What is the convention of numbering carbon atoms in organic molecules? I am in high school and haven't had a formal chemistry course, however I am fascinated with biology and picked up a biology book more advanced than the one the school is using to read on my own. The book is: Biology by Peter Raven (2007) ISBN: 0073227390

I am confused about something in the organic chemistry section.

I don't understand the convention that is used to number carbon atoms in for example, glucose.

In case you don't know what I'm talking about, here is an example: a structural diagram of an a-glucose molecule courtesy of Wikipedia.

The numbers next to the carbon atoms, I don't understand how they are ordered. Can someone please explain? I know this is an elementary question for most of you, but the help is appreciated.

Glucose occurs as the open chain form aldehyde that determines atom numbering, and as the epimeric cyclic hemiacetal(s) you provided,

The "ether" is a hemiacetal.  C-1's -OH can be up or down.  Either way, it opens and reversibly dehydrates to the aldehyde R-C(=O)H.  Number one atom in a chain is the end that bears the highest total mass.  Double bonds count double their other end atom.  The cyclic form's C=1 also has the largest summed mass attached, two oxygens plus a carbon and a hydrogen.


When is the n+l rule followed when working with energy of different orbitals? While going through my book, I came across a problem which gave some options about the order of energy of 3s, 3p and 3d orbitals of hydrogen atom. The correct answer was given as:

3s, 3p and 3d orbitals all have the same energy

I also read somewhere that the order of energy of different orbitals increases as the value of $n+l$ increases where $n$ is the principal quantum number and $l$ is the azimuthal quantum number. So my guess was $3s<3p<3d$ .
What is the reason for this?

3s, 3p and 3d orbitals all have the same energy in hydrogen atoms ( or monoelectron atomic ions ) according to the nonrelativistic Schroedinger model.
Fine measurements and relativistic models of hydrogen atoms shows there is slight difference in their energies, related to the fine structure of spectra  and Lamb shift ( Wikipedia  Hyperphysics )
For multielectron atoms, s, p, d orbitals have different energies, as there come in the play 2 major effects, which affect electron energies:

Mutual electron repulsion
Screening/shielding of the nucleus charge by other electrons.


Why is there an increasing solubility in water for chlorides, chlorates, perchlorates in that order? According to Wikipedia $\pu{100 mL}$ of water dissolve at $\pu{25 ^\circ{}C}$ about $\pu{35 g}$ of $\ce{NaCl}$ (1), but about $\pu{79 g}$ of $\ce{NaClO3}$ (2), and around $\pu{210 g}$ for $\ce{NaClO4}$ (3).

This looks to me roughly like its doubling solubility for every single higher level of oxidation.

Earlier I was under the impression that the % of $\ce{Cl}$ or % of $\ce{Na}$ in the water would still be the same, and the amount of additional oxygen is what could be causing the increase in solubility.... ie. now each molecule has 3 oxygen atoms as opposed to none in case of $\ce{NaCl}$. But this doesn't hold true.

I would also like to know what is the effect of multiple solutes sharing same atoms on each other.  For example does dissolving some chlorate cause the solvent to hold more chloride ?

As you may have guess Chemistry isn't my strong suite, I know basic chemistry, but now when I am looking at it again.. the whole thing has changed atleast w.r.t what I was taught during my school days. (right from how we determine the number of electrons)

Science is all about discovering hidden regularities, patterns, and laws. In that sense your bold generalization is good, but will it hold if we check just one more data point? Let's see: $\ce{KCl}$ has solubility of 25g/100ml, $\ce{KClO3}$ has 8, and $\ce{KClO4}$ has 1.5. Er, well...

What we have here is an interplay between solvatation energy and lattice energy, the latter being determined by the crystal structure, which is a tricky thing, and hence so is solubility. Don't expect it to be predictable from qualitative considerations. You don't know it until you measure it. Chemistry is an experimental science, after all.


Why are so many ionic compounds brittle? Even when ionic compounds are strong (e.g. rubies and sapphires), they are neither malleable nor ductile in the slightest, and if sufficient stress is put on them, they will shatter, not bend. Why is that?

Ionic crystals are hard because of tight packing lattices, say, the positive and negative ions are strongly attached among themselves.
So, if mechanical pressure is applied to an ionic crystal then ions of similar charges may be forced to get closer to each other.
Now, by doing so, the electrostatic repulsion can be enough to split or disorient completely the lattice infrastructure. Thus imparting the brittle character.


How does heptafluoropropane suppress fire? How does heptafluoropropane suppress fire?

It says here that it does so by "inhibiting the chain reaction."  What does that mean specifically? Does $\ce{O2}$ attach itself to the heptafluoropropane?

Curt F. is probably correct in his estimate that heptafluoropropane extinguishes fire primarily by physical means. According to an overview by Choy and Fong, 'An Introduction to Clean Agents: Heptafluoropropane' (Int. J. on Eng. Performance-Based Fire Codes, vol. 5, nr 4, p. 181$-$184, $2003$),

For heptafluoropropane, the contribution of physical mechanisms to the extinguishment of fires predominates over the chemical mechanism. It suppresses  fires  primarily  by

extracting  heat  from  the flame reaction zone,
reducing the flame temperature below that which is necessary to maintain sufficiently high reaction rates by a combination of heat of vaporization [and] heat capacity.

They add that,

Oxygen depletion plays an important role in reducing flame temperature. The energy absorbed in decomposing the agent by breaking fluorine bond[s] is quite important, particularly with respect to decomposition production formation.

The chemical effect does indeed involve tying up active radical species. As an example, consider the mixture $\ce{H2/O2}$. The primary active species are $\ce{OH^\bullet}$ radicals.
$$\ce{H2 + O2 -> 2OH^\bullet}$$
The chain starts to grow.
$$\ce{OH^{\bullet} + H2 -> H2O + H^\bullet}$$
$$\tag{two reactive sp. instead of one}\ce{H^{\bullet} + O2 -> OH^{\bullet} + O^\bullet}$$
$$\tag{two reactive sp. instead of one}\ce{O^{\bullet} + H2 -> OH^{\bullet} + H^\bullet}$$
The bottom two steps show especially clearly why fires are violent, quick reactions (after incubation period is over). Heptafluoropropane's chemical extinguishing effect arises from

[----] the thermal decomposition of small amounts of heptafluoropropane in the flame which form fluorinated fragments such as $\ce{CF3}$ and $\ce{CF2}$. These will then consume the key combustion chain-propagating species $\ce{H}$ and $\ce{O}$, but to a lesser extent on $\ce{OH}$ radicals. The rates of chain-branching combustion reaction will decrease, the chemical flame is inhibited and the flame propagation is halted. (Choy, Fong)

To increase the chemical inhibition, heptafluoropropane (or HFC-227ea) is sometimes mixed with $\ce{NaHCO3}$.

Flame inhibition by sodium species is believed to be due to chemical scavenging of major radical species (e.g., $\ce{OH}$, $\ce{H}$) in the flame.
source: Skaags, 'Assessment of the Fire Suppression Mechanics for HFC-227ea Combined with $\ce{NaHCO3}$', U.S. Army Research Laboratory, link

One commercial source, unlike Wikipedia, groups pentafluoroethane together with heptafluoropropane.

source: Group of Companies "RealSnabService"

Further reading, more in-depth discussion and related topics:

Luo et al.. 'Effect of Hydrofluorocarbons and Perfluorocabons on Suppression Efficiency of $\ce{CH3I}$'. Process Safety and Environment Protection Research Group. link
Williams, et al.. 'Intermediate Species Profiles in Low Pressure Methane/Oxygen Flames Inhibited by 2-H Heptafluoropropane: Comparison of Experimental Data with Kinetic Modeling'. Annual Conference on Fire Research, $1998$. link
Sheinson. 'Heptafluoropropane With Water Spray Cooling System As A Total Flooding Halon 1301 Replacement: System Implementation Parameters'. Navy Technology Center for Safety and Survivability. link
Grosshandler, et al.. 'Extinction of Hydrofluorocarbon Flames with F/H Ratios of Unity and Greater'. Annual Conference on Fire Research, $1998$. link


Is it possible to produce sodium bicarbonate from potassium bicarbonate? I balanced the chemical reaction which is as follows:

NaCl + KHCO3 = NaHCO3 + KCl

I've also gathered solubility values which are the following:

KHCO3
22.4 g/100 mL (20 °C)
NaCl
35,9 g/100 mL (20°)
KCl
34.2 g/100ml (20°C)
NaHCO3
9.6 g/100ml (20 °C)

Since NaHCO3 is less soluble than the other compounds, I could separate it by boiling off some of the water up to a certain point, but I'm not sure how to predict which way the reaction will be going and if it will proceed completely.

How do I find out which way the reaction will proceed, and to which extent it will be complete?

I'd be glad for any help.

Boiling off some water is a decent plan. Perhaps it should be done at reduced pressure so as to keep the temperature low. Magnesium and calcium bicarbonates decompose at temperatures above 60C in water to yield insoluble carbonates, and several data sheets for sodium bicarbonate suggest the same. One said decomposition was complete at 100C. I suppose you could consider that any H2CO3 in equilibrium with NaHCO3 could decompose to CO2 + H2O. Heat would just drive the equilibrium to completion. I suppose that's why I could find no solubility data for NaHCO3 or KHCO3 at temperatures higher than 60C.

One way to go about the preparation is to dissolve 58.5 grams (1 mole) of NaCl in 260 grams H2O, and 100 grams of KHCO3 (1 mole) in 260 grams H2O. Heat both to 60C; they will completely dissolve. Then pour one into the other (it may make a difference which way you do it) and stir. The amount of water is sufficient to dissolve 1 mole (84 grams) of NaHCO3 at 60C (but the other salts could and probably will affect this). So the mixed solution will be clear; all salts will be soluble.

Cooling to 20C will (should!) ppt 34.1 grams of NaHCO3, leaving 49.9 grams in solution (again, ignoring the effect of the KCl). Cooling to OC should ppt another 14 grams of NaHCO3, leaving 35.9 grams in solution. You could then boil off about half of the water before the solubility limit of the KCl is reached, but because of the possible loss of CO2, you may not get much more NaHCO3.


Role of substituents in radical initialized polymerization Here, the answer given is (iv), and I don't see the reason why methoxyethene is not forming an addition polymer with radical initiators. Clearly, the first option will form polystyrene and the second, polyacrylonitrile. But after that, I feel lost. After looking at the answer key, I felt that there would be some role of the OCH3 here as it may react with the peroxide initiator, but I'm not quite sure what. As far as I know, the radical is generated at the double bond which then keeps polymerizing independent of the substituent. Can somebody tell me what is the mechanism of polymerization in the (iv) part?

The compound IV is a great compound for cationic polymerization , the explanation is the oxygen which is giving electron with resonance : The form where the oxygen has a plus charge and the end carbon a minus is the reason why this compound will be able to start a radical polymerization but with a really low constant of polymerization (kinetic constant), and also the reason why the cationic polymerization will have high constant.

Compounds with strong (electron donor and tiny) resonance will not be good for radical mechanism, but if you put a phenyl group behind the ether function it will be better because the resonance effect will be "diluted".


Cause of continuous spectral lines We know different types of atoms have different spectral lines. But what exactly are the spectral lines representing? What causes white light to have a continuous spectra while other atoms show discontinuous spectral lines?

The spectral lines represent transitions between pairs of discrete energy levels in the atom; an electron is excited by absorbing the energy of a photon and a transition from one of these levels to another occurs as the photon is destroyed. Emission lines occur when the reverse process happens.

If enough energy is absorbed the atom can be ionised and an electron with un-quantised energy can be liberated and moves to a position of complete separation where attractive forces between electron and nucleus is insignificant. As the liberated electron's energy is un-quantised the absorption giving rise to ionisation is continuous. The absorption is truly continuous, no matter how high a spectral resolution is used discrete absorption lines will not be observed.

Black body radiation is continuous emission from a hot body. The analysis of this emission lead Plank in 1901 to propose, for the first time, the idea that the photons were quantised, thus avoiding the 'ultra-violet catastrophe'. He did this by allowing ultra-violet photons to have more energy than visible ones which in turn had more energy that infra-red ones. Nowadays we know that he energy is directly proportional to frequency via  the Planck constant, $E=hv$.

In the black body, all the numerous energy levels in the body are highly excited and transitions between them continuously occur as the body is being continuously heated. These transition energies overlap one another in energy so that the spectrum is continuous.

(In molecules there are also discrete energy levels, in the ground state these are due to the vibrational and rotational motion of the molecule.  A molecule may also have electronically excited states and these can also have discrete vibrational and rotational energy levels. Furthermore there are dissociative excited states that have continuous absorption spectra.)


Analytical Chemistry predicting the order of elution of compounds in a reverse phase HPLC Predict the order of elution for a mixture containing caffeine and benzoic acid, using a reverse phase HPLC. Explain your answer.

My try:

Since benzoic acid is more acidic that caffeine, which means it has a lower pH, it is hence more polar. Since for HPLC, the mobile phase is the polar phase, elution of polar compounds will be favoured hence benzoic acid will be eluted first. Is my answer and explanation correct?

The logic is wrong. Lower $\mathrm{pH}$ (actually you mean $\mathrm{p} K_\mathrm{a}$ here) does not mean more polar.

It is more complicated to decide the polarity then you think. For reverse phase HPLC, the more polar and more soluble compound come out first. For reverse phase HPLC, it is common to have acid (TFA for example) in the eluent. It this case, caffeine can be protonated and become much more polar because it is ionic now. Solubility is another factor to consider. So if you want to predict the order, you have to provide more information.


Compare ONO bond angle in nitrogen oxides Which has maximum $\ce{ONO}$ angle?

  $\ce{NO3-}$
  $\ce{NO2+}$
  $\ce{NO2-}$
  $\ce{NO2}$

According to me $\ce{NO2+}$ should have the maximum angle because the structure is linear but in answer it is given $\ce{NO3-}$. We can consider the lone pair in $\ce{NO2+}$ but still also by VSEPR theory I don't think it is possible to decrease angle from 180 to less than 120.

You are right, that's $\mathrm{NO_2^+}$, since it is linear, and whoever says otherwise is wrong.

Also, those (except for NO2) are not compounds, but particles.


How to you show a 3-center 2-electron bond in a Lewis structure? How would you write $\ce{PCl5}$ or other hyper-conjugate molecules as a Lewis structure that indicates 3-center 2-electron bonds?

I have seen depictions [1] for transition metal complexes but not for main group non-metals such as sulfur or phosphorous.

What about the probably "not so good" approach might be the bent bond as sometimes seen for diborane?

(source)

The latest recommendation by IUPAC I know, exemplified in rule "GR-1.9 Multi-center bonds":

(Brecher, Pure Appl. Chem., 80, 277-410, 2008; doi 10.1351/pac200880020277, open access)

stating:

  "As a matter of convention, any such multi-center character is ignored
  when producing chemical structure diagrams, and regular bonds
  connecting pairs of atoms are used instead."

To accommodate Karsten Theis' comment on the initial form of this answer, I would like to add the perspective provided by Robert Grossmann's The Art of Writing Reasonable Organic Reaction Mechanisms.  Starting page 270 (2nd edition, 2003) the chapter introducing reactions with transition metals draws attention about "Conventions of Drawing Structures"; highlighting that there are field dependent conventions about what line and dash represent.  To quote:

  "The conventions for drawing organometallic and inorganic compounds
  differ in subtle ways from those used to draw “ordinary” organic
  compounds. The most important difference is the way in which bonds are
  drawn. In organic compounds, one does not use a line to connect a bond
  to an atom. In organometallic and inorganic compounds, however, a line
  sometimes connects an atom and a  $\sigma$ or $\pi$ bond. In this
  case, the line indicates that the pair of electrons in the $\sigma$ or
  $\pi$ bond is shared with the metal also."

(loc. cit. p. 271)

  "An even more confusing situation arises in complexes in which the
  electrons in a $\pi$ system spread over three or more atoms are used
  to make a bond to a metal. In this case, the usual convention is to
  use a curved line to indicate the $\pi$ system and a single line to
  connect the $\pi$ system to the metal, regardless of the number of
  electrons in the $\pi$ system (the organometallic chemists’
  convention). However, sometimes the curved line is omitted and single
  lines are used to connect the metal to each of the atoms in the $\pi$
  system (the crystallographers’ convention). The representation that
  would make the most sense to organic chemists, in which a single line
  represents a two-electron  bond and a dative bond is used to show
  two-electron bonds between each individual $\ce{C=C}$ $\pi$ bond and
  the metal, is simply never used."

(loc. cit., p. 271)

  "Formal charges are usually omitted in inorganic and organometallic
  complexes. Only the overall charge on the complex is indicated. For
  example, formal charges are usually not assigned in Lewis acid–base
  complexes involving transition metals. The acid–base bond is sometimes
  indicated by an arrow pointing from the ligand to the metal, but more
  often it is indicated by an ordinary line."

(loc. cit., p. 271)

This precedes just the section about electron counting (specifically in organometallic compounds with transition metals). I speculate, however, Robert Grossman aiming to highlight differences between "organic" and "organometallic" convention meant does while accidentally deploying does not in the first paragraph.


Chromium +1 is unknown while Mn+2 is known? Although both $\ce{Cr+}$ and $\ce{Mn^2+}$ both ions have $d^5$ configuration, still only $\ce{Mn^2+}$  is well-known while $\ce{Cr+}$ is not common. What is so special about Cr that it does not form $\ce{Cr+}$?

Even with the stabilization of the $d^5$ electron configuration, chromium is quite electropositive and the (second) ionization energy for removing an additional electron from $\ce{Cr^+}$ is rather low.  It's outweighed by the extra electrostatic attraction or solvation energy available by having two charges on the chromium, therefore you are ordinarily going to go on to $\ce{Cr^{2+}}$.

Chromium can adopt a +1 oxidation state in certain complexes, such as $\ce{Cr(CN)_5NO^{3-}}$.  See https://en.wikipedia.org/wiki/Chromium.


What is the relation between the following two optical isomers? I think they should be diastereomers, but the answer to this question is enantiomers, if it is enantiomers then how?

you can check this out it is not exactly the same but Cl is replaced by OH


Electolysis of H2O to produce O2(g), in a small, portable unit Full disclosure, I'm just a high school chem student, but I've got this question and I couldn't find the answer anywhere online.
I was reading an article about some futuristic "artificial gill", which kinda sounds like BS to me, but it got me thinking about this.

http://www.smithsonianmag.com/innovation/student-claims-have-designed-working-artificial-gills-180949465/

We were just learning in class about electrochemistry, and we just did electrolysis, so I was just thinking, instead of filtering the water, what if they used electrolysis to produce O2 gas?
There would still be hurdles, like diluting the 100% O2 to about 20% gas by volume, storing the necessary electric power to drive the reaction, and making sure that the unit wouldn't blow up, but could it theoretically work if we could overcome these other issues?

Basically, could you produce enough O2 gas to survive, through the electrolysis of H2O, in a unit small enough for a man to use?

In scuba diving the main issue is pressure. To be able to breathe, you need to counteract the pressure of the column of water on your rib cage. This is why divers breathe from compressed gas tanks. Otherwise their rib cage and lungs would be crushed by the pressure. At higher pressures, air becomes too viscous to be breathed without too much effort, and the build-up of dissolved nitrogen in the tissues leads to narcosis ("being drunk"), so the oxygen has to be mixed with helium, that is much less viscous. So, with theses 'artificial gills' or with electrolysis, you still need a tank of helium, and a powerful compressor.

Overall, it is much more convenient to carry tanks of compressed mixtures of gases. Compressing gases is a form of energy storage (that could even power cars). But if you are in a nuclear-powered submarine, then you would be better off doing electrolysis.


IUPAC name for 1,2,3-trichlorocyclopropane? What is the IUPAC name for the two isomers of 1,2,3-trichlorocyclopropane? Similarly what is the name for isomers of 1,2,3,4-tetrachlorocyclobutane, 1,2,3,4,5-pentachlorocyclopentane, etc?

1,2,3-trichlorocyclopropane has two isomers, one of them with all the chlorine atoms facing in one direction and the other with one chlorine facing the opposite direction of the other two. In CIP rule, if two substituents on an atom differ only in their absolute configuration, R takes priority over S. However I have difficulty applying the rule here: two of the substituents are exactly the same. Nevertheless the two molecules are clearly distinct.

In practice the problem could be circumvented by using trans- and cis- prefix. But it does not address the key problem and fails on longer chains --- 1,2,3,4-tetrachlorocyclobutane has 4 possible arrangements, though not necessarily all of them have the problem.

Here are the isomers of cyclobutane and their mirror images. I use + and - to denote a chlorine atom above or below the plane. The number in the parenthesis is the number of distinct molecules.

+ + | - -    (1)
+ + | - -

+ - | - +    (1)
+ + | + +

+ - | - +    (1)
+ - | - +

+ - | - +    (1)
- + | + -

Achiral cyclic compounds like 1,2,3-trichlorocyclopropane may contain pseudoasymmetric centers.
Pseudoasymmetric centers have distinguishable ligands (“a”, “b”, “c”, “d”), two of which are nonsuperposable mirror images of each other (enantiomorphic). The lower case stereodescriptors “r” and “s” are used to designate the absolute configuration of pseudoasymmetric centers, for example (2R,3r,4S)‑2,3,4‑trichloropentane:

Note that the stereodescriptors “r” and “s” describing a pseudoasymmetric stereogenic unit are invariant on reflection in a mirror (for example “r” remains “r”, and “s” remains “s”), but are reversed by the exchange of any two ligands (“r” becomes “s”, and “s” becomes “r”).
The configurations (“r” or “s”) at pseudoasymmetric stereogenic units are determined by application of the Cahn-Ingold-Prelog (CIP) priority system and the sequence rules. In particular, note that an atom or group with descriptor “R” has priority over its enantiomorph “S”. In order to establish the order of precedence of ligands in a stereogenic unit, the atoms of the stereogenic unit are rearranged in a hierarchical diagram, called a “digraph”. Descriptors specified in digraphs may correspond to the final descriptors or to temporary auxiliary descriptors used only for ranking ligands and never appearing as final descriptors.
Similar to the above-mentioned case of (2R,3r,4S)‑2,3,4‑trichloropentane, the structure of (1r)‑1,2,3‑trichlorocyclopropane contains one pseudoasymmetric center, which has two ligands that are nonsuperposable mirror images of each other:

The other isomer of 1,2,3‑trichlorocyclopropane is (1s,2s,3s)‑1,2,3‑trichlorocyclopropane, which contains three pseudoasymmetric centers. Each one has two ligands that are nonsuperposable mirror images of each other:

It may not be obvious how to choose the stereodescriptors ‘r’ or ‘s’ for 1,2,3-trisubstituted cyclopropanes. The ranking is a complex process and requires the use of auxiliary descriptors.
Generally, the stereodescriptors ‘r’ and ‘s’ are assigned by applying the Cahn-Ingold-Prelog (CIP) Priority System as for the stereodescriptors ‘R’ and ‘S’. The order of precedence is reached by the construction of a hierarchical digraph and the application of the Sequence Rules.
The constitutional formula of a cyclic molecule is transformed into an acyclic digraph according to the methodology that is described in Subsection P-92.1.4.3 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) and which has been recommended by Prelog and Helmchen.
In order to obtain the acyclic digraph of a stereogenic unit, the didentate ligand (i.e. the ring) is transformed into two monodentate ligands by leaving intact in each case one bond with the core of the stereogenic unit and cleaving the other bond. At the end of each of the two branches thus obtained, a duplicate atom of the core is attached. (Note that this methodology is similar to the case of multiple bonds.) For example, for the pseudoasymmetric stereogenic unit at the atom numbered ‘1’ of (1r)‑1,2,3‑trichlorocyclopropane:

where $\ce{(C)}$ is a duplicate representation of the core of the pseudoasymmetric stereogenic unit.
As the digraph is acyclic, it is now possible to specify the stereodescriptor ‘R’ or ‘S’ for the stereogenic unit at the atoms numbered ‘2’ and ‘3’ in each branch. These are auxiliary descriptors that are different in each branch.

Since the auxiliary descriptors are different, they allow the branches to be ranked using Sequence Rule 5, which states that ‘R’ has priority over ‘S’.

P-92.1.3.5 Sequence Rule 5

An atom or group with descriptor ‘R’, ‘M’, and ‘seqCis’ has priority over its enantiomorph ‘S’, ‘P’ or ‘seq Trans’.

Therefore, the configuration of the pseudoasymmetric stereogenic unit at the atom numbered ‘1’ of (1r)‑1,2,3‑trichlorocyclopropane can be determined to be ‘r’.

In the same way, the configuration of each pseudoasymmetric stereogenic unit of (1s,2s,3s)‑1,2,3‑trichlorocyclopropane can be determined as ‘s’.


Equation to calculate the triple point Is there any equation that can calculate the triple point of a substance? Because a method for calculating the critical temperature and the Boyle temperature of a substance that I posted in another question I recently found that it also seems to be able to predict the triple point for some substances.

I searched the Internet and couldn't find any other such equation, in fact I came across a site that said there was none. Is this true?  My equation predicts it from the Lennard-Jones parameters. Source:

Here is an excerpt from a document that I'm putting together. It also predicts the reduced triple point temperature to be 0.5

In principle, you can use the integrated Clapeyron and Clausius–Clapeyron equations to calculate $P$ vs. $T$ co-existence lines and find the crossing point which is at the triple point. You need the $\Delta H$ for fusion and evaporation and the densities of the solid and liquid phases.
The equations are $$ p_\text{liq-vap}=\exp\left(\frac{\Delta H_\text{vap}}R\left(1/T_1-1/T\right)\right)$$ and for the solid vapour line use $\Delta H_\text{fus} +\Delta H_\text{vap}$

For the solid-liquid line use $$ p_\text{sol-liq} = P_1 + \frac{\Delta H_{\text{fus}}}{\delta V}\left(\ln\left(T\right)-\ln\left(T_1\right)\right)$$

where $\delta V = M\left(1/d_\mathrm l-1/d_\mathrm s\right)$ where $M$ is the molecular weight and $d$ the densities of liquid and solid. $T_1$ and $P_1$ are temperature and pressure at the triple point. Thus you have to iterate $T_1$ and $P_1$ by fitting (say by least squares) to the calculated curves to whatever data you have to find the triple point. Tricky!

Naively, I am surprised that using Lennard-Jones parameters would work as they describe interaction potential between two atoms and not in a solid or liquid.


Product of intramolecular aldol condensation of 3-methyloctane-2,7-dione What is the major product formed when 3-methyloctane-2,7-dione is treated with sodium hydroxide?

My main issue is where the base deprotonates. Two enolates are possible (I neglected the other two because upon cyclisation, they form seven-membered rings which are less stable than the five-membered rings otherwise formed). Enolate 2 should be more stable than enolate 1 due to the greater degree of substitution:

However, these two enolates lead to different products. So, which product should be formed? I know that if a strong base like LDA was used, a proton will be abstracted from one of the less hindered terminal carbons. But hydroxide ion is small and should be able to abstract more hindered protons.

For the enolates that can be formed, the others have already given comprehensive answers: All four enolates are possible and since we are under thermodynamic control (weak base), we expect o observe them all. Numbering the chain carbons according to IUPAC, and noting the non-carbonyl end of the enolate, I expect the enolate ratio to be $3>6>1\approx 8$.

But your question asked for the reaction’s major product. First of all, the two terminal enolates (carbons 1 and 8) are to be dismissed: The seven-membered ring is much less stable. Then, the reaction mechanism shows us which enolate leads to a thermodynamically much more stable product than the other:

The more stable enolate 2 unfortunately cyclises to give a β-hydroxyketone 5 which has a quarternary carbon between the ketone and the newly-formed hydroxy group. No elimination is possible here. The other enolate aldol product 3, however, can eliminate to the corresponding cyclopentene 4. This should likely be given as the reaction’s major product since all steps up to the elimination are reversible. (The elimination is, too, but it is still most stable.)

Indeed, Stork and Williard report that the cyclopentene 4 is obtained when 3-methyloctane-2,7-dione is refluxed with ethanolic KOH for 2 hours, although no yields are given as it is only mentioned in a footnote.[1]

Stork, G.; Williard, P. G. Five- and six-membered-ring formation from olefinic α,β-epoxy ketones and hydrazine. J. Am. Chem. Soc. 1977, 99 (21), 7067–7068. DOI: 10.1021/ja00463a053.


What does "parallel in-register" mean for a protein structure in a fibril? I read a paper by Ross and Poirier [1, p. S12] which has the following lines (see section Commonalities of amyloid structure) that describe the structure of protein in a fibril:

  The most extensively characterized amyloid fibril is that formed from the β-amyloid (Aβ) peptide implicated in AD. Using SOLID-STATE NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY, the in-register, parallel β-sheet organization of fibrils formed by Aβ10–35, a fragment of the full-length 42-residue Aβ peptide, was first described.

I couldn't understand what "in-register, parallel β-sheet" means.

References

Ross, C. A.; Poirier, M. A. Protein Aggregation and Neurodegenerative Disease. Nat Med 2004, 10 (7), S10–S17. https://doi.org/10/dppb95.

There is a clear explanation and figure showing this in the following reference:

Anton Gorkovskiy, Kent R. Thurber, Robert Tycko, Reed B. Wickner, PNAS 2014, 111 (43) E4615-E4622.

A parallel beta sheet is one where the direction from N- to C-termini on adjacent strands run in a parallel direction (rather than antiparallel). In-register means that each residue in one molecule is aligned with the same residue in the neighboring molecule, producing lines of identical residues along the long axis of the filament.


Where is the acid in DNA/RNA? It is well known that the A in both DNA and RNA stands for acid, but where is the acid in chemical formula for the compound, and it is classified so based on what acid-base theory? Like Arrhenius, Brønsted-Lowry or Lewis.

Just to better explain my question, when I say DNA I am essencially thinking about nucleotides linked together, am I missing something in the structure that contains my answer or this is the way to go?

The phosphate ester groups which connect the nucleotides contain one acidic proton at their OH group, and two at the end of each strand. As you can see here, here and in the picture below (with the groups marked turquoise), most of these groups are deprotonated depending on pH. They act as Brønsted-Lowry acids.


How many stereogenic centres are there in discodermolide? It is generally said that carbon-carbon double bonds cannot behave as stereogenic centres but the above molecule contains three such centres. Can somebody please point out all the stereogenic centres (including the double bonded carbons) and explain how and why this is so?

Summary

The confusion probably arises because of incompletely standardized definitions of at least some subset of the words "stereocenter", "stereogenic", "stereogenic center", "stereogenic unit".

IUPAC definitions

The place to turn for authoritative definitions of these terms is IUPAC's Gold Book, as noted by @jerepierre in the comments.  Here is what it has to say:

  stereogenic unit (stereogen/stereoelement)

  A grouping within a molecular entity that may be considered a focus of stereoisomerism.

  At least one of these must be present in every enantiomer (though the
  presence of stereogenic units does not conversely require the
  corresponding chemical species to be chiral). Three basic types are
  recognized for molecular entities involving atoms having not more than
  four substituents:

  A grouping of atoms consisting of a central atom and distinguishable
  ligands, such that the interchange of any two of the substituents
  leads to a stereoisomer. An asymmetric atom (chirality centre) is
  the traditional example of this stereogenic unit.
  A chain of four non-coplanar atoms (or rigid groups) in a stable conformation, such that an imaginary or real (restricted) rotation
  (with a change of sign of the torsion angle) about the central bond
  leads to a stereoisomer.
  A grouping of atoms consisting of a double bond with substituents which give rise to cis-trans isomerism.

Thus, the unequally substituted C=C moieties in discodermolide are "stereogenic units" according to IUPAC.

Stereocenter

Unfortunately, this term is not defined in IUPAC's Gold Book.

Stereogenic center

Unfortunately, this term is not defined in IUPAC's Gold Book.

Wikipedia

Wikipedia is not as authoritative as IUPAC but it is likely that chem.se readers will encounter wikipedia entries for these terms far more often than IUPAC's in the course of regular web searching.

Stereogenic unit

No Wikipedia entry

Stereogenic center

Redirects to "stereocenter", see below.

Stereocenter

  A stereocenter or stereogenic center is an *atom* bearing groups such that an interchanging of any two groups leads to a stereoisomer.

This definition does not apply to C=C double bonds, because only an atom can be a stereocenter, not a group of atoms.  Supporting this interpretation is a complete lack of reference to cis-trans isomerism on the rest of the page.

Usage by chemists

Discodermolide is referred to in many sources as having 13 "stereocenters" or "stereogenic" centers.  This conforms with the Wikipedia, atom-centric definition of stereocenter.  Here are some example sources:

Wikipedia's discodermolide article: (+)-discodermolide has
...13 stereogenic centers...
A presentation from 2004:
molecule possesses: 13 stereocenters
A paper from 2003: Paper title: Regulated-Stereoselective Construction of Thirteen
Stereogenic Centers...

All these uses comport with both Wikipedia definitions of "stereocenter"/"stereogenic center".  Chemists do not seem to use the IUPAC term "stereogenic unit" very much in actual practice.   I could not find any relevant pages using both the term "discodermolide" and the phrase "stereogenic unit", for example.

Bottom line

It is just possible to pass a thread of consistency through the eye of this needle of confusing terminology.

Discodermolide has thirteen (13) stereogenic centers.
Each of those stereogenic centers is by definition also a stereocenter, and is also a "stereogenic unit".
In addition to those 13 stereocenters, there are three additional "stereogenic units" in discodermolide corresponding to the unequally substituted C=C bonds.


Why green leaf turns brown in Brine solution? I immersed a semi dry green leaf in glass of concentrated NaCl solution . I left it in mild sunlight for around 3 hours . Solution before was clear but afterwards it became reddish brown and the color of leaf turned brown. I want to know why this happened? Please help.

It is most likely the $\ce{Na+}$ is formed a complex salt with chlorophyll (bulky group) subsequently displacing $\ce{Mg+}$ cation, even though $\ce{Na+}$ doesn't have extra electronic shells for complex ion stabilisation, instead the chlorophyll (through the $\ce{N}$ atoms with lone pairs) provided extra stabilisation of the positively charged $\ce{Na+}$ cation.

Chlorine dissociated from $\ce{NaCl}$ to form aqueous chloride ions $\ce{Cl-(aq)}$ forming a typically greenish-brown solution you noticed during your experiment.


How do native sulfur crystals form? I was visiting an elements exhibition in the Ulster Museum (Northern Ireland) and there were these native sulfur crystals on display.

My question is how do sulfur crystals like these form in nature? Does the sulfur precipitate out of a solution? Are they the product of a chemical reaction?
I am confused because of sulfur's negligible solubility in water and I couldn't imagine any other way such perfect crystals could form.

I am really interested in making such crystals myself but as I have said, I don't know by what process or under what conditions this crystallization of sulfur occurs.

Tl;DR: Wikipedia mentions some of the ways in which elemental sulfur is produced naturally:

Through volcanic emissions, including emissions from hydrothermal vents. Some of them found near hot springs and near volcanic deposits. Also, lakes of molten sulfur up to ~200 m in diameter have been found on the sea floor due to submarine volcanoes.
As minerals: sulfide minerals, such as pyrite (iron sulfide), cinnabar (mercury sulfide), galena (lead sulfide), sphalerite (zinc sulfide), and stibnite (antimony sulfide); and the sulfate minerals, such as gypsum (calcium sulfate), alunite (potassium aluminum sulfate), and barite (barium sulfate).
Action of anaerobic bacteria on sulfate/sulfide minerals.

Long answer:
I found a paper1 which discussed #3 point in details. It said that although sulfide minerals can be oxidized to zero-valent sulfur by sulfate reducing bacteria in presence of molecular oxygen, it may pose the following problems:

exposure to oxygen would drastically decrease growth of microbes thereby slowing down sulfide production
on geologic timescales, excess supply with oxygen would convert
sulfide into sulfate rather than native sulfur
enormous amounts of oxygenated water would be needed to produce sulfur in large amounts

So, in order to overcome these problems, scientists proposed four possible mechanisms as a means to form native sulfur:

a modified sulfate reduction process that produces sulfur compounds
with an intermediate oxidation state,
coupling of sulfide oxidation to methanogenesis that utilizes
methylated compounds, acetate or carbon dioxide
ammonium oxidation coupled to sulfate reduction
sulfur comproportionation of sulfate and sulfide.

These reactions are found to be thermodynamically favorable and especially useful in environments containing dissolved sulfide. This provide evidence that microbial species functioning in such environments can produce native sulfur in absence of light and external oxidants such as $\ce{O2}$, nitrate, and metal oxides.
Reference

Labrado, A. L., Brunner, B., Bernasconi, S. M., & Peckmann, J. (2019). Formation of Large Native Sulfur Deposits Does Not Require Molecular Oxygen. Frontiers in microbiology, 10, 24. https://doi.org/10.3389/fmicb.2019.00024


What are the ozonolysis products of phenanthrene? What is / are the product(s) of ozonolysis of phenanthrene?

I am not sure I understand how to proceed. Which of the $\ce{C=C}$ double bonds is most susceptible to this type of reaction?
I am confused given the resonance and aromatic character of the compound which doesn't make me feel comfortable doing ozonolysis the usual way.

Phenanthrene reacts quickly with one mole of ozone at the 9,10-double bond according to this paper here and the OrgSyn procedure here referred to above by @Buttonwood. In chloroform a polymeric mono-ozonide is obtained. In MeOH the reaction takes a different course resulting in the formation of a cyclic methoxy-hydroxy-peroxide and a cyclic dimethoxy-peroxide. Reductive work up (Zn/AcOH or Me2S) of these products gives the di-aldehyde.


pH Indicators for Hoffman apparatus I've read that bromothymol blue can be mixed in with the electrolyte in the demonstration of electrolysis using a Hoffman apparatus. The indicator clearly shows the local differences in pH for the anode and cathode. And the blue color provides good visual contrast between the fluid and the gases that evolve. But I've found this indicator tends to break down quickly into a colorless solution while operating the apparatus, and after awhile you lose the indicating effect as well. Does anyone know of a better choice of indicator to use that will last longer?

I've never used it for that particular experiment, but bromocresol purple has a similar transition range and a very obvious colour change. The trouble with using indicators here is that many are organic compounds and are subject to redox reactions near the potentials at which electrolysis of water is occurring, so they break down or polymerize if they reach the electrodes. Bromothymol blue is known to oxidize at below 1 V and deposit on the electrode. Bromocresol purple tends to polymerize, but it seems to be at a relatively high potential (>1.5 V), so it may fare better.

It may also be possible to limit the diffusion of the indicator to the electrode surfaces. Perhaps using an indicator trapped in polymer beads or a block of agarose gel.


Why is the boiling point of polonium less than that of tellurium? My textbook states that polonium has a lower boiling point than tellurium because it has weaker intermolecular forces of attractions (van der Waals forces).

Why are van der Waals forces of attraction less in polonium as compared to tellurium?

Polonium's atomic size is greater than tellurium, and I believe van der Waals forces are directly proportional to the surface area. Is this the reason?

It's mainly about relativistic effects. Relativistic affects mainly affect heavy nuclei, such as polonium etc, so makes their properties slightly different to the trend.[ref] The outer electron configuration of Polonium could differ so much from its lighter family member tellurium due to the electron constriction, causing an outer electron to fall into a lower shell, that the intramolecular forces are weaker. Thus a lower boiling point.

It explains a lot of things that do not seem realistic with very heavy nuclei. Such as the "golden" colour of Gold. It should be silvery, but due to relativistic effects it is a "golden" colour. It's also like Copernicium behaving more like a noble metal, unlike the other metals of group 12, even being gaseous in room temperature.

Reference: Wikipedia article on relativistic quantum chemistry


Why sugar solution/sugarcane juice becomes sour when kept for certain time? I stir-dissolve some sugar in a cup of water and then keep it open for certain time (say a few hours), when I taste it I feel that it has became sour.
I have observed the same thing many times when I went to juice shop(s) and asked especially for sugarcane juice when I drank it, it tasted a bit sour (sometimes too sour to drink) and I just realized that the sugarcane (or other fruits) juice was extracted some hour(s) ago because I know that a fresh juice doesn't taste sour until kept (open) for a certain time.
From all this experience one thing I know is that the sugar solution or sugarcane (or other fruits) juice becomes more or less sour when kept for a certain time after extraction. It's most probably due to acid (which usually tastes sour) formation but I don't know what chemical changes turn a sugar solution or (sugarcane) juice sour (acidic) when kept for a certain time?
Any explanation will be greatly appreciated. Thanks.

This is not a complete answer but it should give you a head start.

I suspect two things: Fermentation and Caramelization.
In any fermentation process, sugar will turn into acid. It is observed from this answer from cooking.SE that at a temperature of 35-45 °C, fermentation occurs. If any sugar laden food is kept at this temperature, bacterial or yeast colony would flourish giving you the fermentation effects and making your food acrid. Fermentation process is a two-step process. First, glucose is turned to ethanol by Saccharomyces cerevisae, also known as Brewer's yeast. This ethanol is then converted to acetic acid by Acetobacter aceti, giving you the sour effect. This two-step process is called aerobic fermentation.

Caramelization generally occurs at higher temperatures, at 150-170 °C where sugars are broken down to simpler products and turning sugar into a brown-black mixture of products. This reaction is also called Maillard reaction. At this temperature, the reaction happens almost instantly but at lower temperature (35-45 °C), this reaction will takes hours, days or even months to finish it but it will speed up if there is moisture where it will take a few hours. Note that caramelization process is very complex as there are various steps. The finished product is a brown-black goo which is actually a mixture of several products. But if put it in a simplified way, sucrose breaks into glucose and fructose which further breaks into acetic acid, maltol, furan and many other products. This acetic acid contributes to the sour taste. So, in a hot-humid day, sugar can turn sour.


How many crystal forms of HCl have been identified - where to find the unit cells/lattice constants? This blog about stomach acid shows this image of the "zig-zag" form of crystalline HCl (see also Wikipedia where it looks (to me) like five four HCl's per unit cell), but links to this cubic crystal structure (2nd image) data card instead with only one HCl/cell.

I've read crystallographic "data cards" in a past life, but I have no idea how to search for all the forms of HCl on-line (and non-paywalled) now, in order to get the lattice constants, angles, unit cells etc. to build a 3D model of each form of crystalline HCl in Blender for example.

note: I am really looking for just the dimensions and shape of the unit cell, and coordinates of the atoms within the unit cell as a name and numbers, as one would find on a crystallographic data card. I can do all the plotting - no need for plotting software.

below: Here's the Wikipedia image mentioned above, from here. It certainly looks like the Chlorines are at 0.25 and 0.75 along the a-axis (vertically) and 0.00 and 0.50 along the b- and c-axes, so there seems to be enough information here to work out or infer the complete 3D structure, assuming it's cubic, and further infer that a=b=c=521pm, which suggests a density of this solid of about 1.7 g/cm^3. But remember, when you infer, you make an inf out of e and r.

From the first picture depicted, I would refrain from stating there are five HCl molecules per unit cell.  One thing, while there is some choice in the definition of a unitcell (there are several definitions possible, even under the constrain "it should be the simplest and still complete representation possible"), lower multiplicities occur more often; and more frequently seen values are $Z = 1, 2, 4$.

Second remark, again to the first picture: no, it does not look so as there were five molecules per unit cell.  Here, the projection of the $(a,b)$-plane is depicted (the little green and red axis emerging the origin O bear a label).  To graphically determine the actual multiplicity or the number of molecules present in a unit cell, it is better to "walk around" the unit cell; for example by accessing either *.cif (typical for small molecule structures) or *.pdb (more typical for protein structures) in a appropriate program, like CCDC's Mercury (likely used in the first drawing), or Jmol (used for the second drawing), or Avogadro.  Each of the three programs may be deployed under Windows, Linux, or MacIntosh; and in terms of displaying structures and to perform measurements of distances and angles available for free (as free beer).

It is possible, yet not in every case necessary to go from *.pdb or *.cif via Blender to prepare a dataset for 3D print though.  The "just do it" software that comes to my mind for this purpose is Werner Kaminsky's cif2vrml (windows only).  If you have Mercury (which actually may read both *.pdb, *.cif, beside *.mol, *.mol2, *.xyz, too), there is a built-in file -> print 3D menu, too; allowing you to chose between *.vrml or (then colourless) *.stl file.  Doing so found entry in publications, where examples in the Journal of Chemical Education like DOI: 10.1021/acs.jchemed.5b00597 and DOI: 10.1021/acs.jchemed.6b00293 come pretty close to what you intend to do. (As you are not afraid of larger molecules, consider the dedicated NIH 3D exchange as a valuable source of inspiration, too.)  Both jmol as well as Avogadro (referring to 1.2.0; not to the branch of Avogadro2) equally offer VRML export, the former including a dedicated how-to here.

For Jmol (the second figure presented), there are (at least) two options to read *.cif data.  The one by the GUI typically opens the *.cif, representing both the crystallographic motif and the unit cell (the later may be switched off, if necessary / useful).  Note that the GUI in the top row differs slightly from the one accessible by right-mouse-button click inside the window.  Via the toprow gui, you may access the CLI console, too -- allowing you to access many other functions, too.  For example, if you decompressed the *.zip archive, placed your *.cif into the same directory where jmol.jar lives, typing

  load cell test.cif {1 1 1}

will load the data set, test.cif, and will fill the complete unitcell till indices {1 1 1}. For one of the data found at COD, Jmol provided me this representation:

Where to get the *.cif? Literature databases like ACS' Scifinder and Elsevier's Reaxys indicate / link to such primary literature.  Depending on the journals policy, these *.cif data may be freely available in the SI (ACS journals, for example), or not (Tetrahedron family ...).  Clearly, a dedicated database ones like CSD (organic, metalorganic; preferentially small molecules than proteins) is an option.  For the service and potential insight offered, an annual academic subscription is very well invested money (was once around US$ 300-400 per year and full time equivalent).  If you encounter their structure code in a literature reference, but not the *.cif, you may request the data free of charge here; regardless if you / your institution holds a subscription, or not.

For mineral compounds, for example ISCD or the one by AMS.  The baltic COD is much younger than the others (CSD started more than 50 years ago), so the data coverage is not yet this complete, but searchable by keyword and structure, too; yielding by the former (allowing only H and Cl as the only elements present) 5 entries for HCl.

Where to find all the *.cif of HCl? It is not really organic / metallorganic (CSD), nor really a mineral (ISCD).  Perhaps via Scifinder and Reaxys one may retrieve a more round answer.

Addition

One reference found doi:10.1038/213171a0 is based on powder diffraction studies of HCl and DCl, listing these structures known to the authors back in 1967:

It continues with further powder diffraction studies, presenting an orthorhombic model, with four molecules per unit cell:

and a projection very similar to your first figure:

From these two it may be possible to construct (manually) a *.cif file which is then readable to Mercury others.  For this nature paper, Thomson Reuter's Web of Science alone knows about 139 citations, the most recent one in 2015 10.1021/acs.jpcc.5b01653 about predicted supraconducting polymorphs of HCl and HBr in the monoclinic $C2/m$.

Furthermore, a cif file based the Italian 1933 structure report the English wikipedia refers to is accessible in the COD if searching by keyword (H and Cl elements present, excluding O and C, minimal = maximal number of elements = 2).


Why does sodium polyacrylate solidify, on addition of water? Sodium polyacrylate is a white powder with a characteristic odor. But when we add water to this powder, instead of forming an aqueous solution, it instantaneously  solidifies and forms more amount of powder like substance, also called "hot ice".
Why and how this phenomenon occur?

YouTube video link

Sodium polyacrylate is an example of a super-absorbing polymer. It is a cross-linked (network) polymer that contains sodium atoms. It absorbs water by a process called osmosis. When the (sodium-containing) polymer is placed in contact with water, there is a tendency for the sodium to distribute equally between the network and the water.

That means, some of the sodium atoms want to leave the network and move to the water. When these sodium atoms leave, they are replaced with water molecules. Water swells the polymer network to try to keep the sodium concentration balanced between the polymer and the water.

The cross-links that connect the chains together prevent them from dissolving/breaking apart in the water. The resulting product appears very similar to fake snow, as you mentioned, a solidified powder like substance.

Sodium polyacrylate can absorb it’s own weight 300x over making it an amazing chemical to aid in water cleanups.

The question was good.


Is there a specific term for reactions of the form A + B → C + D? I'm working with an abstract model of chemical reaction networks. In this model, some of the reactions have the form $\ce{A + B <=> C}$ and others have the form $\ce{A + B<=>C + D}$. I'm wondering if there is some existing terminology I can use to refer to these two cases.

I know that the terms "unimolecular" and "bimolecular" refer to the number of reactants, but in this case the important thing is the number of products. The ideal thing would be if there's a general term describing whether a reaction has the same number of products as it does reactants.

Alternatively, since the species I'm modelling are abstract polymers, I can use the terms "cleavage" and "ligation" to refer to the $\ce{A + B <=> C}$ case, depending on the direction in which the reaction proceeds. If there is a corresponding term for $\ce{A + B<=>C + D}$ type reactions in polymers, that would also be useful to know.

I suppose at a stretch I could call it an exchange or a substitution reaction, but (as far as I know) that tends to imply that the thing being substituted or exchanged is a single atom, rather than a whole section of polymer. (In my case the reaction might be something along the lines of $\ce{aaabb + ccdddd -> aaadddd + ccbb}$, where $\ce{a}$, $\ce{b}$, $\ce{c}$ and $\ce{d}$ are different types of monomer.)

There are 4 'base-types' of chemical reactions:

Synthesis reactions ($\ce{A + B -> AB}$)
Decomposition reactions ($\ce{AB -> A + B}$)
Substitution reactions ($\ce{A + BC -> B + AC}$)
Metathesis reactions ($\ce{AB + CD -> AD + CB}$)

Given that you talk about polymers that in one case combine to form a new (single) polymer: $\ce{A + B <=> C}$ you could call this a synthesis reaction.

You other example, $\ce{A + B <=> C + D}$, is an exchange of parts of the polymers so that would be a metathesis reaction.

For polymers there are a couple of additional options to name the types of reactions, but these depend a lot on the specific chemistry (e.g. hydrolysis if water is one of the reactants) so that is probably difficult if you have 'abstract' polymers.


Diels-Alder rate of reaction Why is cyclopentadiene more reactive than 9-anthracenemethanol when acting as a diene?

Both molecules are locked in an s-cis conformation, which is favourable. I'm thinking that it's due to the extensive conjugation in 9-anthracenemethanol?

Yes, you are exactly right.  Anthracene is better described by this structure (where there are no fixed single and double bonds, but rather a delocalized aromatic structure)

than a structure like this (where the single and double bonds are fixed and distinct from one another).

So in anthracene, the aromatic character must be disrupted to some degree in order to undergo the Diels-Alder reaction.  In cyclopentadiene, there is no aromaticity to disrupt, the double bonds are basically fixed like in buta-1,3-diene (and, as you point out, locked in the required s-cis conformation).

A final note, benzene won't undergo a Diels-Alder under normal conditions because its aromaticity is completely disrupted and lost during the process.  In anthracene the aromaticity remains in two of the rings, a much smaller price to pay.


Evaluate indicator choice (phenolphthalein) A student wishes to find the concentration of a $\pu{50 mL}$ solution of acetic acid $(K_\mathrm{a} = \pu{1.8e-5}).$ He performs a titration using a solution of $\pu{0.2 M}$ sodium hydroxide. The student selects phenolphthalein as his indicator choice which has a $K_\mathrm{a} = \pu{3.98e-10}.$ Justify the student’s choice of indicator.

My attempt: I found the $\mathrm{p}K_\mathrm{a}$ of the indicator to be $9.4$ and the fact that the titration was weak acid + strong base meant the $\mathrm{pH}$ at equivalence would be around that too, so the indicator was justified. In saying this, my justification isn’t very mathematical.

Easy marks for phenolphthalein usability are:

$99.9\%$ of acetic acid is titrated at about
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{1000} = 7.75$$

$100\%$ of acetic acid is titrated at
$$\mathrm{pH} = 7 + \frac12 \cdot (\mathrm{p}K_\mathrm{a} + \log c) = 9.385 + \frac12 \log c$$

Phenolphthalein color transition is $8.2-10.0.$


Why is the product of the reaction between iron and steam is iron (II, III) oxide and not iron (II) oxide or iron (III) oxide? The reaction between Iron and steam in my textbook is given as this:

$$\ce{3Fe + 4H_2O <=> Fe_3O4 + H_2} $$

My question is why these following reactions are not possible or do not happen.
$$\ce{Fe + H_2O <=> FeO + H2}$$
$$\ce{2Fe + 3H_2O <=> Fe_2O_3 + 3H2}$$

I mean, how do the atoms know to form iron (II,III) oxide instead of Iron (II) oxide or iron (III) oxide in this reaction?

Recall that $\ce{Fe3O4}$ is basically just an equimolar mixture of $\ce{FeO}$ and $\ce{Fe2O3}$. So, in that way, both of the reactions that you mentioned do infact happen, and the result is what was mentioned in your textbook.


What is Meant by Mole Percent? I am attempting to interpret an experimental section on a scientific paper. The paper calls for 0.51 mmol of 3,5-dimethoxy phenol, 1.0 mmol of 3-methyl-3-butenal, 5 mol% of phenylboronic acid, 20 mol% of benzoic acid, and 2.5 mL of heptane. I am trying to figure out what is meant by mole% in this case. There are many different interpretations of this I have seen online, including mole percent being based on a fraction of the limiting reagent (the phenol in this case), the mole percent being based on the fraction of all of the total moles of reactant and catalyst (most fitting of the definition of mole percent), and the mole percent being based on all of the moles of reactant, catalyst, and solvent. It makes most sense that the mole percent for this protocol would be based on the limiting reagent, but I included the calculations based on the total moles of reactants and catalysts for reference. Does anyone know what the term "mole percent" would appropriately based on in this case?

0.51 mmol (phenol der.) + 1.0 mmol (aldehyde) + x mol PhB(OH)2 + y mol PhCOOH = z total mol

0.2 z (total mol) = y mol PhCOOH
0.05 z (total mol) = x mol PhB(OH)2

0.51 mmol (phenol der.) + 1.0 mmol (aldehyde) + 0.2z + 0.05z = z total mol
0.75z = 1.51
z = 2.013

0.2 (2.013) = 0.40 mol benzoic acid
0.05 (2.013) = 0.10 mol phenyl boric acid

If the mole percent was based on the moles of the LR, the moles of catalyst would be as follows:
0.10 mol benzoic acid and 0.026 mol of phenyl boric acid

Eli, you should have read the paper very carefully before asking questions. I think you are very intelligent person, so don't choose easy way out. Now, to answer your question, about mol%. It is right in the paper. As you already figured out and also Orthocresol pointed out in his comment, The amount equals to given mol% is $LR \times \frac{given \ mol\%}{100}$. It is in the General procedure A of given reference:

  Condensations of phenols and naphthols with $\alpha,\beta$-unsaturated ketones. An oven-dried $\pu{25 mL}$ Schlenk tube equipped with a teflon-coated magnetic stir bar was evacuated and purged with argon. The phenol ($\pu{0.2 mmol}$), pentafluorophenylboronic acid ($\pu{8.5 mg}, \ \pu{0.04 mmol}, \ \pu{20 mol\%}$), diphenylphoshinic acid ($\pu{8.7 mg}, \ \pu{0.04 mmol}, \ \pu{20 mol\%}$), $\alpha,\beta$-unsaturated ketone ($\pu{0.4 mmol}$) and heptane ($\pu{1 mL}$) were added to the tube under argon.....

Accordingly, The phenol used: $\pu{0.2 mmol}$ ($LR$), pentafluorophenylboronic acid used: ($\pu{0.04 mmol}$ meaning $0.2 \times \pu{20 mol\%}$), and diphenylphoshinic acid ($\pu{0.04 mmol}$ meaning $0.2 \times \pu{20 mol\%}$).


Temperature relation for isomerism reactions When we have an isotherm isomerism reaction
$\ce{A(g) -> A^*(g)}$  with $\ce{ΔH^0 = 0}$
is the ratio of preassures $\ce{P_A^* / P_A}$ independent from temperature?

The van't Hoff equation would predict that the equilibrium constant is not dependent on temperature, since in this case:
$$\frac{\mathrm{d}}{\mathrm{d}T}\ln K_{eq} = \frac{\Delta H^\circ}{RT^2}=0$$
Strictly speaking the equilibrium constant is a function of fugacities, but it is probably safe to assume that the pressures are also independent of T, so the answer would be yes.


Sodium and potassium produce a strange liquid, why the liquid explode when it gets contact with water? I saw an experiment sodium and potassium mixed together to produce a liquid alloy  but when the liquid get contact with water then some kind of explosion occur.

Why the explosion occur when the liquid  made contact with water?

Same explosion will occur if it would get contact with other form of liquid?

This site provides an explanation about the mechanism of the explosion.

...causes an explosion because the metal dissolves, generating an
extreme amount of heat and transferring electrons to the water. The
dissolution step also generates steam and forms hydroxide ions and
hydrogen, which can be ignited, making the process even more
energetic.
Some researchers have puzzled over how the process can occur so
quickly, though. They recognized that the steam and hydrogen generated
early on in the reaction should form a buffer layer over the metal
surface and impede water from continuing to react.
...within a fraction of a millisecond of making contact with water,
the Na/K droplets form numerous spikes that protrude into the water.
Molecular dynamics analysis indicated that nearly instantaneous
transfer of electrons from the spikes to the water rapidly generates
positively charged alkali ions, which vigorously repel and cause a
so-called Coulomb explosion. It is the speedy manner in which that
process propagates and generates reactive metal surfaces that triggers
the overall explosion.
The researchers have figured out many of the key aspects that enable
this highly exothermic reaction to become explosive, rather than
self-quench.

It does not generally reacts with any other liquids. Hence they are stored in mineral oil.

It is highly reactive with water and is stored usually under hexane or
other hydrocarbons, or under an inert gas (usually dry nitrogen or
argon) if high purity and low levels of oxidation are
required.(Wikipedia)


InChIKey generation from InChI I am trying to find some good resources to understand generation of InChIKeys from the InChI. There seems to be no resource available online. I have gone through the Wikipedia article for InChI, but it hardly contains any information as to how the InChI keys are generated. Can anyone provide me with this information as to how we can generate InChIKeys from InChI. Also, if you happen to find any good resources for the same, please do share them with me.

InChI is the acronym for the IUPAC International Chemical Identifier. It is basically as system, to catalogue molecular information. Its prime advantage is, that it is just a text string. That basically means, that it can easily be understood by machines. The other advantage is, that it is also human readable. It consists of several layers: the molecular formula, connectivity of the heavy atoms, placement of the hydrogen atoms/ stereochemical information. This shall not be confused with the IChIKey, which is a string of 27 characters derived from the hash code of the InChI string.

A quite thorough presentation/ documentation of the software and its algorithms (including the generation of the InChI string and the InChIKey) is available as an open access paper (currently asap): Stephen R Heller, Alan McNaught, Igor Pletnev, Stephen Stein and Dmitrii Tchekhovskoi, Journal of Cheminformatics, 2015, 7:23. Since the abstract is not very comprehensive, I will quote a short passage from the introduction, which outlines the contents of the article:

  We then describe the workflow of InChI generation (normalization, canonicalization, and serialization stages), as well as generation of the compact hashed code derived from InChI (InChIKey); the related algorithms and implementation details are briefly discussed.

In response to R.M.'s comment I am also including the introductory sentence of the linked paper, which describes what an IChIKey is:

  InChIKey is a compact chemical identifier derived from InChI. The InChIKey is always only 27-characters long.

For a more technical description of the algorithms and the scope of the InChI string and the InChIKey, the reader is pointed towards the technical manual: Stein S, Heller S, Tchekhovskoi D, Pletnev I. IUPAC International Chemical Identifier (InChI) InChI version 1, software version 1.04 (2011) Technical Manual. The description of the InChIKey can be found in section V, pp 53-62:

  The InChIKey is a character signature based on a hash code of the InChI string. A hash code is a fixed length condensed digital representation of a variable length character string. Providing a signature derived from an InChI string is helpful for search applications, including Web searching and chemical structure database indexing; also, it may serve as a checksum for verifying InChI, for example, after transmission over a network.

Please note that the InChIKey by itself does not contain enough information to restore the molecular structure. This is also mentioned in the open access paper:

  A disadvantage of the InChIKey is that one loses the ability to algorithmically restore a structure from a textual label: InChIKey is a structure-based registry-lookup identifier [...].

But these are not the only sources. As an IUPAC member, most information relating to the developement and implementation can be accessed freely at iupac.org. Since itself is an non-profit organisation, it also has its own website, www.inchi-trust.org. This includes a very extensive technical FAQ (html and pdf) and since it is open source software also the source code.

The software is still actively developed by the chemical community and the trust itself. As of the current version (1.04, Sept. 2011) there are still several shortcomings, which have bean addressed in Structure that breaks InChI.


Why does the concentration of hydrogen ions increase in the electrolysis of dilute sulphuric acid? In the electrolysis of dilute sulphuric acid, I'm given these two half-equations:

anode: 4OH-(aq) -> 2H2O(l) + O2(g) + 4e-

cathode: 4H+(aq) + 4e- -> 2H2(g)

Why does the concentration of hydrogen ions increase?

Isn't the concentration of hydrogen ions decreasing here, rather than increasing as it is being discharged at the cathode?

N.B: (This question originates from a past AQA GCSE Question: see question 2 part (d) here.)

Replace the anode equation with your new equation. Then find the overall equation for the reaction. You are correct that the net number of $\ce{H+}$ is not changing as a result of this process.  However, something is being consumed. What is it? How does that change the concentrations of the other species present?


Why is there a factor of half in the definition of bond order? I was recently taught about molecular orbital theory and bond order:

$$\text{Bond order} = \frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2}$$

I was wondering, what is the purpose of dividing by 2?

Because there are two electrons in a covalent bond! Say all the electrons you have is two bonding electrons... Would you call that a bond order of two? Or is it just a good old single bond?


How to prepare solutions of different concentrations of potassium from stock solutions with different concentartions? I need to prepare different concentrations of $\ce{K+}$ ($10$, $20$, $30$, $40$, $\pu{50 mM}$) by diluting a solution of $\pu{100 mM}$ $\ce{K+}$ with an isotonic buffer of $\pu{5 mM}$ $\ce{K+}$.
I need $\pu{4 mL}$ of each concentration.

I really don't remember how to calculate this. Because I need to consider the $\ce{K+}$ within the isotonic buffer right, because they will also add amount of $\ce{K+}$? If I would have diluted the $\pu{100 mM}$ $\ce{K+}$ with just water, then I understand how to do.

If you mix two solutions of different concentrations with each other you can easily determine the resulting concentration from the general formula for it:
$$c = \frac{n}{V}\tag1\label{concentration-general}$$

You have one solution with $c_1 = \pu{100 mM}$ and one with $c_2 = \pu{5 mM}$, therefore with $\eqref{concentration-general}$ you can determine the amount of substance for each, and the total amount of substance:
$$n_\mathrm{tot} = c_1\cdot V_1 + c_2\cdot V_2\tag2\label{amount-of-substance-total}$$

Assuming there is not volume contraction, we can simply add them to know the total volume:
$$V_\mathrm{total} = V_1 + V_2 \tag3\label{volume-total}$$

The total concentration therefore is:
$$c_\mathrm{total} = \frac{n_\mathrm{total}}{V_\mathrm{total}}
  = \frac{c_1 V_1 + c_2 V_2}{V_1 + V_2}\tag4\label{concentration-total}$$

For your dilution series you will need to modify it, a $\eqref{volume-total}$ is fixed.
$$V_\mathrm{total} = V_1 + V_2 = \text{const.}$$

Therefore we modify $\eqref{concentration-total}$:
\begin{align}
  c_\mathrm{total} &=
    \frac{c_1 V_1 + c_2 (V_\mathrm{total} - V_1)}{V_\mathrm{total}}
    \tag{4a}\label{concentration-total-mod-a}\\
  c_\mathrm{total} &=
    \frac{c_1 (V_\mathrm{total} - V_2) + c_2 V_2}{V_\mathrm{total}}
    \tag{4b}\label{concentration-total-mod-b}\\
\end{align}

Now you have one remaining unknown, solve for it:
\begin{align}
  V_1 &= \frac{c_\mathrm{total} - c_2}{c_1 -c_2} V_\mathrm{total}\\
  V_2 &= \frac{c_\mathrm{total} - c_1}{c_2 -c_1} V_\mathrm{total}\\
\end{align}

Unless you have very accurate equipment, however, I do suggest using larger quantities than $\pu{4 mL}$.
\begin{array}{rrrrrr}
c_\mathrm{total}/\pu{mM} &
V_\mathrm{total}/\pu{mL} &
c_1/\pu{mM} &
c_2/\pu{mM} &
V_1/\pu{mL} &
V_2/\pu{mL} \\\hline
10 & 4 & 100 & 5 & 0.21 & 3.79 \\
20 & 4 & 100 & 5 & 0.63 & 3.37 \\
30 & 4 & 100 & 5 & 1.05 & 2.95 \\
40 & 4 & 100 & 5 & 1.47 & 2.53 \\
50 & 4 & 100 & 5 & 1.89 & 2.11 \\\hline
\end{array}


Which is more stable? A phenyl carbanion or a disubstituted vinylic carbanion? I came across a question which asks to work out the stability order for the following species-

b.
c. $\ce{R2C=CH-}$

Can the lone pair on phenyl carbanion participate in resonance?  If yes, then how? Also, how is that more stable than the vinylic carbanion?

Can the lone pair on benzene carbanion participate in resonance?

As Jori has pointed out, the carbanion lone pair is orthogonal to the pi system of the aromatic ring.  Therefore it cannot participate in resonance.  Here is a drawing that illustrates this point.

how is it more stable than vinylic carbanion?

The $\mathrm{p}K_\mathrm{a}$ approach discussed in @Jori's answer becomes problematic when we are trying to look at a small difference between two large $\mathrm{p}K_\mathrm{a}$s.  A $\mathrm{p}K_\mathrm{a}$ of $43$ means $1$ ionized molecule exists in every $10^{43}$ molecules.  Certainly difficult to measure experimentally and so a significant error is attached to these estimates. In fact, using the data from Evan's $\mathrm{p}K_\mathrm{a}$ table we find a $\mathrm{p}K_\mathrm{a}$ of $43$ for the benzene $\ce{C-H}$ proton and a $\mathrm{p}K_\mathrm{a}$ of $50$ for an ethylenic proton, suggesting that the phenyl carbanion might be more stable than the vinyl carbanion.
An alternate way to examine this question involves comparing the hybridizations of the phenyl and ethylenic $\ce{C-H}$ bonds.  Since bonds with more $\mathrm{s}$-character will stabilize electrons better than bonds with less $\mathrm{s}$-character (because the $\mathrm{s}$-orbital is lower energy than a $\mathrm{p}$-orbital), whichever bond has more $\mathrm{s}$-character should produce a more stable carbanion.
The $\ce{C-H}$ bond in benzene is exactly $\mathrm{sp^2}$ hybridized as required by the symmetry of the benzene molecule.  Experimentally, we know that the $\ce{H-C-H}$ angle in ethylene is $117^\circ$.  Using this fact along with Coulson's theorem (reference1, reference2, reference3) we can determine that the $\ce{C-H}$ bond in ethylene is $\mathrm{sp}^{2.2}$  hybridized.
The $\mathrm{sp^2}$ hybridized orbital in benzene has more s-character than the $\mathrm{sp}^{2.2}$ hybridized orbital in ethylene.  This would lead us to suspect that the phenyl carbanion is more stable than the vinyl carbanion.
Of course, once we remove these protons we expect the resulting carbanions to undergo further rehybridization in order to further stabilize their lone pair (carbanion) electrons.  It seems like a reasonable first approximation that they would both rehybridize similar amounts and the "relaxed" phenyl carbanion would remain more stable than the "relaxed" vinyl carbanion.  But in any case, the transition state leading to the phenyl carbanion will be more stable than the transition state leading to the vinyl carbanion because of the hybridization \ $\mathrm{s}$-character reasons presented above.
Edit: OP's comment on hybridization
Molecules are not just $\mathrm{sp, sp^{2}}$ or $\mathrm{sp^3}$ hybridized.  While the $\ce{C-H}$ bonds in methane are exactly $\mathrm{sp^3}$ hybridized (as required by symmetry), the $\ce{C-H}$ bonds in $\ce{CH3F}$ are not (see here for a discussion).  Bonds can have any hybridization index (the superscript to the right of "$\mathrm{p}$") between zero and infinity.  Hybridization indices are not limited to integer values, nature can mix $\mathrm{s}$ and $\mathrm{p}$ orbitals in whatever ratio is required to create the most stable bonds and molecules.


What do dashes mean in chemical formulas? I'm working on summarizing some research for an essay; here's the main article.

The basic premise is making nanoparticles from gold and iron(III) oxide that have properties of both gold and iron.

The formula they give is $\ce{Fe2O3-Au}$. The procedure involves chemical reactions, so I'm assuming it is a new molecule or particle of some sort, but what does the dash mean in the formula? Does it relate to the type of bond or something?

Also, and this is probably bonus since the useful part of the question deals with dash notation, but from the procedure I've generated the partial equation:

$$\ce{Fe2O3 + H2O + Na3C6H5O7 + HAuCl4 → Fe2O3-Au} + \text{something}...$$

After the reaction, the samples are put through a centrifuge and some unknown supernate is then removed, leaving the $\ce{Fe2O3-Au}$. Do you have any guesses on what the supernate is composed of? I was thinking something along the lines of $\ce{HCl}$, but I'm not sure what would happen to the sodium or carbon.

This nomenclature refers to a core-shell arrangement of nanoparticles. This is where you have a nanoparticle core consisting of one material that has been coated/decorated/wrapped/overlaid with another material. There are a few accepted forms of nomenclature for describing these types of nanoparticles:

core@shell In your case, it would be Fe2O3@Au
core-shell In your case, it would be Fe2O3-Au
core/shell In your case, it would be Fe2O3/Au

There are plenty of examples of each syntax in the literature, and has been used for eons.

Others may answers for the additional questions you have here.


MolView is unable to generate correct 3D structure for hydroxypropadiene MolView is unable to generate the correct structure of hydroxypropadiene. It instead generates the structure of acrolein (which has the same molecular formula). Why is this so? Is it because hydroxypropadiene is so unstable that the calculations done while generating the structure degenerate it to acrolein?

MolView is using the default GLMol renderer. Switching to JMol or ChemDoodle does not change the structure.

EDIT: The exact workflow I used was open MolView, delete the caffiene molecule, draw this molecule and click the 2D to 3D button. The generated URL is http://molview.org/?cid=7847, which, when pasted in a browser, interestingly redirects to Acrolein and not the compound that was drawn.

I suspect MolView simply fails to parse the compound's name you are trying to input. "Hydroxypropadiene" isn't the proper name, at least not by IUPAC standards. According to Google services, this name doesn't appear in any textbooks and is mentioned only in three research articles from 1980s.

The PIN propa‐1,2‐dien‐1‐ol appears to work correctly with MolView:

When you are trying to get a 3D image from the 2D drawing, there is the following issue which MolView usually reports (not sure why you haven't received the message):

  Failed to load structure from sketcher. This structure may contain allenes, which cannot be represented in the SMILES notation.

MolView relies on SMILES in order to fetch the 3D structure.
The problem is likely that Ketcher generates wrong SMILES because PubChem indeed receives C=CC=O instead of the proper one, C=C=CO.


Electroplating copper onto steel I'm rather unfamiliar with chemistry but I want to electroplate steel with copper.

From Wikipedia:

  The cations associate with the anions in the solution. These cations are reduced at the cathode to deposit in the metallic, zero valence state. For example, in an acid solution, copper is oxidized at the anode to $\ce{Cu^2+}$ by losing two electrons. The $\ce{Cu^2+}$ associates with the anion $\ce{SO4^{2−}}$ in the solution to form copper sulfate. At the cathode, the $\ce{Cu^2+}$ is reduced to metallic copper by gaining two electrons. The result is the effective transfer of copper from the anode source to a plate covering the cathode.

I have also found a YouTube video where a guy uses $\ce{HCl}$ instead of $\ce{H2SO4}$.

What are the requirements for the reaction to work? Does it have to be an acid solution? Can I use vinegar?

You can attempt to use a copper item as the anode in a vinegar solution; the $\ce{Cu}$ is oxidised to $\ce{Cu^2+}$ and the acidity of the vinegar prevents the precipitation of $\ce{Cu(OH)2}$. The copper acetate formed in solution will migrate to the cathode, where you can place your item to be plated.

However, as my own experiment has shown, too high a current will result in a black deposit, not pink as the copper crystals are plated to quickly to form large crystals and thus scatter light. So use a low current!


Kinetics vs Thermodynamics The way kinetics is taught at the undergraduate level (Arrhenius and collision theory) chemical equilibrium is determined governed immensely by activation energy of the reaction. According to thermodynamics, however equilibrium is a function of free energy change. In a way, thermodynamics and kinetics seem to contradict each other. What insight am I missing. Is the energy in energy vs reaction coordinates supposed to be Gibbs free energy instead of enthalpy? Would this imply activation energy changes as a reaction proceeds? Is the way kinetics is taught wrong?

"...chemical equilibrium is governed immensely by activation energy of the reaction."

This sentence is incorrect. The relative amounts of species in a chemical equilibrium are not at all influenced by the kinetics; if anything, kinetics are influenced by the chemical equilibrium (such as in the Marcus theory of electron transfer reactions). A rigorous derivation of the expression for an equilibrium constant can be made with no direct reference to reaction kinetics.

Chemical equilibria are what happens to systems after they've been left without interference for an infinite time; any system must reach a steady state in finite time, otherwise it would essentially be a perpetual motion device. In other words, for a chemical equilibrium to be achieved, the system is given more than enough time for any finite activation energy barrier to be overcome.


Fluorine Preparation:How can this be explained on basis of Lewis Acid-Base Reaction? How can this be explained on basis of Lewis Acid-Base Reaction ?

$\ce{2 K2MnF6 + 4 SbF5 → 4 KSbF6 + 2 MnF3 + F2↑}$

References:

https://en.wikipedia.org/wiki/Fluorine#Industrial

Google Books reference (click me)

$\ce{2 K2MnF6 + 4 SbF5 → 4 KSbF6 + 2 MnF3 + F2↑}$

Occurs in two stages. First there is a (Lewis) acid-base reaction

$\ce{2 K2MnF6 + 4 SbF5 → 4 KSbF6 + 2 MnF4}$

where the Lewis acid $\ce{SbF5}$ accepts a lone pair from a fluorine in the manganese complex, and the adduct then falls apart to leave $\ce{MnF4}$. This is thermally unstable at ambient temperatures (Housecroft and Sharpe) and falls apart in the redox reaction

$\ce{2 MnF4 → 2MnF3 +F2}$


What is the compound called when it is in solution? When a compound precipitates out of solution, for example Calcium Carbonate from a water solution, it is called the precipitate.

What is the same compound called before it precipitates out - once it has dissolved? I found the word "electrolyte", but I'm unsure this applies to all solutions - only ionic, and I'm not sure a CaCO3 solution is ionic.

I believe the term you are looking for is solute; a solute is a substance dissolved in another substance, known as a solvent.


Shouldn't 1 mole of any ideal gas at open space occupy more than 22.4 L volume at S.T.P. due to diffusion? So if diffusion occurs,1 mole of any ideal gas should occupy a much greater volume in open space at STP.

Here is the calculation for the volume occupied by one mole ($n$) of an ideal gas at STP (temperature $T = 273.15\ \mathrm K$, pressure $P = 1\ \mathrm{bar}$):

gas constant: $R = 0.08314\ \mathrm{L\ bar\ mol^{-1}\ K^{-1}}$
ideal gas law: $PV = nRT$

$$\begin{align}
V &= \frac{nRT}{P}\\[6pt]
&= \mathrm{\frac{1\ mol\ \times\ 0.08314\ L\ bar\ mol^{-1}\ K^{-1} \times\ 273.15\ K}{1\ bar}}\\[6pt]
&= 22.71\ \mathrm L
\end{align}$$

Another way of looking at this of course is that any 22.71 liter chunk of particulate-free air, of arbitrary shape, will contain about 1 mole of air. Molecules of the different gases composing the air will diffuse in to and out of this region, but it will always contain about 1 mol of air at STP.


Distinguishing between potassium chloride and sodium chloride I have these very large blocks of salt intended for a water softener in my house, but I'm not sure if they're potassium chloride or sodium chloride, according to what I've read they're both widely used for water softeners. I've thought about distinguishing them by weight but I don't have any sodium chloride with the same granularity as the salt blocks.

Per Wikipedia:

  ..., MgSO4·H2O, can be combined with a solution of potassium chloride to produce potassium sulfate.

So, cooling an aqueous solution KCl in the presence of Epsom's Salt:

$\ce{2 KCl + MgSO4 <=> MgCl2 + K2SO4}$

creates potassium sulfate, which falls out of the solution forming distinct orthorhombic crystals.

If Na2SO4.7H2O is created instead from aqueous NaCl in place of KCl, you will get something that resembles a gluey paste.


Why does cobalt have a higher atomic mass than nickel? I understand that this deviation from the general trend of increasing atomic number and atomic mass is because cobalt's isotopes tend to have more neutrons than nickel's. Why is that the case? How would one go about predicting similar cases like this?

I've read a bit about the shell model. I understand that 28 is a "magic number" of nucleons, i.e. a filled shell. In that case, wouldn't it be the most stable for both atoms to mainly be found with 28 neutrons; and so, cobalt-55 and nickel-56 would be the most stable isotopes of each element. Why is that not the case? Am I missing or misunderstanding something?

Protons in each nucleus repel each other electrostatically.

As atomic number increases, there is more repulsion and an increasing advantage of spreading out the protons with neutrons.

Calcium-40 is the last (greatest atomic number) observationally-stable isotope with an equal or greater number of protons than neutrons.

Cobalt-55 and Nickel-56 are both unstable to beta decay, because it is favorable to emit a positron converting a proton into a neutron.

The natural abundance of isotopes are not necessarily in proportion to the stability, but instead stellar and cosmic processes.  As explained in the 1963 article Computer Guides Professor Hoyle to New Hypotheses:

  Working with Professor W. A. Fowler of the California Institute of Technology, he had begun with an analysis of the distribution of isotopes of the common metals, and found that nickel presented an unusual case; in the natural metal on Earth, nickel-58 is some three times more abundant than nickel-60, which is contrary to what one would expect from considerations of nuclear stability. He found the solution to this puzzle in the effect of neutrino emission from highly evolved stars, leading to extremely short life times, in the range of 20 minutes to 10 hours, and also in the phenomenon of electron-positron pair production. Nickel-58 could be made in an explosion in a large star which occurred so rapidly that the process of synthesis was cut short.

So, while the details of that article may be out of date, the reason cobalt has a higher natural abundance atomic mass than nickel is a combination of nuclear stability and stellar processes.

See On the e-process - Its components and their neutron excesses for further discussion of natural abundance ratios of Co and Ni isotopes.


Energy of the n-th level for an atom Going through the Bohr's model and his assumptions, I came across with this formula to find the energy of the n-th level of any atom:

$$E = - \frac{Z k_e e^2}{2r_n} = -\frac{Z^2(k_e e^2)^2m_e}{2\hbar n^2} \approx \frac{-13.6Z^2}{n^2}~\pu{eV}$$

Now, let's say we take as an example the atom of hydrogen, the energy of the $n = 1$ and only level is ($\pu{-13.6 eV}$). Now, according to Wikipedia (not really a reliable source) this can be interpreted as

  An electron in the lowest energy level of hydrogen ($n = 1$) therefore has about $\pu{13.6 eV}$ less energy than a motionless electron infinitely far from the nucleus.

I don't really understand why there is a "minus" in the result. Why does it say that it has less energy than a motionless electron when he, himself, suggested that the electron's acceleration does not result in radiation and energy loss?

I also understand this formula only worked for the atom of hydrogen (that's why I took it as an example) and due to this problem and that's why the Sommerfeld model was postulated.

It's like an asteroid falling toward Earth. Potential energy converts to kinetic, until it crashes, then the energy is found in whatever aftermath there is.  The total energy is still zero. The "falling" from a great distance is a change from high potential energy to low. By convention, that higher level is assigned zero, so the lower energy must therefore be negative.

For an electron and a bare nucleus far apart, and not counting any nuclear or non-chemical phenomena, we assign zero as the total energy.  When the electron is allowed to fall, it falls faster and faster, potential energy converting to kinetic, until "crashes" - it enters the 1s orbital, n=1, releasing a photon of 13.6eV. (In constrast, an asteroid makes a big crater, stirs up debris, and creates tremendous heat.)  The photon carries 13.6eV, and so the newly formed hydrogen atom is -13.6.  This is  called binding energy, and is negative by the convention physicists and chemists have followed since long ago.

BTW, that formula is only for any atom with exactly one electron, and Z not too big.  With two electrons it no longer applies, and for the heavier elements such as uranium, there are relativistic effects and finite nucleus size effects requiring more sophisticated math and at the very least, some sort of correction terms to that formula.


Will aluminum continue to corrode after its surface is oxidized? After doing some reading, it is common to read that aluminum is corroded by oxygen, but that the thin layer of aluminum oxide protects the internal metal from further damage. How much does this layer of aluminum oxide protect the underlying layers? Does it completely seal off the aluminum or is there still a slow degradation compared to a metal like iron?

No. Aluminium stops corroding once a strong oxide layer is formed.

Aluminium is a fairly reactive metal. The reason why it usually survives in air and water under normal conditions is because the aluminium oxide layer on the surface is very strong and normally protects the metal from further attack. Even if the layer is disrupted, it reforms quickly preventing any further reaction. Aluminium oxide is a very strong compound and impervious to water and oxygen. In bulk form it is known as corundum and appears in nature as ruby and sapphire, just about the hardest non-diamond minerals.

But there are substances that can disrupt that layer and expose aluminium to attack from air and water. Mercury and its salts and gallium metal both do this. Once they penetrate the surface layer, aluminium will be destroyed fairly rapidly and will crumble to a weak bulky mess. See examples on youtube from NurdRage and NileRed.


Why don't Gringard Reagents react with carbonyls at the alpha position? According to many texts, the reason why Gringard reagents cannot be used with alcohols is because the Gringard reagent is a strong base, which would deprotonate the alcohol. ($p\pu{K}_b$ of Gringard $\approx 50$).

However, carbonyls also have a fairly acidic hydrogen at the alpha position ($p\pu{K}_a$ of enol $\approx 20$), so why does the Gringard reaction react with the electrophilic carbon, rather than as an acid-base reaction with the alpha position to form an enolate?

It actually happens sometimes and you end up with the enolate being formed in appreciable amounts. But it depends on other adjacent substituents that can withdraw or donate electron density to the alpha carbon.

If you draw the mechanism for the alpha attack you can see that the enolate can react further with the aldehyde and produce aldol condensation adducts. Furthermore, the carbon atom of the aldehyde is way more electrophilic than the hydrogen atom bound to the alpha carbon, that is the main reason why the enolate end up being formed in very small amounts and sometimes you don't even detect it.


Will propane liquefy under these conditions? Consider a 1:1 mixture of propane and nitrogen at $60°\mathrm{F}$ and $75\:\mathrm{psi}$. Will the propane liquefy if pressure is increased to $200\:\mathrm{psi}$? I am experimenting in the oilfield with a shallow heavy oil lease.

First of all, you might want to consider using proper SI units.

If you know the saturation temperature (boiling point) of propane at a given pressure:

At a pressure of p = 75 psi (0.52 MPa), the saturation temperature of pure propane is 37.13 °F (2.85 °C); thus, at a temperature of 60 °F (16 °C), propane would be gaseous.

At a pressure of p = 200 psi (1.38 MPa), the saturation temperature of pure propane is 104.54 °F (40.30 °C); thus, at a temperature of 60 °F (16 °C), propane would be liquid.

However, a 1:1 mixture of propane and nitrogen contains only 50 % propane; i.e., the mass fraction of propane is wpropane = 0.5. The corresponding amount-of-substance fraction (mole fraction) of propane is xpropane = 0.39.

For ideal gas mixtures, the ratio of partial pressures equals the ratio of the number of molecules:
ppropane = xpropane • p

Therefore, the partial pressure of propane is approximately ppropane = 78 psi (0.54 MPa).

At a pressure of p = 78 psi (1.38 MPa), the saturation temperature of pure propane is 39 °F (4 °C); thus, at a temperature of 60 °F (16 °C), propane would be gaseous.

If you know the vapour pressure of propane at a given temperature:

At a temperature of 60 °F (16 °C), the vapour pressure of propane is 107.71 psi (0.74 MPa).
Thus, at a pressure of p = 75 psi (0.52 MPa), pure propane would be gaseous. At a pressure of p = 200 psi (1.38 MPa), it would be liquid.

However, once again, a 1:1 mixture of propane and nitrogen contains only 50 % propane; i.e., the mass fraction of propane is wpropane = 0.5. The corresponding amount-of-substance fraction (mole fraction) of propane is xpropane = 0.39.

For ideal gas mixtures, the ratio of partial pressures equals the ratio of the number of molecules:
ppropane = xpropane • p

Therefore, the partial pressure of propane is approximately ppropane = 78 psi (0.54 MPa). Since this partial pressure is below the vapour pressure of 107.71 psi (0.74 MPa) at a temperature of 60 °F (16 °C), propane would be gaseous.


Raman Spectrum of Water This is the Raman spectrum of water. There is $\pu{1635 cm-1}$ Raman peak corresponding to $\ce{HOH}$ and $\pu{3410 cm-1}$ Raman peak corresponding to $\ce{OH}$. But why is there general nonzero background in intensity? What caused it?

This is taken using 532nm wavelength raman system:

This is taken using 785nm wavelength raman system up to 2750 cm-1 only (versus 4000 cm-1 of the first spectrum):

In the above picture. Why is the background dimishing to zero towards 2750 cm-1. My guess is that as wavelength gets higher, water absorbs more. But can it absorb the noise too?

This answer is based on the educated guess that the green excitation Raman spectrum was acquired with a dispersive Raman instrument*.

  But why is there general nonzero background in intensity? What caused it?

For any real-life instrumentation, you should never assume to measure 0 signal for 0 analyte concentration (see also Does a calibration curve of absorbance vs concentration need to pass exactly through the origin?) nor expect the baseline to be at zero unless corrected for, even if there really were no Raman scattering outside those two bands, the instrument and/or sample preparation procedures (electronics, optics, masking, cross-sensitivity, ...) will often lead to some non-zero signal.

The CCD in the spectrograph has a so-called dark current. So the raw signal you observe is dark current plus photon signal. Dark current is influenced by settings of the CCD (you trade off sensitivity and dark current) and by temperature.

The constant part of the dark current shouldn't trouble you too much: you can measure and then subtract it (which is in fact what spectrometer software does where you see the baseline at zero). However, as subtracting spectra sums their noise** (error propagation). As Raman measurements typcially suffer from low signal-to-noise-ratio, you may want to avoid this, so Raman spectrometer software usually either outputs raw spectra or offers you the choice whether you want to subtract the dark spectra.

For SNR of Raman spectra, you'll have to take into account shot noise (Photon noise): the noise is directly proportional to the signal intensity, and that includes optical background such as fluorescence or elastically scattered or reflected light as well as your Raman signal. (And then come the other sources of noise such as flicker noise and the various sources of noise related to your application).

For a thorough discussion of the noise properties of Raman spectrometers, there's a whole chaper of McCreery: Raman Spectroscopy for Chemical Analysis, ISBN: 978-0-471-23187-5 on the topic
(for FT instruments, see Zhao & McCreery: Multichannel FT-Raman Spectroscopy: Noise Analysis and
Performance Assessment, ApplSpec, 1687 - 1697, 1997.)

About the 785 measurement in comparison:

The green excitation y-axis looks like raw counts (the data itself would reveal more: raw counts are usually integer numbers). In contrast the red spectrum's y-axis goes from 0 to slightly above 1.0. Which means that some transformations have been done (which is impossible to say from the data alone - you'll have to talk to the manufacturer what they do).

   water absorption

Absorbance of in the range below 1100 nm is < 0.2/cm. So yes, you loose about 1/3 of the intensity through 5 mm water and back - but you can avoid that by focusing not as deep into the water.
What hurts your signal more is that the quantum efficiency down there is pretty awful for many CCDs (around 10 %, there's a reason why their spectral range ends in the mid of the -CH stretching vibrations).

* A rather safe guess, as FT instruments for such short wavelength AFAIK exist at the moment only as "home-built" instruments developed in some research facilities - whoever uses one of those is going to be fully aware that they are using something special.

** more precisely: the variances of independent random variables are additive