What is the molecular formula of the nitrate of a metal Q that forms a carbonate of the form QCO3? A metal $\ce{Q}$ forms a carbonate and a nitrate. The molecular formula of the carbonate of $\ce{Q}$ is $\ce{QCO3}$.  What is the molecular formula of the nitrate of $\ce{Q}$?

The answer says it is $\ce{Q(NO2)3}$ which I think is incorrect; I think the answer should be $\ce{Q(NO3)2}$. Am I right?

The carbonate anion is $\ce{CO3^{2-}}$. This tells us that the metal cation is doubly positively charged:
$$ \ce{Q^{2+} + CO3^{2-} -> QCO3} $$

The other two anions mentioned in your question are

the nitrite anion $\ce{NO2^-}$
the nitrate anion $\ce{NO3^-}$

Since the question you're concerned with asks for the formula of the nitrate of $\ce{Q}$, the correct answer would be $\ce{Q(NO3)2}$.

However, the compound $\ce{Q(NO2)2}$ is, at least from a charge perspective, also correct. But that is the nitrite of $\ce{Q}$.


What gas law is involved when your ears "pop" when you are on a plane high in the air? When you are up in the air, the air pressure decreases and the altitude increases. What gas law is involved in the situation wherein your ears "pop" because of the difference in pressure.

The temperature and the volume of the inner ear are constant. When your ears pop during descent, air from the cabin goes into the ear, increasing the pressure. The law is the following:
$$n / P = const$$
You can derive this from the ideal gas law. It has no special name.


What is IUPAC name of this compound? Should the benzene group be taken as a substituent, so as to select the longest C atom chain?
In what cases is it taken as as a substituent?

The correct answer is

  (d) none of these

In the usual substitutive IUPAC nomenclature, simple halogen compounds are always expressed by prefixes (‘bromo’, ‘chloro’, etc.). Hence, the senior parent structure of such compounds is not determined by a principal characteristic group (a characteristic group chosen for citation at the end of a name by means of a suffix or a class name, or implied by a trivial name).

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the ring or the chain can be the senior parent structure; however, for the preferred IUPAC name, the ring is always selected as the senior parent structure.

  P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)

  Two methods are recognized to name systems composed of rings and chains (exclusive of linear phanes).

  (1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.

  (2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.

  (…) For selection of a preferred IUPAC name, see P-52.2.8.

  P-52.2.8 Selection between a ring and a chain as parent hydride

  Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

Therefore, the benzene ring is selected as the senior parent structure for the preferred IUPAC name of the compound given in the question.

According to P-29.1.2, The chloromethyl ($\ce{-CH2Cl}$) substituent is a compound substituent group, i.e. a substituent group consisting of two parts, a simple substituent group (the parent substituent group, here: methyl) to which is attached one or more simple substituent groups (here: chloro).

According to P-16.5.1.1, parentheses are used around such compound prefixes. Therefore, the preferred IUPAC name for the compound that is given in the question is (chloromethyl)benzene.


How can I separate sodium ions from brine or molten NaCl? Is possible to separate sodium ions from an aqueous solution of NaCl? If so, how do I go about it?

You can't simply produce isolated sodium ions, you also need to have an equivalent amount of counter ions of the opposite charge. So, in this case, you are going to get sodium chloride unless you introduce some other negatively charged species.

So you could just evaporate water which will give an increasingly concentrated solution of sodium chloride and, eventually, solid sodium chloride (which will contain all your sodium ions alongside an equal number of chloride ions.

This sort of separation is commonly used for a different reason: to produce pure water. Sea water is too salty to drink but there is far more of it than fresh water. So distillation (retaining the fresh water) or reverse osmosis (which produces more concentrates salt on one side and fresh water on the other). These techniques produce both fresh water and more concentrated salty water (which is one way of getting more sodium ions in your product).


How to determine which gas would have the lowest pressure? If the volume of each sample of gas was reduced to one-tenth of its original size and the temperature remained the same, which gas would have the lower pressure. I'm getting confused because my options are $\ce{C6H4}$, $\ce{C3H4}$, $\ce{N2}$, $\ce{O2}$, and $\ce{Cl2}$. Our teacher told us to pick the polar compound but the problem is all of them are non polar. Would the answer be $\ce{C3H4}$? I have a feeling I might be drawing the lewis structure wrong because I'm positive the other four are non polar.

Not all of them are polar. The gases  $\rm N_2$, $\rm O_2$,  $\rm Cl_2$ are not polar.

The $\rm HC\equiv C-CH_3$ is polar, but $\rm H_2C=C=CH_2$ should not be and $\rm C_6H_4$ is also polar, if it is right chemical formula you wrote. But I'm not sure which is more polar, but I will say that it is $\rm C_6H_4$

.

In more polar compound there will more interaction between molecules and that is why pressure will be smaller, simply because they "want" to be liquid form - not in gas.


Dehydration of primary alcohol with a cyclobutadiene substituent Note: Primary alcohol will follow E2 mechanism, therefore, removal of protonated water and proton (RDS) shall proceed in a single step in the above mechanism.
My questions are:
1- According to wikipedia "Although the parent compound has not been isolated, some substituted derivatives are robust and a single molecule of cyclobutadiene is quite stable."
As cyclobutadiene is not a stable compound, Is the primary alcohol (reactant) a stable compound?
2- Is the ring expansion step and the final product (benzene) correct?

You have misinterpreted what Wikipedia has said. I'd put whole quote here:

Cyclobutadiene is an organic compound with the formula $\ce{C4H4}$. It is very reactive owing to its tendency to dimerize. Although the parent compound has not been isolated, some substituted derivatives are robust and a single molecule of cyclobutadiene is quite stable. Since the compound degrades by a bimolecular process, the species can be observed by matrix isolation techniques at temperatures below $\pu{35 K}$. It is thought to adopt a rectangular structure.

These somewhat stable cyclobutadiene derivatives are not just derivatives with a simple alkyl substituent. They are with bulky substituents, e.g., tetrakis(tert-butyl)cyclobutadiene. Therefore it is safe to assume that given compound (2-cyclobutadienylethanol) does not exist. However, as Aniruddha Deb has said, we can treat it like it is existing only for this theoretical question (we don't have other choice):
Since it is a primary alcohol it can't undergo usual $\mathrm{E1}$ elimination with an acid. The $\mathrm{E1}$ pathway through formation of a primary carbocation is not the most likely pathway here, because primary carbocations are extremely unstable. Therefore, more likely, this reaction undergoes through an $\mathrm{E2}$ mechanism where the transition state should be lower in energy. Accordingly the protonation of the alcohol group to give the good leaving group $\ce{-OH2+}$:

And then, a weak base (which could be either $\ce{H2O}$ or $\ce{H2PO4-}$, or even another molecule of the alcohol) could abstract $\beta-\ce{H}$ from nearby $\beta$-$\ce{CH2}$ group, leading to give the conjugated alkene with cyclobutadienyl group (vinylcyclobutadiene):

We really don't know how stable vinylcyclobutadiene is, but it should be more stable than original alcohol since it has additional resonance contribution with external vinyl group. This compound can undergoes either Diels-Alder or Cope rearrangement to give a bicyclo[2.2.0]hex-1,5-diene derivative. This derivative should be unstable due to severe strain, and expected to rearrange to Dewar benzene derivative, which then rearrange to benzene upon heating (Ref.1).
Reference:

L. Watts, J. D. Fitzpatrick, R. Pettit, "Cyclobutadiene," J. Am. Chem. Soc. 1965, 87(14), 3253–3254 (https://doi.org/10.1021/ja01092a049).


Why does microwave heat up things so much more quickly than visible light? I wonder why microwave has more "energy" to heat things up than visible light does. Microwave has a much longer wavelength, therefore should supposedly has less energy than visible light.

But does wavelength and frequency really have to do with how fast a certain type of electromagnetic radiations give an object heat, or it is something to do with how a microwave oven is built?

Microwaves don't have more energy, they just resonate at the frequency that causes molecular bonds to rotate. This specifically applies to dielectric molecules, molecules like water that have electric dipoles.
From Wikipedia:

Water, fat, and other substances in the food absorb energy from the microwaves in a process called dielectric heating. Many molecules (such as those of water) are electric dipoles, meaning that they have a partial positive charge at one end and a partial negative charge at the other, and therefore rotate as they try to align themselves with the alternating electric field of the microwaves. Rotating molecules hit other molecules and put them into motion, thus dispersing energy. This energy, when dispersed as molecular vibration in solids and liquids (i.e., as both potential energy and kinetic energy of atoms), is heat. Sometimes, microwave heating is explained as a resonance of water molecules, but this is incorrect such resonances occur only at above 1 terahertz

For reference, energy is related to wavelength by the formula:
$$E=\frac{hc}{\lambda}$$
and
$$\lambda=\frac{c}{f}$$
where

$\lambda$ is wavelength
$c$ is the speed of light
$f$ is frequency
$h$ is Plank's constant
$E$ is energy

You can see by the first equation that as $\lambda$ gets smaller $E$ gets larger.


Are all (pure) alkanes, alkenes and alkynes - tasteless, odourless and colourless? non-pure alkanes have a taste, but

Are all pure hydrocarbons, tasteless, colourless and odorless?
Are there hydrocarbons that do have a taste, colour, odor?

If so, what makes these pure hydrocarbons different, even though they are only made of carbon and hydrogen?

Why would a mixture of two odourless alkanes have an odour?

As already pointed out by Greg E and Uncle Al, numerous hydrocarbons without any (heteroatom-based) functional groups do show characteristic smells.

Naphthalene is always remembered for the typical smell of grandma's mothballs.

More pleasant smells are found in the group of monoterpenes, i.e. in those molecules with the sum formula $\ce{C10H16}$ that are derived from the combination of two isoprene units.

These volatile compounds were/are used as fragrances for perfumes. Typical examples are myrcene (1), $\alpha$-terpinene (2), limonene (3), $\alpha$-pinene (4), $\beta$-pinene (5), $\alpha$-thujene (6), sabinene (7), and camphene (8).

As far as the colour of hydrocarbons is concerned:

A sufficiently conjugated system is far from being colourless, take lycopene (9), the deep red coloured dye in tomatoes as an example.


Why don't 1,2-substitued alkenes polymerize as much as mono or disubsituted alkenes? Ethylene, propylene, PVC, acrylates. All of these monomers that are industrially produced seem to only have substituents on one of the carbons, rather than both. Why don't we usually see 1,2-substituted alkenes undergoing polymerization? Sterics? Electron cloud interference?

It is true that we do not see the products made by the polymerization of 1,2-substituted alkenes or 2-olefins (even natural rubber does not fall in this category). Although a number of 2-olefins such as 2-butene in large amounts is produced by petroleum industry (naphtha cracking), they have not been utilized as monomers to obtain commercial polymers in daily life because of their low reactivity toward commonly used polymerization catalysts, probably due to steric effects. Perhaps, if 2-olefins are reactive as much as ethylene towards polymerization to give high molecular weight polymers, there would be a large number of new polymer materials in daily use.
Nonetheless, a few studies on the polymerization of 2-olefins by novel transition metal complexes, specifically using trans-2-butene as the monomer, have been reported (Ref. 1-3). According to all these three references, the monomer-isomerization has been occurred during polymerization process, indicating the steric effect. For instance, both cis- and trans-2-butene give high molecular weight polymers with a recurring unit of the corresponding 1-olefins (after monomer-isomerization by metal catalyst prior to the polymerization), which consists of $\ce{–CH2–CH2–CH(CH3)\!–}$ units instead of $\ce{–CH(CH3)–CH(CH3)\!–}$ units (Ref.1 and 3). Thus, it is unlike that highly condensed polymer can be produced from 2-2-olefins until the discovery of a novel metal catalyst, which prohibit the in situ monomer-isomerization.
References:

Mark D. Leatherman, Maurice Brookhart, "Ni(II)-Catalyzed Polymerization of trans-2-Butene," Macromolecules 2001, 34(9), 2748–2750 (DOI: https://doi.org/10.1021/ma002142j).
Mark D. Leatherman, Maurice Brookhart, "(E)−(Z) Selectivity in the Polymerization of 2-Butene Promoted by Ni(II) Brookhart-Type Catalysts," Macromolecules 2005, 38(6), 2072–2075 (DOI: https://doi.org/10.1021/ma048373y).
Kiyoshi Endo, Yoshiyuki Kondo, "Polymerization of trans-2-Butene with ($\alpha$-Diimine)Ni(II) Complex in Combination with $\ce{Et2AlCl}$," Polymer Journal 2006, 38(9), 1160–1164 (DOI: https://doi.org/10.1295/polymj.PJ2006058).


Relation between size of ion and hydration energy Suppose I take 1 tablespoon of an ionic crystal and put it in excess water. We know that shall the hydration energy of the crystal be more than the lattice energy, the crystal will separate in ions and water molecules will surround each ion, making it soluble.
I wonder if the relation is direct or inverse in:
i) Anion's size and hydration energy
ii)Anion's size and lattice energy
iii) Cation's size and lattice energy
iv) Cation's size and hydration energy

I think the Cation's size is directly proportional to the hydration energy, since the bigger the atom, more easily shall it form the bond with the water molecules. The Anion's size should be inversely proportional to the hydration energy as the bigger the atom, more difficult would it be to form a strong bond between the water molecules. Similar reason goes for the other two.

PS: I'm new in chemistry so please try to explain a little.

We can look at solubility as a competition between lattice energy and hydration.

$\Delta H_\text{solution} = \Delta H_\text{hydration} + \Delta H_\text{lattice}$

If the enthalpy of solution is exothermic or slightly endothermic, the compound will be soluble (we are ignoring entropy here).

If we take a simplified view of both lattice energy and enthalpy of hydration, we can see that they are both generally proportional to both the size of the ions as well as the charges, that is:

$$E \propto\frac{|q_1q_2|}{r} $$

Where $q_1$ and $q_2$ are the charges on the species in question and $r$ is the inter-nuclear distance.  So, ions with higher charges are more attracted to each other AND form strong ion-dipole interactions with water.  At the same time, larger ions experience weaker attractions to each other AND form weaker ion-dipole interactions with water.

It is important to remember that this simplified analysis does not take lattice structure and crystal packing into account at all.

The differences in solubility between different ionic compounds follows two general trends.

Mismatch in Ion Size
In general, when there is a difference in the size of the ions in a compound, the compound is more soluble.  This probably arises from poorer packing in the solid state.  For example at room temperature, $\ce{MgSO4}$ has a solubility of 1.16 g/100mL while $\ce{BaSO4}$ has a solubility of $2.45*10^{-4}$ g/100mL$^{[1]}$.  The barium ($159$ pm) and sulfate ($242$ pm) ions are much closer in size than magnesium ($86$ pm) and sulfate$^{[2]}$.
Covalent Character
In general, the more covalent character present in an ionic compound, the lower it's solubility.  We can estimate covalent character by looking at the difference in electronegativities of the ions in the compound.  The larger the difference in electronegativities, the less covalent character in the compound.

For example, $\ce{AgF}$ has

$\Delta \mathrm{EN} = (4.0 - 1.93) = 2.07$

Solubility = $\dfrac{180}{100} \mathrm{~g~mL^{-1}}$

While $\ce{AgCl}$ has

$\Delta \mathrm{EN}= (3.5 - 1.93) = 1.57$

Solubility = $\dfrac{0.193}{100}\mathrm{~g~mL^{-1}}$

In summary, the solubility of ionic compounds is greater when there is a significant size difference between the ions and when there is a greater difference in their electronegativities.

Footnotes

[1] Tro et al, Chemistry, A Molecular Approach, 2nd Canadian Edition

[2] https://en.wikipedia.org/wiki/Ionic_radius


Why does electrolysis of aqueous concentrated sodium bromide produce bromine at the anode? It is often stated that electrolysis of any aqueous concentrated halogen compound will produce the halogen at the anode, despite it having a more postitive standard electrode potential than the oxidation of hydroxide, while when it is dilute oxygen will be produced, but what is the reason for this?

Why does the principle of concentrated solutions during electrolysis only apply for halogens and not any other element. For example, why isn’t sodium produced at the cathode when concentrated sodium chloride is electrolysed?

There is important to know the actual potential depends on actual activities of reagents ( approximately concentrations for dilute solutions).

Additionally, the electrode potential is the thermodynamic quantity. If some net reaction kinetic is involved, other potentials affects the actual potential at the solution border, like the diffusion potential and dielectric layer potential.

We consider reactions
$$\begin{align}
\ce{Br2 + 2e- &<=> 2 Br-}\\
\ce{O2 + 2 H2O + 4e- &<=> 4 OH-}\\
\end{align}$$

The initial bromine activity is very low.

Activity/concentration ratio of ions typically decreases with concentration, as seen in the Debye-Huckel equation,
but for high salt concentrations it often jumps up the roof.

Therefore, activity of bromides is very high and the actual potential gets low.

$$E_{c_{\ce{Br2}}/c_{\ce{Br-}}}=E^{\circ}_{c_{\ce{Br2}}/c_{\ce{Br-}}} + 0.059 \log \left( \frac{a_{\ce{Br2}}}{a_{\ce{Br-}}}\right)$$

OTOH,
The potential for hydroxide oxidation  is pushed up for neutral solutions to the middle between standard redox potentials for acidic and alkaline solutions $$(+0.401 + +1.229)/2=+0.815$$

$$E_{c_{\ce{O2}}/c_{\ce{OH-}}}=E^{\circ}_{c_{\ce{O2}}/c_{\ce{OH-}}} + 0.059\log \left( \frac{a_{\ce{O2}}}{a_{\ce{OH-}}}\right)$$

Very important thing is the oxygen overpotential. Its value is listed in range +0.56 V on Ni to +1.02 V on Au, in contrary to the maximal value for chlorine ( on graphite)  +0.12 V.

For $\ce{NaCl}$ solutions, difference of standard potentials is too high to produce metallic sodium, unless one uses mercury as the cathode, where happens the hydrogen, related to its reaction kinetics. This is used for production of sodium hydroxide, where sodium amalgam reacts with water.


How are atomic masses determined? The atomic mass (ma) is the mass of 1 atom of an element.
Elements have isotopes.
Therefore, there must be an atomic mass (ma) associated with each isotope of an element.
How do we know the atomic masses of the isotopes of an element?
https://www.ciaaw.org/atomic-masses.htm

The link above shows atomic mass (ma) data.
How was this data generated?

Also, I have come across the atomic mass constant (mu). It has a value of 1.6605402(10) x 10^-27.
How do I obtain this value? i.e. How was this number derived?
Thank you.

The mass of a particular isotope is determined by mass spectrometry. Hopefully you know this technique. In order to obtain a high degree of precision, the beam must go through several mass spectrometers coupled in series, some of which are quadrupoles or octupoles. It would take hours to explain the developments of this technique.


There's an absolute zero, is there an "absolute hot"? This might be more of a physics question, but is there a ceiling on how hot things can get?

What happens at this temperature?

In the actual theories of physics the highest temperature which has a physical meaning is the Planck's temperature.

$$T_\mathrm{P} = \frac{m_\mathrm{P} c^2}{k} = \sqrt{\frac{\hslash c^5}{G k^2}} \approx \pu{1.4e32 K}$$

For the moment no theory predict higher temperature because of the limit of our theories.

There is a Wikipedia article about absolute hot with some references. You must have a look.

  Contemporary models of physical cosmology postulate that the highest possible temperature is the Planck temperature, which has the value $\pu{1.416785(71)e32}$ kelvin [...]. Above about $\pu{10^{32} K}$, particle energies become so large that gravitational forces between them would become as strong as other fundamental forces according to current theories. There is no existing scientific theory for the behavior of matter at these energies. A quantum theory of gravity would be required. The models of the origin of the universe based on the Big Bang theory assume that the universe passed through this temperature about $10^{−42}$ seconds after the Big Bang as a result of enormous entropy expansion.


How do I take chemistry notes on Mac? I am learning Chemistry now, and I want to take some notes that involve huge amount of organic chemistry on Mac. How can I do that? Pages does not seem support scientific use.

If you know LaTeX and you want to see what you are writing you can use lyx, and some TeX package like mhchem and chemfig. So you can write all the things you want and add comments. This is what I have done with my Mac just now:

Here is the code, you can make these lines as a default template:

\documentclass[12pt,a4paper]{article}
\usepackage[utf8]{inputenc}
\usepackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry}
\usepackage{chemfig}
\usepackage{tikz}
\usepackage[version=4]{mhchem}
\setlewis{4pt}{0.8em}{black}

From here you have to write! The date is automatically added!

\title{Introduction to chemistry notes}
\begin{document}
\maketitle
\section{mhchem}
You know that \ce{ CH4 + 2O2 -> CO2 + 2H2O}     this is a simple equation\ldots
Okay maybe you can do this with word, but if you are familiar with Chemistry S.E. you will be very fast with this notation.

\ce{Zn^2+
<=>[\ce{+ 2OH-}][\ce{+ 2H+}]
$\underset{\text{amphoteres Hydroxid}}{\ce{Zn(OH)2 v}}$
<=>C[+2OH-][{+ 2H+}]
$\underset{\text{Hydroxozikat}}{\ce{[Zn(OH)4]^2-}}$
}

\section{Chemfig}
\begin{figure}[h]
\setatomsep{1.8em}
\schemestart
\subscheme[90]{\chemfig{*6((-[@{a1}]H_2@{a2}\Lewis{26,N})=[@{r1}]-[@{r2}]=[@{b2}](-[@{b1}]N =[@{c1}] N -[@{c2}]*6(=[@{d1}]-[@{d2}] =[@{e1}](-[@{e2}] OH)-=-))-=-=)}\chemrel{<->}
\chemmove[-stealth,shorten <=1pt, shorten >=1pt]{%
    \draw(a2)..controls +(90:5mm) and +(135:5mm)..(a1);% first arrow
    \draw(r1)..controls +(225:12mm) and +(300:12mm)..(r2);% ring arrow
    \draw(b2)..controls +(10:3mm)..(b1);% ring arrow
    \draw(c1)..controls +(340:8mm)..(c2);% ring arrow
    \draw(d1)..controls +(225:12mm) and +(300:12mm)..(d2);% ring arrow
    \draw(e1)..controls +(20:4mm)..(e2);% ring arrow
    }
\chemfig{*6((=[@{a1}]H_2@{a2}\chemabove{N}{\oplus})-[@{r1}]=[@{r2}]-[@{b2}](=[@{b1}]N-[@{c1}] N =[@{c2}]*6(-[@{d1}]=[@{d2}] -[@{e1}](=[@{e2}] \chemabove{O}{\ominus}H)-=-))-=--)}
}
\schemestop
\caption{This is a coniguated system\ldots}
\label{fig:con}
\end{figure}
Okay the lesson will end and in one hour you'll draw this figure only, maybe you should use an external software\ldots %what the professor means whit this last phrase?!
\end{document}

Sincerely I think that is best to take handwrite notes because I think that writing and drawing chemical formula allow you to memorize better but of course this is only a thought.


What does it mean to 'fuse' ores and why do we use alkali metals in the process? For the extraction of chromium, fusion of chromite ore with sodium or potassium carbonate in excess of air is done. What exactly happens in this 'fusion' process?
According to my textbook (class XII chemistry, NCERT): d- and f-block element, extraction of chromite ore
According to the dictionary: To melt or fuse (ores) in order to separate the metallic constituents.
Now,  while this makes sense and is sort of (implied) by my textbook? how exactly is this process done? I want to learn the technicalities related to it! Further I've heard we use alkali metals in fusion, why do we use them ?

"To fuse" is another word for "to melt" (e.g. "heat of fusion"). Specifically, if you say you want to fuse two materials, you melt them in the hope that they will mix.

In this case, you melt the carbonate, and hope that the chromite will dissolve in it. Because e.g. $\ce{Cr2O3}$ has a melting point of $\pu{2435 °C}$, chromite ($\ce{Fe(II)Cr2O4}$) typically around $\pu{2140 °C}$. That is out of reach with a common Bunsen burner, and even if you did reach it, it would still be the same insoluble mess after cooling down.

When your chromite is dissolved in alkali carbonate, the mixture easily dissolves in $\ce{HCl}$, and you can properly run your wet analytics.

Soldering for example is a very similar process: Your soldering iron cannot nearly melt copper, but the molten tin/lead/etc. fuses (partly dissolves/melts) with the copper surface, and you get a continous metallic connection.


Major product in reaction of butan-2-one + selenium dioxide When reading about $\ce{SeO_2}$ I read this following reaction:
$\ce{CH3CH2COCH3 ->[SeO2] CH3CH2COCHO + CH3COCOCH3}$
In the following reaction it was stated that $\ce{CH3CH2COCHO}$ was the major product. But I fail to understand why ?
The two possible reasonings I can think of are that the terminal hydrogen is either less sterically hindered or that it is more acidic. But I am not quite sure of either.
What is the reason for the same ?

To understand the major factor we have to consider the mechanism of the reaction first. The reaction you mentioned is popularly know as Riley oxidation. The mechanism is as follows:

As you can observe the RDS of the reaction is the tautomerization of the keto form into an enol. Thus the extent of enolization determines the major product in this reaction.
As the reaction takes place in an neutral environment the acidity of the alpha-hydrogen plays the major role in deciding the major enol formed and so the corresponding oxidized product is major.
In short, the major product in Riley oxidation is the product associated with the major enol content which in the given conditions is determined by the acidity of respective alpha-hydrogens.


how to avoid bubble formation in hexane I have faced this problems when I was working in lab with hexane for the first time.

I was trying to guide dispersion of nano-particles in hexane solution to a fluid cell (vacuum chamber) with a syringe and a flexible tubing (about 2mm or 3mm diameter). All the materials that I use are hexane resistant.
The fluid cell is kept on very low pressure, so the goal is not to have any bubble formed in it.
When I inject hexane fluid, midway before it reaches the chamber, the vapor bubble are formed ( I think because of low boiling point of hexane ) , which is bad I don't want any bubbles go in the chamber.
However, I figured that if the flow is not horizontal or parallel to the ground (i.e when i inject vertically up ward to chamber) I can avoid the bubble flow.

I also thought to do the injection process fast, however the whole flow has to be very slow for the experiment.
 I would like to know what exactly causing it and what are your suggestions to avoid the bubble formation in a horizontal flow.

I was thinking maybe adding some other solution with higher boiling point would be a case to avoid it. But I don't know exactly what solution. And also it is important to keep in mind that hexane solution contains magnetic nanoparticles and I don't want any effect (like aggregation and so on) of added solution into the dispersion nanoparticles.
Thank you,

As you state, hexane has a low boiling point. If the bubbles are hexane vapor, as opposed to an air leak, then cool the fluid and apparatus more. You might pre-chill the hexane in a dry ice bath. The sublimation temperature of $\ce{CO2}$ at 1 atmosphere is about 195 K, and the melting point of a mix of hexane's various isomers is about 179 K, so there should be no problem with crystals blocking the flow.

Be careful with the cold -- frostbite is not painful at first, but is as damaging as a burn!


Is there an error on Wikipedia's page on characteristic rotational temperature of HF? I am referencing this article: https://en.wikipedia.org/wiki/Rotational_temperature
So, I see that $\theta _R = \frac{(h/2\pi)^2}{2Ik_b}$.
I took the value of $I$ from https://cccbdb.nist.gov/mieigen2x.asp.
However, when I do the calculations for $\theta _R$ by plugging in the numbers, I get
$$\theta _R = 46.8324 K$$
not $30.2K$.
Here is my Mathematica snippet

My question is, has Wikipedia gone wrong, or am I messing this up somewhere?

The units of the inertial constant I are reported in $\pu{amu \cdot Å^2}$. One unified atomic mass unit (or Dalton) is approximately $\pu{1.660e-27 kg}$. Using the correct formula in MATLAB:
>> amu=1e-3/6.022e23;
>> h=6.626e-34; kb=1.3806e-23; tr=@(I) ((h/(2*pi))^2)/(2*I*kb*amu*1e-20);
>> tr(0.86)
ans =

 2.820249699307124e+01

Using instead $\pu{I= 0.80 amuÅ^2}$ from the NIST website returns a value closer to that on the Wikipedia table.


Difference between basicity and reducing character My understanding of basicity and reducing character:

Reducing character is the ability of a substance to reduce something else. So it itself must get oxidized. Since $\ce{BiH3}$ has a large radius, it will loose hydrogen atoms easily. Hence I conclude $\ce{BiH3}$ has more reducing character than $\ce{NH3}$.
Basicity is how easily can the compound give up hydrogen atoms/electrons.

Now I end up thinking basicity and reducing character are the same thing. And that $\ce{BiH3}$ should have more basicity than $\ce{NH3}$. But, this is not the corrrect answer.

So what's the difference between reducing character and basicity?

On mostly all Google searches I find that reducing character is directly proportional to basicity. But that doesn't say why they are different in the first place.

Basicity is how easily can the compound give up hydrogen atoms.

Isn't this the definition for acidity instead? (assuming your "hydrogen atoms" = proton) Basicity instead refers to how easily a compound can accept protons on its lone pair, not give them up.

$\ce{BiH3}$ is a weak base, because the lone pair on bismuth is dispersed over a large surface area, as compared to that on the nitrogen in ammonia. (details here) Being diffused, the lone pair is unable to attack a proton, and hence, is a weak base.

$\ce{BiH3}$ is a better reducing agent, because on getting oxidised (losing a hydrogen), the negative charge formed on the bismuth atom is dispersed over a much larger surface area. Hence, the negative charge is stabilized much more on the bismuth atom than it would be stabilized on the nitrogen atom in ammonia.

Notice that the dispersion of the lone pair makes it more diffused, hence, less basic. While dispersion of the excess negative charge stabilizes the specifies formed after loss of a hydrogen, hence, a better reducing agent.


How thermochromism happen? I learned that when $PbO$ is hot, it is orange. When $PbO$ is cold, it is yellow. I know this's called thermochromism. But how do that happen? Is it because the ions inside doing something?

In the case of lead oxide and many other inorganic thermochromic materials, the color change is a result of a phase change that slightly alters the distance separating the inorganic ions.

In ordered materials, the energy levels present in atoms and molecules are replaced by energy bands.  Said differently, in materials with extended order there are so many energy levels clustered around one another at a certain energy point and separated from one another by such small amounts of energy, that the distinct energy levels of the molecules "fuzz" out into a band.  Instead of a highest occupied and lowest unoccupied molecular orbital (HOMO and LUMO), a valence band and a conduction band results.

Just as the HOMO-LUMO separation will determine where in the energy spectrum an organic molecule will absorb light, the energy separation between the valence band and the conduction band (called the band gap) will determine where a material with extended order will absorb light.

In the case of lead oxide and other crystalline materials, slight changes in the crystal structure (phase change) that can be brought about by heating or cooling the material, will change the size of the band gap.  This change in the energy separation between the valence band and conduction band will result in a change in the optical (and electronic) properties of the material.  If the change in optical absorption occurs in the visible region, then the thermochromic properties of the material are apparent to the eye.

Here is a link to a nice paper that starts simply, but then goes into greater depth, on this subject.


Why is Cyanidine soluble in water? Cyanidine is not polar due to its make-up of only hydrogen atoms and carbon atoms, correct? But "like dissolve like", which means it would HAVE to be polar, or no? Is it then soluble solely because of its OH-groups being able to create hydrogen bonds with the water molecules, thereby making it miscible?

Cyanidine actually looks to me to be a rather highly polar compound. It's essentially a $\ce{C15}$ hydrocarbon having 5 hydroxyl groups and a formal positive charge.

The ratio of hydroxyl groups to the number of carbons is the same as that of water-miscible isopropyl alcohol. (Of course smaller molecules like isopropyl alcohol also tend to be more soluble than larger molecules.)

The most important feature indicative of it's rather high polarity and water solubility is it's formal positive charge. This is analogous to placing a positive charge on a large amine by protination in order to turn the amine into a polar, water soluble compound.


Are all solutions of weak acid/bases buffers? I am having a difficult time understanding what makes a buffer a buffer.

Buffers in my textbook are defined as a solution of a weak acid or base and their conjugate acid/base. So if I were to just dissolve acetic acid (a weak acid) in water why would this not be defined as a buffer. Acetic acid incompletely dissociates into an acetate ion (the conjugate base) and a Hydrogen ion. Why is it that for a base to be made the acetic acid would have to be mixed with something like Sodium acetate if acetic acid incompletely dissociates into a conjugate base in the first place? Could someone please help me understand why this is the case?

I totally agree with @DavePhD and I want to show you how the pH dependence of the degree of acid-base ionisation provides the background of the buffering mechanism. Titration curves effectively illustrate the influence of the pH on the ionisation of weak acids and bases. The Figure below shows the change in the pH of a 0.1 M acetic acid solution on its titration with a strong base. If you just dissolve acetic acid in water, yes it is true that there will be a small fraction that ionizes to yield the conjugate base, acetate. However if you add the slightest amount of a base, for example, it will result into a steep increase in the pH (look at the beginning of the titration curve) and the solution is not a buffer.

That initial steep increase in the pH slows down quickly, and a pH plateau appears in the range of 4.0-5.8. As in this range $[\ce{CH3COOH}]$ is comparable to $[\ce{CH3COO-}]$ (i.e. their ratio is within a tenfold range), we will have a two-component system containing both a weak acid and a weak base—i.e. a buffer. The centre of the plateau is at 50 % titration where $[\ce{CH3COOH}]=[\ce{CH3COO-}]$. Therefore, according to the equation $\mathrm{pH}=\mathrm{p}K_\mathrm{a}+\log{\frac{[\ce{CH3COO-}]}{[\ce{CH3COOH}]}}$, the $\mathrm{pH}$ will just equal the $\mathrm{p}K_\mathrm{a}$. Following the plateau (upon the exhaustion of the buffer capacity), the pH will again increase steeply, and the top of the titration step will be reached, i.e. the point of neutralization at which the degree of titration of the acetic acid will be 100 %

From the figure above, the region marked inside the rectangle is a buffer. Because it can provide the most efficient pH stabilization against both acidic and basic shifts. In this region, the ratio of the two components (acid and conjugated base) is one or close to one. Accordingly, a buffer is effective in the $\mathrm{pH}$ range of its $\mathrm{p}K_\mathrm{a} \pm 1$ $\mathrm{pH}$ unit


Is the reaction between a carboxylic acid and alcohol endothermic? I would say that it is endothermic, and then by increasing the temperature, one could get more product, disturbing the equilibrium.

Is this true, though? During an esterification, some bonds are broken, releasing energy. The opposite reaction splits the ester and adds the atoms back, which I suppose releases energy which is then bound in the bonds adding back the water molecule.

When in doubt, go find data. Here is a typical such reaction - the esterification of ethanol and acetic acid to make ethyl acetate.

$$\ce{CH3CH2OH + CH3CO2H -> CH3CO2CH2CH3 + H2O}$$

The following thermodynamic values are from the NIST Chemistry Webbook.

$$\begin{array}{|c|c|}\hline
\text{compound} & \Delta _\pu{f} H^\circ \ (\pu{kJ mol^-1}) \\ \hline
\text{ethanol} & -276 \\
\text{acetic acid} & -484 \\
\text{ethyl acetate} & -480 \\
\text{water} & -286 \\ \hline \end{array}$$

Now for some Hess's Law:

$$\begin{aligned}
\Delta_\pu{r} H^\circ &= \sum_\mathrm{products} \Delta_\pu{f} H^\circ - \sum_\mathrm{reactants} \Delta_\pu{f} H^\circ \\
&= (-480\ \pu{kJ mol^-1} + -286\ \pu{kJ mol^-1}) - (-276\ \pu{kJ mol^-1} + -484\ \pu{kJ mol^-1}) \\
&= -766\ \pu{kJ mol^-1} - (-760\ \pu{kJ mol^-1}) \\
&= -6\ \pu{kJ mol^-1}\end{aligned}$$

Hey! This one is exothermic (slightly).

Here are the enthalpies of reaction for some other esters (by no means a complete or even representative list):
$$\begin{array}{|c|c|}\hline
\mathrm{ester} & \Delta_\pu{r} H^\circ \ (\pu{kJ mol^-1}) \\ \hline
\text{methyl acetate} & -9 \\
\text{ethyl acetate} & -6\\
\text{methyl benzoate} & -5 \\
\text{ethyl crotonate} & +1 \\ \hline \end{array}$$

I would thus guess that formation of most carboxylic esters from carboxylic acids and alcohols would be slightly exothermic or slightly endothermic depending on structure.

The reasons esterifications are often heated are 1) to increase the rate of reaction and 2) to distill off water as it forms.


Number of optically active isomers of [Co(NH3)2Cl2(NO)(OH)]? My Attempt:

Since Compound is of the form M[a2b2cd]

I got three optically active isomers of the form M[(ab)(ab)(cd)] , M[(ab)(ac)(bd)] , M[(ab)(bc)(ad)]
and since NO is a bidentate ligand, total number of optically active isomers should be multiplied by 2 due to linkage isomerism.

Hence i was getting the answer as 2*3=6 but answer given was 4.

Can you please help me with this?

Try drawing out [M(ab)(ab)(cd)]: you'll see that it's actually optically inactive!

So, there are 2 optically active isomers, which along with their respective enantiomers adds up to 4 (M[(ab)(ac)(bd)] , M[(ab)(bc)(ad)],  as you've mentioned).


Determining Kohn-Sham and Hartree Fock virtual orbitals: The underlying field In Frank Jensen Introduction to Computational Chemistry from 2007, I stumbled upon this paragraph about whether to assign meaning to the KS orbitals or not

  Another difference is that the unoccupied orbital energies in Hartree–Fock theory are determined in the field of N electrons and therefore correspond to adding an electron, i.e. the electron affinity. The virtual orbitals in density functional theory, on the other hand, are determined in the field of N − 1 electrons and therefore correspond to exciting an electron, i.e. unoccupied orbitals in DFT tend to be significantly lower in energy than the corresponding HF ones, and the highest occupied molecular orbital–lowest unoccupied molecular orbital (HOMO–LUMO) gaps are therefore much smaller with DFT methods than for HF. This also means that orbital energy differences in DFT are reasonable estimates of excitation energies, in contrast to HF methods where excitation energies involve additional Coulomb and exchange integrals.

I can't seem to understand why the Kohn-Sham equations are solved in the presence of N-1 electrons and not N. From my understanding the Kohn Sham and the Hartree Fock equations are almost identical. Where does this difference come from?

That is true: in the Kohn–Sham model electrons both in occupied and in virtual orbitals are "moving" in the field $n-1$ electrons, while in the Hartee-Fock model electrons in occupied orbitals are "moving" in the field $n-1$ electrons, but electrons in virtual orbitals are "moving" in the field all $n$ electrons. And this is simply "by construction": you just have to look at the very definitions of these models. So, in short:

In the Kohn–Sham model, it is the exchange-correlation hole that by definition "excludes" a single electron from equations describing each and every orbital.

If you look at the Kohn–Sham equations
$$
\left(-{\frac  {\hbar ^{2}}{2m}}\nabla ^{2}+v_{{{\rm {eff}}}}({\mathbf  r})\right)\phi _{{i}}({\mathbf  r})=\varepsilon _{{i}}\phi _{{i}}({\mathbf  r}) \, ,
$$
where
$$
v_{\rm {eff}}(\mathbf {r} )=v_{\rm {ext}}(\mathbf {r} )+e^{2}\int {\rho (\mathbf {r} ') \over |\mathbf {r} -\mathbf {r} '|}d\mathbf {r} '+v_{{{\rm {xc}}}} \, ,
$$
it is last term of the Kohn–Sham effective potential, the exchange-correlation potential,
$$
v_{{{\rm {xc}}}}({\mathbf  r})\equiv {\delta E_{{{\rm {xc}}}}[\rho ] \over \delta \rho ({\mathbf  r})} \, ,
$$
that causes the above mentioned feature of Kohn–Sham orbitals.

For the case of the Hartree-Fock equations
$$
  \hat{F} \phi_i({\mathbf r}) = \epsilon_i \phi_i({\mathbf  r}) \, ,
$$
the summation in the Fock operator by the very definition of the model runs over all occupied orbitals only,
$$
  \hat{F}
  =
  \hat{H}^{\text{core}}
  +
  \sum_{j=1}^{n}[\hat{J}_{j} - \hat{K}_{j}] \, .
$$
For occupied orbitals, in the Hartree–Fock equation defining some particular (occupied) $i$-th orbital the Coulomb and exchange contribution from this orbital itself perfectly cancel each other, i.e. $\hat{J}_{j} \phi_i = \hat{K}_{j} \phi_i$ when $j=i$. Thus, in the Hartree-Fock model an electron from any occupied orbital is also moving in the field $n-1$ electrons. And this is of course physically more than reasonable: at the end of the day electron should not interact with itself.

But for virtual orbitals the situation in the Hartree-Fock model is different from that in the Kohn-Sham one: for any virtual orbital $\phi_k$ the sum in the Hartree-Fock equation defining it still runs over occupied orbitals only, but none of the terms $\hat{J}_{i} \phi_k$ and $\hat{K}_{i} \phi_k$ this time cancel each other since $k$ is simply grater than $n$.


Boiling points of HF, HCl, HI I'm really confused over how to weigh the molecular mass with intermolecular forces such as hydrogen bonding in HF in terms of figuring out the melting points in relation to each other.

I know that HF has the highest melting point out of HCl and HI, but then HCl and HI both have significantly larger molecular mass.

So how does one properly explain why HF has the higher melting point out of HCl and HI? (Not just saying that generally hydrogen bonds out weigh the molecular mass when considering boiling point)

Is there any mathematical basis?

In HF there is a hydrogen bond which always outweighs all other factors when deciding properties related to stability. H-Bonded molecules are very stable


What is the difference between cross-linked bond and secondary bond in polymers? In solid materials, there are two kinds of bonds between polymers: primary and secondary. Primary bonds can be also called cross-linked bonds.

What is the difference between the two (strength, type, response to stress/strain, etc)?
How do they compare to bonds inside a polymer?

This reference indicates that primary bonds in a polymer are the intramolecular bonds (the covalent bonds) that hold the polymer together. Cross-linking different chains of polymer together does occur through a covalent bond, such as a disulfide bond.

The secondary bonds that help to give the polymer its physical properties are intermolecular forces, such as hydrogen bonding for polymers that contain hydroxyl or primary or secondary amine groups. These secondary bonds can be disrupted by changing solvent conditions or heating, for example, which wouldn't disrupt the primary bonds.

So, I suppose that you could say, loosely, that there are "two kinds of bonds between polymers in solid material" if you mean between the polymer chains before they have an opportunity to cross-link (by a change in pH, for example). After they cross-link, the physical properties of the bulk sample of the polymer will still be affected by secondary bonds, the intermolecular interactions, between the cross-linked chains.

The primary bonds will be stronger and can only be affected by chemical changes, the secondary bonds (intermolecular forces) can be affected by changing the physical conditions.


Why is O2 a biradical? $\ce{O2}$ has a double bond in its normal form. That is $\ce{O=O}$. There are no unpaired electrons in this case are there since there are 2 lone pairs on each oxygen.

However 1 resonance structure would be $\ce{O-O}$ (result of homolytic cleavage of double bond) where each $\ce{O}$ is a free radical (a negatively charged one at that). If you have this in hydrogen it is likely going to form hydrogen peroxide.

You could also have $\ce{O-O}$ where 1 is positive and the other is negative and this is also 2 free radicals.

And finally there is $\ce{O#O}$ where both oxygens are positively charged and are free radicals. Why are both positively charged? It is because 3 bonds already to oxygen means 1 lone pair and 5 electrons around oxygen is +1.

Is it because of these resonance structures giving 2 free radicals in $\ce{O2}$ that $\ce{O2}$ is considered a biradical?

We can draw the 3 Lewis structures (or the corresponding resonance structures) pictured below for $\ce{O_2}$

Since an oxygen atom has 6 electrons,

A would correspond to a structure with a single bond between the oxygen atoms, 2 lone pairs on each oxygen and an unpaired electron on each oxygen; however A does not have an octet around each oxygen, in fact, each oxygen would only have 7 electrons
B would correspond to a structure with a double bond between the oxygen atoms, 2 lone pairs on each oxygen and no unpaired electrons on each oxygen; B does have an octet around each oxygen, but it is not a biradical
C would correspond to a structure with a triple bond between the oxygen atoms, 1 lone pair on each oxygen and an unpaired electron on each oxygen; however C does not have an octet around each oxygen, in fact, each oxygen would have 9 electrons and this would be impossible for oxygen

So while structure A would indicate a biradical, we wouldn't "expect" it to count for much since the oxygens do not have octets.  This inability to clearly predict the biradical nature of $\ce{O_2}$ illustrates one of the failings of both Lewis structures and resonance theory.
In order to correctly predict the biradical nature of $\ce{O_2}$ we must move up to molecular orbital theory.  Below is the molecular orbital diagram for $\ce{O_2}$.  As you can see it does predict that $\ce{O_2}$ should be a biradical with an unpaired electron in each of its degenerate, highest occupied molecular orbitals.

Edit: response to OP's comment

When I think of triple bond I don't think of 2 3 electron bonds(which
is what you drew). Rather I think of 3 2 electron bonds(1 sigma bond
and 2 pi bonds)

Structure C does represent 3 two-electron bonds (not 2 three-electron bonds), that's just how you draw the Lewis structure.

This type of triple bond would make the oxygen positive with 5
electrons around it.

No, the formal charge on the oxygen in structure C is
Z = 6 - 3 unshared - (1/2 * 6 shared)= 0,
there is no formal charge on oxygen in the "triple bond" structure and as I noted above, there are 9 electrons around it (not 5), which is impossible for oxygen.

I am assuming electrons are shared equally with half around 1 atom and
half around the other(which is the basis for formal charge

Yes, that's correct.


Sulfur trioxide structure What resonance structure is the most abundant in sulfur trioxide: the one with one double and two single, the one with two double and one single or the one with all three double bonds? Is it actually correct to write an expanded octet when it is not necessary?

The answer depends on which teaching you follow.

Traditional teaching, dating back to Linus Pauling, would prefer three double bonds to minimise charge separation, requiring an expanded octet.

Even older traditional teaching would favour the validity of the octet rule and take charge separation as a tolerable by-product.

Both sides have their advantage and disadvantage:

The Pauling method

correctly shows that all three bonds are equal without resorting to mesomery
incorrectly assumes a bond order of $2$ rather than the actual $1.4$

The charge-separation method

predicts a bond order of $1.33$ which is much closer to $1.4$
requires mesomery of three structures to explain the three equal bond lengths.

I for my part prefer the mesomeric charge-separation description, because we have no trouble drawing benzene in the same way and because I like the octet rule. Others will disagree and prefer the three double bonds.

I wouldn’t know anybody who prefers a description with two double bonds and one single bond. It is just between the two models making it all the less correct.

Please also see this excellent answer to a related question by Martin which includes molecular orbitals for $\ce{SO3}$


Storage of the Halogens in PTFE Bottles I'm looking to buy some Iodine for various reactions, and noticed that the person who sells it has it in a glass jar with a plastic lid of unknown type.

I'm looking up chemical resistance charts and it says that Iodine, Chlorine and Bromine both work absolutely fine when placed inside of a PTFE bottle, assuming that the Chlorine is not in a gaseous form.
As for Fluorine, one website states that it attacks the plastic severely, whereas the other states that the PTFE is resistant until around the 100°C mark.

These sites are both here and here.

Now when I go look at people asking the question, "How should I store elemental Iodine?", everyone seems to respond along the lines of keeping it in a glass jar with a ground glass joint or PTFE/Teflon lined cap.

However since my research suggests that PTFE alone is capable of storing Iodine, is this a viable option? I would love to know from past experiences, because no-one seems to recommend PTFE on its own on any site I checked. Of course however I will also be storing said PTFE container inside of another non-reactive material container.

I understand that I'm probably better off just storing it in glass containers but I have PTFE bottles lying around and would like to use them if at all possible.

Since this question revolves around the Halogens themselves, could the same be potentially done for Bromine and Chlorine?

Though PTFE is not attacked by the halogens, they may discolor PTFE (particularly iodine, and at elevated temperature), since the plastic is somewhat permeable to halogens (end even to water).


Why doesn’t aniline undergo Friedel-Crafts alkylation? Even though aniline is an activated benzene derivative, it still doesn’t undergo Friedel-Crafts alkylation. Why?

Can it undergo Friedel-Crafts acylation?

The answer lies in the fact that aniline is a Lewis base and $\ce{AlCl3}$ is a Lewis acid. The reaction between aniline and $\ce{AlCl3}$ hampers the catalytic activity of $\ce{AlCl3}$ required to perform the Friedel-Crafts alkylation and acylation.

Despite the activation of the $\ce{NH2}$ group, Friedel-Crafts alkylations and acylations fail because the $\ce{NH2}$ group acts as Lewis base and interacts with the Lewis acid catalyst. ... A way to overcome this problem is to convert the $\ce{NH2}$ group into an amido group prior to Friedel-Crafts reaction. ...The amide group can be hydrolysed back to aniline group after alkylation.
(Pg. 1167 An Introduction to Organic, Inorganic and Physical Chemistry. 4th edition. Catherine E. Housecroft, Edwin C. Constable. Pearson Education Limited)

The aniline forms a Lewis acid base adduct with $\ce{AlCl3}$ :
This prevents the interaction between alkyl or acyl chloride with $\ce{AlCl3}$.


Why can Neutral Red (indicator) not test values below its pKa? It is the only indicator I have found that does not test below its pka, and can't think of any reason for this.

Neutral Red has a pKa value of 6.8.
(
Reference 1,
Reference 2
),
but looking at its pH testing range, it only shows from pH 6.8-8.0 (Reference 3)

What would be the reason for this?

Thank you

The colour of neutral red in aqueous solution changes from violently red (protonated cation) to yellow/mostly colourless (the uncharged amine).

If the $pKa$ is 6.8, then the dye is still very much red at pH 6.8, and you won't be able to see the colour difference to, say, pH 5.

You have to keep adding base up to a pH of eight until the concentration of the colourful cation is so low you cannot see it any more.

Other indicators change from one strong colour to another, there the $pKa$ will be more in the middle of the transition range.


Conversion from t-butanol to alkene How can I accomplish the above conversion?

My thoughts:

One way of introducing the double bond, would be using the appropriate Grignard Reagent after oxidation of t-butanol (somehow)
I think that initial conversion to carbonyl compound is necessary - only then can we follow up with nucleophilic addition.

P.S. The question was in the organometallics section of my assignment, possibly hinting that reagents apart from the above species may be used. However, won't it be great if we could design a mechanism using only the starting species and commonly available reagents?

t-butanol + allyltrimethylsilane + Lewis acid

This is called the Hosoni-Sakurai Reaction (see J. Org. Chem. 2006 71 8516)

Can only find an image for the more common addition to carbonyl compounds but there are examples of additions to alcohols (see the JoC reference)


Radial probability for orbitals Questions

Question F-2:

I think question f1 requires one to relate $\mathrm{\psi^2}$ or. Probability density with radius but I am just in high school and have not learnt about Hamilton operator

Question F-1:
In second I think A should be 1s and B should be 3p

For the 1st one, the maximum probablity of finding the electron occurs at the Bohr's radius,i.e. at $\ce{r=r_0}$. This can be guessed by thinking this result classically. But if you want to know the exact method, the wave function for $1s $ orbital is, $$\ce{$\psi$_{1s} = 1/$\sqrt{\pi a_0^3}$ e^{-r/a_0}}$$ So, the radial probablity distribution for the $1s$ orbital will be,$$\ce{P(r) = |$\psi$|^2 4$\pi$r^2dr}$$ which will be eqal to , $$\ce{P(r) =4/a_0^3 r^2e^{-2r/a_0}}$$ To find the value of $r$ at which $\ce{P(r)}$ is maximum , you just have to find the solution of $\ce{dP/dr =0}$. You can verify very easily that the solution is $\ce{r=r_0}$.For the next one, the graph of B) is definitely for $2s$. Now, according to judgement of no. of radial nodes of $2s$ and $3p$, they  both have 1 radial node each. So, that is inconclusive. But if you would have known the wave functions, it would be easier.The wave-function of $2s$ is $$\ce{$\psi$_{2s} = c. (2- r/a_0)e^{-r/2a_0}, where c is a constant.}$$But, wave function of $3p$ orbitals are, $$\ce{$\psi$_{3p_z} = c_1(6r/a_0 - r^2/a_0^2)e^{-r/3a_0}cos$\theta$}$$ $$\ce{$\psi$_{3p_x} = c_2(6r/a_0 - r^2/a_0^2)e^{-r/3a_0}sin$\theta$cos$\phi$}$$ $$\ce{ ($\psi$_{3p_x} is actually an additive linear combination of $\psi$(r)Y_1^1 and $\psi$(r)Y_1^{-1})}$$ and, $$\ce{$\psi$_{3p_y} = c_2(6r/a_0 - r^2/a_0^2)e^{-r/3a_0}sin$\theta$sin$\phi$}$$$$\ce{ ($\psi$_{3p_y} is actually an subtractive linear combination of $\psi$(r)Y_1^1 and $\psi$(r)Y_1^{-1})}$$So, you can see at $r= 0$, all the $3p$ orbitals have value of wavefunction (i.e. |$\psi$ |) = $0$ (simply check by putting $r$ as $0$ in all given $3p$ wave functions). But at $r =0$,$\ce{ $\psi$_{2s}}$ has non-zero finite value so, $\ce{|$\psi$|^2}$ does not vanish at origin for$ 2s$ orbitals which the given graph is showing. So, it can't be $3p$ as the value of the wave function doesn't match at origin. If you consider higher orbitals like $4d $ also, their wavefunction will also vanish at origin. So, the figure B) corresponds to $2s$.


Regioselectivity in Electrophilic Aromatic Substitution of Pyrrole and Indole If we look into the structures of Pyrrole and Indole, they are very similar. Only Indole has one extra Benzene ring attached to it, i.e. if we attach one Benzene ring to Pyrrole, what we'll have is nothing but Indole.

But if we delve into the preferable positions of Electrophilic Aromatic Subsitutions for these compounds, it's pretty well known that Pyrrole undergoes EAS preferentially at the $2$-position, whereas for Indole it's the $3$-position which is favoured.
A typical example of an Elcetophilic Aromatic substitution reaction where this distinct feature between Pyrrole and Indole can be clearly visible is the Vilsmeier-Haack Reaction. The major outcomes of the reaction in the two cases clearly show the difference.

Now, the preference of $2$-position in case of Pyrrole can be justified easily by considering the greater stability of the $\sigma$-complex intermidiate formed in case of substitution at $2$-position due to more number of resonating structures as clearly been described in the answer to the question Regioselectivity in Electrophilic Substitution of Pyrrole
Now my question is, applying the similar logic Electrophilic substitution at the $2$-position should be even more favoured in case of Indole, as the $\sigma$-complex in that case should be further more stabilised due to resonance with the Benzene ring as well. Therefore, the preference of substitution at $2$-position should be exceptionally higher than at $3$-position in case of Indole. But in reality, it's actually the opposite. Why is this anomaly observed in case of Indole ?
The resonating structures in case of two different substituent positions in Indole are shown below,

P.S. Thoughtful answers are highly appreciated. Any kind of Quantum Mechanical Calculations or Computations or Orbital Analogy are also welcome.

The traditional explanation for the reactivity of indole at C–3 is that attack at C–3 does not disrupt the aromaticity of the benzene ring in the cationic intermediate (see any organic chemistry book; for heterocycles a good book is Joule and Mills as has been suggested in the comments):

You have drawn many of the resonance structures for the intermediate, including the two shown above. So I should mention that the major resonance contributor should be the one where the positive charge is borne by the nitrogen. In this case, the positively charged nitrogen has four bonds and a complete octet; whereas if the positive charge is on carbon, that's formally a six-electron, electron-deficient carbocation. Many thanks to Buttonwood for pointing this out.
To a first approximation, one could consider the reactivity of indole to be as if it were a benzene ring plus an enamine, and enamines are very, very happy to react β to nitrogen and not α. This is of course not an enamine, but the idea is the same.
For pyrrole, there is no "other benzene ring" to worry about, so this consideration does not apply.


In what ways does the transfer of heat affect bond stability? It is to my understanding that certain amount of joules may change an element's/compound's physical state given a certain temperature. How does a certain added number of joules affect the orbital of a bonded electron ( I realize my question's ambiguity by not stating a specific type of bond).

How does a certain added number of joules affect the orbital of a
  bonded electron

If you pump enough energy into an element\molecule you can promote an electron from one orbital to the next highest.  For example, the hydrogen atom has a single 1s electron, put enough energy in and you can promote that electron to the 2s orbital.  That means that, on average, the electron would spend more time further away from the nucleus.  In molecules, there are numerous "sub-levels" between each (molecular) orbital.  These "sub-levels" relate to rotational and vibrational energy.  So as you pump energy into the system the molecule will rotate and vibrate more violently.  If you put enough energy in, the bond will break and the atoms connected by that bond will fly apart.


Expectation value of an observable in quantum system In quantum systems, sometimes if the wave function is not an eigen value of an operator then it is expanded in terms of bases states, then we apply the operator over the bases states and we get different eigen values for an observable with different probabilities corresponding to the projection coefficients of the bases states.

  In a book, I read that after making a measurement we get an observation, and the probabilities of other observations shrink down to zero. The observation remains the same for howsoever times we make it after the measurement. Like if a coin is tossed, if we get head after first measurement it remains head, even if we make any number of observations.
  My question is that what is the need of expectation value of an observable (given by formula as summation of the product of probability associated with an observable and the eigen value of observable) as after measurement we get only one eigen value of the observable with 100% certainty. I think knowing only the possible values of the observable is sufficient. What is the significance of expectation value and then does the formula for expectation value given in books is for before the measurement?

Roughly, the significance is that before measurement the system can be in a superposition of states, the wavefunction of which can be represented by a linear combination of the individual eigenfunctions

$\Psi = \sum_i c_i{\psi_i}$

However the very act of observing the system reduces the wavefunction to a single eigenstate due to interaction with the external world

For most cases, the expectation value is just the quantum mechanical way of calculating properties. Remember, the wavefunction is not observable! Only properties like density, dipole moment, etc are and the expectation value allows us a way to go from wavefunctions to properties via operators.

These other answers may also be helpful

https://physics.stackexchange.com/questions/67756/quantum-collapse
https://philosophy.stackexchange.com/questions/48425/is-collapse-of-the-quantum-state-conceptually-misnamed
https://physics.stackexchange.com/questions/151457/how-does-a-wavefunction-collapse


How does H2SO4 dissociate? Apologies for this extremely basic question, I'm just beginning with Chemistry so please don't be too harsh on me.

My book says that sulfuric acid, $\ce{H2SO4}$, dissociates in its ions following this reaction: $$\ce{H2SO4 -> H2^+ + SO4^{2-}}$$

My question is, why can't the dissociation reaction happen like this:
$$\ce{H2SO4 -> 2H^+ +SO4^{2-}}$$

I know hydrogen is a diatomic gas, but here I don't know if H will dissociate as a gas or as a liquid (since $\ce{H2SO4}$ is a liquid, not a gas).

I'm trying to learn, thank you for your understanding and your time.

$\ce{H2SO4}$ is one of common strong acids, meaning that $\ce{K_{a(1)}}$ is large and that its dissociation even in moderately concentrated aqueous solutions is almost complete.

Arrhenius dissociation:

$$\ce{H2SO4 <=> H+ + HSO4-}~~~~~~~~~~\ce{K_{a(1)}}=\ce{large}$$

Brønsted-Lowry Dissociation:

$$\ce{H2SO4 + H2O <=> H3O+ + HSO4-}~~~~~~~~~~\ce{K_{a(1)}}=\ce{large}$$

This accounts for the vast majority of protons donated by the acid. However, since it is diprotic, you may want to take into account the second dissociation, which is technically weak but has a larger $\ce{K_a}$ than many weak acids.

Arrhenius 2nd Dissociation:

$$\ce{HSO4- <=> H+ + {SO_4}^2-}~~~~~~~~~~\ce{K_{a(2)}}=1.2\times10^{-2}$$

Brønsted-Lowry 2nd Dissociation:

$$\ce{HSO4- + H2O <=> H3O+ +{SO_4}^2-}~~~~~~~~~~\ce{K_{a(2)}}= 1.2\times10^{-2}$$

This second dissociation may need to be taken into account for some calculations, but it is negligible in concentrated solutions.


Is zero-point vibrational energy an intensive or extensive property? The zero-point vibrational energy (ZPVE) is the non-zero amount of non-electronic energy that molecules have even at zero kelvin and is purely quantum mechanical in nature. For a molecule with $c$ countable normal modes, it is defined as

$$
E_{\text{ZPVE}} = \frac{1}{2} \hbar \sum_{c} \omega_{c}
$$

where $\omega_{c}$ is the energy or frequency of each normal mode.

Intensive and extensive variables have always confused me. This reference gives the following concise bulleted definition:

  Extensive:

  Additive
  State variables (only sometimes)

  Intensive:

  Not additive
  Field variables
  Point variables (i.e., same at each point in a system at equilibrium)

with further examples of each (also on Wikipedia):

  Extensive: $Q$, $W$, $U$ (and other free energies), $N$, $V$, $M$, $C$, $n$
  Intensive: $P$, $T$, $\mu$, $\xi$ (electric field), any field variable, ratio of extensive/extensive

To use an example, the total energy of a system is extensive. If I have one molecule of caffeine, calculate $U$, and then take two caffeine molecules (even at infinite separation), and calculate $U$, they'll be different.

Now consider the ZPVE. For a nonlinear molecular system, there are $3N-6$ normal modes, where $N$ is the number of atoms. Take caffeine again, which has 24 atoms. There are 66 normal modes for one molecule and 138 for two molecules. If the ZPVE is considered to be a bulk or macroscopic property, then clearly it increases as the total number of normal modes increases, independent of whether or not the resulting normal modes are localized to one molecule. If the ZPVE is considered to be a single-molecule property (I hesitate to call it microscopic), then a single molecule is a single molecule, and it would make sense to "isolate" it in a way that doesn't make sense for other properties; that is, it doesn't make sense to only consider the temperature of a single molecule assumed to be in isolation that would in fact be part of some bulk phase.

So, perhaps a better wording of the question is: is ZPVE a microscopic (ugh) or macroscopic quantity? A property like polarizability also exists in a weird gray area for me as well.

Mathematically, an extensive property $f$ is one for which $f(\lambda x,\lambda y,z) = \lambda f(x,y,z)$ for extensive variables $x$, $y$, and intensive variable $z$. For example, the Helmholtz free energy $A(T,\lambda V,\lambda N) = \lambda A(T,V,N)$ is extensive.

Following this definition, the zero-point energy $$E_\text{ZPVE}(\{\omega_c\},\lambda n) = \lambda E_\text{ZPVE}(\{\omega_c\},n),$$ where $n$ is the number of normal modes, is extensive in the thermodynamic limit $N \to \infty$.

The polarizability should be able to be treated similarly.

It bears noting that a lot of macroscopic extensive properties are not extensive in the microscopic regime. For example, if we consider two identical systems, each with an energy $U$, then the combined system has energy $2U$ plus an additional energetic term $U_\text{sys-sys}$, accounting for superficial energetic interactions between the two subsystems. The surface, and hence the superficial energy term, scales as one dimension removed of the volume, and likewise vanishes in the thermodynamic limit.

Furthermore, it's perfectly valid to have properties that are neither extensive nor intensive, like the number of microstates that a system possesses.


What's the difference between ΔG° and ΔG°'? In biochemistry contexts, one often sees $\Delta G^{\circ\prime}$ ("delta G naught prime"), rather than the normal standard free energy change $\Delta G^{\circ}$ ("delta G naught").

What's the difference between the two quantities?
Is there a formal definition of the two terms?

The prime usually denotes a standard free energy that corresponds to an apparent equilibrium constant where the concentration (or activity) of one or more constituents is held constant.

For example, for $\ce{HA <=>A- + H+}$ the equilibrium constant is $K=\frac{[\ce{A-}][\ce{H+}]}{[\ce{HA}]}$ and the corresponding standard free energy change is $\Delta G^\circ=-RT \ln(K)$. If you know the value of $K$ and the start concentration of $\ce{HA}$ then you can compute the equilibrium concentrations of $\ce{HA}$, $\ce{A-}$, and $\ce{H+}$.

However, if the pH is held constant then $[\ce{H+}]$ is no longer a free variable and the apparent equilibrium constant is $K'=\frac{[\ce{A-}]}{[\ce{HA}]}$ and the corresponding standard free energy change is $\Delta G'^\circ=-RT \ln(K')$. So $\Delta G'^\circ=-RT \ln(K/[\ce{H+}])=\Delta G^\circ+RT\ln[\ce{H+}]$

In biochemistry there is often several important constituents in addition to  $\ce{H+}$ that are held constant such as $\ce{Mg++}$, phosphate, etc.


How to get back my beautiful crystals made from iron and sulphuric acid I have put some old nails and screws in the 30% sulfuric acid to prepare an electrolyte for plating. After some bubbling and horribly irritating smell (probably $\ce{SO2}$), the reaction slowed down. Overnight, beautiful green/blue crystals formed:

I just love them. After another day, the reaction almost stopped. I suppose the acid was all bound to iron. To continue the process, I poured some concentrated sulfuric acid into the solution. That was a big mistake - the crystals reacted with concentrated acid and formed ugly white dust:

What did I create? What are the crystals and what's that dust? How to reverse the reaction? Does the dust dissolve to be electrolysed?

I realize this question is over two years old, but I have to add the correct answer here: the iron (II) sulfate you had at the beginning was in the heptahydrate form.  SEVEN molecules of water for every iron (II) sulfate ion pair.  Now, what do you think would happen to that crystal structure when you add one of the most powerful dehydrating agents in existence?  That's right, it sucks all the water out of your crystals, and you're left with a powdery, noncrystalline form.  You could have just added less concentrated acid, that was diluted enough to not be such a strong dehydrating agent.

Incidentally, iron is a surprisingly strong reducer of sulfates.  I did this reaction recently and noticed gold flecks floating around in the solution.  I was shocked to see gold, since I obviously never added any, until I realized that the iron had reduced the sulfate all the way down to sulfide, and what I was seeing was iron pyrite -- fool's gold.  Adding more acid eliminated that almost immediately.


Hybridization and bonding in ethenone H2C=C=O? My book says ethenone is a planar molecule, so I was trying to understand its hybridization,I feel that like allenes ethenone too will not be planar. Supposing oxygen and the carbon near it to be in the xz plane and assuming that the pi bond between them involves py orbital of carbon, this implies the bond between that carbon and the adjacent carbon involves the pz orbital of that carbon. This implies that the end carbon's hydrogens will be in xy plane. What is wrong with my understanding?

You are right.the hibrydization is the same with allenes but in allenes there are 2 CH2 groups but in ethenone there is a CH2 and there's a O with two none bonding pairs. The hydrogens will be in the same plane as C=C=O but  the electron pairs on oxygen are not in the same plane


Reaction to form amino acid $$\ce{CH2=CH-CH=O ->[CH3SH][(1,4 addition)] Product (P_1) ->[(i) NaCN][(ii) NH3\\(iii) H3O+] Q_1}$$
Correct statement regarding $\ce{Q_1}?$

$\ce{Q_1}$ is essential amino acid
$\ce{Q_1}$ is acidic amino acid
$\ce{Q_1}$ is nonessential amino acid
$\ce{Q_1}$ on Lassign test gives prussian blue colour with $\ce{FeSO4}$ and $\ce{FeCl3}$ solution

By 1,4 addition (as mentioned in question) carbon-3 (taking carbon as carbon-1 of aldehyde group) will have $\ce{SCH3}$ added to it and oxygen must have $\ce{H}$ added to it.
After $\text{P}_1$ is formed, it can again tautomerize from an alcohol to an aldehyde.
The $\ce{^-CN}$ can now attack the aldehyde. Looking at the answer we can ay that the product must be an amino acid and so the $\ce{NH3}$ group and $\ce{^-CN}$ must be present on same carbon because then the $\ce{^-CN}$ will hydrolyze to give a $\ce{-COOH}$ group.
But how is $\ce{NH3}$ supposed to attack a carbon having $\ce{^-CN}$ and a hydroxyl group. Can we make the alcohol a leaving group somehow?
Or is the number of carbons in chain 1 less than what it should be? What should be the correct answer and approach for this problem?

The starting compound is an organic compound known as acrolein. $\ce{CH2=CH-CH=O}$
Now your first reaction would be:
$$\ce{CH2=CH-CH=O ->[CH3SH][1,4 addition]CH3-S-CH2-CH=CH-OH} $$
Now, this compound tautomerizes to form methional
$$\ce{CH3-S-CH2-CH=CH-OH <-> CH3-S-CH2-CH2-CH=O}$$
Now, the second step in your reaction is a named reaction known as Strecker Amino Acid Synthesis. This converts an methional (in this case) into methionine.

Now, onto the question at hand, it asks about certain properties of methionine. Methionine is an essential amino acid. So A is correct. Since methionine doesn't have any extra acid or base groups, it is neutral.
As for the Lassaigne Test, if the compound had only nitrogen, it would have shown a prussian blue color. However since there is sulphur as well, it shows a violet colouration. So, the answer is A.


Why change in enthalpy is negative? I noticed that the enthalpy of the products of an exothermic reaction (bonds formed are stronger than bonds broken) is lower than the enthalpy of the reactants. Enthalpy equals internal energy plus $PV$. In a constant pressure environment, change in enthalpy equals to change in internal energy as $PV$ in constant. Internal energy consists of $E_k$ (kinetic energy) of particles and $E_p$ (potential energy) of bonds between particles. When stronger bonds are formed, $E_p$ is bigger. And the energy of the heat released to environment equals to the energy gained in form of potential energy, so why is the internal energy lower?
It would be helpful if someone shows me my mistake :)

Bonds actually have negative potential energy. Think about it this way: in order to break a bond you have to expend energy so when a bond is created, energy must therefore naturally be released. Hence internal energy is lower.


Pt[BrCl(PR₃)₂] cis, trans isomer nmr - Which is cis and which is trans? the question is,
The two square planar isomers of $\ce{Pt[BrCl(PR_{3})2]}$ have a different $\ce{^{31}P}$-nmr spectra. One shows single group of lines, the other shows two distinct resonances, each one being similar to the single resonance region of the first one. Which isomer is cis and which is trans?
Thanks.

Look at the two structures drawn below

In the trans-isomer each phosphorous atom is in the exact same chemical environment.  Each one has a chlorine on one side, a bromine on the other side and a phosphorous diagonally across.  Therefore only one signal is observed for the phosphorous.

In the cis-isomer, each phosphorous is in a different chemical environment.  One phosphorous has a chlorine diagonally across from it, the other has a bromine diagonally across from it.  In this case we observe two phosphorous signals, one for each of the two different phosphorous environments.


What is the IUPAC name of K4[Fe(CN)6]? The IUPAC name of $\ce{K4[Fe(CN)6]}$ is:
A. Potassium hexacyanido ferrate(II)
B. Potassium hexacyanido ferrate(4-)
C. Tetrapotassium hexacyanoferrate
D. All are correct

Actually I had been going through the IUPAC exercise of my book when I encountered just this question.
I know that the correct answer is option A but in the answer key the answer is given as option D.
Can this really happen ? Can a compound have more than one IUPAC name?
Our teacher told that whichever name can be used to deduce the structure of the coordination compound can be termed as the IUPAC name of that compound but I am not at all convinced with this fact.

All three would appear accepted by IUPAC, though A is preferred generally.

Quoting from the red book (the IUPAC guide for nomenclature of inorganic compounds):

IR.5.4.2.2:

  It is possible to provide information on the proportions of the constituents in names by using one of two other devices: the charge number, which designates ionic charge, and the oxidation number, which designates oxidation state. In nomenclature, the use of the charge number is preferred as the determination of the oxidation number is sometimes ambiguous and subjective. It is advisable to use oxidation numbers only when there is no uncertainty about their assignment.
  The charge number is a number whose magnitude is the ionic charge. It is written in parentheses immediately after the name of an ion, without a space. The charge is written in arabic numerals, followed by the sign of the charge. Note that unity is always indicated, unlike in superscript charge designations (which are used in formulae). No charge number is used after the name of a neutral species.

  The oxidation number (see Sections IR-4.6.1 and IR-9.1.2.8) of an element is indicated by a Roman numeral placed in parentheses immediately following the name (modified by the ending ‘ate’ if necessary) of the element to which it refers. The oxidation number may be positive, negative or zero (represented by the numeral 0). An oxidation number is always non-negative unless the minus sign is explicitly used (the positive sign is never used). Non-integral oxidation numbers are not used for nomenclature purposes.

Potassium hexacyanidoferrate(II) and potassium hexacyanidoferrate(4–) are specifically given as examples based on this rule, making A and B both valid IUPAC names.

Section IR.5.2.1:

  The proportions of the constituents, be they monoatomic or polyatomic, may be indicated in generalized stoichiometric names by multiplicative prefixes, as was the case for the constituents of binary

Tetrapotassium hexacyanidoferrate is given as an example based on this rule, making C also a valid IUPAC name.

As an aside, the red book specifically makes provision for the fact that, at present, there are multiple allowed IUPAC names:

  A planned future project on inorganic PINs will need to face the problem of choice between the equally valid nomenclature systems currently in use.

This differs from organic nomenclature when one preferred IUPAC name is given (known as the PIN), even where multiple valid names may exist (and are often preferred in everyday use for historical reasons).


Extraction of monosodium glutamate (MSG) from kelp What is the extraction process of monosodium glutamate (MSG) from kelp?
How could one do this at home?
Is there a specific kelp species to use?

Monosodium glutamate was originally discovered by Kikunae Ikeda using a simple aqueous (hot water, namely his soup) extraction of kelp that was later refined and patented.  That particular species was Laminaria japonica. This along with a more detailed description of the original process and the history of MSG processing is given here.

The sulfuric acid treatment and precipitation process described therein certainly cannot be simply and safely done at home. The main purpose of that is to convert other compounds (e.g. glutamine) into glutamate to give higher yields of pure MSG for commercial usage. As stated in this Wikipedia article:

  "MSG is freely soluble in water",

so, as Ikeda enjoyed the water extract, I would suggest trying that first.  If you need a more concentrated extract, simply let the water dry.  I wouldn't use more than mild heat, maybe an oven on very low setting with a fan, to prevent degradation of your product.

An interesting side note, stated in both sources above, is that Ikeda's discovery of MSG was also recognized as the discovery of a new flavor at that time (in addition to sweet, salty, sour and bitter).


what is the difference between excitation and emission spectroscopy? I know about absorption spectroscopy but really confused between excitation and emission spectroscopy. If they are different, then how?

Absorption:

Light with a specific frequency is selected
This frequency is shined (radiated) onto a molecule (sample)
The amount of light that shines through the molecule is measured, ie. the detector is behind the sample (180 degrees).
Step 1 is repeated with a different frequency.

When the sample absorbs a specific frequency, less light of that frequency will shine through to the detector. The machine records this and a peak appears in the spectrum for that frequency.

Frequency is usually measured in wavelength, and a wide range of wavelengths are applicable. In UV-VIS, this can be eg. 300-800 nm, so 800 nm is measured, then 799 nm, then 798 nm...down to 300 nm. In IR, you will go into micrometers. In X-ray, you will be around 10 nm I think.

Emission:

A light with a specific frequency is selected
The sample is irradiated with this frequency.
The amount of light that shines from the molecule is measured, ie. the detector is to the side of the sample (90 degrees).
The emitted light is measured for a range of frequencies
Step 1 is repeated with a different frequency.

In both processes, the sample absorbs light. The major difference here is the detector being located to the side of the sample. This allows it to ignore the light being shined onto the molecule, and instead measure light that is emitted from the sample. Think of the incident light as a column. In absorption spectroscopy, that column is absorbed by the sample and continues to the other side, where the detector can measure it. In emission spectroscopy, you want to avoid the column of light. Think of the sample like a light bulb that shines light in all directions. The detector in fluorescence is placed to the side, avoiding the column of light and picking up the light from the sample instead. The advantage to fluorescence is you can often selectively measure a moiety (part of the molecule) with superior sensitivity, whereas in absorption impurities or stray light are more prone to interfere with the signal and also leads to higher concentrations usually being necessary.

The reason the emission wavelength detected is different than the frequency used to irradiate it is due to Stoke's Shifts.

Stoke's Shift

Stoke's Shift requires basic knowledge of the difference between electronic and vibrational states. A molecule has electronic states, S=0, 1, 2, 3...etc. Within each electronic state is a set of vibrational states, v=1,2,3.... For example, two electrons can be in the 0 electronic state (termed ground state), but one electron can be in the v=2 vibrational state, while the other electron in v=4.

This is easier to picture if you look up Stoke's Shift, or specifically the diagram on Wikipedia's page for Fluorescence.

For example, in emission spectroscopy, an electron absorbs light, being excited electronically from S=0, v=1, to S=1, v=2. Vibrational relaxations will occur before electronic relaxations. That means the electron will first relax from S=1, v=2 to S=1, v=0. After this, it relaxes to S=0, emitting a photon in the process for detection. Because it relaxed its vibrational state first, energy was lost not in the form of light (thermodynamically). As a result, the photon finally emitted is at a lower energy than the photon that had been absorbed. That's the rudimentary explanation.


How to know when dissolution is complete? In a given solubility equilibrium, I know that if $ Q >K_s$, then the salt will precipitate until $ Q =K_s$. If $ Q <K_s$, however, would the precipitate dissolve until $ Q =K_s$ too?

You've described the general tendency for any equilibrium process, including solubility. So, the short answer is yes, you will dissolve more until the reaction quotient is equal to the equilibrium constant.


How to separate alloys? How do you separate metal alloys into their constituent elements, especially industrially? I don't think you can always use melting points, because alloys, depending on the composition (looking at phase curves) can have one common melting point.

Here's a phase diagram for a copper-nickel alloy, which I think is a system where both components are completely soluble in one another in all ratios.

This composition cannot be separated by cooling it down, as I understand it (even though, as mentioned here, the composition would not be uniform if the molten alloy is not cooled down slowly enough). As described here, the solid that would start to separate (the alpha phase) is itself an alloy.

EDIT: What about electrolysis? Could that work here?

You seem to approach the separation of alloys much like the separation of water-soluble salts, probably based on the fact that both involve the liquid/solid phase diagrams, which may even look alike. This is wrong; these systems are fundamentally different.

Say, you have a water solution of some well-soluble salt which you need to recrystallize. You cool it down, so the salt precipitates. You filter the crystals out of the solution, then you probably rinse them with something which would remove the traces of solution without damaging the crystals, or just let them dry.

Now suppose you have a metal alloy, maybe not even of the kind you mentioned above; let it have a simple eutectic diagram, like Bi-Cd.

You melt it, then carefully cool it down to a certain point, and end up with a mixture of some (supposedly) pure metal crystals with a still-liquid alloy. What good does it do? You can't really filter it. You can't rinse the crystals. You can't let them "dry". It is useless.

With metal alloys, your best bet is to dissolve both metals and separate them chemically, based on their different properties. In some relatively rare cases, you might be able to melt the alloy and perform chemical conversion of one component without affecting the other; examples include cupellation or Bessemer process referenced earlier in the comments. In yet more exotic cases, you might be lucky enough to chemically etch one component out of solid alloy; this is how Raney nickel is made. Unless you are that lucky, go back to the start of this paragraph.


Why is 11-oxabicyclo(4.4.1)undeca-1,3,5,7,9-pentaene aromatic? I have been told by my instructors that for a compound to be aromatic, it must be planar. In this given molecule the oxygen seems to be out of the plane of the conjugated system: how can it still be aromatic, then?

The aromatic system should be planar. The oxygen is not part of the aromatic system.
Please note, of course, that your drawing does not capture the 3-dimensional geometry of the system. You would still have to analyze the carbon skeleton to see if it satisfied the planarity constraint enough to be aromatic.


Are carbocations that bad? My professor has an anti-fetish for carbocations. In any mechanism that we draw with a carbocation, the maximum earnable credit is automatically halved. In any case, he draws all mechanisms involving carbocations as concerted reactions; he never wants us to draw an carbocation by itself. If a carbocation could be drawn (as in the case of the formation of an acetal; specifically the step in which water leaves) then we must have electrons flowing from somewhere else to give the carbon an octet.

Now, is this valid? Are all mechanisms involving carbocations concerted? Excluding the obvious case of SN1; I'm pretty sure he'd accept that carbocation (and he hasn't touched on SN1 ever).

Are all mechanisms involving carbocations concerted?

Many carbocations are stable, observable entities.  Many large, delocalized carbocations like trityl perchlorate form very stable salts (reference see p. 196), other carbocations are easily observed by nmr.

If free carbocations didn't exist as discrete intermediates how could you explain the stereochemical results from an SN1 reaction where a chiral starting material generates a planar, long-lived carbocation that can be attacked from both sides to produce an achiral product?  Any type of concerted process would be expected to proceed with inversion of configuration (SN2).

(image source)


Reaction of 1,2-diketones with periodic acid $\ce{HIO4}$ can oxidise alcohols. But does $\ce{HIO4}$ undergo any reaction/act as oxidant with 1,2-diketones?

Yes it does, under basic conditions to give two carboxylic acids, according to this paper by Bunton & Shiner J. Chem. Soc., 1960, 1593-1598  here

Abstract of the paper:

  The oxygen atom of the acetone from oxidation of pinacol and 2-methyl- propane-1,Z-diol by periodic acid comes from the l,Z-diol, in agreement with the suggestion that an intermediate is formed by electrophilic attack of the periodate. In contrast to this, the additional oxygen atom of acetic  acid, formed by oxidation of biacetyl by periodate in basic solution, comes from the periodate. The mechanism is a nucleophilic attack by periodate upon the carbonyl-carbon atom,  and does not seem to involve the hydrates of the ketone as intermediates. In the oxidation of methylacetoin by periodic acid the oxygen of the acetone comes from the hydroxyl group of the hydroxy- ketone, and the additional  oxygen atom of the acetic  acid comes from the periodate. The mechanism  probably  involves  nucleophilic attack by the periodate  ion  upon the carbonyl  group, and co-ordination of the hydroxyl group on to the iodine atom of the periodate.


What is the typical range for the molar absorptivity of paint? For example, $\ce{TiO2}$ is used a lot as pigments in paint, what is the range of molar absorptivity for that? Would I be able to determine this from this graph:

Ideally I could estimate a range for acrylic paints. I really just need a range, and it only needs to be as accurate as a few orders of magnitude.

Well, I cannot say much from the above. Estimation of molar absorptivity experimentally would involve some measurements with different concentrations. I think single measurement may not give that information.

$TiO_{2}$ if present in paint then it is only one component ( ~10-40% or more maybe). Other components may affect the molar absorptivity too.
 If you want to find molar absorptivity of paint (with several components combined) you can just measure paint highly diluted in a solvent  and measure UV-spectra at different known concentrations and use the Beer-lambert law.

For molar absorptivity of only $TiO_{2}$ I have different proposal.

If you have some
$TiO_{2}$ then you can weigh 5 different amounts of it. Then dissolve it in acid (sulphuric acid or mixture with hydrochloric acid or you can check yourself which is better ) to make 5 solutions (should be dilute solution ). You know the concentration for these solutions.

Then, measure absorption with UV spectrometer.
Finally, use the Beer-Lambert law to get the molar-absorptivity based on the absorption peak (~350 nm).

Anyway this is a weird idea. (I am not sure if absorption of acid species, say sulphate ions will affect this measurement.)
And if you really try, let us know.


Do most chemical reactions happen at constant pressure? While studying enthalpy $H=U+PV$ and its changes, I realized I am not clear on the following: chemical reactions happen with the external pressure being constant and equal to the atmospheric pressure $P_{atmos}$. But, in general, the system's internal pressure $P_{gas}$ would seem to be different from $P_{atmos}$ and variable during a reaction...
The change in enthalpy $\Delta H= Q$, where $Q$ is thermal energy, only when the gas internal pressure matches the environment external pressure: $P_{gas}=P_{external}=constant$ Under those circumstances, the emitted or absorbed thermal energy $Q$ goes into change #H# while the gas expands/contracts. The work $PV$ done by the gas is perfectly matched by the $PV$ work done on it by the environment.
How common is for the gas internal pressure to be constant and equal to $P_{atmos}$ during chemical reactions to justify $\Delta H = Q$?

What happens during a reaction is important insofar as reversibility is concerned (a reversible path being a special way of performing a reaction in which equilibrium is sustained throughout), but not when computing $\Delta H$. The condition $\Delta H = q$ is derived from the basic law of conservation of energy and the definition of enthalpy. Because H is a state function, to compute the change in its value you need to know only the initial and final states of the system (and of course the values of H corresponding to those states). For $\Delta H = q$ to hold requires that (1) only expansion (pV) work be performed and that (2) the initial and final pressures of the system equal each other.
You can dream up scenarios in which you start at mechanical equilibrium (balanced pressure) between surroundings and system but end up at a constrained system volume (rigid). For instance, if an explosion moves a piston to a constrained setting within a cylinder. In that case it is not true that the enthalpy change and heat are equal. On the other hand, if the explosion is constrained such that the volume can expand until pressures balance (and external constraining pressure is constant), then enthalpy change and heat are equal.


Carbohydrate conformations I have to questions regarding the conformations of carbohydrates.

There are 38 different conformations for an aldohexose or ketohexose, including $\alpha$ and $\beta$ anomers, namely 2 chairs, 6 boats, 6 skew-boats, 12 half-chairs and 12 envelopes.

Each of these 38 conformations can have 729 different isomers ($5~\ce{OH}$ and $1~\ce{CH2OH}$) because of dihedral and tetrahedral angle ($3^6$ isomers, 3 (gg, gt, tg for each dihedral angle)).

For example, glucose has $2~(\alpha~\text{and}~\beta~\text{anomers}) \times 38 \times 3^6 = 55404$ conformations.

Is this calculation correct?

The nomenclature of the 38 different conformations is still not clear to me, which leads me to my second question.

For chair form, the literature uses $\ce{^4C_1}$ and so on. But I'm not sure, which 4 carbon constitutes the reference plane.

Similarly other used terms are $\ce{^{3,O}B}$ for boats, $\ce{^3S_5}$ for skew conformations.

Can anyone please explain how to interpret these symbols?

After going through several literature, finally I am able to answer these questions. It might be important starting point for someone wanting to know carbohydrate chemistry, that's why I'm posting my answer.

Yes, I'm right. For α- and β-glucose 27702 configuration are possible. So, in total $27702 \times 2 = 55404$ possible structures for glucose [1].

  For xylose, 81 rotamers were optimized per puckering conformation for a total of 3078 conformations. For α-glucose, β-glucose, and β-mannose, 729 rotamers were generated for each pucker, resulting in 27 702 conformations for each of these six-carbon monosaccharides. For GlcNAc, 972 conformations per ring pucker were generated for a total of 36 936 geometries. Thus, the initial set of conformations considered for all molecules in this study comprised 123 201 geometries.

Actually these nomenclature is known as Cremer-Pople (Dr. Pople is the Noble prize winner chemist for his outstanding contribution to formulate DFT) nomenclature. They defined puckering parameters and generalized ring structure conformations for 3/4/5/6 or more carbon species. That's a real interesting world!
Courtesy: Shinya Fushinobu, University of Tokyo
It might not be clear from this picture. So I will try to explain them separately.

Chair: Two chair conformation possible. (O-C2-C3-C5 forms the reference plane). If C-4 is on top then $^4C_1$ and vice versa.

Boats: Boats forms when two plane intersects each other. The bottom two corner atoms are used to distinguish. If two bottom corners are upside down then they are used as superscript on the left and vice versa.

Skew-Boats: Same as chair. Reference plane is formed by (O-C2-C3-C5). If C5 is on the top and C1 is in bottom then it's named as $^5S_1$

Reference

Mayes, H. B.; Broadbelt, L. J.; Beckham, G. T. Journal of the American Chemical Society 2014, 136 (3), 1008–1022 DOI: 10.1021/ja410264d.