Why is the melting point of propane lower than that of both methane and ethane? I've seen their properties in my textbook and multiple online sources, but I can't seem to find a straight answer as to why. Excluding propane, the trend is increased melting point as the molecule gets bigger, but propane is the weird exception here.

Ok, let's first look at the trend...

$$
\begin{array}{c|c|c|c}\hline
\text{Chemical} & n \text{ in }\ce{C_nH_{2n+2}} & \text{m.p. / K} & \text{m.p.}(n) - \text{m.p.}(n-1)\text{ / K} \\
\hline
\text{Methane} & 1 & 90.7 & \\
\text{Ethane} & 2 & 90.4 & -0.3 \\
\text{Propane} & 3 & 85.5 & -4.9 \\
\text{Butane} & 4 & 133\mathrm{-}139 & +50.5 \\
\text{Pentane} & 5 & 144.1 & +8.0 \\
\text{Hexane} & 6 & 177\mathrm{-}179 & +36.9 \\
\text{Heptane} & 7 & 182.2\mathrm{-}183.0 & +4.6 \\
\text{Octane} & 8 & 216.0\mathrm{-}216.6 & +33.7 \\
\text{Nonane} & 9 & 219.0\mathrm{-}220.0 & +3.2 \\
\text{Decane} & 10 & 242.7\mathrm{-}243.9 & +23.8 \\
\hline
\end{array}
$$

Using the mean for the alkanes with melting point ranges given (e.g. 136 for n-butane and 178 for n-hexane), it isn't clear there is any "great" linear trend. In fact it seems that the even number n-alkanes are on one curve and the odd n-alkanes are on another.

There is a "general trend in that with increasing MW the MP increases. However the exceptions would seem to be both methane and ethane, not propane. Also it is obvious that a simple correlation with MW alone is not sufficient to characterize the trends. All in all I think one must be careful not to read too much into such trends without a very careful analysis.


What is made at the copper cathode when electrolyzing water? I just electrolyzed water with $\ce{NaHSO4 -> HSO4- <=> H+ + SO4^2-}$ with copper electrodes.
At the anode this reaction happened:

$\ce{Cu -> Cu^2+ + 2e-} \hspace{26 mm} E_0 = -0.34V$
$\ce{Cu^2+ + 6H2O -> [Cu(H2O)6]^2+}$

Can anyone confirm this last reaction? There was a blue-greenish substance falling to the bottom.
At the cathode this reaction happened:

$\ce{2H+ + 2e- -> H2} \hspace{27 mm} E_0 = 0V$

$\ce{2H2O + 2e- -> H2 + 2OH-} \hspace{5 mm} E_0 = -0.83V$
$\ce{OH- + H+ -> H2O}$

However, additionally, the cathode ($\ce{Cu}$) turned black. What reaction could have happened there? I electrolyzed at $12V$ using a car battery charger.

Look at your redox potentials for the half reactions.  12V is tremendous overkill.  At the cathode (reducing) you will make hydrogen to which copper is inert.  Could you reduce sulfate to sulfide or sulfur and have that react with copper?  CuS is black.

At the anode (oxidizing) you will oxidize water to oxygen - but one atom at a time.  It will be very reactive to the copper surface.  CuO is black.  You will also oxidize copper to Cu(II).  It coordinates four waters square planar.  The apical waters are distant and weakly bound given the Jahn–Teller effect,

http://chemwiki.ucdavis.edu/Inorganic_Chemistry/Coordination_Chemistry/Coordination_Numbers/Jahn-Teller_Distortions
http://www.adichemistry.com/inorganic/cochem/jahnteller/jahn-teller-distortion.html

One possibility is to add a series resistor to lower the working voltage.


How to determine the most stable carbanion? Which of the following carbanions is the most stable?

I think the answer is either a or c, as in d there are no chances for the carbanion to have its charge become less. In b I think there is a + inductive effect due to methyl group which gives it more negative charge.
The ones left are a and c.  Nitrogen is an electronegative element so it might lessen the charge on the carbanion, but in both a and c the carbanion is same distance from it. I am unable to reason out it from here.

Wikipedia says:

The stability and reactivity of a carbanion is determined by several factors. These include
1)The inductive effect. Electronegative atoms adjacent to the charge will stabilize the charge;
2)Hybridization of the charge-bearing atom. The greater the s-character of the charge-bearing atom, the more stable the anion;
3)The extent of conjugation of the anion. Resonance effects can stabilize the anion. This is especially true when the anion is stabilized as a result of aromaticity.

You can see that point 3) is not relevant here. We do not have a conjugated system, nor is there any chance for resonance.
Keeping in mind point 1), the $\ce{N^+}$ has a $\ce{-I}$ effect, but inductive effect reduces over distance. The negative charges in options a), b) and d) are either in the meta or para position.
Also, remember that $\ce{+I}$ effect destabilises the carbanion.
In c), the carbanion is primary, but in a), it is secondary. Primary carbanions are more stable due to lesser $\ce{+I}$ effect.
So your answer is c).


Which has more metallic character: aluminium or magnesium? Which element has more metallic character - aluminium or magnesium?

I know that metallic character decreases along a period (from left to right) and increases down a group.

Aluminium comes after Magnesium in the same period (period 3) in the periodic table. So, I thought Magnesium would have greater metallic character.

However, I read in a book that Aluminium has greater metallic character. Is this true? If this is true, why?

Two types of metallic character
In fact, there are two type of metallic character if you look at the metal from the chemical point of view or if you look at the metal from the physical point of view:

So it really depends on how do you define metallic character

Chemical metallic character
Since metallic character in chemistry is defined as:

the tendency of an atom to lose electrons.

We can understand that it is directly related to electronegativity.

chemical property that describes the tendency of an atom to attract
electrons (or electron density) towards itself

Basically, it is the opposite (that would be Electropositivity...)! An so it increases from right to left. The picture below shows the electronegativity but can be used for the purpose.

I think the reasoning that the book wants you to do is not remembering all the subshell but just the basic chemistry principles.
So from the chemical point of view $\ce{Mg}$ has a greater metallic character compared to $\ce{Al}$.
Physical metallic character
The physical metallic character is defined taking in account the physical properties of the metal. A metal, from the physical point of view, has a high electrical conductivity, high thermal conductivity, and high density.
These properties are harder to predict:

In this case, however, $\ce{Al}$ has a more metallic character compared to $\ce{Mg}$.


What bond exists in "red oxygen", and can something similar happen with hydrogen? "Red oxygen" is the nickname for O$_8$. It only exists at extremely high pressures.

What kind of bond does it have? Can hydrogen theoretically have something similar at very high pressures too?

That is what I found on wikipedia. It has a cubic structure and it was not predicted theoretically.
here the pressure is not very high but is high(10 GPa)
High-pressure studies of hydrogen typically employ a diamond anvil cell (DAC) made of two diamonds holding a few-micrometer-thick gasket containing a tiny sample of hydrogen.
These are some more images I could found of hydrogen at high pressures, but I do not think any of them are like red oxygen.

Sources

Bonding, structures, and band gap closure of hydrogen at high pressures

This is a must read: Solid Hydrogen at High Pressure


Acid hydrolysis of ethyl acetate uses I was wondering if the above mentioned reaction has any uses? I looked up applications and significance and couldnt find anything. I came across Saponification but that involves a base and is a completely different reaction. Then I thought the products maybe useful but i realised that ethanol and acetic acid are used to produce said ester. then what exactly is the point of this reaction?

The products of the reaction are perfectly miscible with water and are harmless to environment in low concentrations. My guess that the leftovers of ethyl acetate can be turned to ethanol and acetic acid before disposal.

For example here you can see that ethyl acetate is forbidden from sewer disposal while ethanol and acetic acid are allowed.

https://www.vumc.org/safety/waste/chemical-waste-sewer-disposal


Crystallization of solutions and liquids The process of crystallization of a liquid or any solute occurs when its concentration exceeds the maximum concentration which the solvent can carry or when the temperature drops below the freezing temperature in cases of liquids. Then why is it that these materials require a small "nuclei" or a jagged edge to start forming crystal on, why cant it form crystal anywhere within the bulk of the matter. (eg cloud formation from condensation of water)

Crystallization is rightly called by you a process.  Simplified, there are two steps to be considered: Nucleation and crystal growth.  For the sake of simplification, let me consider the crystallization of a molecular compound in absence of ionic charges.

Nucleation refers to the initial stage, where once isolated molecules of the dissolved substance assemble to a first crystal germ. This is energetically hindered because these particles of especially small size have unfavourably high ratio of particle surface energy per molecule in these just formed particle (or perhaps better: aggregates). This energetic drawback has to be overcompensated. In addition, molecules loose degrees of freedom accessible in solution, at first: dynamic freedom, and secondly, depending how strictly rules are followed to build the crystal, a number of orientations in space otherwise available and accessible--In a perfect crystal, the structural close environment of one molecule is going to be repeated for all molecules, the very next ones and for their next ones, over ranges of Angstroms to Centimetres.

Once such a germ is formed,

Crystal growth more or less refers to simply attachment of additional molecules according to these "(energetically justified) previously determined rules" to yield a bigger entity.  The need of oversaturation of the solute is less prominent than during nucleation, yet still existent to energetically favour this process. In this description, crystal growth is again simplified, too.  It may happen, in the (sub)process of crystal growth, conditions are met that allow crystallisations in realisations even more energetically favourable than at early stages of crystal growth so that a compound preferentially forms a polymorph A initially, then ones in polymorph B. And often a crystallization does not yield only one (perhaps even single)crystal, rather than multiple ones.  So nucleation may occur in one spot of the solution while simultaneously crystal growth already takes place on germs already formed in other parts of the solution.

Complementary, think this way: it is every time easy to dissolve a perfect crystal (already considering entropy), but easily difficult to bring molecules back into a perfectly ordered crystal.

The science and art (like craftsmanship) of crystallization is of high importance in our daily life, spanning areas of electronic devices (semiconductors) till pharmaceutics. Journals like Crystal Growth & Design or CrystEngComm are examples concentrating on this subject.


Are there any good examples of commonly ingested molecules that contain particular toxic individual elements? In the house on my own and an article popped up on my feed with the following statement:

  While the Government has insisted the chemical is safe, cadmium is
  recognised as a cause of lung cancer and during the Second World War
  was considered by the Allies as a chemical weapon.

The proceeding paragraphs are about zinc cadmium sulphide.

full article

I think this is a flawed argument, essentially inferring that any given molecule can be said to be toxic just because one of it's component parts is toxic but I can't think of a good example to back this up.

What is a good example of a common compound that is ingested often (daily?) but contains a particularly toxic element if that element were ingested in some way on it's own?

Table salt! What would be worse than putting sodium (it can spontaneously combust if you get it wet) and chlorine (used as a war gas in WWI) all over your food?

Then there's water which always contains some $\ce{OH-}$ (an active ingredient in Drano) and $\ce{H+}$ (that stuff eats through metal, man). Although those aren't elements.

And hiding inside every $\ce{O2}$ molecule are two deadly oxygen atoms; once freed, they'll wreak havoc on just about anything they come into contact with.


What does CkC mean in chemical bonding? I'm working on a chemistry question where I'm supposed to use average bond energies to estimate enthalpy changes for a reaction. It gives me these bond energies to work with:

$$\begin{align}
\ce{C-H}&=&413\\
\ce{C-C}&=&348\\
\ce{C-Br}&=&276\\
\ce{H-Br}&=&366\\
\ce{CkC}&=&614\\
\end{align}$$
I know how to do the question, but I don't what $\ce{CkC}$ is. I can't find it online or in my textbook. What does the "k" mean?

$\ce{C=C}$ average bond energy is somewhere around $\pu{614 kJ/mol}$ so "k" probably indicates a double bond


Naming unsaturated hydrocarbons $\ce{CH2=CH-CH=CH-CH2-CH3}$

Is this hexa-1,3-diene or hex-1,3-diene?
And why?
When do you add an extra a and when do you not?

According to the general methodology described in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), a double bond is expressed by changing the ending ‘ane’ of the name of the corresponding saturated parent structure to ‘ene’.

  P-31.1.1.1 The presence of one or more double or triple bonds in an otherwise saturated parent hydride (…) is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. (…)

For example, the name ‘hexane’ is changed to ‘hex-1-ene’.

If the compound contains multiple double bonds, a multiplicative prefix is placed before the ending ‘ene’ and the letter ‘a’ is inserted between the root of the name of the parent structure and the multiplicative prefix.

  P-31.1.1.2 The multiplying prefixes ‘di’, ‘tri’, etc., are placed before endings denoting unsaturation to indicate the number of multiple bonds of each kind, as required, for example, ‘diene’ and ‘triyne’. For euphonic reasons, when the endings ‘ene’ and ‘yne’ are preceded by a multiplying prefix and a locant the letter ‘a’ is inserted. (…)

See also:

  P-16.8.2 For euphonic reasons, the letter ‘a’ is inserted between the root of the name for polyenes, polyynes, and polyenynes, and the numerical multiplicative prefix ‘di’, ‘tri’, etc., preceding the ending ‘ene’ or ‘yne’.

Therefore, the correct name for the compound given in the question is hexa-1,3-diene (not hex-1,3-diene).


Citric acid and ammonium nitrate Soil based question. I have soil with to much Ammonium Nitrate in it. I am wondering if I use citric acid and water mixed will that flush out the extra Ammonium Nitrate that has built up in the soil.

Will citric acid bond to Ammonium Nitrate?

Any incite on these to compounds will be greatly appreciated.

Ammonium nitrate is very soluble in water, so watering the soil will remove it, though, depending on the amount and dilution, it could then pollute streams, lakes and other bodies of water, possibly causing an algae bloom.

Your best choice might be to experiment with nitrate-tolerant plants, such as lettuce, ornamental cabbage and rhubarb, or perhaps bell peppers.

Citric acid will not help in any way, and would make the soil less likely to support life.

In nature, denitrifying bacteria will slowly reduce the $\ce{NH4NO3}$. In an anaerobic environment, it might be converted to $\ce{NH4+}$.


Can't decide between E1 and E2 mechanisms for this reaction I can see that due to the substrate being bulky, an elimination reaction will take place and not substitution. The reagent is a strong base. E1 reactions are independent of the base chosen, and E2 requires strong base. There is no information about the geometry of the substrate (i.e., whether the beta hydrogens are anti to the halide). How do I decide between the two mechanisms?

You have a tertiary halide. These dissociate rather easily into the relatively stable tertiary carbocation and the free anion. That in and by itself should be indication enough that the reaction proceeds via E1. Alcoholates are only moderately strong bases so the mechanism cannot be $\mathrm{E1_{cb}}$.

You could consider E2 because the substrate is non-cyclic which means that it can always orient itself in such a way that a hydrogen atom is anti with respect to the halide. The only thing to take care of is steric hindrance; just draw the appropriate Newman projections to determine the likelyhood.

Once you have the tertiary carbocation, any of the β hydrogens could dissociate. You would likely initially get some kind of mixture based on the corresponding $\mathrm pK_\mathrm a$ values weighted by the number of hydrogens. Looking at the three different carbons, one is obviously much more acidic because it is benzylic. This explains the major product.

Even if initially another hydrogen is deprotonated, you can assume that equilibration takes place towards the conjugated styrene system given catalytic excess base.


Why are antibonding orbitals higher in energy? I found an explanation on Quora.com

https://www.quora.com/Why-are-antibonding-orbitals-more-destabilized-or-higher-energy-when-theyre-between-two-atoms-of-very-different-electronegativities

I wasn't satisfied with their reason about electrons being inside the two nuclei region.

Why exactly does electrons being shared between atoms reduce the energy? What's the quantum mechanical and mathematical argument? What's the quantum mechanical and mathematical explanation of the octet rule?

Electrons being shared sounds like repulsion forces would increase. Anti bonding sounds like a way to reduce electron repulsion by the coulomb force.

You shouldn't stop asking "why," but really explaining why is really, really complicated. It sort is like learning to run. First you learn to waddle around, then you can walk, and then you can run. Starting off such an explanation with quantum mechanics isn't throwing you in the pool it is more like throwing you in the middle of the Pacific Ocean.

Start with this. I assume that you're happy with the shell structure of atoms. Now we want to make molecules.

Consider the reaction: $$\ce{H + H -> H2}$$

Two hydrogen atoms each with a $\ce{1s^1}$ electron bond to form a "molecule" with 2 electrons in some sort of orbital. There is no shell model for a molecule, so how do we conceptualize the two hydrogen orbitals combining? A straightforward approach is the linear combination of atomic orbitals (LCAO). So we have a $\ce{1s}$ orbital from the first hydrogen and a $\ce{1s}$ orbital for the second hydrogen. Remember that each of these orbitals can hold two electrons. We now, with some nativity, just assume that the energy of atomic orbitals must be conserved. So how much in energy the molecular orbital is lower than the hydrogen orbital then the other orbital must exist with that much "antibonding" energy. But for $\ce{H2}$ there are only two electrons, thus the antibonding orbital is empty and the net result is that a $\ce{H2}$ molecule has less energy than two lone H atoms and the molecule will stay together.

Now consider the reaction:
$$\ce{He + He -> He2}$$
Both He atoms have two $\ce{1s}$ electrons. Thus when we try to form a $\ce{He2}$ molecule both the bonding and the antibonding orbitals are occupied. The net result is that the atoms in the $\ce{He2}$ molecule have no net affinity to stay bonded together. Notice also that this would not be true for a $\ce{He2^+}$ ion. With two electrons in the bonding orbital and only one in the antibonding orbital the ion would be "stable" in free space - at least until the ion bumped into some other atom or molecule in interstellar space.

Now using LCAO is naive. However it works well for a lot of chemical reactions. The really nice thing is frankly that the mathematics is fairly simple. Think of Newton's gravity equation. It worked well enough to put men on the moon, but it isn't exact. Now with GPS satellites the GPS clocks must be corrected for relativity. Scientists knew about relativity before we went to the moon of course, but the expanded mathematical treatment just wasn't necessary.

So does that answer your question in enough detail so that you can accept the explanation?

Edit - adding information about quantum mechanics

The easiest treatment of atomic orbitals is to assume hydrogen like orbitals and tweak the value of the orbitals based on the interactions of electrons. The gist here is that the simplest mathematical treatment assumes that the electrons move around the nucleus in defined orbits like the planets move around the sun. It was quickly realized that this sort of modeling was woefully inadequate.

Even with this simple treatment there is a massive fundamental problem that bedevils all such calculations. The three body problem in physics. If our solar system just had the sun and the earth, then it would be possible to use Newton's gravity formula to solve the equation which predicts the movement of earth exactly. However when the rest of the solar system is added, then an exact solution is no longer possible. We can solve the movement numerically by essentially calculating where the planets will be after some time $t$. So based on where all the bodies are in the solar system we calculate a vector for earth and all of the other planets and advance the time by an hour. We repeat the calculation 525,960 times and we can predict where all the planets will be exactly one year from now.

Now skipping several other mathematical models which offered better solutions than the planetary model, we arrive finally at the Schrödinger equation which gives quantum mechanical calculations for orbitals. The high level results are threefold. First the Schrödinger equation is subject to the same three body problem as before so an exact solution is only possible for a system with one electron and the nucleus. Second the electron orbitals are not discrete paths through 3D space but rather smeared out probability functions. Third different types of suborbitals (s,p,d and f) have different shapes.

Now with the more realistic model for atomic structure we can again combine atomic orbitals to get molecular orbitals, with the new constraints introduced by quantum mechanics. But as we did before, when we create a molecular bonding orbital from two hydrogen atoms, then we get an antibonding orbital as well. Thus the same explanations used for hydrogen and helium bonding as before still apply, however the actual values of the bonding energies do change.

I'd be remiss if I didn't point out something else. Going back to the the motions of the planets in the solar system we also stumble into something known as chaos theory in mathematics. Because of all the interactions small errors in the mass or velocities of the planets would result in much different results for the calculations. This is important in chemistry because the ethane molecules in a gas aren't a static entities, but rather dynamic ones. The molecules are rotating and the atoms are vibrating. Molecules are bumping into one another causing rotations and vibrations to change.

Now trying to model all of this in one model is too complex for us to handle yet computationally. So chemists build models for different aspects of the problem and then try to piece together an overall picture.

In the olden days, AAA would print a route map for someone wanting to drive from Augusta, Maine to Tampa, Florida. They didn't draw the whole route on a massive roll, but rather pieced the route together. Each page showed the pertinent decisions relative to that stretch of road. So a page in a large city might be a few miles, but out in the country a page might be a 100 miles.

So rather like AAA, professional chemists don't have one massive bonding model that we keep hidden from new students. Rather we describe atomic orbitals and molecular bonding in a number of different ways depending on the situation. By piecing the overall model of a molecule together we can understand predict overall behavior.


Mixing two ideal gases I had total blackout with these ideal gas equations as I haven't needed these in years. If I have two gasses, A and B which both have same volume V and quantity N. We also have temperatures T1 and T2.

Now, I let energy change between these two ideal gases. How do I get the final temperature out using these informations?

I realise we take the ideal gas equation $p = \frac{NRT_1}{V}$ and get the answer from there, right?

Because you aren't told, there are three possible cases to consider (well there's more, but these are the natural cases):

Two monoatomic Gases
One monoatomic and one diatomic gas
Two diatomic gases

We should just note that the equilibrium for this system will be the state at which the temperature of each system is equal. We want to find where that temperature ends up between $T_1$ and $T_2$. Let's call this $T_f$. So we have,$$T_1\le T_f\le T_2.$$

Also note that this a constant volume process not a constant pressure one.

$\textbf{Two Monoatomic Gases}:$

The heat capacity of a monoatomic ideal gas is simply $C_v=\frac{3}{2}nR$.

Then, the internal energy of the system at a given temperature is,$$U=\frac{3}{2}nRT.$$
Then, the change in internal energy of one system is the negative of the change in internal energy of the other,$$\Delta U_1=-\Delta U_2$$so that,$$\frac{3}{2}nR(T_f-T_1)=\frac{3}{2}nR(T_2-T_f)$$
Upon rearranging this yields,
$$T_f=\frac{T_2+T_1}{2}$$which gives the (perhaps) intuitive result that the final temperature the gases is simply the average of the two initial temperatures for the fairly restrictive conditions we have here.

$\textbf{One Monoatomic and One Diatomic:}$

The only thing different here is that the heat capacity of a diatomic gas is $C_v=\frac{5}{2}nR$. That is, we pick up an extra $\frac{1}{2}k_bT$ for each rotational degree of freedom per molecule.

It is still gonna be true that changes in energy are equal and opposite so we can skip straight to that step and write,
$$\frac{5}{2}nR(T_f-T_1)=\frac{3}{2}nR(T_2-T_f)$$but we see now that it actually matters now whether the hotter or colder bath had the diatomic molecules. For the expression above, let's say $T_1$ was colder than $T_2$. This gives,
$$T_f=\frac{3}{8}T_2+\frac{5}{8}T_1$$
And by the symmetry of the situation, if the diatomic gas started at the hotter temperature $T_2$, we just switch the coefficients and get,
$$T_f=\frac{5}{8}T_2+\frac{3}{8}T_1$$

This basically tells us that the system with the larger heat capacity needs more energy input than the other.

$\textbf{Two Diatomic Gases}$

Well, this actually ends up being exactly the same as the monoatomic gas case as I'm sure you can convince yourself. So, here the final temperature is simply the average of the initial temperatures as well.

$\textbf{Further Cases}$

Note that I used a linear diatomic which only has two rotational degrees of freedom, but one could easily imagine using water vapor which has three rotational degrees of freedom and combining this with a diatomic or a monoatomic. The form of the answer is always gonna be the same as above though. Just two coefficients in front of the initial temperatures which sum to one.

I'm pretty sure I haven't gone astray, but if I have I'm sure someone will promptly inform me :)

Edit:

It just dawned on me that the question title and the body of the question don't say the same thing. That is, letting two systems of gases equilibrate by heat exchange is different than having them separated and allowing them to mix. I haven't really thought about it, but because the initial volumes of the containers are the same, the answer might be the same as what I write above, but the answers probably aren't the same in general.

Maybe someone else can answer the gas-mixing case.


Why are crystalline solids anisotropic? Why crystalline solids are anisotropic whereas amorphous solids are isotropic?

Is it because in crystalline solids, X, Y and Z dimensions vary, but so does in amorphous.

Please write as simple as possible.

To qualify as «crystalline», there must be a minimal unique pattern (unit cell) to construct the solid only by applying regular, periodic translations. You may predict the outcome of the translation both in proximity, as well as in far distance reliably.  In an amorphous state, there is no such predictability.

For two, anisotropic means that a vectorial property depends on the orientation within the sample.  In the case of an amorphous sample (as contrasting to crystalline) like glass there is no orientational difference.  As an example, the speed of light in glass is independent on the direction you look at the sample; thus, glass is an isotropic material.

In a crystal, however, the speed of light depends on the relative orientation of the sample, thus crystals generally are anisotropic.  Depending on the symmetry of the underlying Bravais lattice, there may be two, or three special directions of observation along those vectorial properties are equal to each other (e.g., uni- and biaxial crystals in birefringence).  Then, these directions of observation are isotropic to each other; allong all other directions of observation, however, the crystal's vectorial properties are anisotropic.


How can you predict relative Rf values of cis-trans isomers in thin layer chromatography? This is a "homework" question.

I'm writing up a lab in which we synthesized 1,4-diphenyl-1,3-butadiene. We analyzed the products using TLC with 4:1 Hexane:Ethyl Acetate as the solvent.

One of the isomers has a larger Rf value than the other (0.83 and 0.95).

I'm having trouble understanding how to figure out which of the isomers is which on the TLC. I know more polar compounds are going to have a lower Rf, like cinnamaldehyde.

My best guess is that, although the all the double bonds are conjugated, the phenyl rings have a higher electron density than the rest of the molecule. Since the trans,trans isomer is essentially symmetrical, it has little or no molecular dipole. The cis,trans isomer has a small molecular dipole because it is asymmetric. Therefore the cis-trans isomer has more affinity for the stationary phase (giving it the Rf value of 0.83).

Am I on the right track here?

Well, apparently I was wrong.

In fact, the trans,trans isomer is adsorbed more strongly by the stationary phase because it is entirely planar. The cis,trans-isomer cannot adopt a planar conformation because of steric hindrance between the two phenyl rings. The non-coplanar geometry lowers the efficiency of interactions with the stationary phase, making it more mobile.

So the cis,trans isomer has a higher Rf, and the trans,trans isomer has a lower Rf.


Let us say that height of HCP is h then what is the distance of tetrahedral voids In a hexaoal cose-packedstructue et $h$ be the spacing between equivalent basal planes.  Then what are the "heights" of the tetrahedral voids lying between any two adjacent, equivalent basal planes, as distances from either plane?

The book gives the answer $h/8$ and $(7h)/8$. How is this derived?  My (the original asker's) expectation, based on how I understand the hcp structure, was that the heights would be $h/4$ and $(3h)/4$.

Actually $h/8$ and $(7h)/8$ are correct ... but not the whole truth.  TL, DR the correct set is $\{h/8,(3h)/8, (5h)/8, (7h)/8\}$.

Here is how to get $h/8$ and $(7h)/8$:

Think of the unit cell as consisting of a "base" having parts of seven atoms, then three middle atoms entirely inside the cell, then the top layer having parts of seven more atoms.  Define $h$ as the height from the middle of the base atoms to the middle of the top atoms.  Then each of the three atoms at height $h/2$ above the base is the common vertex of two tetrahedea.  One has its base at the base of the larger cell, and since the center of the tetrahedron is one fourth of the way from the base to the apex it's height is $(h/2)/4=h/8$.  The second of the two tetrahedea is upside down, with its base at the top of the cell, and the height of its center is $h-(h/8)=(7h)/8$.

Now for the curveball: these cannot be all solutions.  We forgot to include the possibility of voids involving atoms from surrounding cells!  If we include these surrounding atoms we discover that there is a layer of tetrahedral voids at $h/8$ above the mid-level, matching those at $h/8$ above the base, and another layer at $h/8$ below the midlevel matching those just below the top.  These additional voids are hard to "see" because they involve atoms in surrounding cells.  Nevertheless they are there, the complete solution set is $\{h/8,(3h)/8, (5h)/8, (7h)/8\}$, and an equal number of tetrahedral voids (three per cell) lie at each of these heights.


Helicity in Molecular motors and topological isomers I'm working on a project to understand the work of the 2016 Nobel laureates and I've come across a problem of not being able to find the right sources to understand basics of helicity (more specifically right-handed
helicity (M) and lefthanded helicity (P)) to understand things like the diagram below (I understand cis, trans, I don't know much about the (P,P) and (M,M)).

I would like some suggestions for books or articles that would help me learn more about these concepts.

Thanks in advance!

Any pretty non-old organic stereochemistry book should work.

M/P notation

You present some organic compounds, so let's see definition from the Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013:

  P-92.1.1.2 The helicity rule: stereodescriptors ‘M’ and ‘P’

  Helicity is the chiral sense of a helical, propeller, or screw-shaped molecular entity (see ref. 37). The ‘helicity’ rule is expressed by Prelog and Helmchen (see subsection 5.1, ref. 36) as: “Depending on Whether the identified helix is left- or right-handed it is designated ‘minus’ and marked ‘M’ or ‘plus’ and marked ‘P’ ”.

  The application of this system to the description of conformations and configurations considers the torsion angle between two specified (fiducial) groups are attached to the atoms linked by that bond. The sign of the smaller torsion angle between the fiducial groups defines the sense of chirality sense of the helix (see torsion angle, P-94).

  P-92.1.1.2.1 The chirality of hexahelicenes is denoted by the stereodescriptors ‘P’ and ‘M’.

  (P)-hexahelicene (PIN)                  (M)-hexahelicene (PIN)

  (…)

The description follows further, the rule can be applied also for biphenyls, allenes, or similar structures, I'd say where along the axis some two approx. perpendicular groups are in clockwise or anticlockwise rotated positions.

The arrows show the helicity orientation in the direction from the observer; but may be confusing(opposite), when one applies common hand and thumb methods for determining a screw thread handedness, but results in the same correct helicity orientation/descriptor.

Δ/Λ notation

In the question you're mentioning molecular motors. These can be also propeller like structures. If they were from inorganic chemistry world, they could use different notation Δ/Λ; from IUPAC Red Book 2005:

  IR-9.3.4.11 The skew-lines convention

  Tris(bidentate) complexes constitute a general family for which a useful, unambiguous convention has been developed based on the orientation of skew lines which define a helix.
  Examples 1 and 2 represent the delta (Δ) and lambda (Λ) forms of a complex such as $\ce{[Co(NH2CH2CH2NH2)3]^{3+}}$.   (…)

            2.
  delta (Δ)                      lambda (Λ)

This one is the Δ isomer:

Δ/Λ likely corresponds to organic's P/M respectively (however, in some structures, there might be doubts which parts of the structure should be interpreted as the propeller blades.. see e.g. Q. Molecular knots absolute configuration)

The structures in your ask, 3,3′-dimethyl-2,2′,3,3′-tetrahydro-1H,1′H-4,4′-biphenanthrylidene, do have a combination of a geometrical isomerism at the connector double bond, and two chirality axes (perpendicular to the double bond connector).


Bleach in water only good for one day? In this time of COVID-19 our establishment is being told by government sources to spray down surfaces with a 9:1 water bleach solution, and then told to throw it out within 24 hours. What makes it less effective when we have over the counter products with bleach in them that obviously sit on the shelves for months and years?

Commercial liquid chlorine bleaches, used in the laundry, are stabilized with sodium hydroxide at high pH. When you dilute the bleach in water, the alkali concentration also decreases. As a result, the active ion (hypochlorite $\ce{OCl-}$) slowly begins to decompose to the chloride ion (the common salt).


Bonding of the complex when both weak and strong field ligand are present In the complex $\ce{[Ni(H2O)2(NH3)4]^2+}$ the magnetic moment ($\mu$) of $\ce{Ni}$ is –

Zero

$2.83~\mathrm{BM}$

$1.73~\mathrm{BM}$

$3.87~\mathrm{BM}$

It is easy to find the unpaired electrons and hence magnetic moment if we know the bonding strcture.
$\ce{NH3}$ and $\ce{H2O}$ are strong and weak field ligands respectively according to spectrochemical series. I understand that pairing of electrons occur in the case of strong field ligands and it does not occur in the case of weak field ligands.
Nickel has $\ce{d^8}$ configuration here. I dont understand whether I should pair this electron or not. If I pair this electron, I get this:

If I pair the electron then I have to use one $\ce{3d}$ orbital and one $\ce{4d}$ orbital. I have not used both orbitals for overlap in any other question before so I don't know if that's right. Well, I think it should be because anyway these orbitals will make hybridized orbitals of equal energy for bonding. In that case, we get $\ce{d^2sp^3}$ (octahedral geometry).
If I do not pair the electron, I get $\ce{sp^3d^2}$ hybridization.
I am not sure but is this logic correct that since there are more number of strong field ligands than weak field ligands, pairing will occur?
Should I pair the electrons or not and why?

$\ce{NH3}$ and $\ce{H2O}$ are strong and weak field ligands respectively according to spectrochemical series.

No, this is not correct. I keep saying this time and time again. Ammonia is a medium-field ligand at best, and water is the defined ‘zero point’ below which weak field ligands begin.

  I understand that pairing of electrons occur in the case of strong field ligands and it does not occur in the case of weak field ligands.

This is simplified at best. The transition from low spin to high spin depends on both the metal and the ligands (and also the geometry). In fact, some complexes are known both in a high-spin and in a low-spin state with identical central metal and ligands. You always need to consider the big picture to determine whether a high spin or low spin complex will result.

I am also going to completely ignore the stuff you wrote about hybridisation. You should not consider hybridisation in transition metal complexes as I have posted numerous times.

To answer this question you should ask yourself how many ligands you have around the nickel centre and which geometry that implies. The complex being tetraamminediaquanickel(II) results in six ligands and thus an octahedral configuration. The next step should be to consider the molecular orbital scheme of an octahedral complex (although the d-orbitals alone carry all the necessary information so even the crystal field model can answer it).

Figure 1: simplified MO scheme of an octahedral $\ce{[ML6]}$ complex ignoring π $\ce{L\bond{->}M}$ interactions. Originally taken from Professor Klüfers’ web scriptum of his coordination chemistry course at the LMU Munich.

The important bit to note is that the d orbitals split into a $\mathrm{t_{2g}}$ and an $\mathrm{e_g}$ part. $\mathrm{e_g}$ corresponds to the $\mathrm{d}_{z^2}$ and $\mathrm{d}_{x^2- y^2}$ orbitals. Thus, a perfect octahedral geometry will generate two degenerate orbitals which should be filled with parallel spins by Hund’s rule. Even if you start to assume Jahn-Teller distortions here (which are not likely, since a high-spin $\mathrm{d^8}$ octahedral configuration is not susceptible to Jahn Teller because there are no unevenly populated degenerate orbitals), that would not lead to an energy difference large enough to induce spin pairing. Remember that a typical 3d-metal octahedral complex is high-spin even in $\mathrm{d^6}$ cases; the energy difference between $\mathrm{t_{2g}}$ and $\mathrm{e_g}$ is generally much larger than what can be achieved by Jahn-Teller distortion.

Now you may decide to argue using square planar complexes. However, the complex in question cannot be square planar since there are six ligands.

Therefore, a magnetic moment of zero is not an option in the case presented; $\ce{[Ni(H2O)2(NH3)4]^2+}$ cannot reasonably adopt a low-spin configuration. The high-spin complex is paramagnetic and in a triplet state ($2S+1 = 3$).


"Orthogonality" of chemical elements Chemical elements are said to be immutable at normal energies, but not at the energies that are orders of magnitude higher.

Given that nucleus at normal energies gives us the Periodic Table and at higher energies becomes unstable, can we say that (with fusion) we will be able to transform an arbitrary chemical element into some other chemical element (if we have enough energy)?

Certainly nuclear fusion and nuclear fission can convert elements into other elements. Indeed, this is how different elements form in the first place.

The one issue I have is with your "arbitrary chemical element." You can use fusion to put together lighter nuclei or elements into heavier ones - up to iron, or use fission to divide more massive nuclei into lighter ones.. up to about iron.

The limit here is the binding energy of the nucleus which maximizes with $\ce{^{56}Fe}$.

In general, though, your point is a good one. Given enough energy (e.g., in a supernova), you can create lots of different elements.


Conductance and Conductivity of acid base titration? I have a question with $\ce{NaOH}$ initially present with given concentration and conductivity, then an equal volume of given concentration of HCl is added and the final conductivity is given, again a same volume of HCl of same concentration is added, the conductivity of which is again given. Now I need to find molar conductivity of $\ce{HCl ,NaOH, NaCl, H2O}$.

Molar conductivity($\lambda_m$), conductivity($\kappa$), Molarity($M$)

Using the formula $\lambda_m=\kappa/M$,firstly I found the molar conductivity of NaOH, afterwards when HCl was added, there will be only NaCl remaining, the rest being used up, so I again found molarity of NaCl and put in the formula and found molar conductivity, but in the last part I am confused, there is only HCl and NaCl remaining and I think one of these must be true:
$$\kappa_{solution}=\kappa_{\ce{NaCl}}+\kappa_{\ce{HCl}}\\\lambda_{solution}=\lambda_{\ce{NaCl}}+\lambda_{\ce{HCl}}$$
Probably the second one due to Kolrauch's Law? Also I am not aware of what is going on here because I'm just using a pretty formula. Please explain in detail the physical situation and the processes involved.

Edit:
The question is:

  The conductivity of $0.1$ M $\ce{NaOH}$ is $0.0221\;\Omega^{-1}cm^{-1}$. When an equal volume of $0.1$ M $\ce{HCl}$ solution is added, the conductivity decreases to $0.0056\;\Omega^{-1}cm^{-1}$. A further addition of $\ce{HCl}$ solution, the volume of which is equal to that of first portion added, the conductivity increases to $0.017\;\Omega^{-1}cm^{-1}$. Calculate $\Lambda_m(\ce{NaOH}),\Lambda_m(\ce{NaCl}),\Lambda_m(\ce{HCl}),\Lambda_m(\ce{H^+,OH^-})$

As the concentrations of $\ce{HCl}$ and $\ce{NaCl}$ are equal, that concentration can be regarded as the "concentration" of the solution. If you divide both sides of your first equation (which includes $κ$) with that concentration you would get your second equation (which includes $λ$). So both equations are valid. The problem would arise if the concentrations of $\ce{HCl}$ and $\ce{NaCl}$ are different, you couldn't define the concentration of the solution, and you couldn't calculate it's molar conductivity, therefore only the first equation would be usable. Note that the Kolrauch's Law can also be expressed through $κ$.


Naming a molecule with both alcohol and aldehyde functional groups How would you name this molecule as it has an alcohol and aldehyde group. The aldehyde surely has the higher priority but if I start numbering there, I'm not sure how to write alcohol.

Best try I have is 3-methylpent-2-ene-1-hydroxy-2-al.

You have correctly identified the principal characteristic group as well as the principal chain. Thus, the parent structure of the compound is pent-2-enal. You also correctly included the ‘3-methyl’ substituent.

The missing $\ce{-OH}$ group cannot be expressed as suffix ‘ol’ since it is neither the principal characteristic group nor situated on the principal chain. Therefore, it has to be indicated by the prefix ‘hydroxy’. The corresponding rule in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows.

  P-63.1.4 Substitutive nomenclature, prefix mode

  Hydroxy groups are indicated by the prefix ‘hydroxy’ when:

  (1) a group having priority for citation as the principal characteristic group is present; or

  (2) a hydroxy group cannot be denoted by a suffix.

In this case, the hydroxy group is placed on a methyl side chain, which results in the compound substituent group ‘hydroxymethyl’.

  P-29.4 COMPOUND SUBSTITUENT GROUPS

  P-29.4.1 Compound substituted substituent groups

  A compound substituted substituent group is formed by substituting one or more simple substituents into another simple substituent that is considered as the principal chain. The choice of the principal chain is fully discussed in Section P-44.3. The first criterion to be applied is that the principal chain is the longest chain; it will be applied in the following examples of acyclic compound substituent groups.

  (…)

Parentheses are used around such compound prefixes in order to avoid ambiguity.

  P-16.5.1 Parentheses (also called curves or round brackets)

  P-16.5.1.1 Parentheses are used around compound (see P-29.1.2) and complex (see P-29.1.3) prefixes; after the multiplicative prefixes ‘bis’, ‘tris’, etc.; to enclose a multiplied parent compound consisting of a parent hydride and a substituent suffix or a multiplied substituent prefix even though preceded by a basic numerical prefix, such as ‘di’, ‘tri’, etc.

Therefore, the complete systematic name for the compound that is given in the question is (2Z)-2-(hydroxymethyl)-3-methylpent-2-enal. (Also note the stereodiscriptor for the double bond.)


What are standard controllers for automatic pH control? My background in control theory comes from electrical engineering (robots, motors, etc.) and I want to build a control system for automatic pH control.

Can I directly apply my knowledge from classical control to chemical control? E.g. can I use a PID controller in a similar way as I am use to, or are there special pitfalls/best practices in chemistry that I should be aware of?

Short system description: My plan is to have two pumps that can add either base or acid to a solution under constant agitation (shaker plate), using a standard glass electrode sensor for measurements.

Automatic pH control is not that easy as is shown by the fact that automatic titrators, the equipment that you want to build, are relatively expensive.
Besides titration, automatic titrators usually have also a pH stat function, which is for maintaining a certain fixed pH.

In first place, note that to  have a good control, typically you must know which type of system you are working with.

Normally, to maintain a constant pH only one reagent is added, either acid or base.
This is because chemical systems typically become more acidic or more alkaline when they proceed, but do not oscillate.
When you add both reagents, they usually start to compete: a bit of acid is added, it passes the set point, base is added, it passes the set point in the other direction, etc.
This happens especially when the response of the system is slow and the mixing is not good (shaker plate is not a good idea, use a stirrer instead).
The way to solve this is by adjusting very well the concentrations of the reagents, the delay times, addition rates, etc., and this is different in every system.

You also should know in which part of the pH curve you are working to set the concentration of the reagents and the rate of addition.
The pH curve has three very different zones, a vertical part and two horizontal parts.

Depending on where you are, the conditions must be different.
Note also that the inflection point becomes less obvious when weak acids or bases are involved.

Finally, you might need to integrate also a temperature compensation since, depending on the $\Delta\epsilon$ you have set, the pH change with temperature will be to high.

In short, a PID might work in some specific system, but for a general purpose system you will probably need some more sophisticated control like a PLC.


Finding the mystery ester strucuture using NMR I have an unknown ester with the chemical formula $\ce{C9H10O2}$ that is used as a flavoring agent in candies. It exhibits the following H-NMR and C-NMR (Attached file: unknown ester NMR)

Based on the NMR peaks I believe the unknown is benzyl acetate, as the C-NMR and the first two peaks of H-NMR are identical to the unknown NMR (Attached file: benzyl acetate NMR).

Knowing that aromatic protons are not equivalent, as shown in the unknown's H-NMR at 7.4 and 7.5 ppm, I do not know why the aromatic protons for benzyl acetate are integrated and shown as a singlet.

Is this just due to the sensitivity of the prediction algorithm (I am using Chembiodraw)? Or benzyl acetate is not the unknown?

Your help is greatly appreciated!

Prediction software always has its limitations, and there is always a degree of error in the calculation. For the ChemDraw predictions, you will see that for the 3 aromatic environments, it has done 3 independent calculations, and has happened to arrive at the same chemical shift. This simply means that these shifts are coincident, not equivalent.

Remember, prediction software is like any tool — only as good as the person using it, and should not replace a proper assessment of the data.

Your assessment here should first look at your proton environments; all 10 protons are accounted for. We have a $\ce{CH3}$, a $\ce{CH2}$, and a mono-substituted benzene. The $\ce{CH3}$ and $\ce{CH2}$ are not directly connected to any other group that is causing obvious splitting. Secondly, the $\ce{^13C}$ spectrum shows 9 peaks, consistent with our $\ce{CH3, CH2}$ and a mono-substituted benzene. We also have a peak at ~δ170, which is a $\ce{-C(O)-{}}$

There are only a few ways you could put these groups together. Rather than using ChemDraw to predict the spectrum and see which is identical to your question, you should rationalise why each possibility is or isn't the correct answer.

Trying to rationalise a bunch of peaks around δ7.4–7.5 is the of minimal interest, and remember that a real spectrum will almost always look different to a simulation. The peaks you should focus on willjustify the chemical shifts for the $\ce{-CH3}$ group and the $\ce{-CH2}$ group in the proton spectrum, and the $\ce{-CH3, -CH2}$ and $\ce{-C(O)-{}}$ in the carbon spectrum. There is only one possible correct answer.


How to simulate chemical reactions in GAMESS? I am rather newb in GAMESS field but finally, learned to make input files for most molecular simulations and could even use TD-DFT for excited states (special thanks to Geoff Hutchison for his help).

Now I want to start simulating simple chemical reactions. I looked at the GAMESS manual and think I need to make transition states and find least energy pathways but have no idea how to make input file for that.

Can anyone give me a simple input file example for a simple reaction?

As in geometry optimization, you are searching for stationary point on the potential energy surface (PES). Not for local minimum, but for saddle point, therefore in GAMESS, you specify RUNTYP=SADPOINT. You would also need the correct (non-guess) Hessian matrix, which you can calculate separately with RUNTYP=HESSIAN. In transition state (TS), you should have zero gradients (stationary point) and one imaginary eigenvalue of Hessian (the reaction coordinate in TS).

For working example, see e.g. SN2 transition state. In principle you need very good guess at TS geometry, which you obtain by scanning along some chosen coordinate. The point highest in energy should be reasonably similar to the TS, and you optimize to saddle point starting from there. Already at the guess geometry you must have one imaginary Hessian eigenvalue (resembling the reaction you are looking for, not methyl rotation somewhere else).

But beware, finding the transition states is one of the most difficult topics in computational chemistry and you are not guaranteed to find one, no matter how hard you try. One of the biggest problem is solvation, as you often wish to find reaction barrier in solvent (where majority of chemistry happens), but you can calculate only in vacuum. Just think of how carefully the workbench chemist finds the correct solvent for given reaction, in computer you have only very crude approximations to it (and most often just vacuum).

That said, it is great fun and rewarding to find the transition states. For best results, use the simplest theory available, as it is better to have smooth and well behaving PES, even though it is slightly wrong, so it is no harm to start with HF or BLYP (or B3LYP). To get some ideas, start with simple reactions, i.e. where no charged species are involved, so electrocyclic reactions are probably best choice, e.g. Diels-Alder.


Backbonding in molecules like BF3 Can anyone please briefly explain to me the concept of backbonding along with necessary conditions for backbonding with some example? As I understand, backbonding is a donation of electron from electron-rich to vacant orbital, but I don't know anything else.

$\ce{BF3}$ has trigonal plannar structure all the three $\ce{B-F}$ bonds lie in plane and thus p-orbitals of boron and fluorine become parallel.

Boron has empty p-orbital and p-orbital of fluorine contains lone pair, and hence  boron acts as Lewis acid and fluorine as Lewis base. Fluorine donates its lone pair to boron and this bonding is called backbonding.


What is the mechanism for the reaction of alkenes with diborane and chloramine? $$\ce{CH3-CH=CH2 ->[B2H6/THF][NH2Cl/H2O, Δ] Product (P)}$$
What is $\ce{Product (P)}?$

First of all, in this reaction I performed hydroboration to form an alkylborane. But I don't have any ideas what to do next. I tried a lone pair attack of nitrogen on the boron atom of the alkylborane, but what should I do next? What is the product and by what mechanism is it formed?

Organoboranes react with chloramine to give the corresponding amine as reported by Brown et al. [1], so P is n-propylamine (propan‐1‐amine).

The mechanism is similar to the oxidation of boranes using $\ce{H2O2}/\ce{NaOH}$, but instead of migration to an oxygen atom, the alkyl chain migrates to a nitrogen atom. The $\ce{N-B}$ bond is cleaved in the same way.

Reference

Brown, H. C.; Heydkamp, W. R.; Breuer, Eli.; Murphy, W. S. The Reaction of Organoboranes with Chloramine and with Hydroxylamine-O-Sulfonic Acid. A Convenient Synthesis of Amines from Olefins via Hydroboration. J. Am. Chem. Soc. 1964, 86 (17), 3565–3566. DOI: 10.1021/ja01071a036.


Reduction of glucose to hexane with hydroiodic acid My book (NCERT India) states that n-hexane is formed when glucose is heated in the presence of hydrogen iodide. That is, the following reaction takes place:
$$\ce{C6H12O6 + HI ->[{\Delta}] C6H14}$$
This seems a bit strange to me because I only know of three possibilities that can occur when a compound containing hydroxyl groups is treated with a strong acid:

A substitution reaction, where all the $\ce{-OH}$ groups are replaced by $\ce{-I}$
A dehydration reaction, where $\ce{H2O}$ molecules would be eliminated and it would end up with few double bonds.
Formation of an ether.

There are also other sources which mention this reaction (thanks to Karan Singh, Tyberius, and user55119 for pointing these out). A Michigan State University webpage states that:

  Hot hydriodic acid (HI) was often used to reductively remove oxygen functional groups from a molecule, and in the case of glucose this treatment gave hexane (in low yield).

Also, Dietary Sugars (edited by Preedy, V. R.) states (p 80) that

  Glucose is reduced with concentrated hydriodic acid in the presence of red phosphorus to form a trace amount of n-hexane. This indicates an unbranched chain of six carbon atoms [...]

It goes on to cite the 5th edition of Principles of Biochemistry by Nelson et al. (published 2008), but I could not find this statement in any recent edition of this book.

Lastly, a similar reaction was carried out by Kiliani (Ber. Dtsch. Chem. Ges. 1885, 18 (2), 3066–3072) where the cyanohydrin of fructose was reduced to to 2-methylhexanoic acid using hot HI and phosphorus.

I would like to know:

Is phosphorus required for this reaction, or not?
What is the mechanism by which this reaction proceeds, with or without phosphorus?

Hydroiodic acid is a reducing agent.  As Wikipedia says:

  Although harsh by modern standards, HI was commonly employed as a reducing agent early on in the history of organic chemistry. Chemists in the 19th century attempted to prepare cyclohexane by HI reduction of benzene at high temperatures, but instead isolated the rearranged product, methylcyclopentane (see the article on cyclohexane). As first reported by Kiliani,[10] hydroiodic acid reduction of sugars and other polyols results in the reductive cleavage of several or even all hydroxy groups, although often with poor yield and/or reproducibility.[11] In the case of benzyl alcohols and alcohols with α-carbonyl groups, reduction by HI can provide synthetically useful yields of the corresponding hydrocarbon product (ROH + 2HI → RH + H
  2O + I2).[8] This process can be made catalytic in HI using red phosphorus to reduce the formed I2.[12]

When $\ce{HI}$ acts as a reducing agent, the oxidized product is $\ce{I2}$.  To regenerate $\ce{HI}$ from the $\ce{I2}$, phosphorus can be used.  Thus, phosphorus is not strictly required for the reaction, as long as you are OK with using $\ce{HI}$ in superstoichiometric amounts and/or getting very low yields.
The overall scheme for $\ce{HI}$ reduction of an alcohol to an alkane is first displacement of the (protonated) alcohol by iodide, leading to formation of alkyl iodide.  Then, $\ce{HI}$ catalyzes the reduction of the alkyl iodide to the alkane, leading to $\ce{I2}$ as a byproduct.  The reduction step apparently follows a radical mechanism, at least under some conditions.
I'm not 100% sure how $n$-hexane would form from glucose.  It is unclear to me how $\ce{HI}$ would catalyze the reduction of the glucose aldehydic (or anomeric) carbon to an alcohol.  But, that is only one carbon out of six.  The other carbons have bona fide hydroxyl groups, and as such can be deoxygenated and reduced by $\ce{HI}$ to form alkyl carbons.  For example, Lv et al. report that hydrocarbons can be obtained from the five-carbon sugar alcohol xylitol in yields above 60%.  Unlike glucose, xylitol has no aldehydic/anomeric carbon, and only has alcohol groups.  This process did use phosphorous acid as an in situ iodine recycling agent.  Also, Yang et al. were able to obtain 5-methylfurfural from fructose by using $\ce{HI}$, without using an in situ agent to recycle $\ce{}$ to $\ce{HI}$.


What is the characterstics of additives that it must posses to make any mixture hypergolic? This is about hypergolicity of rocket propellant.Turpentine gave hard-start with RFNA and WFNA but excellent with mixed acid(mixture of HNO3 & H2SO4).Similarly,Furfuryl alcohol & WFNA(also RFNA) become hypergolic mixture.So,what properties must additives posses to ignite particular mixture at any time without requiring oxygen?

The ideal formulation of rocket fuel is a balanced mix of  oxidizer and fuel. A strong oxidizer and a highly flammable compound or strong reducer, which is termed the "fuel", Is the Ideal combination for a fuel that will combust hypergolically when mixed. The compounds in hypergolic fuels are ideally liquids, because fluidity allows for easier mixing.

As oxidizers, Nitrates work well because the large quantity of oxygen atoms attached to the single nitrogen atom makes for a strong oxdizer. Other compounds, such as Sulfates, Phosphates, Chromates & Dichromates, Chlorates & Perchlorates, Peroxides, and Permanganates also serve as fairly good oxidizers for this reason. However, strong oxidizers do not always require high oxygen content. For example, Chlorine Trifluoride is a very strong oxidizer and is commonly used in many different hypergolic rocket fuel combinations

As far as the rest of the mixture, the fuel, most flammable liquids work well. Aniline, Liquid H2, Hydrazine and it's analogues, Kerosene, Gasoline, and most short-chain alkenes work well with the right oxidizer. An ideal choice when it comes to Hypergolic fuels is fairly small alcohols and ethers, like ethanol, diethyl ether, furfuryl alcohol, and especially Glycerol. It seems that the occasional oxygen atom in the fuel makes it easier and faster to oxidize when the time comes.

In conclusion, Adding a small amount of a stronger oxidizer in solution with your main oxidizer should help spark the reaction if it has trouble getting started. Also, while adding a fuel booster doesn't always work, sometimes adding a little ethanol or glycerol can help get the reaction in motion as well as help it run more smoothly. This is similar to how ethanol is often added to gasoline to increase the efficiency of the burn.

I kind of have an interest in the chemistry of rocket fuels, so if you ever find yourself looking for another answer, feel free to mention me directly.


Fractional vs Simple distillation After performing both simple and fractional distillations for a 50/50 mixture of methyl acetate/ethanol, i discovered that my total recovery (in mass) of substances after distillation was lower for the fractional distillation than for the simple distillation. Is this to be expected, or would it be because of an error on my behalf? (To clarify, my initial mass for F. distillation was 17.39g and my recovered mass was 14.95g. My initial mass for S. distillation was 16.54g and my recovered mass was 16.04g.) What would be some reasons in explaining this result?

In practice distillation is not 100% efficient: You never get a perfect separation of a mixture.  But fractional distillation is more efficient, so you would expect the product to be of higher purity than that produced by simple distillation.

E.g., if you have a 50/50 mixture by mass of two liquids, you would expect a higher concentration of the higher-boiling-point liquid at the end of a fractional distillation apparatus, which means a smaller fraction of the original mass.


What chemical reaction happens on worn out metal handles? I randomly came across a picture of this. It's the comparison between right handed and left handed people. As more people use the door handle, the metal becomes more faded and turns gold color. What is this gold color underneath and why does friction between your fingers and the door handle cause this reaction?

It looks like a brass ( a class of alloys, containing mostly copper and zinc, being golden-like ), which surface was covered by protective plating , possibly by chrome or nickel(*). It was probably worn out mechanically and/or by corrosion due acidic sweat.
(*)Nickel plating could be worn out easier than hard and resistant chrome.


What is the chemistry behind heat produced when blood is bleached? When I discard blood (20-40 mL) in a beaker containing bleach (400mL), I can feel that heat is produced in the beaker. What is the chemistry behind it? Will heat be produced due to Lysis of haem protein in RBCs? Or any other mechanism?

This reaction produces heat for the same reason as burning methane in oxygen. There's a lot of chemical energy stored in the reactants, and comparatively little in the products.
The hypochlorite is oxidising the organic content of the blood - not just breaking proteins but then breaking down the amino acids too, as well as anything else that happens to be in the sample.


Equalization of pressure in heat exchanger I was solving numericals on Kinetic Theory of Gases when I came across this question

  Two closed vessel of equal volume contain air at 105 kPa, 300 K and are connected through narrow tube. If one of the vessel is now maintained at 300 K and other at 400 K what will be the pressure in the vessel?

I don't want the solution of this question but I want to know why the pressure will be same in both the vessel when one is maintained 300 K and other 400 K. The molecules in  the vessel at 400 K will have greater kinetic energy and will exert greater pressure on the walls than the molecules of the other vessel.

Now one will say that some of the molecules of vessel at higher temperature will move to the other vessel so that the pressure remains same in both the vessels, but why is it so. Why can't the pressure in both the vessels be different?

Macroscopics

Imagine the tube is initially closed. Then one vessel is heated up. As $V$ and $m$ are constant:
$$p=k\,T$$

So if $T$ is increased, $p$ is increased. As $p$ is increased, there will be a net flow of mass, until they are balanced. That's what we observe. Now, we can think of it microscopically.

Microscopics

Initially, there are the same number of particles in each vessel. When heating up there is an increase in velocity. So, in the high-temperature vessel the same number of molecules are with higher velocity. Imagine a plane in the middle of the narrow tube, the number of collisions per time, because of velocity, will be higher in $400\ \mathrm K \to 300\ \mathrm K$ direction. It indicates a net flow of mass. This flow of mass will equal pressures on both sides.


Chemical kinetics of a reversible reaction I am having trouble with the following question:

  Consider the following reversible reaction in which the reaction is first order in both
  directions: $$\ce{[A] <=> [B]}$$ $k_\mathrm a$ is the rate constant for the forward reaction and $k_\mathrm b$ is the rate constant for the reverse reaction.

  If only reagent $\ce{A}$ is present at $t = 0$, such that $[\ce{A}](t = 0)$ = $[\ce{A}]_0$ (and likewise for $[\ce{B}]_0$), then at all subsequent times $[\ce{A}]+[\ce{B}] = [\ce{A}]_0$. Write down a differential equation for $\frac{\mathrm d[\ce{A}]}{\mathrm dt}$ where $[\ce{A}] \equiv [\ce{A}](t)$ is the concentration of the reagent $\ce{A}$, and hence show that:

  $$[\ce{A}]=[\ce{A}]_0\frac{k_\mathrm b+k_\mathrm a \exp[-(k_\mathrm a+k_\mathrm b)t]}{k_\mathrm a+k_\mathrm b}$$

Here's my attempt:

$$\frac{\mathrm d[\ce{A}]}{\mathrm dt}=-k_\mathrm a[\ce{A}]+k_\mathrm b[\ce{A}]\tag{1}$$

Then using the identity in the question:

$$\frac{\mathrm d[\ce{A}]}{\mathrm dt}=-k_\mathrm a[\ce{A}]+k_\mathrm b([\ce{A}]_0-[\ce{A}])\tag{2}$$

Rearranging...

$$\frac{\mathrm d[\ce{A}]}{\mathrm dt}+(k_\mathrm a+k_\mathrm b)[\ce{A}]=k_\mathrm b[\ce{A}]_0\tag{3}$$

I believe that this is a first order linear differential equation with the integrating factor: $\mathrm e^{\int (k_\mathrm a+k_\mathrm b)\,\mathrm dt}=\mathrm e^{(k_\mathrm a+k_\mathrm b)t}$ Thus:

$$\frac{\mathrm d}{\mathrm dt}\left([\ce{A}]\mathrm e^{(k_\mathrm a+k_\mathrm b)t}\right)=k_\mathrm b[\ce{A}]_0\mathrm e^{(k_\mathrm a+k_\mathrm b)t}\tag{4}$$

$$[\ce{A}]\mathrm e^{(k_\mathrm a+k_\mathrm b)t}=\int k_\mathrm b[\ce{A}]_0\mathrm e^{(k_\mathrm a+k_\mathrm b)t}\,\mathrm dt\tag{5}$$

This gives me:

$$[\ce{A}]\mathrm e^{(k_\mathrm a+k_\mathrm b)t}=k_\mathrm b[\ce{A}]_0\frac{\mathrm e^{(k_\mathrm a+k_\mathrm b)t}}{k_\mathrm a+k_\mathrm b}+c\tag{6}$$

But this doesn't give me the final result if I put in the boundary conditions. Where have I gone wrong?

The simplest and most general way to solve this type of scheme is to calculate the effect of an amount x that reacts with initial amounts $A_0$ and $B_0$, thus
$$\ce{A~~~ <=>[k_a][k_b] ~~~B}  \\  A_0-x  ~~~~  B_0 +x $$
which produces the rate equation
$$ \frac{dx}{dt} = k_a(A_0-x)-k_b(B_0-x)$$

the variables can be split to give a straightforward integration instead of having to solve a complicated pair of differential equations.
$$\int _0^x \frac{dx}{k_aA_0-k_bB_0-(k_a+k_b)x} = \int _0^t dt$$

which, after some rearrangement, gives
$$\ln\left(1-\frac{(k_a+k_b)}{k_aA_0-k_bB_0}\right) = -(k_a+k_b)t$$
and this can be rearranged to
$$x= \left( 1-e^{-(k_a+k_b)t}  \right)\frac{(k_aA_0-k_bB_0)}{k_a+k_b}$$

with the initial condition that at $t=0$, $B_0=\ce{[B]=0 }$ this becomes
$$x= \left( 1-e^{-(k_a+k_b)t}  \right)\frac{k_aA_0}{k_a+k_b}$$
The concentration of A is $A_0-x$ which is

$$ \ce{[A]}_t =A_0 \frac{k_b+k_ae^{-(k_a+k_b)t}}{k_a+k_b}$$

This expression has the correct form at all times. $\ce{[B]}_t$ can be calculated similarly from $B_0+x$.


Mimic a volcano eruption with household chemicals? How can one mimic a volcano eruption using household chemicals safely?

We've tried it in a testing tube with baking soda and vinegar. This wasn't quite as impressive as expected.

On the web we've found recipes that add food coloring (red) and dish washing liquid to the mix. I guess the latter is for some bubbling effect.

What we have:

A volcano made from paper-mâché (classic volcano cone form, perhaps a bit steeper than natural)
Center hole big enough to fit a 1.5 liter water bottle

Note: this is supposed to be done with kids, so please don't propose any method that would generate poisonous fumes or similar.

Meanwhile tried a soda ($\ce{NaCO_3}$, i.e. not baking soda $\ce{NaHCO_3}$) solution poured into a dish washing solution, then adding a citric acid solution. It was pretty good, but a bit too quick. Vinegar seems to work better than citric acid, too. Also I used warm water to dissolve the reagents.

The 'elephant's toothpaste' experiment with hydrogen peroxide decomposition works well to produce large amounts of foam and might be suitable for what you are looking for.
Low concentration hydrogen peroxide solutions are readily available and not dangerous provided you don't get it in your eyes or drink it or anything silly.

Fill the bottle with some hydrogen peroxide solution, add a few drops of orange food colouring and some dish washing liquid and then add something to catalyse the decomposition. You could use something like potassium iodide or manganese dioxide to give you a very rapid decomposition but if you want a slightly less violent and more sustained reaction then apparently dried yeast dissolved in warm water works quite well (yeast contains good amounts of catalase, an enzyme which catalyses the decomposition of hydrogen peroxide).

Having never done this experiment, I can't tell you exactly what quantities to use so you should probably test it before doing it as a demonstration. However, there are some descriptions of the experiment on the internet to give you a good starting point.

http://www.sciencebob.com/experiments/toothpaste.php
http://www.stevespanglerscience.com/lab/experiments/elephants-toothpaste


What dictates the lifetime of a solvated electron in a given solvent? Solvated electrons have a long lifetime in ammonia solutions, but their counterparts in water (called hydrated electrons) have a much smaller lifetime, of the order of microseconds in very pure water.

(image from: Boero et al, Phys Rev Lett 2003, 90, 226403)

What properties of the solvent account for these very different lifetimes? Both are polar protic solvents, so what other(s) factor(s) could be involved here?

For those who have never had the pleasure of personally doing this, see this video. It has been known since 1807 that dissolving sodium in liquid ammonia results in a beautiful color. It was originally thought the color was due to some familiar complex instead of a solvated electron. A similar phenomenon happens with other alkali metals in ammonia.

Research in the field says that it takes, at least, forty some ammonia molecules to solvate a given electron. Metstable cavities form and their stability could depend highly on electrostatic interactions to solvate our electron. Ammonia will slowly react by evolving hydrogen gas,

$$\ce{2 NH3 + 2 e- -> H2 + 2 NH2-}$$

Differences in solvent can play a huge role in the stability of our solvated electron. An analogous decomposition occurs in water,
$$\ce{2 H2O + 2 e- -> H2 + 2 HO-}$$

Famously the latter occurs faster than the former, see here in comparison to my earlier link. We could say that these differences in rate reflect different stabilities of our solvated electron. Although, the addition of an appropriate catalyst to our ammonia will result in rapid evolution of hydrogen.

A more quantitative description for the difference in energies is obtained by measuring the UV-vis spectrum for a solvated electron in both water and ammonia. One will quickly notice that the band appears at higher energies in water than it does in ammonia and there are considerable differences in the band shapes (the band is wider in ammonia).

So Why Do They Differ

So here I have to make the disclaimer that no current theories quantitatively reproduce the observed phenomena, such as the absorption spectrum for a solvated electron. This question has no established/accepted answer as it is an ongoing area of research.

The self-ionization of water has an equilibrium constant on the order of $K = 10^{-14}$ and that of ammonia is on the order of $10^{-30}$. Hydronium formation occurs in the case of water and ammonium in the case of ammonia. Hydronium has a considerably lower $\mathrm{p}K_\text{a}$ than ammonium and so it’s reasonable to see why a reaction in water would be more likely with an electron. (Ammonia rarely produces a weakly acidic species, but water often produces a strongly acidic complex.)

A slightly deeper reason may be that water forms more ordered local domains/structures in solution than ammonia and this influences the rate by resulting in a larger entropy of activation when the hydrolyzed electron breaks up these larger structures, recall that $k \propto \exp (\Delta S ^\ddagger /R) $.

Just speculation though.


IUPAC name of a molecule with a cyanide group and a double bond This is the molecule:

After identifying the functional group, what should we prefer? A longer carbon chain or the one which is short but has a double bond in it.

I personally think that we should go for longest carbon chain. So the name should be 2-ethenylpentanenitrile.

The relevant simplified criteria for the choice of a principal chain are:

greater number of substituents corresponding to the suffix (principal characteristic group)
longest chain
greater number of multiple bonds

The corresponding wording of the rules taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) is as follows.

P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES
When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.
P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).
(…)
P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].
(…)
P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.
The senior ring, ring system, or principal chain:
(a) has the greater number of multiple bonds (P-44.4.1.1);
(b) has the greater number of double bonds (P-44.4.1.2);
(…)

You have correctly identified the suffix (“nitrile”) for the principal characteristic group. Thus, according to the first criterion, the parent structure could be pentanenitrile or but-3-enenitrile since both include the nitrile group.
The next criterion for the principal chain is the greater number of skeletal atoms (i.e. the longest chain). Thus, the principal chain corresponds to the pentanenitrile part and not to the but-3-enenitrile part since pentanenitrile has a longer chain than but-3-enenitrile. Therefore, the correct name is 2-ethenylpentanenitrile.
Many people would probably still call it 2-vinylpentanenitrile; according to current IUPAC recommendations, however, the prefix “vinyl” is retained only for general nomenclature, “ethenyl” is recommended as the preferred prefix.


Way to detect lead in gasoline Is there any way that i can detect lead in gasoline, not by sending of to a lab but to be able to do this by myself?

As pointed out in my earlier comment, you won't find tetraalkyl lead compounds in current petrol used for cars. In most countries, it was banned decades ago and replaced by MTBE.

However, the situation seems to be different for fuels used in aviation,   such as AvGas 100 where lead concentrations up to $\mathrm{1\,g\,L^{-1}}$ can be found.

Even the low lead variant, AvGas 100LL contains $\mathrm{0.56\,g\,L^{-1}}$ lead, see product specifications from TOTAL.

In an older technical note by R. E. Mansell and T. A. Hiller, published in Anal. Chem., 1973, 45, 975-975 (DOI), a possible approach for the decomposition of tetraethyl lead and subsequent detection of $\ce{Pb^{2+}}$ is outlined, referring to the ASTM D526-56 method:

  […] recommends refluxing 50 ml of gasoline with 50 ml of $\ce{HCl}$ for 30 minutes to decompose the tetraethyllead.[…]

In the aqueous layer, $\ce{Pb^{2+}}$ can be detected by complexometric titration with EDTA, either directly, or by back titration with a solution of a $\ce{Mg^{2+}}$ salt.


How are glycosidic linkages formed? Consider glycosidic linkages in disaccharides.

I know a condensation reaction occurs, with the loss of water.

But which OH is lost? The O in the linkage belongs to which monosaccharide?

Example: take sucrose, the O in the linkage belongs to fructose or glucose?

I couldn't find any literature regarding this. Any help is appreciated .

In sucrose molecule the  glycosidic linkage is formed by the OH group on the anomeric carbon of α-D-glucose and the OH group on the anomeric carbon of β-D-fructose.

source :

https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(McMurry)/Chapter_25%3A_Biomolecules%3A_Carbohydrates/25.08_Disaccharides

As seen, in either alpha d-glucose or beta d-fructo furanose may react to give sucrose.

Let us examin.

Alpha D-gluco Pyranose gives an intermediate 1 and 2 by protonation at anomeric carbon.

Similarly  Beta D-fructo furanose gives an intermediate 3 and 4 by protonation at anomeric carbon.

If you examine , structure 3 is more stable then structure 4.

The negative inductive effect of CH2OH in fructofuranose destablizes this intermediate 4.

Hence structure 2 , an intermediate of aplha D-gluco pyranose (more stable) attacks beta D-fructo furanose togive sucrose.

The above diagrams are author drawn in CHEM DRAW


How to determine the magnetic character of heteroleptic complexes? I was asked to find out the magnetic behavior (and essentially electronic configuration) of heteroleptic complexes using Valence Bond Theory and Crystal Field Theory.

But I am confused about how to do it. Especially in the ones which have both strong field and weak field ligands together; how am I to decide whether pairing of electrons will take place or not ?

For a more concrete example consider the complex $\ce{[Cr(NH3)_4Cl_2]^{+}}$

Based on the available information, we would start by figuring out the 3D geometry of the complex. $\ce{Cr^3+}$ should be octahedral, with two possible "combinations", cis or trans. Let's start with the trans version. The symmetry of this system is $D_{4h}$ so we call this "tetragonal".

With a tetragonal geometry, the MO's would be arranged in such a way that there will be two degenerate orbitals at the lowest energy, and three higher-energy orbitals. The MO diagram typically looks something like this:

Chromium(III) has an [Ar] $3d^3$ electronic configuration so we need to fill up three levels. Now the question is whether or not the splitting caused by the ligands are large enough to cause pairing.

We know for a fact that the trans version of the example is a green solution, so it absorbs red light (700 nm or ~14,300 $cm^{-1}$). Pairing energy is typically in the range of 20,000 - 30,000 $cm^{-1}$ so the 'gap' between yz and xy is probably about only ~14,000 $cm^{-1}$. As such, I would say the trans isomer is magnetic and not paired, because it is energetically more favorable to promote the 3rd electron to the higher energy level than to force it to pair.

Similarly, we know for a fact that the cis version is violet. This means it absorbs yellow light, about 570 nm or 17,000 $cm^{-1}$. This is still lower than the pairing energy, so I'd say that The tetraamminedichlorochromium(III) is magnetic.


Improving Elephant Toothpaste experiment we are trying to perform the Elephant Toothpaste experiment involving the reaction of yeast and Hydrogen Peroxide for the students at school.

However, the results have been very underwhelming, likely because we have to use the 3% version of Hydrogen Peroxide.

My question is: What are some other ways to help create a stronger propulsion or thicker foam?

Some things I have been experimenting with are reducing the water mixed with the yeast, and swirling the bottle to mix the yeast and Hydrogen Peroxide. Any advice is much appreciated.

You might try horseradish as a catalyst, rather than yeast. Use the fresh root, not the prepared sauce, and test a few different plants - some are more effective than others.
Cosmetic supply stores stock 6% (20 volume) $\ce{H2O2}$, but that concentration is more likely to cause skin irritation or bleach fabrics.
The most effective catalyst for decomposing  $\ce{H2O2}$ that I've used is potassium permanganate, $\ce{KMnO4}$, but it has a few liabilities:

It's a strong oxidant and can start a fire when mixed with organic chemicals.
It stains skin and fabric anything from pinkish-purple to black. The stain can be removed with sodium bisulfite, $\ce{NaHSO3}$, sold as an iron-stain remover. That can be a fun demonstration but is not suitable for young students to use.


describing in words a primitive heated water vapor lift The primitive lift illustrated in the left hand side of the following figure operates using heated water vapor ("gas") that pushes on a sealed piston. During its operation, the vapor undergoes the rectangular cyclic process shown in the right hand side of the figure.
I'm trying to describe in words how one might realize each of the four steps.
My idea is : We assume the system is isolated entirely from its surroundings such that it obeys the ideal gas law. First, heat is added through an external source to the gas while the piston is held fixed such that Temperature increases while the volume remains constant, therefore the pressure increases from $p_{1} \to p_{2}$ (A). Then, the piston is released and the gas is no longer heated. The volume will increase from $V_{1} \to V_{2}$ and the pressure will remain constant. When the piston reaches the top it is stopped and held still.
Here I'm blocked, I don't know how to explain a situation in which pressure decreases while volume is held constant .
Any help would be appreciated.

To carry out the B transformation, you have to heat a lot. If you don't heat, the volume will of course increase, but the pressure will decrease. B is a process difficult to carry out, because it is not easy to  heat enough a gas who is inflated to as to maintain the inner pressure. Then, at the end of B, the temperature is very high. You block the position of the piston, and you cool the system : the pressure will decrease from P2 to P1, during the process C. The last process D is also difficult to carry out. Because in D you must compress the gaz so as to decrease its volume, but the pressure must not increase. It means that the system should be cooled during D. At the end of D, the temperature will be extremely low.


Force Constants for Angles From Gaussian To AMBER: which value should I put in the matrix equation? I am trying to make a new force field in AMBER for a novel compound including boron (which does not exist in GAFF). For this, I calculated freq and I found bond force constants but I couldn't find force constants for angles.

For example I tested $\ce{H2O}$: I calculated $\ce{H2O}$ frequencies in Gaussian and I got force constants from the log file. In the log file, the $\ce{O-H}$ stretching force constant was $\pu{9.1465 mDyne/\require{mediawiki-texvc}\AA{}}$ and I converted this to approximately $\pu{1315.65 kcal mol^{-1} \AA{}^{-2}}$. In AMBER, $v = 0.5k(l-l_{o})$, so I multiplied this by 0.5 and I found

ow-hw 657.825

which is very close to the GAFF2 forcefield:

ow-hw  612.98

which can be found on line 87. However, I couldn't find a way to calculate force constants from Gaussian for angles. In GAFF2 for $\ce{H2O}$, the values are

hw-ow-hw   43.276     104.520       AMBER              1    TIP3P_water

I searched and I found a lot of answer like this, but in this answer, for the Seminario approach I do not know what value to put in the matrix equation. Could you provide an example on a Gaussian log file?

I really don't know exact answer but I have an idea for you it may help: You can use some "force field developer toolkits". One of them is: VFFDT. It parametrizes (for amber forcefield *.frmcd files) bond and angle force constant from gaussian out/log file.


On stretching a rubberband does the entropy increase or decrease? It seems that on stretching a rubberband, entropy or the degree of disorder is likely to increase. Is this the correct answer?
My text says that on stretching, the arrangement of particles become more 'ordered' and hence entropy 'decreases'. How?

Think of it this way, when an elastic polymer is in its relaxed state, these molecules are all tangled up with each other and have no particular direction to them, but when you apply a force, i.e stretch the rubber band, you end up disentangling some of the polymer molecules.

We are not stretching the actual polymer molecules, just sort of changing the way they are organised. This disentanglement corresponds to a reduction in micro-states that the system occupies and thus, a decrease in entropy.

In simpler words, there's more ways for a polymer chains to jumbled up with each other (more microstates) than for them to be aligned.

More simply-- If the chain were totally stretched out, it would only have one possible conformation. By coiling itself, it increases the number of
conformations it can have, and therefore, its entropy.

A more rigorous treatment can be found if you read up about the ideal chain, which serves as a model for such systems.

Thus, assuming that no other energy exchanges takes place, the tendency of the elastic band to shrink back to its original conformation is due to something called an entropic force, i.e a tendency of the system to move to a configuration that maximises entropy.

Some Details

Note: I am not quite confident with this material and might've made some mistakes--if that be the case, I am sure someone here will correct me

Consider the following model system: The chain is made of $n$ rigid segments of size $b$; Thus, the length of the chain is $L = nb$. Also, assume that the chain has one fixed end, and one free end and the distance between the two is denoted as $r$.

Then, one can derive the probability of finding the chain ends a distance $r$  apart is given by the following Gaussian distribution:

$$ P(r,n)\mathrm dr = 4\pi r^2 \left(\frac{2nb^2\pi}{3}\right)^{\frac{-3}{2}}\exp\left(\frac{-3r^2}{2nb^2}\right)\mathrm dr$$

The classical entropy is approximately $ S = k_\mathrm b\ln(P(r,n))$

From hereon, one can go on to show that for an ideal chain, maximising its entropy means reducing the distance between its two free ends (i.e reducing $r$)


What is the oxidation state of all the atoms in this compound, W(CH3CN)3(CO)3? $\ce{W(CH3CN)3(CO)3}$

I'm thinking:
$\ce{CO}$ is neutral so $\ce{O}$ is -2 and $\ce{C}$ +2

$\ce{CN}$ is negative one so $\ce{C}$ is +2 and $\ce{N}$ -3

$\ce{CH3}$ is positive one so $\ce{H}$ is +1 and $\ce{C}$ is -2

$\ce{W}$ is neutral so $\ce{W}$=0

is this right?

HINT:

Compound is neutral, So the sum of oxidation state of all the element should be ZERO.

In what you have got, is sum of oxidation state of all the element is zero? This way you can check whether you are right or wrong.


Why are metal oxides generally more stable than their corresponding elemental forms? Why is the oxidation of certain metals such as iron favorable from a thermodynamic standpoint? What is it about these metal oxides that makes them inherently more thermodynamically stable than their starting, elemental forms? The only reason I can imagine is that iron, being a d-block element, has a low electronegativity and cannot stabilize valence electron density as well as oxygen, the second most electronegative element on the table.

Also, is oxidation always thermodynamically favorable?

The oxidation of gold is not favorable.  For many other metals the metallic bonding is weaker than the bonding with oxygen.

Let's look at the steps in oxidation:

Breaking metallic bonds (endothermic).

Ionizing the metal atoms (endothermic).

Electron affinity of oxygen (exo for the first, endo for the second).

Lattice energy of the oxide (exothermic).

The lattice energy is so high that the process is favorable.


Free software for calculating the net charge of an amino acid as a function of pH? Does a free software for calculating the net charge of amino acids as a function of pH exist?

I like MarvinSketch. You can run it in your browser or on your computer. Under Insert > Template, you can find all the amino acids. Once you've inserted one, select Calculations > Protonation > Isoelectric point and click OK. You'll get a window that looks like this showing the net charge versus pH (I picked a pretty boring one, but other amino acids produce more complex curves):


Why do we need to add Al in excess for the reaction between a nitrate and a base to give ammonia? I don't know if why we add Al for the reaction between Nitrate and a base so it would release NH3 gas.

So what I want to know is if what is the role of Al here?
And why do we have to add it in excess?
Can we do this with every nitrate? ( in my case I came across the reaction of Sodium Nitrate and Sodium hydroxide)

And thank you for your time!

Hydroxide and nitrate alone do not react  with each other.

Their reaction with aluminium is the part of the classical quantitative determination of nitrate content by distillation of produced ammonia and by the back acid titration.

Nitrite + ammonium ions interfere, but the former can be eliminated due their high reactivity, the latter can be determined and/or eliminated by prior distillation without aluminium.

Aluminium acts as a reduction agents, reducing nitrates to ammonia, forming alkali aluminates.

$$\ce{3 NO3-(aq) + 8 Al(s) + 5 OH-(aq) + 18 H2O(l) -> 3 NH3(g) + 8 [Al(OH)4]-(aq)}$$

Generally, aluminium need not to be in an excess. But if there is the desire the all nitrate ions react quantitatively, aluminium and hydroxide must be in excess.

Usually, the Devarda's alloy is used, instead of pure aluminium, containing also copper and zinc.


Electron pairing and repulsion If two electrons that get paired occupy the same orbital, then wouldn't there be a heavy amount of repulsion between the two?
As is, since electrostatic force ${\displaystyle \propto }$  $\ce{ 1/ (distance)^2}$   so since the distance between two electrons is way smaller than between the electrons and nucleus, I don't really assume it has any role here.

Does this have anything to do with the strong nuclear force? As far as I know its range is too short and exists only between protons and neutrons.

Summing up, what causes two paired electrons to stay in the same orbital rather than get thrown away due to "the force"?

No strong nuclear force is involved. Instead, tt's the same force as always: electromagnetic interaction. By staying in the same low orbital as the other electron, the electron in question will continue to be subject to a stronger attractive force from the nucleus. Note that the MO diagrams (see e.g. wikipedia) typically already account for the intra-orbital electron-electron repulsion, and you still have HOMO-LUMO gap.

It is also helpful to remember that electrons are "spread out" over the orbital, so once you do the math, you realize that the average distance between electrons does not have to be small compared to the average electron-nucleus distance.

Let's get semi-quantitative: $\ce{H_2}$, 1.4 bohr, HF/STO-6G($\zeta = 1.20$). After the SCF is completed, I transformed the integrals from the AO basis to the MO basis, which yields the following, relevant integrals in Mulliken notation:

(1|h|1) -1.24964
(2|h|2) -0.49918
(11|11)  0.65720
(22|22)  0.68091
(11|22)  0.64686
(12|12)  0.17440

Nuclear repulsion is 0.71429, all values in hartree. We can then calculate the low-spin, ground-state energy as:
$$
E_\text{gs}^\text{RHF} = E_{nn} + 2\cdot(1|h|1) + (11|11) = -1.12778,
$$
wherein we add the nuclear repulsion energy, the one-electron energy of the lower MO (once for each electron) (one-electron meaning that this is the sum of the electron's kinetic energy and its nuclear attraction) and the relevant two-electron integral for the Coulomb repulsion. Since the electrons are of opposite spin, no exchange term (as is responsible for the spin-pairing energy, cf. Hund's rule) is needed.

The energy of the high-spin, "excited" state can be calculated according to:
$$
E_\text{triplet}^\text{RHF} = E_{nn} + (1|h|1) + (2|h|2) + (11|22) - (12|12) = -0.56207,
$$
where an additional exchange term enters the equation to account for partly-correlated electrons due to Pauli-repulsion.

This calculation is semi-quantitative because RHF in such a small basis set does not yield quantitative results. One would have to go to natural orbitals obtained through full CI in a larger basis set for fully quantitative results. However, for illustrating the relative size of the quantities involved, this level of calculation is sufficient.


What is the correct order of chemical reactivity in terms of oxidising property for the following elements: F, Cl, O, and N This is a textbook question. I am confused as to which factors i should consider while sorting out these elements. If I take electron gain enthalpy, then Cl should be greater than F, but if i consider electronegativity F is greater than Cl . I'd be grateful to anyone who provides  some idea..

See, F is the strongest oxidising agent in the entire periodic table. Also, according to periodic table trends, oxidising character decreases down the group and increases across a period. You should check the reactivity of any element according to the question given, here, according to the oxidising character. If nothing is given, usually IE and Electronegativity is considered.
Therefore, I think the order should be F>O>Cl>N. Cl>N because Cl is more oxidising than N.
I must add that a simple Google search would have yielded the same result.


How can I recover PbNO3 from PbI2? I'm a chemistry school teacher.

I want to know how to recover lead Nitrate from the ppt of PbI2.

I know I can do it using conc HNO3 but I do not know if NO or HI will form. Also I would need a guide on the conditions of the reaction.

I don't want to dispose the lead if I can avoid that.

Thank you

The reaction is given here.

$$\ce{3PbI2 + 8HNO3 ->[\Delta] 3Pb(NO3)2 + 3I2 + 2NO + 4H2O}$$

  Lead(II) iodide react with nitric acid to produce  lead(II) nitrate,
  iodine, nitrogen monoxide and water. Nitric acid - 30% solution. The
  reaction takes place in a boiling solution.


How is the avagadros constant derived? İ heard that :
Charge of a mole of electrons/charge of single electron =avagadros constant
How is this derived ?

I believe it is the number of carbon atoms in 12 grams.


Can anyone tell me what reaction is this? I came across this reaction, but I have no idea what reaction is this. Can anyone please tell me?

What will be the products?

My first guess is that a carbene will be formed which will attack on the alkene, but I don't know what to do after that.

Diazomethane under the action of light or heat releases nitrogen leaving a  neutral, divalent carbon species with a lone pair and an empty orbital called a carbene. This carbene then reacts with the double bond of alkene breaking it to form a cyclo- species. in your example, cis-2-butene will react with diazonmethane to form cis-1,2-dimethylcyclopropane with no detectable amount of the trans isomer.

However, do note that if the same reaction is carried out via a triplet carbene, generated via triplet sensitization, then a 1.3:1 mixture of trans to cis dimethylcyclopropane is observed. [3]

Source

https://www.masterorganicchemistry.com/2011/11/05/reagent-friday-diazomethane-ch2n2/
[Source for pic. 1]
http://www.ochempal.org/index.php/alphabetical/c-d/diazomethane/ [Source for pic. 2]
http://reag.paperplane.io/00000832.htm
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Diazomethane%2C_Carbenes%2C_and_Cyclopropane_Synthesis