iron the coating forms a spongey surface easily penetrated by further oxygen.


How is the formula of mean activity coefficient derived? The mean activity coefficient is defiend as follows:
$$\gamma_\pm = (\gamma_+\gamma_-)^{1/2}.\tag{1}$$
If Debye-Hückel equation
$$-\log\gamma_i = 0.5z_i^2\mu^{1/2}\tag{2}$$
is used, then the mean activity coefficient has the form
$$-\log\gamma_\pm = 0.5|z_+z_-|\mu^{1/2},\label{eqn:3}\tag{3}$$
where $\mu$ is the ionic strength, $z_i$ is the charge of species $i,$ $\gamma_i$ is the activity coefficient.
But I'm confused about the equation \eqref{eqn:3}. It can be found as equation \eqref{eqn:3-38} in Snoeyink and Jenkins' Water Chemistry [1]:

[…] developed for ionic strengths of less than approximately $\pu{5E-3}$ and can be stated as
$$-\log\gamma_i = 0.5Z_i^2\mu^{1/2}.\tag{3-34}$$
Because anions cannot be added to a solution without an equivalent number of cations (and vice versa). it is impossible to determine experimentally the activity coefficient of a single ion. Therefore, Eqs. 3-34, 3-35, and 3-36 cannot be verified directly. However, it is possible to define, and measure experimentally, a mean activity coefficient, $\gamma_\pm,$ as,
$$\gamma_\pm = (\gamma_+\gamma_-)^{1/2}.\tag{3-37}$$
The Debye-Hückel and Güntelberg relationships can be extended to the mean activity coefficient thus:
$$-\log\gamma_\pm = 0.5|Z_+Z_-|\mu^{1/2},\label{eqn:3-38}\tag{3-38}$$

Below is my derivation process:
$$
\begin{align}
-\log\gamma_+ &= 0.5z_+^2\mu^{1/2};\tag{4.1}\\
-\log\gamma_- &= 0.5z_-^2\mu^{1/2},\tag{4.2}
\end{align}
$$
so
$$
\begin{align}
-\log\gamma_\pm &= -\log[10^{-0.5z_+^2\mu^{1/2}}\times 10^{-0.5z_-^2\mu^{1/2}}]^{1/2} \tag{5.1}\\
  & = -\log[10^{-0.5\mu^{1/2}(z_+^2+z_-^2)}]^{1/2} \tag{5.2}\\
  & = 0.25\mu^{1/2}(z_+^2+z_-^2). \tag{5.3}
\end{align}
$$
This result is different from $0.5|z_+z_-|\mu^{1/2}.$ Can someone tell me which step is wrong in my process?
Reference

Snoeyink, V. L.; Jenkins, D. Water Chemistry; Wiley: New York, 1980. ISBN 978-0-471-05196-1.

The problem is you were starting with an expression for the mean activity coefficient for salts where both the cation and anion were monovalent, and then attempting to derive, from this, a general expression for salts of all empirical formulas. To correct this, you need to start with the general expression for the mean activity coefficient of salts of any emprical formula.
If the empirical formula of the salt is of the form $\ce{A_pB_q}$, where A is the cation and B is the anion, then the general formula for the mean activity coefficient is:
$$\gamma_{\pm} = \sqrt[^{p+q}]{\gamma^p_+\gamma^q_-}$$
[Adapted from: https://en.wikipedia.org/wiki/Activity_coefficient, where I've taken the Wikipedia expression and substituted $\gamma_+$ for $\gamma_A$, and $\gamma_-$ for $\gamma_B$, to correspond to the nomenclature you are using.]
Charge balance dictates that  $|z_+| = q$ and $|z_-| = p$. [E.g., in $\ce{A_2B_3}$, the charge on A must be 3+, and that on B must be 2- (or some integer multiple of those).]
Substituting, we have:
$$\gamma_{\pm} = \sqrt[^{|z_-|+|z_+|}]{\gamma^{|z_-|}_+\gamma^{|z_+|}_-}$$
If you use that in place of eqn. (3-37) in your post, and work through your calculations again, you will get the result show in eqn. (3-38), which is a general result for any values of p and q.


Deuteration of enone - mechanism? What happens when the above compound is mixed with D2O (deuterated water)?
As a result of tautomerism, I'd expect the 4H atoms, one alpha to carbonyl carbon, and three at gamma position; to be replaced with deuterium.

However, I'm not clear about the mechanism that'd be followed to achieve the same.

My thoughts:
The first step should involve formation of carbanion by abstraction of hydrogen gamma to carbonyl group. The carbanion is stabilised by extended conjugation, and has three resonance structures.

Now, how does the reaction exactly proceed? Could someone please provide a detailed mechanism, it'd be really helpful. Thanks a lot.

As TAR86 mentioned, you seem to understand where deuteration occurs in crotonaldehyde. The process is usually conducted under alkaline conditions. All of the steps are reversible and need not occur in the same order as in the diagram. For instance, enolate 2 can deuterate at the α-position prior to γ-deuteration. Hope this helps.


Ring chain tautomerization of hydroxyacetone My chemistry textbook says that hydroxypropane-2-one and 1,2-epoxy-2-hydroxypropane are ring chain tautomers, but it did not mention the steps for conversion of one into the other. I tried to figure them out but got stuck.

My attempt:

At first, even though this is not related to the question, I want to say that you're confused between double-headed arrow between the two resonance forms and double harpoon in reversible reactions. (The harpoons should be written as below.)

Your incomplete mechanism has two errors:

The hydrogen atom bonded to carbon is far less reactive than the one bonded to oxygen.
The positive carbon, which is bonded to oxygen, a very electronegative element, is very, very unstable, which is not likely to happen.

This is the right mechanism for the reaction:

Step 1: Oxygen is a good nucleophile, so it is most likely to take the most acidic hydrogen.
Step 2: Then, the negative charged oxygen will attack the electrophile carbon atom, forming epoxy ring.


Why doesn't the particle in a ring model have a potential energy term? I was thinking of the "particle on a ring" problem as a subdivision of a bigger problem, i.e. the hydrogen atom. While in the hydrogen atom problem, the Hamiltonian contains a potential energy term, the Hamiltonian in the "particle on a ring" problem does not.

Why is it so? What are its implications in the mathematical solution to the problems?

The short answer is that the particle in a ring is defined to have no potential energy term (at least, not beyond what Karsten Theis has pointed out). Once you introduce something else into the system, then it is no longer really the particle on a ring problem any more.

However, if you want to relate it to the hydrogen atom, then note that in the hydrogen atom, the Coulomb potential is a function of only $r$, i.e.

$$V(r) = -\frac{1}{r}$$

in atomic units. In the particle on a ring model system, the particle is already constrained to be at a particular value of $r$, so even if we were to hypothetically introduce a Coulomb-type potential $\propto 1/r$, this would just be a constant. The effect of having a constant potential energy is just to shift every eigenstate up in energy by the same amount; it has no real physical effect.

What are the differences in the mathematical solutions? Well, you already know it, surely; the eigenstates of the hydrogen atom are atomic orbitals, whereas the eigenstates of the particle on a ring are simply $\exp(\mathrm im\phi)$ with $m \in \mathbb{Z}$ to satisfy the boundary condition.

Of course, this $\exp(\mathrm im\phi)$ term is indeed part of the mathematical form of the atomic orbitals. This is because when you solve the Schrödinger equation for the hydrogen atom, you can successively separate out different degrees of freedom. Firstly you separate the bit which depends on $r$, which goes on to become the radial wavefunction $R(r)$. Then you need to solve for the angular wavefunction $Y(\theta,\phi)$, and the way of doing this is to again use separation of variables, i.e. assume $Y(\theta,\phi) = f(\theta)g(\phi)$. That gets you to a differential equation which looks like $(-1/g)(\mathrm d^2g/\mathrm d \phi^2) = m^2$, which has $g = \exp(\mathrm im\phi)$ as its solutions. I'm glossing over some details here, but that's the general idea; consult e.g. Griffiths' Introduction to Quantum Mechanics for the nitty-gritty bits.


Hydrolysis products of chlorides of group 15 $\ce{NCl3}$ on hydrolysis produces $\ce{NH4OH}$ and $\ce{HOCl}$.

$\ce{PCl_3}$ when hydrolyzed produces $\ce{P(OH)3}$ i.e. $\ce{H3PO3}$.

$\ce{AsCl_3}$ when hydrolyzed produces $\ce{As(OH)3}$.

However when $\ce{SbCl3}$ or $\ce{BiCl3}$ are hydrolyzed they produce $\ce{SbOCl}$ and $\ce{BiOCl}$ respectively.

Why are the products so widely varying ? How can we explain the reactions theoretically ?

Nitrogen trichloride

Nitrogen trichloride is quite unstable and hydrolyse easily:

$$\ce{NCl3 + 3 H2O → NH3 + 3 HOCl~~~~~~~~(1)}$$

$$\ce{NH3 + H2O -> NH4OH~~~~~~~(2)}$$

$$\ce{NCl3 + 4H2O -> NH4OH + HOCl~~~~~~~(1) +(2)}$$

Phosphorus trichloride

$$\ce{PCl3 + 3H2O → H3PO3 + 3HCl}$$

Arsenic trichloride

$$\ce{AsCl3 + 3 H2O → As(OH)3 + 3 HCl~~~~~~~~(1)}$$

This is the first and main reaction of hydrolysis of arsenic chloride to form arsenous acid. But other species (arsenite ions) like $\ce{[AsO(OH)2]−}$ and $\ce{[AsO2(OH)]^2−}$ also exist in solution and are the conjugated bases.

$$\ce{AsCl3 + 4H2O ⇄ H[As(OH)4] + 3HCl~~~~~~~(2)}$$

$$\ce{AsCl3 + 2H2O ⇄ HAsO2 + 3HCl~~~~~~~~~(3)}$$

And at last aresnic trichloride fully hydrolyse to form arsenic(III) oxide

$$\ce{2AsCl3 + 3H2O → As2O3 + 6HCl~~~~~~~~~~(4)}$$

Antimony trichloride

$$\ce{SbCl3 + H2O → SbOCl + 2HCl~~~~~~~~(1)}$$

With more water it forms $\ce{Sb4O5Cl2}$ which on heating to 460° forms $\ce{Sb8O11Cl2}$.

$$\ce{SbCl3 + H2O ⇄ Sb(OH)Cl2 + HCl~~~~~~~(2)}$$

$$\ce{4SbCl3 + 5H2O ->[50 C] Sb4O5Cl2 + 10HCl~~~~~~~(3)}$$

Final reaction to form antimony trioxide.

$$\ce{2SbCl3 + 3H2O ->[\Delta] Sb2O3 + 6HCl~~~~~~~~~~(4)}$$

Note that speculation on the composition of antimony oxychloride has been raised. Some say it is a mixture of antimony trioxide and antimony trichloride.

Bismuth tricloride

$$\ce{BiCl3 + H2O -> BiOCl + 2HCl~~~~~~~~(1)}$$

Main reaction to form bismuth oxychloride. Further hydrolysis yield and intermediate monohydrate $\ce{BiCl3.H2O}$.

$$\ce{BiCl3 + H2O(vap) ->[50 C] BiCl3.H2O~~~~~~~~(2)}$$

$$\ce{$$BiCl3 + 2H2O → Bi(OH)2Cl + 2HCl~~~~~~~~~(3)}$$

Explaination

Now, you may been be wondering that hydrolysis of nitrogen and phosphorus yield one product but that of arsenic, antimony and bismuth yield 3 or 4 products. This is because of presence of vacant d or f orbitals where electrons can reside easily. For this reason, $\ce{Sb^3+, Bi^3+}$ solvate in aqueous solution to form antimonyl or bismuthyl ($\ce{SbO+, BiO+}$) and other species like $\ce{[Bi6O4(OH)4]^6+, [Bi6(OH)12]^6+ ,  [Bi(H2O)9]^3+}$ and similar antimony ions. Other factors also play role like relativistic effects, lanthanoid contraction, inert pair effect etc.


How can I know which molecule is more stable when bond order it's equal? So, I was wondering, I have these molecules $\ce{O^+2}$, $\ce{O^-2}$, $\ce{O2}$ and $\ce{C^+2}$.
When I try to find their bond order and their magnetic nature I get this:
$\ce{O^+2} = \frac{1}{2} \left( 10 - 5 \right) = 2.5$ - Paramagnetic
$\ce{O^-2} = \frac{1}{2} \left( 10 - 7 \right) = 1.5$ - Paramagnetic
$\ce{O2} = \frac{1}{2} \left( 10 - 6 \right) = 2$ - Paramagnetic
$\ce{C^+_2} = \frac{1}{2} \left( 7 - 4 \right) = 1.5$ - Paramagnetic
You can note that $\ce{O^-_2}$ and $\ce{C^+_2}$ have the same bond order and are also paramagnetic, but how can I know which one is more stable so I can arrange them in stability order if $\ce{O^-_2}$ and $\ce{C^+_2}$ have the same orders?
Thanks!

To recognize the more stable molecule you may extend the description with bond orders is the one considering how the mathematical concept to mix atomic orbitals by LCAO and visualizing the results in Molecular orbital diagrams and compute the overall energy of such a molecule.  Eventually, you compare the total energy of $\ce{O^-_2}$ with $\ce{C^+_2}$ using the same external reference.
As about the hydrogen molecule shown below:

Two important details about symmetric diatomic moleculs to retain are:

For $n$ atomic orbitals, LCAO yields $n$ molecular orbitals.
The split between the lowered bonding molecular orbital and the higher antibonding orbital is not symmetric.  The difference between the energy level of the initial atomic orbitals and the the antibinding molecular orbital is higher, than the difference between the initial atomic orbitals and the binding molecular orbital.

So you draw the diagram including the energetic levels, and populate the levels with electrons from the bottom up.  If there multiple molecular orbitals of same energy, you fill the orbitals first with one, and later with the second electron. You then compute the the overall energy of the molecule and later compare the results about the molecules in question against a reference in common, e.g. vacuum.
Note that the notation $\sigma$ and $\sigma{}^*$ is about the rotation symmetry around the bond between the molecule (contrasting to $\pi$ and $\pi{}^*$ as perpendicular to this), and not if the molecular orbital in question is conceptually formed from $s$, or $p$ atomic orbitals.  Equally, the energetic consecution of levels of the molecular orbitals may change, e.g. between nitrogen and oxygen:

(using material from here)


Symmetry lost in orbitals? I've always thought that orbitals lead to a loss of symmetry, and have never been able to give myself a satisfactory answer to this.

I'll explain via an example:

Let's take an $\ce{N^3+}$ atom. It's perfectly spherical, and has no distinguishing 'up' and 'down'. There is no set of 'preferred coordinate axes' for it since it has spherical symmetry (except the nucleus, but I doubt that matters).

Now, let's give it three electrons. They arrange themselves in the $2p$ orbitals, one in each (by Hund's rule). Now, suddenly, the atom has lost its spherical symmetry — we have a distinct triplet of orthogonal directions separate from the others.

This leads to these questions:
How can symmetry 'break' this way? Are the directions of the axes 'hidden' in the atom beforehand? Are they themselves wavefunctions (though a wavefunction of wavefunctions sound odd to me, this explanation makes sense-random events can break symmetries)

So, I'd like a clear explanation of how/why the symmetry breaks.

What you describe is unfortunately a very common misconception. Defining orbitals does not break spatial symmetry. It is still completely arbitrary which directions you define as $x$, $y$ and $z$, and thus it is entirely arbitrary how you orient the $2p$ orbitals in space. Therefore rotational invariance is still preserved.

The other thing to remember is that all the $p$ orbitals are (essentially) degenerate and so you can take any linear superposition of them you want. For example you can write down an orbital that looks like

$$\frac{1}{\sqrt{3}}\left(\phi_{2p_x}+\phi_{2p_y}+\phi_{2p_z}\right) $$

and placing an electron in this orbital does not break rotational symmetry.


Recrystallisation of dibenzylideneacetone Recently, I carried out the synthesis of dibenzylideneacetone. I had a lot of problems with the purification process, especially the part involving recrystallisation. I was instructed to carry out the recrystallisation using an "ethanol/water" pair. That means to say, I was supposed to dissolve the crude product in ethanol until just enough dissolved, then add a small volume of water until some of the non-polar dibenzylideneacetone precipitated out of the solution and lastly, add another small volume of ethanol to re-dissolve the precipitate that had just formed. Once done, the beaker is removed from the hotplate and left to cool at room temperature. Subsequently, the crystals were washed with cold ethanol to finally obtain the final product. It is similar to what was done in this video.

During the practical I had to repeat the recrystallisation step multiple times for a variety of reasons. Firstly, there were some instances when the dibenzylideneacetone melted to form a yellow oily liquid. Secondly, there were also some instances when the crystals "crashed out" of solution upon cooling. This resulted in poor-quality crystals which clumped together. I am just wondering if anyone has any tips on how I could have improved my procedure so that I don't have to perform recrystallisation multiple times and yet still obtain poor-quality crystals.

Armarego's Purification of Laboratory Chemicals suggests using dichlormethane–hexane pair instead of ethanol–water; it also lists crystallization from warm ethyl acetate, ethanol or petroleum ether as alternative [1, p. 342].

  Dibenzalacetone [DBA, bda, trans-trans-1,5-diphenyl-1,4-dien-3-one, 1,5-(bisphenyl)-penta-1E,4E-diene-3-one] [538-58-9] $\ce{C17H14O}$, $M~234.3$, $m~107°$, $111°$, $112°$, $113°$, $120-122°$.
  Purify the ketone by flash chromatography (150 mesh $\ce{Al2O3}$ deactivated with $6\%$ v/w $\ce{H2O}$) using  petroleum ether $40-60°/\ce{EtOAc}$ ($4/1$, v/v), and the yellow solid is recrystallised by layering a concentrated $\ce{CH2Cl2}$ solution with hexane (i.e. $\ce{CH2Cl2}$/hexane, $1:4$, v/v).
  Also recrystallise the ketone from hot ethyl acetate $(\pu{2.5 ml/g})$ or $\ce{EtOH}$.
  It has been also crystallised from petroleum ether b $30-60°$.

References

Armarego, W. L. Purification of Laboratory Chemicals; Elsevier: Boston, MA, 2017. ISBN 978-0-12-805457-4.


Is nitrogen of aniline sp² or sp³ hybridized? I found a thread which discussed this which totally confused me. (refer to the part of the link which talks about $\ce{sp^2/sp^3}$)
The question "How to rationalise the resonance structures and hybridisation of the nitrogen in a conjugated amine?" is helpful, but doesn't have aniline specifically. I understand that if the lone pair is in resonance, it becomes $\ce{sp^2}$.
Can someone help me out?

The nitrogen in aniline is somewhere between $\mathrm{sp^3}$ and $\mathrm{sp^2}$ hybridized, probably closer to the $\mathrm{sp^2}$ side.  We are correctly taught that the nitrogen in simple aliphatic amines is pyramidal ($\mathrm{sp^3}$ hybridized).  However in aniline, due to the resonance interaction between the aromatic ring and the nitrogen lone pair, considerable flattening of the nitrogen occurs (if it were completely flat it would be $\mathrm{sp^2}$ hybridized).

We can assess the nitrogen hybridization by measuring its barrier for pyramidal inversion. If a trigonal nitrogen is $\mathrm{sp^2}$ hybridized, the barrier will be zero. On the other hand, in aliphatic amines where the nitrogen is $\mathrm{sp^3}$ hybridized the inversion barrier is typically around $\pu{4-5 kcal/mol}$.

(pyramidal inversion diagram)

In aniline this barrier is very low, somewhere around $\pu{1-2 kcal/mol}$. This indicates that the nitrogen in aniline is not quite planar, but is much closer to being  planar ($\mathrm{sp^2}$) than pyramidal ($\mathrm{sp^3}$).


Choice of equipment for distilling I need to distill some reagents, and the book Purification of laboratory chemicals suggests first refluxing over $\ce{P2O5}$ for several hours before distilling across.

The obvious choice is a solvent still head like the ones used to distill THF, but I don't need very much of this so it seems wasteful.

The way suggested that I do it is to set up a normal reflux, let it cool, and then quickly change the condenser for a side arm. This will obviously work but seems like a lot of effort. I think there should be some other piece of glassware to allow this process. Is this the case?

http://www.sigmaaldrich.com/catalog/product/aldrich/z516333?lang=en&region=US

This condenses, and depending on what you have the knob turned too will either send it back to it's origin or to a collector. (I think, I have something similar in my lab but the link doesnt have a great drawing)


Size of target substance in IR and Raman Spectroscopy Usually what is the size (in nanometer) of the sample or target substance in IR and Raman Spectroscopy?

Which one has largest size of sample it can scan, how many molecules or atoms are involved? If you can make the size bigger or scan coverage bigger, then the more intense would be the IR or Raman stroke signal? Or no difference in intensity in the case of Raman even if sample size is increased? Please answer separately for IR and Raman since they are different. One uses IR light, the other laser. I guess IR light needs to be focus at a spot just like a laser too.

This all depends on the exact geometry of your spectrometer light path, if you measure in transmission or reflection, using ATR, etc. AND of course on your sample concentration.

If you want to ask about the minimal size: As small as you can get your focal point or laser spot. A few dozen microns, maybe less. Local sample heating can be a problem. You can buy scanning IR and FT Raman microscopes.

In transmission, you can measure IR on a cm sized sample, given that is is translucent. With Raman, that would be tricky, because you should focus the scattered light again.

Minimum number of molecules for IR is very low, but that also depends on how strong the absorption line in question is. Raman used to be very insensitive, that has become a lot better with FT Raman.


Synthesis via Sn2/Sn1, E2, E1: Choice of solvent I uploaded pictures on http://www.physicsforums.com/showthread.php?t=765106

I was looking at the answers in my textbook. They seem to make incongruent choices of solvents.

In e) and j), why choose EtOH and H2O/CH3OH? I would rather choose an aprotic solvent like DMF to better allow Sn2 to occur. Especially with j), there is a lot of competition with E2 because OH- is a strong base and because we are attacking a secondary carbon.

I see this again in k). Secondary carbon attacked: why not add DMF to prevent E2 and get best yield?

Your thinking is not wrong. For (e) DMF is a better choice of solvent - this doesn't mean that MeOH won't work just that DMF is better and is typically used in these reactions. For even better results you can use DMSO. If you dare, as DMSO/cyanide is a very toxic cocktail.
(j) I don't think it is a practical synthetic reaction since it can possibly give you elimination and/or $S_N1$ (ie no inversion of stereochemistry) products.


Where does the RT term come from in the derivation for the activation enthalpy from the Eyring equation? It is easy to show an Arrhenius-like equation from the Eyring equation, but if you do this, you get that the activation energy is about equal to the activation enthalpy. However, the real approximation is that $E_\mathrm{a} = \Delta H^\ddagger + RT$. Why is this?

Using vibrational partition functions to define the reaction rate constant produces an equation of the form $\displaystyle k=aT^be^{-\Delta U_0^\mathrm{O}/(RT)}$ where $a,b$ are constants independent of $T$ and $\Delta U_0^\mathrm{O}$ is the difference in zero point energies at the transition state compared to reactant, this is related to what is commonly called the activation energy. This approach allows molecules to have discrete vibrational levels and so a thermal population of these. The constants $a,b$ are found from the partition functions.

The (classical) Arrhenius equation is often written as $k_A=Ae^{-E_A/(RT)}$. Taking the log of each equation and differentiating wrt. $T$ gives

$$\frac{d\ln(k)}{dT} =\frac{bRT+\Delta U_0^\mathrm{O}}{RT^2}\equiv \frac{E_A}{RT^2}$$

thus the Arrhenius activation energy is $E_A\equiv bRT+\Delta U_0^\mathrm{O}$.

(Texts on statistical thermodynamics gives details of partition functions, some book on kinetics also give this detail.)


Can I determine purity of isopropanol with a hydrometer? Assuming I start with a known 99.8% purity of isopropanol, after recovery with a distillation apparatus, could I use a hydrometer to test for water content? If so, is there a special one I should use?

There is an image of a table provided by Chu and Thompson [1] listing isopropanol water solution densities:

The essential data part is (weight % vs density):

  77.87 0.8361
  89.84 0.8069
  99.91 0.7808

There is density gradient -0.0026 / 1 w%, resp. -0.00052 / 0.2 w%, determining purity by density via hydrometers is rather illusory.

Some success could be gained by pycnometers — using special glass flasks with a glass cap with a hollow. You would weight the flask, the flask with water and the flask with isopropanol and you would calculate the density.

In any case, as differences are minimal and as liquids have considerable thermal dilation, one must carefully watch the liquid temperature.

In case of the pycnometry measurements, the glass and both liquids must have the same temperature, the best the one for which density is tabelized.

References

Chu, K.-Y.; Thompson, A. R. Densities and Refractive Indices of Alcohol-Water Solutions of n-Propyl, Isopropyl, and Methyl Alcohols. J. Chem. Eng. Data 1962, 7 (3), 358–360. https://doi.org/10.1021/je60014a011.


Molecular formula for 4-methyl-1-pentanol? According to Wikipedia 4-methyl-1-pentanol is $\ce{C_6H_{14}O}$.

But here is what I got, which is slightly different.

Methyl = Think of methyl group, which is $\ce{CH_3}$.

Pentanol = Penta means we have 5 carbons, -ane ending means we have $2n+2$ hydrogens, i.e. 12 hydrogens and -ol ending means we have a hydroxyl group $\ce{-OH}$.

So the molecular formula that I arrived at is: $\ce{CH_3C_5H_{12}O}$ = $\ce{C_6H_{15}O}$.

How did I end up with an extra hydrogen compared to Wikipedia answer?

Actually, the formula you arrived at is $\ce{CH3C5H12OH}$ = $\ce{C6H16O}$.  In any case, in your "2n+2" step, you need to reduce that number of hydrogens by 2.  One is eliminated where the methyl group is attached, and a second is eliminated where the $\ce{-OH}$ group is attached.  So removing 2 hydrogens from $\ce{C6H16O}$, gives you the desired $\ce{C6H14O}$.


Do atoms react to fill outer shell or 8 valence electrons? I recall in middleshool chemistry we simply said electron shell configurations were 2,8,8,many which I would have no problems with.

But now I'm learning that shells contain $2n^2$ electrons where n is the shell number.

Then why is "full outer shell" and "8 valence electrons" still used interchangeably when describing stability? i.e we are taught that atoms react in ways that achieve an outer shell of 8 valence electrons. But at the same time, they react to form full outer shells.

Excellent question I must say.
To middleschoolers they really don't teach advanced concepts regarding orbitals. But since this question came to your mind, I guess you're capable of handling what I have to say next.
Unlike what's taught to you, shells don't exist just like that, they exist as a collection of subshells. Subshells are given names such as $\mathrm{s}$, $\mathrm{p}$, $\mathrm{d}$, and $\mathrm{f}$. The number of subshells a shell has is given by it's principal quantum number, or simply the shell number. In this way, the 1st shell has only one subshell, which is $\mathrm{1s}$. The second shell has two subshells; $\mathrm{2s}$ and $\mathrm{2p}$. The third has three subshells; $\mathrm{3s}$, $\mathrm{3p}$, and $\mathrm{3d}$.
Every subshell is further composed of orbitals. Orbitals are different from orbits, orbits have electrons in a precisely defined path, while orbitals show a cloud of probability of where the electrons could exist. Each orbital holds two electrons.The orbitals also have exciting shapes:
The $\mathrm{s}$ subshell consists of just one orbital, the $\mathrm{s}$ orbital. It's a spherical cloud of electron probability:

The $\mathrm{p}$ subshell consists of three orbitals, $\mathrm{p}_x$, $\mathrm{p}_y$, and $\mathrm{p}_z$. They are dumb-bell shaped, oriented along the $x$, $y$, and $z$ axes respectively:

The $\mathrm{d}$ subshell has five orbitals, $\mathrm{d}_{xy}$, $\mathrm{d}_{yz}$, $\mathrm{d}_{zx}$, $\mathrm{d}_{x^2+y^2}$, and $\mathrm{d}_{z^2}$. They have interesting shapes.

The $\mathrm{s}$ subshell can hold two electrons, the $\mathrm{p}$ subshell can hold six electrons,the $\mathrm{d}$ subshell can hold ten electrons, and the $\mathrm{f}$ subshell can hold fourteen electrons.
In middleschool, they teach you to fill the lower electrons first, and move to the top, if you want to find the electronic configuration of a given element. Unfortunately, that's not entirely correct. Infact, considering the subshell concepts, you must use the Aufbau Principle. The rule tells us to fill the electrons in the order given in the picture below:

(source: boundless-cdn.com)
If the picture wasn't clear, the order is $\mathrm{1s}$, $\mathrm{2s}$, $\mathrm{2p}$, $\mathrm{3s}$, $\mathrm{3p}$, $\mathrm{4s}$, $\mathrm{3d}$, $\mathrm{4p}$, $\mathrm{5s}$, $\mathrm{4d}$, $\mathrm{5p}$, $\mathrm{6s}$,...
Do you notice that $\mathrm{4s}$ comes before $\mathrm{3d}$? Thats why you don't get those ten extra electrons of $\mathrm{3d}$, since the $\mathrm{4s}$ subshell is being filled for the elements potassium and calcium.


Difference of Ascorbic acid dissolved or precipitated in Glycerol? Difference of Ascorbic acid (AA) dissolved or precipitated in Glycerol?
Referring to pat US20110008445A1
Solution 1
At 25C, AA was dissolved in glycerol and alcohol (subsequently removed by distillation).
The AA was measured using scattered light (JIS K0101 using formazine as ref).
5 µm sized particles of AA were found to be dissolved in the glycerol.

Solution 1a
Solution 1 was aged at 28C for 10 days.
10% AA remained dissolved
90% AA precipitated
25 µm sized particles of AA were found to be suspended in the glycerol.

Solution 1b
Solution 1 was aged at 15C for 10 days.
10% AA remained dissolved
90% precipitated
80 µm sized particles of AA were found to be suspended in the glycerol.

Analysis
A key aspect of the experiment focusses on AA being in two different states:
Dissolved and Precipitated
The implication is that the AA is somehow different, otherwise why mention dissolved and precipitation?
Also, the AA was initially dissolved in glycerol and alcohol.
Therefore the AA was

Dissolved in Glycerol and Alcohol
Dissolved in Glycerol
Precipitated and suspended in Glycerol

Question
Does Ascorbic Acid remain in its usable form, in all three scenarios?
('Usable' relates to applying this concept, of dissolving the AA in Glycerol and Alcohol, as a precursor to AA encapsulation in lecithin emulsion).

To acknowledge OP's request, I am putting this as my answer:

What OP's initially referring to (US20110008445A1) was a patent application. Since it was a 2011 application, I searched for the patent and found it (Ref.1). It's abstract states that:

  A Suspension of ascorbic acid in glycerol or in glycerol comprising diglycerol, in which the content of ascorbic acid is 13% by mass or greater, and further in which 8 to 12% by mass of the ascorbic acid is dissolved in glycerol or in glycerol comprising diglycerol, and the rest of ascorbic acid is precipitated in the form of microcrystals having a particle diameter of $\pu{25 \mu m}$ or smaller and is uniformly dispersed in the suspension. The suspension of ascorbic acid in glycerol is useful as a base material for cosmetics containing ascorbic acid which exhibits excellent feel in the use (spreadability and Smooth feel on application to the skin).

The patent described the preparation of ascorbic acid suspentions in alcohol or glycerol. Based on my knowledge in ascorbic acid, all forms of suspensions contain ascorbic acid in  in its usable form. However, one should be aware that ascorbic acid tends to have browning effect due to slow formation of related furan derivative on long standing (Ref.2).

References:

Mikiharu Okumura, Takanori Ohta, “Suspension of Ascorbic Acid in Glycerin and Process for Production Thereof,” United States Patent 2012, 8,211,472 B2 (https://patentimages.storage.googleapis.com/62/29/bf/f95fde91d869cd/US8211472.pdf).
T. Lalikainen, M. A. Joslyn, C. O. Chichester, “Browning Reactions, Mechanism of Browning of Ascorbic Acid-Citric Acid-Glycine Systems,” J. Agric. Food Chem. 1958, 6(2), 135–139 (DOI: 10.1021/jf60084a011).


Is this asking for the equivalence point? Is this question asking for the equivalence point of the titration curve?

  At what point does the concentration of acetic acid equal the concentration of the acetate ion? What is the pH of the equimolar, ideal buffer solution at this point.

At first, I thought this question was asking about the point right before the titration curve shoots up. However, I don't recall ever learning about such a point.

If you were titrating acetic acid with a base, then the equivalence point would be where as much base has been added (i.e. moles of base vs moles acid) to the solution as there was acetic acid. This is essentially the point where all the acetic acid has been neutralized.

In general the "pKa" is the term used to define the point at which the protonated and unprotonated forms are equal. The pKa is from the negative log of the acid dissociation constant as given by the reaction:
$$\ce{HA <=> H+ + A-}$$
and the equation: $$ K_\mathrm{a} = \dfrac{\ce{[H+] [A- ]}}{\ce{[HA]}}$$

where $\ce{[H+] = [A- ]}$ then:
$$K_\mathrm{a} = \dfrac{\ce{[H+]^2}}{\ce{[HA]}}$$

and $$ \ce{pKa} = -\log \ce{Ka} = -\log \dfrac{\ce{[H+]^2}}{\ce{[HA]}}$$


Calculating enthalpy of dissolution In a polystyrene cup calorimeter, $4.3\ \mathrm g$ of ammonium nitrate, $\ce{NH4NO}$, was added to $60.0\ \mathrm g$ of water and stirred to dissolve the solid completely. The initial temperature dropped from $22.0\ \mathrm{^\circ C}$ to a final temperature of $16.9\ \mathrm{^\circ C}$.
Calculate the enthalpy change in $\mathrm{kJ\ mol^{-1}}$ for this dissolution process, as represented by the chemical equation below:
$$\ce{NH4NO3(s) -> NH4NO3(aq)}$$
Assume that the calorimeter does not absorb any heat, that the density of the solution is the same as that of water $(1\ \mathrm{g\ ml^{-1}})$ and that the specific heat capacity of the solution is also the same as that of water $(4.18\ \mathrm{J\ g^{-1}\ K^{-1}})$.
$$M(\ce{NH4NO3})=80.05\ \mathrm{g\ mol^{-1}}$$

My attempt:
$$Q=mcT = 64.3\ \mathrm g \times 4.18\ \mathrm{J\ g^{-1}\ K^{-1}} \times -5.1\ \mathrm K = -1.37\ \mathrm{kJ}$$
But this doesn't seem right because I haven't used all the information given in the question, such as the molar mass of ammonium nitrate.

The question requires you to put your answer in kJ $mol^{-1}$. This means kilo joules per mole. So you have calculated the energy in kJ correctly, to get it per mole you need to first calculate the number of moles of ammonium nitrate you used.

Use the equation: $$n=\frac{m}{M_r}$$

n = moles

m = mass in grams

$M_r$ = molar mass

The mole calculation turn out to be 0.05371642723.

Then its just a matter of division, which I calculated to be -25.50 kJ $mol^{-1}$ (2.d.p).

That answer is using your rounded answer for the kJ, so the answer may be off. When i do calculations, i round right at the end to get the most accurate answer.


Does the acid strength of hydrochloric and hydrofluoric acid correlate with the electronegativity of the halides? Is this statement true or false?

  HCl is a stronger acid than HF because fluorine is more electronegative than chlorine.

I'm thinking this is true, because since fluorine is more electronegative than chlorine, it has  a higher ionic energy and therefore a stronger bond between hydrogen than chlorine does with hydrogen. In result, leading HCl to dissociate  more easily. Can you please clarify.

You are correct, HCl is a stronger acid than HF. Fluorine is both more electronegative and smaller than chlorine. Because fluorine is more electronegative, the bond between it and the hydrogen is more polar, meaning that the proton would need to overcome a larger coulomb force to separate from the fluorine. In addition, fluorine is a smaller atom, meaning that the addition of an electron has a larger effect on the net charge of the atom. This both makes the fluoride anion (the conjugate base) less stable and it also - once again - increases the coulomb force between the proton and the fluorine. In addition, the coulomb force is even stronger because there is a smaller distance between the two atoms. As a general rule of thumb, as you move down the halogens, the acids become stronger and they form more stable conjugate bases.


Volume occupied by real gases Molecules of an ideal gas are assumed to have zero volume, the volume available to them for motion is always the same as the volume of the container. In contrast, the molecules of a real gas have small but measurable volumes. At low pressures, the gaseous molecules are relatively far apart, but as the pressure of the gas increases, the intermolecular distances become smaller and smaller . As a result, the volume occupied by the molecules becomes significant compared with the volume of the container. Consequently, the total volume occupied by the gas is greater than the volume predicted by the ideal gas law. Thus at very high pressures, the experimentally measured value of PV/nRT is greater than the value predicted by the ideal gas law.

This is what I read in a website. In the above statements, what is the volume predicted by the ideal gas law? The volume predicted by the gas law, should be the volume of container. Now if volume occupied by real gas is less than the volume of container, how can the volume occupied by it be greater than that predicted by ideal gas law ?

Also, compressibility is given by $Z = \frac{PV}{nRT}$.  In this equation is P the measured pressure or the ideal pressure?

When you are "containing" a gas in a container then you are by definition restricting its volume. Lets assume you heat the gas. If the volume is confined then the pressure will increase. Similarly, two different gases will have different pressures if confined to the same volume.

If you somehow control the pressure (there is a movable piston) then a real gas will occupy less volume because of its intermolecular interactions.

Also see this: Volume of different gas in a container


Stability of Carbanions For comparing stability of carbanion, we use the fact that "negative charge is stable on more electronegative atom." So, The stability order is, {electronegative order: $\mathrm{sp \gt sp^2 \gt  sp^3}$} and therefore, $\ce{HC#C^- \gt H2C=C^-H \gt CH3-C^-H2}$.
My doubt is that, we generally say, carbanion is stable when there is less electron density on $\ce{C^-}$.
In $\ce{HC#C^-}$, $\ce{C}$ is $\mathrm{sp}$ hybridized that means it is more electronegative, so it will pull electrons more as compared to other two. So, electron density order will be reverse of above order. And, $\ce{HC#C^-}$ will be least stable.
By, my thought, answer is completely opposite.
Please help me to get out of this doubt and please correct me if I said something wrong.

$\mathrm{sp}$ hybridized carbanions are more stable as the lone pair of electrons is placed in an orbital consisting of high s-character. In general, electrons are more stable in s orbitals rather than the p orbitals of the same shell. This can be understood in a simple way, by thinking that the s orbitals are closer to the nucleus, and can also be verified by the n+l rule.
Now, when we say $\mathrm {sp}$ hybridized species are more electronegative than $sp^2$ hybridized species, we mean that the $\mathrm{sp}$ orbital are more accomplished in holding an electron within them when compared to their $\mathrm{sp}^2$ counterparts. As electrons are more stable in an $\mathrm{sp}$ orbital, electrons are pulled more towards them rather than the $\mathrm{sp}^2$ orbitals, which makes us say that $\mathrm{sp}$ orbitals are more electronegative than $\mathrm{sp}^2$ orbitals.


How pigments actually function as colorants? Pigments are insoluble colorants that have no affinity to the substrate. So how can they function as colorants if they "can't stick"? Are pigments always applied together with a "binder"?

Consider natural indigo, formerly used to color blue jeans. It is insoluble; reduction converted it into water-soluble "indigo white". Fabric was soaked in indigo white solution; on air oxidation, it became insoluble blue indigo again, with indigo molecules now presumably mechanically entangled in the fiber and held by van der Waals forces.

More often "binders" called mordants are used. They form an insoluble "lake" when they coordinate to the pigment; the lake can bind to fabric fibers. You can read more about the chemistry of the process here.


Decarboxylation of cyclohex-1-ene-1,2-dicarboxylic acid by electrolysis method What is the major product when cyclohex-1-ene-1,2-dicarboxylic acid is electrolysed?

My try:

I got 2 cases but I am unable to decide which product is major, and need help here.

Turns out a similar reaction has been performed,   the reaction proceeds via the following mechanism

  SOURCE

The cyclohexyne dimerizes to 1,2,3,4,5,6,7,8-octahydrobiphenylene. Then further rearranges. So I would say that electrolysis is feasible with rearrangement as depicted in the picture .
 The second compound(1,2,3,4,5,6,7,8-octahydrobiphenylene) which you've drawn is more stable since very unstable compounds (cyclohexyne) do not form even more unstable compounds. The second compound will not be the final product/major product owing to its high instability due to anti aromaticity.

PS
 I haven't delved into the details regarding the instability of cyclohexyne and 1,2,3,4,5,6,7,8-octahydrobiphenylene since they have been well covered in Gaurang's answer


Redox: Oxidation and Reduction I don't get the difference between the gain or loss of electrons in a reaction and changes in oxidation state. Based on my notes and some research:

Gain or loss of electrons

A substance is oxidised if it loses electrons in a reaction.
A substance is reduced if it gains electrons in a reaction.

Changes in oxidation state

A substance is oxidised if the oxidation state of the element increases.
A substance is reduced if the oxidation state of the element decreases.

Whenever I see a chemical equation, I am confused as I do not know if it is a gain or loss of electrons or changes in oxidation state. I have an assignment to do which involves redox reaction, but it is the holidays and I can't contact my teacher. Could someone explain so that I could understand? For example,

$$\ce{2 Ca (s) + O2 (g) → 2CaO (s)}$$

I thought there is a change in oxidation state as on the reactants side, the oxidation state of both $\ce{Ca}$ and $\ce{O2}$ are $0$. As for the product, the oxidation state for $\ce{Ca}$ is $+2$ and $\ce{O}$ is $-2$. Therefore, $\ce{Ca}$ is oxidised and $\ce{O2}$ is reduced.

But one of the questions on my worksheet asked about electron transfer for this chemical equation, and I do not understand why it is a electron transfer.

Look this as:

$$\ce{2Ca + O2 -> 2CaO}\qquad\text{(complete reaction)}$$

then we can this split into two sub-reactions:

$$\ce{Ca - 2 e- -> \overset{+2}{Ca}}\tag{1}$$

Here "$+2$" means that $\ce{Ca}$ lost $2$ electrons (or two negative units) and therefore it was oxidized. Since it lost something negative it has a higher number of protons (positive particles) remained. In other words the atoms is now positively charged. The initial state was $0$ (the same numbers of protons and electrons, or positive and negative particles/units) and it was increased by loosing two electrons (the number increased).

$$\ce{O + 2 e- -> \overset{-2}{O}}\tag{2}$$

The oxygen got two electrons or two negative particles. Now oxygen has two negative particles more than positive particles (protons) and therefore it is $-2$. In other words, the number (oxydation state) decreased ($0 \to -2$).


Rate Law with Rate Determining Step The reaction of tert-butyl bromide with azide ion in aqueous solution is proposed to proceed by the following mechanism:
  $$
\begin{align}
\ce{(CH3)3CBr(aq) &<=>[k$_1$][$k_{-1}$] (CH3)3C+(aq) + Br–(aq)}\\
\ce{(CH3)3C+(aq) + N3–(aq) &->[$k_2$] (CH3)3CN3(aq)}
\end{align}
$$
  $k_i$ is the rate constant.

  Assuming that $\ce{(CH3)3C+(aq)}$ achieves a steady-state concentration, but making no further assumptions about the relative magnitudes of the three rate constants, what is the rate law for this reaction?

The answer is

$$\mathrm{Rate} = \frac{k_1k_2[\ce{(CH3)3CBr}][\ce{N3–}]}{k_{–1}[\ce{Br–}] + k_2[\ce{N3–}]}$$

But how would I obtain that answer using multistep reaction mechanisms and not differential equations?

Since $(CH_3)_3C^+(aq)$  achieves a steady-state concentration, the equations when directly added give the overall reaction whose rate constant is the product of the rate constants of individual reactions. Hence the rate law comes out to be as it is written.


State of a Racemic Mixture Is a racemic mixture always an equi-molar solution of two enantiomers, both in the same state? Is it necessary for both isomers to be in the liquid state for racemisation to occur, or does racemisation also occur in the gaseous and solid states?
Thirdly,can the two enantiomers still form a racemic mixture if they are initially at different temperatures?

Is a Racemic Mixture always an equi-molar solution of two enantiomers, both in the same state?

Yes and yes. A racemic mixture is equimolar in both enantiomers. Otherwise you would call the reaction enantioselective with a so-and-so selectivity.
A racemic mixture contains both enantiomers in the same state, as energy differences between enantiomers are so minuscule that you would never get one solid and one liquid at the same time.

  Is it necessary for both the mixture to be liquid in nature, or does racemisation also occur in the gaseous or solid state?

Racemisation is the process where you start out with a pure enantiomer and in the end get a mixture. For complete racemisation you end up with an equimolar mixture as discussed above.

It is thus possible that racemisation occurs in gaseous and solid phases, not only in solution or liquid. However, the reaction times will differ significantly. I could imagine that the solid racemisation reaction will take a long time.

  Can the two enantiomers still form a Racemic mixture if they are initially at different temperatures?

Higher temperatures just mean that the racemisation process proceeds faster. In the end, the warmer enantiomer will have racemised further on than the colder one, but for $t\to\infty$ the result will be the same.


Electron vs electron density I recently read somewhere that electrons do not exist. it's just the electron density. Is it true that there are no electrons (particles) but only in the form of electron density? And does this contradict the fact of dual nature of electron?

Of course, electrons do exist. We have a myriads of observations confirming their existence. What does not exist is working classical picture of an electron. Electrons can behave in certain circumstances like waves, in some other - like particles. That is already nonsense from the classical physics point of view with its universal dichotomy: something is either a particle or a wave.

But the trickiest part of the wave-particle duality is that despite the fact that electrons can behave as waves and as particles, their are, strictly speaking, neither of them. Our language developed on the basis of observations of macroscopic surroundings with its well-established classical notions of waves and particles is simply inappropriate for the description of the microscopic phenomena. Thus, whenever one tries to use any language except for that of mathematics to describe microscopic world he/she quickly runs into apparent paradoxes.

With respect to the desire of pictorial classical description of an electron I feel like the only thing I have to do is to quote Paul Dirac:

  […] the main object of physical science is not the provision of
  pictures, but is the formulation of laws governing phenomena and the
  application of these laws to the discovery of new phenomena. If a
  picture exists, so much the better; but whether a picture exists or
  not is a matter of only secondary importance. In the case of atomic
  phenomena no picture can be expected to exist in the usual sense of
  the word 'picture', by which is meant a model functioning essentially
  on classical lines. One may, however, extend the meaning of the word
  'picture' to include any way of looking at the fundamental laws which
  makes their self-consistency obvious. With this extension, one may
  gradually acquire a picture of atomic phenomena by becoming familiar
  with the laws of the quantum theory.


How can Arrhenius bases also be Brønsted-Lowry bases? According to ChemGuide,

  The Brønsted-Lowry theory doesn't go against the Arrhenius theory in any way - it just adds to it.

According to Study.com:

  All Arrhenius acids and bases are also Brønsted-Lowry acids and bases.

This makes sense for Arrhenius acids, such as hydrochloric acid. Hydrochloric acid produces hydrogen ions when dissolved in water. Additionally, it donates a proton to water.

However, sodium hydroxide is an Arrhenius base, since it produces hydroxide ions when dissolved in water. However, sodium hydroxide cannot accept a proton to form $\ce{HNaOH+}$. So, how can it be a Brønsted-Lowry base? Similar reasoning can be applied to other Arrhenius bases, such as potassium hydroxide.

Sure it can.  You're just incorrect about the products when $\ce{NaOH}$ accepts an $\ce{H+}$ in aqueous solution.  The result is $\ce{H2O(l) + Na+(aq)}$ instead of $\ce{HNaOH+(aq)}$.  That is, the $\ce{NaOH}$ doesn't stay together.

Now mind you, most people would point out to you that the $\ce{NaOH}$ doesn't stay together as soon as it's dissolved, meaning $\ce{NaOH(s) -> Na+(aq) + OH-(aq)}$ happens right away, and then as soon as you have some $\ce{H+(aq)}$ around it reacts only with the $\ce{OH-(aq)}$ and the $\ce{Na+(aq)}$ doesn't get involved, probably isn't even anywhere near.  But this doesn't change the main point, which is that if you add $\ce{NaOH}$ to water, you have a solution that can react with $\ce{H+(aq)}$, which means you added a Brønsted-Lowry base.


Research-backed articles on cast iron seasoning I've always seasoned my cast iron pans with vegetable oil because that was what I was taught. I'd done some research before, and even came across this StackExchange post on asking the best oil for the job.
To my dismay however, when I tried doing some research myself, all sources pointed to blog posts, particularly the Sheryl Canter one. There is not a single source on her post and she doesn't seem to be very near a materials scientist. The closest paper I could find to the topic was this one. Other than that, it was all anecdotal evidence, many people claiming many different things.
It felt almost absurd that I couldn't find a paper on something that seemed relatively common to me. This will bug me till the end of time unless I figure it out, so I need to know - does anybody have any real sources on cast iron seasoning or what factors play a part in the polymerization of different oils? Thanks in advance.

There is this research on Chinese equivalent of seasoning process called "Kitchen God's Blessing" [1].
They argue that the non-stick property of seasoning comes by iron oxide $(\ce{Fe3O4})$ nanoballs.
If they are correct then the oil or its polymers is not exactly what is causing the non-stick property but the those nanoballs on the surface. Their seasoning method is somewhat different however. It uses temperatures up to 450 °C.
Reference

Gao, C.; Yang, N.; Li, C.; Wang, X.; Yu, X.; Zhang, L.; Wei, Z. Seasoning Chinese Cooking Pans: The Nanoscience behind the Kitchen God’s Blessing. Nano Materials Science 2020. DOI: 10.1016/j.nanoms.2020.06.001.


How to change pH of CuSO4? I'm currently very desperate to know this. I'm doing an experiment about adsorption/chelation of Cu ions in CuSO4 using banana peels using the UV-Visible spectroscopy. How do I change the pH of 0.1M or 0.05M of CuSO4? My main problem is with making the CuSO4 more alkaline/basic or whatever u call it. My base pH of the CuSO4 is about 4.0 or 4.1. I've tried making a pH 5 and pH 6 buffer solution then adding it to the CuSO4 solution, but so far it's only made it more acidic.
These are the calculations:
pH 5
Weigh 5.10575g of potassium hydrogen phthalate(C8H5KO4)
Dissolve it in a 100cm3 beaker
Transfer the solution into a 250cm3 volumetric flask and fill it with distilled water
Measure 5cm3 of 2M NaOH and add it to a 100cm3 volumetric flask and fill it with distilled water
Add 100cm3 of 0.1M potassium hydrogen phthalate and 45.2cm3 of 0.1M NaOH into a 250cm3 beaker
pH 6
Weigh 3.40225g of KH2PO4
Dissolve it in a 100cm3 beaker
Transfer the solution into a 250cm3 volumetric flask and fill it with distilled water
Measure 5cm3 of 2M NaOH and add it to a 100cm3 volumetric flask and fill it with distilled water
Add 100cm3 of 0.1M KH2PO4 and 11.2cm3 of 0.1M NaOH
But I realised the phosphate buffers will cause a precipitate. So I tried using a citric acid + sodium citrate buffer. I think the calculation was something like 88.5cm^3 of citric acid + 11.5 cm^3 of sodium citrate (or it could be the other way around). But I don't think that was a buffer solution, cus when I added the slightest amount of deionized water, the pH increased. I got it from https://www.sigmaaldrich.com/life-science/core-bioreagents/biological-buffers/learning-center/buffer-reference-center.html#citric2.
Someone please help this is experiment is basically graded, so I'm really desperate.
edit: I made the copper sulfate using CuSO4.5H2O if that helps

Alas, your problem has no solution, because of the risk of producing $\ce{Cu(OH)2}$ at pH $5$ and $6$. The solubility product of $\ce{Cu(OH)2}$ is $2.2·10^{-20}$. At pH $5$ and $6$, the concentrations of the ion $\ce{OH-}$ are respectively $10^{-9}$ and $10^{-8}$ M. This means that, at pH $5$ and $6$, the maximum molar concentrations of the ion $\ce{Cu^{2+}}$ are $2.2·10^{-2}$ M and $2.2·10^{-4}$ M respectively. So you will never be able to obtain a $0.1$ M solution of $\ce{CuSO4}$ at pH $5$ or $6$.


TLC plate and Rf values I have calculated the Rf values for 6 samples. How do I determine which is really Sriracha ?

Tl;DR - You do not have enough information.

I cannot help but think that this question is not answerable. You have four samples: Huy Fong Sriracha and three "knock-offs". There are six lanes on the plate. All six samples contain potassium sorbate.

If the question was something more like: You have five mysterious hot sauces and one of them is Sriracha... Then the TLC plate makes sense. One of those poorly labeled lanes is sriracha and the others are the mystery sauces. You just need to pick the one that gives the same pattern of spots with the same Rf values.

You don't have that. You have a mismatch between number of lanes and number of samples. You don't know which one is the authentic standard. All of the sauces contain the potassium sorbate (and they should - its a common preservative). You've determined the Rf values and that is all you can do. Well, maybe not all.
I can tell that lane 1 and lane 4 probably contain the same sauce and likewise lanes 2 and 5. Other than that, Jan's comment is correct. The potassium sorbate should always be at the same Rf (and since it is ionic, that likely means at the bottom of the TLC plate).


How can I identify whether a substance is KOH or NaOH? I've bought $\ce{NaOH}$ or $\ce{KOH}$ (I don't remember which one!) for my electrolyzer and I've tested it. Now I want to improve its performance and I was searching for a better electrolyte when I realized I've forgotten what my current electrolyte is! It is white and has crystal shape, but which one is it, $\ce{KOH}$ or $\ce{NaOH}$? If they are both white how can I know?

A flame test should tell you pretty quickly. Make a small wire loop (preferably steel or nichrome), heat in a bunsen burner or propane torch flame until red hot and all the contaminants are burned off, then dip it in your solution and put it in the flame. If sodium is present, you should get a bright yellow flame, and if it is potassium, the flame will be a dim purple (lilac) color. But a small amount of sodium present can mask that color.


EI-MS: M-28 peak in aldehydes I've noticed a phenomenon in the EI-MS spectra of aliphatic aldehydes. Many of them show a significant peak (sometime it's even the base peak) at $M-28$.

To my best understanding, the only thing that can explain this – and is common to all aldehydes – is removal of the carbonyl $\ce{C=O}$ group.
However, I fail to understand the chemistry behind this. How can the $\ce{C=O}$ disconnect, while the aldehyde hydrogen stays – somehow replacing the entire group (forming a new $\ce{C-H}$ bond)? Or am I missing something and this is not actually removal of the $\ce{CO}$?

For example, here is the EI-MS spectrum of (2Z)-3-methylpent-2-en-4-ynal:

Molecule drawing from www.chemspider.com:

This is indeed a quite intriguing question. Thus I am afraid that my answer will be mostly incomplete and quite speculative on some aspects.

First of all,  book and litterature searches lead to find that this M-28 loss for aldehdyes is mostly attributed as a loss of ethene and not CO. This can be traced back to the 1957 article by Gilpin and McLafferty which describes the EI fragmentation of straight chain aliphatic aldehydes. In the article, they demonstrate, through $\ce{^{18}O}$ labelling of the aldehyde, that it is a neutral ethene loss and not a $\ce{CO}$ loss. I did not find a clear proposal for a mechanism leading to this ethene loss, but as it occurs on long straight chain aldehydes, I would tend to favor a mechanism occuring through multiple $\ce{H^{\bullet}}$ migration within a series of isomeric distonic ions following the scheme, taking hexanal as an example:

But of course this is not sufficient in the case of the example that is the initial question, as there is no terminal ethene loss possible. I wondered if it would apply for other aldehydes, and I was not long to find in the NIST database terminal branched aldehydes (such as for example 5-methylhexanal, for which the EI spectrum is available here) that also lead to a strong M-28 peak, which obviously cannot be obtained through the above mechanism: one would expect a M-42 peak which happens to be present, but with a similar intensity to that of the M-28 peak.

And now we enter the speculative part. Someone with access to a high resolution mass spectrometer and a collection of aldehydes could easily identify if the M-28 is loss of $\ce{CO}$ or of ethene. I will assume that in the case of the question, as well as in the case of terminally branched aldehydes it is a loss of $\ce{CO}$, but this should be experimentally checked. (If someone also finds a reference to high resolution spectra showing loss of CO, please edit this answer accordingly.)

In the case of the (2Z)-3-methylpent-2-en-4-ynal, as well as other aklene aldehydes, one could probably explain the loss of CO as a $\ce{H^{-}}$ (hydride) migration from the carbonyl carbon towards an ionized double (or triple bond), followed by a loss of $\ce{CO}$, following the mechanism:

This mechanism leads to the formation of the methyl-methylenecyclopropene ion. According to Shay et.al, the ionized methylenecyclopropene ion is the most stable structure for these types of ions, but it could also produce any of the isomers, so I chose to start with a 1,5 hydride shift, although other hydride transfers could work out fine.

Although satisfying for the initial question, this does not address the case of the branched aliphatic aldehydes, for which no hydride transfer can lead to the formation of a carbonyl group before cleaving the carbonyl - carbon bond. Considering that the final product is in this case an ionized alcane, the mechanism for the rearrangement cannot proceed through a final bond cleavage which would lead to a bond deficient (unsaturated) system. Somehow, the system has to go through a partially bonded state at one stage. The following proposal could work out:

Initially, the system proceeds though standard α bond cleavage which would usually lead to formation of the m/z 29 $\ce{HCO^{+}}$ ion (present in small abundance in the original question as well as in most aldehyde spectra). An ion-neutral intermediate complex is formed in some instances (although eventually the two steps might be not absolutely separated), which can lead to a proton transfer between the $\ce{HCO^{+}}$ ion and the alkyl radical. This transfer is thermochemically favorable and would lead to loss of $\ce{CO}$.

Point in favor of this hypothesis are:

The thermochemistry for final proton transfer is favorable. Not many radicals have thermochemical data, but one takes the propyl radical (Enthalpy of formation for the propyl radical 100 kJ/mol; Ionization energy of propane 10.94 eV; Enthalpy of formation for propane -104.7 kJ/mol; Enthalpy of formation for $\ce{H^{+}}$ 1530 kJ/mole) one can evaluate its proton affinity at 680 kJ/mol. In comparison, proton affinity of $\ce{CO}$ (at C) is of 594 kJ/mol, thus the proton transfer is exothermic by more than 80 kJ/mol. For ketones, α-cleavages lead to the formation of $\ce{RCO^{+}}$ ions, with R an alkyl group, which cannot as easily transfer a proton (proton affinity of the neutral ketenes on the order of 825 kJ/mol) which also explains why similar rearrangements would be specific of aldehydes.
Fragmentation of the ionized alcane should occur and be present in the mass spectrum. Indeed, as shown in the figure below, some ions (notably m/z 71, 43 and 27) are present in the fragmentation of ionized 2-methylpentane and also in that of 5-methylhexanal, and can harldy be explained by α-cleavage of the aldehyde or by a McLafferty rearrangement alone. Nevertheless, this is a rather weak point: loss of water is also typical of aldehyde and leads to ions which could further fragment into these ions also.

A concern with this hypothesis is that a proton transfer to a radical is not a well established process. Indeed, formation of an ionized alcane leads to an ion which cannot be directly represented using standard Lewis notation, which makes writing mechanisms involving this species a bit worrisome.

The second problem is that although ion-induced dipole interactions are not negligible, the binding between $\ce{HCO^{+}}$ and the alkyl radical is likely to be weak compared to other ion neutral complexes. Thus its lifetime might be too short to allow a proton transfer step.

Further work on this subject should, in my opinion, try to focus on, (i) experimental validation of the loss of CO, and ideally, using D-labelled compounds validate the hydrogen transfer hypothesis. MS/MS fragmentation of the M-28 ion would also allow validation that an ionized alcane is indeed formed in the process. Finally, the mechanism should be validated through some theoretical calculations, at least for the formation of an ion neutral complex and for the barrier (and ideally RRKM kinetics) for the proton transfer within the ion neutral complex.


Why total heat of hydrogenation of 1,3-cyclohexadiene is more than that of benzene? I'm told that heat of hydrogenation (HOH) is directly proportional to number of π bonds and inversely proportional to stability. So, is the aromaticity responsible for this?

Also, what is the general approach to the problems like this? Say, I encountered napthalene and some hydrocarbon which contains about 2 rings and have π bonds less than naphthalene, just like stated above – then how shall I decide?

Is there any specific rule like if a compound will be aromatic then it's HOH will be reduced by this amount or something like that?

Yes. The relatively smaller Heat of hydrogenation (HOH) for benzene as compared to that for 1,3-cyclohexadiene is due to the aromaticity of the first. Analyzing the thermochemistry is indeed among the first and perhaps more intuitive ways to present and quantifies aromaticity itself.

A) Cycloexene HOH = -120 kJ/mol

B) 1,4-Cycloexadiene HOH = -240 kJ/mol

C) 1,3-Cyclohexadiene HOH = -232 kJ/mol

D) Benzene HOH = -208 kJ/mol

While for B the HOH is about double as compared to that of cyclohexane and thus is according to the assertion in the question, the conjugation in C does results in a "lower than expected" value.

The 8 kJ/mol difference is the energy of resonance of the two conjugated double bonds.

The situation is even more striking for D as this difference from "an expected value" amounts now to about 152 kJ/mol. This is again a measure of stability and its distinctively high value is a manifestation of  aromaticity.

The HOH of benzene is even less than that of cyclohexadienes. This shouldn't come as surprise though, as for there are not really double bonds.

In other words, while hydrogenation of individual double bond is an exothermic reaction, hydrogenation of benzene to cycloexadienes is an endothermic process.

Once you disrupt aromaticity, then you can think again in terms of individual double bonds (although delocalisation may occur to little extent, as in C). And this fact should answer the question "why HOH of benzene is higher than that of cyclopentene?", too.


Hydrolysis under basic medium The first compound is said to be hydrolyzed more easily than the 2nd compound below under Alkaline medium . How is this possible ?

I thought due to resonance, -NO2 acquires negative charge which faces repulsion from the solvation effect by basic medium. But it will be relatively easier for oxygen in -CH3O( in 2nd compound) to donate its electrons to have it in resonance with benzene. But this contradicts with answer. Where am I going wrong?

Going by the mechanism of hydrolysis under basic medium, the lone pair of hydroxide ion will attack on C=O carbon.

Due to the negative mesomeric (-M/-R) effect and negative inductive effect of the nitro group, the C=O will act as a more electrophilic substrate as compared to the one which is has ether substituent,

(which due to its +M and +I effect will reduce the electrophilic character of the C=O ),
will be easier to hydrolyse.

The effect of solvation on substituents is less likely to tell which compound will hydrolyse easily as it is not the Rate Determining Step of the reaction. So although the observation is not wrong, its still not the deciding factor.


Reduction of α,β-unsaturated nitro compounds Would it be possible to reduce a α,β-unsaturated nitro compound to a saturated amine using for example $\ce{NaBH4}$? I know these kind of compounds will be reduced by hydrogenation using a $\ce{Pt/C}$ catalyst, $\ce{LiAlH4}$ or freshly prepared Raney Nickel, but I'm looking for a reducing agent that is easier to handle than those.

The book from which I'm learning, Clayden et al., does not really go into much detail on reduction of nitro compounds or unsaturated alkanes, other than that hydrogenation is a good approach to both and that the nitro group is one of the most easily reduceable functional groups. But how about a α,β unsaturated compounds then? How would I asses whether some reducing agent is able to reduce both simultaneously?

I'm not aware of a $\ce{NaBH4}$-based reduction of nitroalkenes to saturated amines, such as in:

It is however possible to reduce nitroalkenes to saturated hydroxylamines using $\ce{BH3*THF}$ in the presence of catalytic amounts of $\ce{NaBH4}$. (DOI)

If there's a chance to perform the desired transformation in one step via catalytic hydrogenation at normal pressure: go for it!

The handling of these reactions is nothing to be afraid of.

Typically, I performed hydrogenations in a standard 2- or 3-neck flasks and used ordinary balloons (think childrens' birthday) as gas storage. Advice: Stuff a piece of rubber tube into the inlet of the balloon and secure it with teflon tape.

Either attach this to the glas olive of a valve attached to your flask
or take a plastic syringe, cut off the base, stuff the the plastic cylinder into the rubber tube. Fill the balloon and attach a needle. Hydrogen may now be fed to your reaction mixture through a rubber septum cap.

UPDATE

I still doubt that the intended reduction by $\ce{NaBH4}$ is possible.

There is however another established route from $\alpha,\beta$-unsaturated nitroalkenes to amines which is worth to be mentioned:

Condensation products of aromatic aldehydes and nitroalkenes are reduced by iron in hydrochloric (or acetic) acid to the corresponding arylalkanones.

These ketones can be converted to primary amines by reductive amination in methanol (or ethanol) in the presence of (excess) ammonia and, of course, hydrogen over Raney Nickel or Urushibara Nickel.

In the course of this reaction, the imine intermediate is in situ reduced to the desired primary amine.

The resulting amines are valuable precursors in the synthesis of natural products: Their conversion to imines or amides and subsequent Pictet-Spengler or Bischler-Napieralski cyclisations furnish tetrahydro- and 3,4-dihydroisoquinolines.


Chemical properties of isotopes I've always heard that different isotopes of the same element have exactly the same chemical  properties.

But for example, I've read that some compounds which have hydrogen atoms interchanged by deuterium become odorless (this is one of the arguments for the vibrational theory of olfaction).

  A 2001 study by Haffenden et al. showed humans able to distinguish benzaldehyde from its deuterated version. In addition, tests with animals have shown fish and insects able to distinguish isotopes by smell

So, are there any situations in which  neutrons do matter (besides nuclear decays)?

Addendum 2013/10/28

Another curious effect is that water is slightly blue, due to vibration. But heavy water is not.

  Heavy water is colorless because all of its corresponding vibrational transitions are shifted to lower energy by the increase in isotope mass. For example the H2O band at 760 nm is shifted to approximately 1000 nm in D2O.

(Left: tube if filled with (light) water. Right: empty tube.)

Short answer: nuclear neutrons greatly affect chemistry involving hydrogen.

Longer answer

Hydrogen is the lightest element, and deuterium is just less than twice as heavy. This makes a pretty massive (pun intended) difference to any vibrational modes involving moving hydrogen/deuterium. This means that, for example, the deuterated equivalents of weak acids are less acidic, and any reaction involving the loss of a proton will be significantly slower with the deuterated version.

Since helium is pretty unreactive, and by the time you're at lithium an extra neutron is only increasing the mass by about 14%, the effects on reaction energetics are much, much smaller for anything else.

I'd suggest looking up the "Kinetic Isotope Effect" for more information and more complex effects the mass change on substituting isotopes can cause to chemical activity.

As Eric Brown said in his answer, different isotopes may also have very different spectroscopic properties if the spectroscopic technique depends either on vibrational or nuclear properties.

As a minor curiosity, I'm given to understand this also means that heavy water is mildly hazardous to your health, since deuterated water and deuterated analogues of various vital chemicals have a nasty tendency to react mostly irreversibly or very slowly with vital enzymes that are "tuned" for hydrogen.


Stripping zinc electroplating with phosphoric acid - hydrogen danger? I'll start by saying that I'm not a chemist, more of an artist. I'm stripping zinc electroplating from some regular hex bolts, nuts, and washers. I'm using a product called Ospho. It's basically phosphoric acid with a surfactant and a couple other additions, from what I understand.

I've read that zinc and phosphoric acid react to produce zinc phosphate and hydrogen gas.

My unknown is whether this hydrogen gas is emitted in great enough quantities to be dangerous / explosive when this reaction is carried out indoors. I don't know how much zinc is being reacted, but electroplating leaves an extremely thin coating and I'm talking about small quantities of fasteners.

Yes, mineral acids like $\ce{H3PO4}$ react with electropositive metals like zinc to produce dihydrogen gas ($\ce{H2}$).

The quantities generated of course depend on how much metal is being dissolved that way. About $65\ \mathrm{g}$ of zinc would produce about $44.8\ \mathrm{L}$ of hydrogen (in Standard conditions).

This may seem like a lot but your nuts and bolts are covered with only a very thin layer of zinc.

Hobby chemists dissolve metals in mineral acids all the time, with minimal danger. Just avoid open flames (use an electrical hot plate) and make sure your work place is well ventilated and you should be fine. Consult any safety advice supplied with the Ortho.


Stereochemical dichloride treatment In this question there is no inversion in A to B, but there is one in C to D. I feel like this must be due to the lone pairs on the $\ce{S}$ (in A) being able to open the ring, whereas this isn't possible in C to D so this can only occur via an $\mathrm{S_N2}$ reaction, but I just can't seem to make the mechanisms work for either.

Because of the absence of a lone pair, C undergoes usual $\mathrm{S_N2}$ transformation under the conditions. However, the availability of a lone pair in A undergoes the nucleophilic substitution of cyanide ion via neighboring-group participation (see above comments by Waylander and J. Deans) by the sulfur atom under the conditions (Ref.1). A very reliable mechanism is given in Ref.1 for the substitution of azide ion (work the same way for cyanide ion):

Reference:

D. D. Díaz, A. Converso, K. B. Sharpless, M. G. Finn, “2,6-Dichloro-9-thiabicyclo[3.3.1]nonane: Multigram Display of Azide and Cyanide Components on a Versatile Scaffold,” Molecules 2006, 11(4), 212–218 (doi: 10.3390/11040212).


Explicit differential equation based model of protein folding I am an applied mathematician interested in the dynamics of potential systems - i.e., systems with multiple unique energy minima. One of the best examples of such systems are protein folding potentials. From sources like This one I know that the potential,
$$
V(x)
$$ exists and is a function of the positions of each residue in space and their interactions with each other. Since such a function exists, there should (mathematically) be a set of ordinary differential equations that capture the dynamics described by the potential.

However, although I have seen definitions of $V(x)$ in the literature, for the life of me I can't figure out exactly what the set of differential equations associated with this system is - i.e., the system is usually written out in a generalized form that applies to all proteins, but the exact information required to build such a potential for one such protein is unclear.

Does anyone have an example of of a particular $V(x)$ (for a particular, short, set of amino acids, say) where all the parameters are known and the system can be solved numerically as a set of ordinary differential equations?

What you are looking for is a force field: http://en.wikipedia.org/wiki/Force_field_(chemistry)

Standard force fields for proteins include CHARRM and AMBER. These have relatively simple, well-defined expressions for defining the various types of interactions between bonded and non-bonded atoms. Parameters are available for all standard atom types, and are fit to experimental and/or electronic structure calculations. These force fields are considered "classical" in that they do not explicitly treat electronic interactions, but rather try to capture these interactions in a coarse-grained way. Typically, there are expressions for describing the following interactions:

bonded interactions (bond stretching)
angle interactions (bond bending)
torsion interactions (dihedral bending)
12-6 Lennard-Jones interactions (non-bonded, dispersion forces)
Short-range charge-charge interactions (Coulombic interactions between charged or partially charged atoms)
Long-range charge-charge interactions (see Ewald summation method for more details)

Note that in these classical models, partial charges are generally statically assigned to each atom. The functional form of these different terms varies depending on the force field and sometimes implementation. These interactions are typically pairwise additive, so to get the total potential energy of the system you simply add together all the various interactions between the appropriate pairs of atoms.

There are certainly more rigorous models that attempt to capture other electronic structure affects, however for large molecules such as proteins, relatively simple force fields are generally applied.

Please note that finding the global minimum in potential energy of a protein is not a trivial problem. People have been working on this for years, it's extremely difficult to develop algorithms that do not get trapped in local minima.


Why does the energy required to break a bond decrease as the number of bonds increase? I noticed that the energy to break a single C bond is $348\ \mathrm{kJ}$, to break a double C bond would be $612\ \mathrm{kJ}$, and to break a triple C bond would be $837\ \mathrm{kJ}$. How come the energy required per bond breakage decreases as the number of bonds decrease?
It seems that for double bonds, the energy required to break a bond is $(612/2) = 306\ \mathrm{kJ}$. Why is this so?

Thank you.

You are considering all the bonds equivalent and that's where you are making mistake.The energy required to break second bond will depend upon several factors such as repulsion, configuration, also note that there are different types of bond formed (sigma and pi bonds) and we know that energy required  to break a sigma bond is greater than that of pi bond.


How is the association/ dissociation constant derived? I came across the following formula to find the association of a solute in particular solvent. Suppose the concentration of the solute given in water (or in some other solvent in which it does not associates or dissociates) is $C_1$ and that in some other solvent ( like benzene ) is $C_2$ then the association factor $n$ is determined by $$K=\frac{C_1}{{C_2}^\frac{1}{n}},$$
where $K$ remains constant.

On what concept is this formula based on? Can we derive it? Is it an empirical formula ?

This looks a lot like a normal equilibrium constant. There is very little background information provided, but I assume that this formula applies to an experimental procedure in which the same total amount of the solute is dissolved into each solvent and the concentrations of free (dissociated) solute are measured. In this case that $C_1$ and $C_2$ should be the other way round. In the following, I assume that $C_1$ is the measured concentration in benzene and $C_2$ is the measured concentration in water.

In order for the formula to work we also need to assume that in benzene (or whatever the other solvent is), the solute associates almost completely.

With these assumptions the derivation of the formula is practically identical to the standard derivation of the formula for an equilibrium constant.

Let's call a single molecule of the solute A and a complex of $n$ associated molecules B.

The association reaction then is

$$n\ce{A <=>[k_1][k_2] B},$$

where $k_1$ and $k_2$ are the rates of association and dissociation reaction respectively.

The speed of the dissociation reaction is proportional to the concentration of B, so we can write $k_2=b\ce{[B]}$.

The association reaction requires $n$ units of A coming together and from probability theory we can tell that the probability of that happening is proportional to the concentration of A multiplied by itself $n$ times, so we can write $k_1=a\ce{[A]}^n$.

At equilibrium the rates of association and dissociation reactions are equal, so we can write:

$$k_1=k_2 \Rightarrow a\ce{[A]}^n=b\ce{[B]} \Rightarrow \frac{b}{a}=\frac{\ce{[A]}^n}{\ce{[B]}}$$

Now we raise both sides to the power of $\frac{1}{n}$ in order to get

$$\left(\frac{b}{a}\right)^{\frac1n}=\frac{\ce{[A]}}{\ce{[B]}^{\frac1n}}$$

Now, $\ce{[A]}$ is the concantration of dissociated solvent in benzene, so we can substitute $\ce{[A]}=C_1$. On the other hand $C_2$ is the concentration of the solute in water, which is the same as the concentration of total concentration of both associated and dissociated forms in benzene, so we can write $C_2=\ce{[A]}+n\ce{[B]}\approx n\ce{[B]}$ if $\ce{[A]}$ is much smaller than $\ce{[B]}$ as we assumed in the beginning.

By substituting we can write

$$\left(\frac{b}{a}\right)^{\frac1n}=\frac{C_1}{\left(\frac1n C_2\right)^{\frac1n}} \Leftrightarrow \left(\frac{b}{na}\right)^{\frac1n}=\frac{C_1}{C_2^{\frac1n}}$$

Because $n$ is independent of $C_1$ and $C_2$, we can substitute $K=\left(\frac{b}{na}\right)^{\frac1n}$ and we get

$$K=\frac{C_1}{C_2^{\frac1n}}.$$


Reactions involving trimethylaluminium Al2Me6 I need to write balanced equations describing the following reactions:

one mole of $\ce{Al2Me6}$ with two moles of water
excess of $\ce{Al2Me6}$ with silicon dioxide
excess of $\ce{Al2Me6}$ with tin(IV) chloride

My answer so far:

a) $\ce{2H2O + Al2Me6 -> Al2Me4(H2O)2}$?
b) $\ce{SiO2 + Al2Me6 -> SiMe4 + Al2Me2O2}$?
c) $\ce{SnCl4 + Al2Me6 -> SnMe4 + Al2Me2Cl4}$?

For a): Many organometallic compounds are easily hydrolyzed to the respective hydrocarbons and metal (hydr)oxides. In this case, the products are $\ce{CH4}$ and more likely $\ce{Al2O3}$, because the formation of $\ce{Al(OH)3}$ requires more moles of water.

Reactions b) and c) should be transmetallations where the methyl groups are transferred from aluminium to the other metal. The excess of $\ce{Al2Me6}$ helps to shift the equilibrium of the reaction to the product side. $\ce{SiMe4}$ and $\ce{SnMe4}$ would be the correct products which you have identified, but aluminium is also converted into its respective oxide/chloride.


What is the benefit of high surface area electrodes? Some electrolytic apparatus is equipped with high surface area electrodes. The number of reduced/oxidized atoms is proportional to the total charge flowing through the cell. Does increasing surface area somehow enable use of greater currents?

A current is simply the rate of flow charge (and hence the rate of flow of electrons); the more electrons flowing, the higher the current.

Anions migrate across the potential gradient between electrodes, carrying the electrons to the anode and the anions are then oxidised.  The oxidised species (often uncharged) must then diffuse across a concentration gradient to move away from the electrode (via Fick's laws).  Until the oxidised species moves away from the electrode surface, they are blocking the surface for more anions to migrate in, and hence the current is eventually limited when the rate of migration of anions is equal to the rate of diffusion of the oxidised species.  This exchange happens at a fixed rate per unit area at the surface of the electrode.

When we use a larger surface area electrode, there is more area for this exchange to happen, therefore more species involved in the exchange, therefore more electrons can be deposited in a given timeframe, hence permitting a higher current to flow.


Bibenzimidazole nomenclature: Indicated H and locants I am trying to understand how the locants work for the following molecule (locants added by me).

I added the locants following IUPAC Blue Book rules from 2013, P-14.4. Basically, first numbering heteroatoms in the ring, assigning the highest locant to the one with an indicated hydrogen, and then number clockwise around the ring (red locants).

I assumed the same would go for the second ring, so the name would be:

$1^1H,2^1H-1^2,2^7-\text{bibenzoimidazole}$

However, Chemdraw says the name is:

$1H,3'H-2,4'-\text{bibenzo[d]imidazole}$

I get the differences between the prime and composite locants (P-14.3.1), however it shows a numbering more like:

Is Chemdraw wrong or am I missing something here? If Chemdraw is right then I'd assume that for some reason the ring assembly has preference for a lower locant, but I can't find the rule that states this in the Blue Book.

Your understanding of the numbering of the individual cyclic systems is correct. If we consider the following arbitrarily simplified separate compounds, the numbering is in accordance with your expectations; low locants are given first to the indicated hydrogen and then to the substituent according to the usual rule P-14.4.

In ring assemblies of two identical cyclic systems that are joined with a single bond, however, low locants are given first to the junction according to Rule P-28.2.1.

  P-28.2.1 Ring assemblies with a single bond junction

  (…)

  Each cyclic system is numbered in the traditional way, one with unprimed locants, the other with primed locants. Lowest possible locants must be used to denote the positions of attachment. These locants must be cited in preferred IUPAC names; they can be omitted in general nomenclature when no ambiguity would result.
  (…)

After that, indicated hydrogen is named and numbered according to P-28.2.3.

Therefore, the systematic name is 1H,3'H-2,4'-bibenzimidazole and not 1H,1'H-2,7'-bibenzimidazole since the locants 2,4' for the junction are lower than 2,7'.


Kc(equilibrium constant) and Kp Its given in the book that at 12.18kelvin Kc=Kp. How is that equated from the equation Kp= Kc(RT)^∆ng. I am unable to sort it out, please guide me in to the topic....

From the ideal gas equation, we get
$$PV=nRT \implies P=\frac{n}{V}RT=cRT$$
where $c$ represents the concentration.

Consider an arbitrary reaction
$$n_AA\rightarrow n_BB$$

$$K_c=\frac{c_B^{n_B}}{c_A^{n_A}}$$
$$K_p=\frac{P_B^{n_B}}{P_A^{n_A}}=\frac{(c_BRT)^{n_B}}{(c_ART)^{n_A}}=K_c \times (RT)^{\Delta n}$$

Most probably, the reaction you are talking about has $\Delta n=0$.


How to synthesize 6-methylhept-1-ene What methods are there to produce this compound? What are its properties as a component of automobile petroleum? How does it impact on detonation resistance of automobile petroleum?

I think that the title compound might be produced from catalytic cracking of petroleum. But I don't have a clue of what methods are there to produce exactly this specific compound. I also think that it helps to increase the detonation resistance of automobile petroleum but I can't explain why. So I'm seeking for the explanation. Surprisingly, there's nothing on the Internet, so this site seems to be the only place where I can possibly find some information.

There are a few widespread misconceptions about this topic. The original question, edits, and comments give the impression that such misconceptions may also contribue to the confusion in this case. Therefore, we should like to clear up some of the misconceptions first.

High-octane fuel does not contain a high concentration of octane.

The composition of gasoline varies widely; anyway, typical gasoline contains less than 1 % octane.

The octane number of gasoline does not describe the composition of gasoline. It refers to the measurement of the knock resistance of gasoline, which is usually expressed by their research octane number (RON) or motor octane number (MON). Both parameters are determined in single-cylinder four-stroke test engines with an adjustable compression ratio according to standardized test procedures, e.g. EN 25164 or ASTM D 2699 for RON, and EN 25163 or ASTM D 2700 for MON. During the measurement, the compression ratio is raised until a certain standardized level of knocking is reached. The corresponding octane number is determined by finding a reference mixture of heptane and “octane” that gives the same knock level. The octane number is the fraction of “octane” (in %) in the “octane”-heptane mixture. For example, gasoline with an octane number of 95 has the same knocking performance as a mixture of 95 % “octane” and 5 % heptane; however, this does not mean that the gasoline contains 95 % “octane”.

Nevertheless, the wrong idea that gasoline would contain a high concentration of octane is widespread. For example, in the first relevant Google hit, Ritter, S. Chem. Eng. News 2005, 83 (8), 37:

  Octanes are an important component of gasoline because they help provide smoother combustion in the car’s cylinders and prevent knocking.

Also remarkable is the fact that many textbooks and lectures select the combustion reaction of octane of all the possible combustion reactions of gasoline to show the principle of combustion:

  $$\ce{2C8H18 + 25 O2 -> 16 CO2 + 18 H2O}$$

The reference substance for the octane number is not octane.

The octane number refers to a mixture of heptane and “octane”. However, the “octane” used for this mixture is not the unbranched alkane octane. It is actually a branched isomer of octane with the traditional name “isooctane”. Thus, gasoline with an octane number of 95 actually has the same knocking performance as a mixture of 95 % “isooctane” and 5 % heptane; however, this also does not mean that the gasoline contains 95 % “isooctane”.

The octane number of pure octane is actually very low. Its knock resistance is worse than the performance of pure heptane, which has a octane number of zero by definition.

The “isooctane” that is used as a reference substance for the octane number is not isooctane.

The name isooctane was in accordance with older nomenclature rules. The corresponding systematic name according to current IUPAC recommendations is 2-methylheptane. However, the “isooctane” that is used for the determination of the octane number is not isooctane, i.e. it is not 2-methylheptane. This “isooctane” actually is 2,2,4-trimethylpentane, which is another isomer of octane. Thus, gasoline with an octane number of 95 actually has the same knocking performance as a mixture of 95 % 2,2,4-trimethylpentane and 5 % heptane; and once again, this does not mean that the gasoline contains 95 % 2,2,4-trimethylpentane.

By way of comparison, the octane number of pure isooctane (2-methylheptane) is only about 23.

Irrespective of which compound is really meant in the original question (octane, 2-methylheptane, or 2,2,4-trimethylpentane; the question and comments also mention the similar alkenes 6-methylhept-1-ene and 2,4,4-trimethylpent-1-ene, which might be a mistake), such components of gasoline are usually not selectively synthesized in a pure form.

Gasoline consists of hundreds of different compounds, and the exact composition can vary widely. It is produced by blending of different product streams of oil refining. Today, only a small part (so-called straight-run gasoline) is directly obtained by distillation of crude oil. Other important product streams are thermally cracked gasoline, catalytically cracked gasoline, catalytic reformate (so-called platformate), isomerate, alkylate, polymer gasoline, and oxygenates.