Use of cryolite and fluorspar in electrolytic reduction of pure alumina by Hall-Héroult process Cryolite $\ce{(Na3AlF6)}$ and fluorspar $\ce{(CaF2)}$ are added to alumina to lower the temperature of the mixture from $\pu{2000 °C}$ to about $\pu{900 °C}$ and also to increase the conductance.

But my question is about the specific effect these compounds are producing. I mean, out of the two mentioned effects, which compound is doing what?

According to the CRC Handbook of Chemistry and Physics, melting point of $\ce{CaF2}$ is $\pu{1418 ^\circ C}$, of $\ce{Al2O3}$ (in its corundum form) is $\pu{2053 ^\circ C}$, and of $\ce{Na3AlF6}$ is $\pu{1009 ^\circ C}$ (lowest of the all three). Thus the cryolite should melt first.
I have read mentions of the problem you asked in the book Nonaqueous Electrochemistry (you can find information about it here. For the best of my summarising ability, I can conclude that the role of the cryolite and other salts added to the reaction mixture is that they decrease the melting point of alumina. As cryolite's melting point is much lower than that of alumina as cited above, when the cryolite melts it acts as a solvent to "dissolve" the $\ce{Al2O3}$, by the process of complexation. Recall how silver chloride dissolves in aqueous ammonia or lead iodide dissolves in excess iodide. The book states that:

Cryolite is used as solvent for alumina because it has high chemical stability and a high solubility for alumina.

In detail:
When pure cryolite ($\ce{Na3AlF6}$) melts, it dissociates into fluoride ions and various lower fluoroaluminates:
$$\ce{[AlF6]^3- -> [AlF4]- + 2 F-}$$
$$\ce{[AlF6]^3- -> [AlF5]^2- + F-}$$
(etc.)
However, in the presence of alumina, the dissociation process is complicated by the fact that the products arise from $\ce{Na3AlF6}$ dissociation can react with $\ce{Al2O3}$ creating oxofluoroaluminates, and it is those complexes that directly get electrolysed (cathodically reduced) rather than the $\ce{Al2O3}$ itself. By various means of calculation and experiments (Raman spectroscopy etc.) they proposed some following reactions (or equilibria):
$$\ce{Al2O3 + 4 [AlF5]^2- -> 3 [Al2OF6]^2- +  2 F-}$$
$$\ce{Al2O3 + [AlF5]^2- + F- -> 3/2 [Al2O2F4]^2-}$$
$$\ce{2 Al2O3 + [Al2O2F4]^2- + 4 F- -> 2 [Al3O4F4]^3-}$$
...
When you add calcium fluoride (or magnesium fluoride) into the mixture, the following reactions also occur:
$$\ce{MgF2 + n F- + (2-n) [AlF5]^2- -> [MgF_{(2+n)}(AlF5)_{(2-n)}]^{(4-n)-} }$$
$$\ce{CaF2 + (3-n) [AlF5]^2- -> [CaF_{n}(AlF5)_{(3-n)}]^{(4-n)-} + (2-n)F-}$$
The electrolytes in the electrolysis process is obviously "$\ce{Na3AlF6}$", but in fact they always add an excess of $\ce{AlF3}$ alongside other salts such as $\ce{CaF2}$ ($\pu{3-5 \%}$), $\ce{LiF}$ ($\pu{0-4 \%}$) and $\ce{MgF2}$ (also $\pu{0-4 \%}$). In fact, those salts decrease the solubility of $\ce{Al2O3}$ rather than increase it; that's why they only add a little of those salts.
Then why adding those salts?
There are a couple of reasons for this. Firstly, addition of those salts increase the density of the electrolyte. Because aluminium is already produced in the liquid state (melting point $\pu{660 ^\circ C}$), if the density of the two components are too similar, it is impossible to achieve a good separation-by-weight. Kvande and Rorvik had derived an equation of the melt density as a function of temperature and salt composition:
$\pu{d = 2.64 - 0.0008T + 0.18(wt. ratio NaF/AlF_{3}) - 0.008(wt. \% Al_{2}O_{3}) + 0.005(wt. \% CaF_{2}) + 0.008(wt. \% MgF_{2}) - 0.004(wt. \% LiF)}$
As you can see from the above equation density of the mixture increases with CaF2 content.
The other reason is that those salts increase the interfacial tension at the liquid Al-electrolyte boundary. The higher interfacial tension is, the less does aluminium re-dissolve in the electrolyte (yes it can!), thus better current efficiency. For example: at the boundary this reaction occurs: $\ce{Al + 3 NaF -> AlF3 + 3 Na}$. Adding more $\ce{AlF3}$ (or other fluorides) helps surpressing this reaction, and also may increase the $\ce{Na+}$ concentration in the electrolyte - the researchers found that $\ce{Na+}$ ions are the primary charge carriers.


What is the softening temperature of styrofoam? Does anybody know the softening temperature of styrofoam (expanded polystyrene)?

By styrofoam I mean the solid made of expanded polystyrene balls used in packaging.

There are a number of different temperatures one could consider the softening point of atactic (randomly distributed phenyl groups) polystyrene, used in making expanded polystyrene :

Softening point: ~90 °C
Heat Deflection Temperature: 95 °C
Typical Injection Molding Temperature: from 170 °C at back to 230 °C at nozzle


Equivalent mass in case of disproportionation reaction I came across a question where the equivalent mass of hypophosphorus acid is to be found in the given reaction:
$$\ce{2H3\overset{+1}{P}O2 -> H3\overset{+5}{P}O4 + \overset{-3}{P}H3}$$

If $M$ is the molecular mass of $\ce{H3PO2}$ then $E_\text{oxidation}=M/4$ (for change in oxidation state from $+1$ to $+5$) and $E_\text{reduction}=M/4$ (for change in oxidation state from $+1$ to $-3$)

Equivalent mass of $\ce{H3PO2}$ in this disproportionation reaction

$$= E_\text{oxidation}+E_\text{reduction}= M/4+M/4= M/2$$

Now, I am not able to comprehend that, why these two types of equivalent mass has been added to arrive the equivalent mass of $\ce{H3PO2}$.

My take: The concept of equivalent mass collapses in this situation!

I like and dislike the question. I assume this is from your Indian university entrance exams. The good part is that it is tricky, the sad part is that the concept of equivalents is so outdated that any North American student will barely recognize it. They should abolish teaching equivalents exactly because this problem you are encountering.

Fact number no. 1, unlike molar equivalents of a given molecule equivalent weights are reaction dependent. Hence addition of equivalent weights is irrelevant. Equivalent weights are useful in analytical chemistry because they simplify calculations. The case of disproportionation reactions and their equivalents is rarely discussed because nobody uses disproportionation reactions in titrations, gravimetry etc.

In your case, you will have two equivalent weights one with respect to oxidation, and one respect to reduction. Solve the disporptionation reaction below and see what you get as an equivalent weight for $\ce{Br2}$ in each case.

$\ce{3Br2 + 6OH^- -> 5Br^- + BrO3^- + 3H2O}$


In Parent chain choosing should one consider longest chain or chain with more substituents? Should one take the longest chain or the one containing the prefix/suffix with most priority.Moreover if a shorter chain has two units of the main prefix will it be taken as parent?

examples:

2-ethyl butanoic acid or
3-carboxy pentanoic acid

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

greater number of substituents corresponding to the suffix (principal characteristic group)
longest chain
greater number of multiple bonds
lower locants for suffixes
lower locants for multiple bonds
greater number of prefixes
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

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

  P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES

  When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.

  P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).

  (…)

  P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].

  (…)

  P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.

  The senior ring, ring system, or principal chain:

  (a) has the greater number of multiple bonds (P-44.4.1.1);

  (b) has the greater number of double bonds (P-44.4.1.2);

  (…)

  (h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);

  (…)

  (j) has the lower locant(s) for endings or prefixes that express changes in the level of hydrogenation, i.e., for ‘ene’ and ‘yne’ endings and ‘hydro/dehydro’ prefixes (P-44.4.1.10);

  (…)

  P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.

  P-45.2.2 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.

  P-45.2.3 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than ‘hydro/dehydro’ prefixes) in their order of citation in the name.

  (…)

You have already found the principal characteristic group, i.e. the carboxylic acid group $\ce{{}-(C)OOH}$ that is expressed as suffix ‘oic acid’.

According to Rule P-44.1.1, the principal characteristic group (suffix) has to be considered first in order to decide the parent chain for this compound. Only the second rule (P-44.3.2) is looking at the longest chain.

Therefore, the name of the first example that is given in the question is 2-ethylbutanoic acid rather than ‘3-carboxypentane’ since the principal characteristic group has to be considered before the longest chain.

Note that, according to Rule P-44.1.1, the parent chain has the maximum number of substituents corresponding to the principal characteristic group (suffix). Also note that the second example that is given in the question contains two carboxylic acid groups.

Therefore, the name of the second example is 2-ethylbutanedioic acid rather than ‘3-carboxypentanoic acid’ since the maximum possible number of substituents corresponding to the principal characteristic group has to be considered before the longest chain.


Finding initial pressure in a chemical equilibrium Consider the equilibrium below. If $K = 4$ and initial pressures of $\ce{CO}$ and $\ce{H2O}$ are equal, giving a total pressure of $1.5~\mathrm{bar}$ at equilibrium. What was the initial pressure of $\ce{CO}$ or $\ce{H2O}$?

$$\ce{CO(g) + H2O (g) <=> CO2 (g) + H2 (g)}$$

I don't know how to get the initial pressures in this situation. Do I have to set up an ICE chart? I'm not sure how to start, any help would be appreciated. The answer is apparently $0.75~\mathrm{bar}$.

If the initial pressure of $\ce{CO}$ and $\ce{H_2O}$ are equal, then let $p_{\ce{H_2O}}=p_{\ce{CO}}=x$.
Since their total is 1.5 bar, i.e. $\ce{p_{H_2O} + p_{CO}}=1.5$
Can you solve this equation?

Edit: This is under the assumption that there was originally no carbon dioxide or hydrogen, I have not yet managed to generalise it to any case.
It is given that $K=4$. From the equation for the equilibrium constant, $K=\frac{\ce{[CO2][H2]}}{\ce{[CO][H2O]}}$.
It is also given that initially, $\ce{p_{H2O} = p_{\ce{CO}}}$; from the ideal gas law,
$PV=nRT \implies P=\frac{nRT}{V}=kC  \tag{1}$
where $k$ is some constant and C is concentration.
So initially, $[\ce{H2O}]=[\ce{CO}]=y$, where $y$ is a constant. Let the concentration consumed by the equilibrium shifting to the right be x.
At equilibrium, $\frac{\ce{[CO2][H_2]}}{[\ce{CO}][\ce{H2O}]}=\frac{x^2}{(y-x)^2}=4$.
Solving this equation gives $x=\frac{2y}{3}$.
Now then, if you look back at (1), pressure is proportional to concentration. Try to use that relationship to solve the problem.

 Now, from $(1)$, we see that pressure is directly proportional to concentration. We have worked out from our algebra that: $[\ce{CO2}]=[\ce{H2}]=\frac{2y}{3}$, therefore $[\ce{CO}]=[\ce{H2O}]=\frac{y}{3}$.

So then $p_{\ce{H2}}= p_{\ce{CO2}} = 2p_{\ce{CO}} = 2p_{\ce{H2O}}$ at equilibrium.
So from substitution into Martin's equation, $\ce{p_{H_2O} + p_{CO} + p_{\ce{CO2}} + p_{\ce{H2}}}=1.5$, we find that $p_{\ce{H2O}} = p_{\ce{CO}} = \pu{0.25 bar}$.
But they are now only a third of their original concentration, and so their pressure has also decreased by a factor of 3. So their original pressure is $\pu{0.75 bar}$ each.


What is the Structure of FeSO₄ • NO? What is the structure of $\ce{FeSO4 \cdot NO}$ that is formed when $\ce{NO}$ is passed through ferrous sulfate solution? If it is a brown ring complex then why does the complex sphere break upon heating to yield $\ce{NO}$? In general, when do the complex spheres break?

What is the structure of $\ce{FeSO4 \cdot NO}$ that is formed when $\ce{NO}$ is passed through ferrous sulphate solution?

The structure is octahedral.

The Fe ion is at the center of the octahedron.

Five water molecules and the NO molecule occupy the vertices of the octahedron.

Sulfate is a separate spectator ion.

The overall charge of the iron coordination complex is 2+.

  If it is brown ring complex, why does the complex sphere break on heating to give $\ce{NO}$? In general, when do the complex spheres break ?

"Brown ring" refers to the macroscopic appearance of the product in a tube.  This is unrelated to the chemical structure.  The binding of NO to Fe is an exothermic reaction.  In accordance with the Van't Hoff Equation, the equillibrium constant shifts in favor of the reverse reaction as temperature increases.

For detailed infomation see Kinetics, Mechanism, and Spectroscopy of the Reversible Binding of Nitric Oxide to Aquated Iron(II)

Interestingly, for many years it was thought the iron was reduced to Fe(I), the NO being oxidized to NO+; however, according to the above reference, the current thinking is that the iron is oxidized to Fe(III), and the NO is reduced to NO-.

In other words, the complex is $\ce{ [Fe^{III}(H_2O)_5(NO^{-})]^{2+} }$


Enthalpy of mixing of a non-ideal solution I have this equation of Flory–Huggins model:
$$\Delta_\mathrm{mix}G = RT(n_1\ln\phi_1 + n_2\ln\phi_2) + \chi n_1\phi_2,$$
where $\phi$ is the volume fraction, $n$ is the amount of substance and $\chi$ is the parameter of solvent-polymer processing that depends only on temperature.
I need an expression for $\Delta_\mathrm{mix}H.$

There are two ways (at least) to solve this:
(1) Evaluate $\Delta S = -\left( \frac{\partial \Delta G}{\partial T} \right)_{p,n} $
and add $T\Delta S$ to $\Delta G$.
or
(1) Evaluate directly $\Delta H = -T^2 \left( \frac{\partial (\Delta G/T)}{\partial T} \right)_{p,n} $
Either way I obtained the result
$\Delta H = \chi n_1 \phi_2 - n_1\phi_2 T \left( \frac{\partial \chi }{\partial T} \right)_{p,n} $


Super-cooled liquids I understand that it is possible to cool a liquid past its freezing point, and still maintain its liquid state because crystallisation would not happen in the absence of a seed crystal.

My question, was marked as a duplicate and I was linked to to Can all substances be supercooled?

One of the answers over there does address my question but only in part.
Here's what it says:

  "However, when there is only liquid present, you will first have to form a small crystal nucleus which is not as stable as the bulk solid i.e. has a higher free energy. This is basically a kinetic barrier which prevents forming the solid.
  The formation energy of such a nucleus will depend on the interfacial energy of the crystal-liquid system. In a substance where this interfacial free energy is very little there won't be too much supercooling."

My question still remains. Why is the lack of a nucleus a kinetic barrier to the formation of a bulk solid? If you continue past the glass transition temperature, we form an amorphous solid anyway. How do the mechanics of vitrification and crystallisation differ in this regard?

When a liquid is supercooled its structure doesn't change significantly, only the particles have less kinetic energy available. As you cool even further, it could occur that particles get caught in local minima and eventually you'll end up with a glass, which can be thought of as a frozen liquid structure. As particles are trapped it will resist shear in an elastic fashion so you have a solid. In this regard it is a barrierless transition. There is no interface whatsoever.

Crystal formation is fundamentally different as the local structure changes first to form a nucleus. This has both energetic and entropic consequences and the aforementioned kinetic barrier is observed. I would also like to add that crystal nuclei are formed all the time but they don't reach the critical size if the temperature is too high.


pH of sodium metasilicate anhydrous vs pentahydrate when dissolved in water I'm not a chemist and don't remember much chemistry from school, so please forgive a naive question (I've read the Wikipedia article on sodium silicate though).

I'm researching this topic to make a positive photo resist developer for making PCBs.

Will dissolving $\ce{Na2SiO3}$ (the anhydrous form) in water produce essentially the same solution as dissolving $\ce{Na2SiO3*5H2O}$ but in different concentration or are these intrinsically different?
Will $\ce{Na2SiO3}$, if left in contact with open air, eventually hydrate to $\ce{Na2SiO3*5H2O}$?
Also, the Wikipedia article talks about the $\ce{SiO2:Na2O}$ ratio affecting the pH of the water solution. Is the information about this ratio already present in the chemical formula so if I buy $\ce{Na2SiO3*5H2O}$ it is always the same ratio or is this a different parameter that is specified in addition to the chemical formula? (I understand that different concentrations of the solution will produce different pH, but consider equal concentrations)

I believe I need the developer to be an alkaline solution so I need to understand what will affect the pH when dissolving sodium metasilicate in water (e.g. if I need $\ce{Na2SiO3}$ or $\ce{Na2SiO3*5H2O}$, if I need to consider other specifications besides the chemical formula, if I need to consider the shelf life of the chemical if it changes with time e.g. gets hydrated from $\ce{Na2SiO3*5H2O}$ into $\ce{Na2SiO3*9H2O}$).
 Searching electronics forums on the internet I haven't found a clear answer. For some people using this sodium metasilicate as a developer works for some it doesn't develop at all, some add $\ce{NaOH}$ to it but there are places that advertise their sodium metasilicate developer as sodium hydroxide free.

As explained in the comment by permeakra, due to the structural differences between anhydrous and hydrated Na2SiO3, their solutions will also be intrinsically different, at least for a while.
Whether anhydrous salt will turn to hydrate or vice versa depends on your ambient humidity and temperature. Anyway, I'd not recommend leaving either of them in contact with open air, for fear they might react with CO2. Silicic acid is so weak that it can be displaced even by carbonic acid.
Well, theoretically, the information about the SiO2:Na2O ratio is already present in the chemical formula Na2SiO3 (with or without water). However, the industrial-grade product may not quite match that formula. In fact, it may be a wild mixture of different silicates, where each individual compound cannot and need not be characterized. For many large-scale applications it suffices to know just the said ratio. Not sure about your application, though.


Is this home-made pesticide from cigarette butts and battery acid safe? An old friend once told me that a good way to kill bugs in my garden was to soak lots of cigarette butts in water to make a gallon of the brown liquid. Then add one ounce of battery acid and mix well. He said to spray it on the garden and wait two to three weeks before harvesting because it would take that long for the mixture to become Niacin. I thought it sounded cool but never tried it.

Now 40 years later I would like to, but I want to know if this is really true or would it ruin my vegetables? It's been 22 years since my chemistry classes and tried to make sense of this link.
It seems like the nicotine/acid thing would become neutral, but again, I have forgotten so much that I'd feel safer running this past a few more knowledgeable folks first.

This seems like a bad idea to me.
The use battery acid (sulfuric acid) seems really terrible. In the past there was an awful problem with coal power plants releasing sulfur oxides into the air which formed acid rain (sulfuric acid). You're proposing making a brew that is orders of magnitude worse than acid rain to put on your garden plants. I think such a strong solution of acid would get rid of the bugs by killing your plants. ;-)
If you goggle nicotine "bug spray" you can find a number of websites that discussing seeping tobacco in water, then filtering the brew to make a spray. In such a solution the nicotine is so dilute that it would be safe to handle. However concentrated solutions of nicotine like used for various e-cigarettes could be dangerous to handle. Lookup nicotine MSDS in google (here is one such MSDS sheet).

(1)So no battery acid is needed.
(2) The various websites state that the sprayed nicotine solution decomposes fairly quickly - a couple of days, not weeks.

Frankly I don't smoke or chew tobacco. Since I'd have to buy the tobacco for such a use, I'd rather just buy a commercial pesticide and follow its use instructions. It is absolutely crazy to me that so many folks so berserk when they see the word "chemical." If you showed a strawberry to most folks they would just identify it as such. But it is nothing but a sack of hundreds of chemicals. So to me all strawberries are "organic strawberries." To someone who is allergic to strawberries one can be deadly. Did you know that there are varieties of lima beans which have a high enough concentration of cyanide that eating them can kill? (dangerous foods) So spraying a nicotine solution may seem "natural" but it is still a chemical.


Estimating magnitude of percentage and absolute error Can someone at least give me a hint to push me in the right direction of how to estimate the magnitude of the absolute and percentage errors when using a smaller amount in mL of a substance for a lab measurement?

When you take a small amount of a liquid in mL, you you usually use a pipette. On every pipette, at the top of it, the absolute error is written. As for the relative error, it is the ratio of the absolute error to the measured value. The percent error is the relative error multiplied by 100.


What happens to air at 1000 degrees Celsius I was looking at the graphs for the density of air at different temperatures (link below) and I was wondering if air would not react with itself at high temperatures? I read that oxygen and nitrogen react at 1600 degrees Celsius, but even below that average temperature (because of the Maxwell Boltzmann distribution) I’m guessing the reaction takes place? Also would it just catch fire with contact with anything in the environment like carbon / almost anything? I’m wondering about the hypothetical use in hot air balloons with very high temperatures and low volumes.

https://www.engineeringtoolbox.com/air-density-specific-weight-d_600.html

At $\pu{1000^oC}$, oxygen and nitrogen will not react substantially or probably even noticeably if at all. Besides that, whatever organic material you are using for a hot air balloon will most definitely ignite immediately. The average temperature that most organic materials char and eventually ignite is $\pu{270^oC}$. At almost double the kinetic energy at $\pu{1000^oC}$, most things will almost instantaneously ignite. Note that in air, temperature diffuses rather quickly so even if the fire or whatever you are using to get to $\pu{1000^oC}$ will probably be significantly colder when it reaches anything flammable. In fact, most fuel used in normal hot air balloons burn at over $\pu{1000^oC}$, but it diffuses very quickly that the average temperature of the air in the balloon is around $\pu{110^oC}$


Meaning of the Term "Heavy Metals" in CofA? I've always wondered, what is the meaning of the following terms on a certificate of analysis:

"Heavy Metals"
"Heavy Metals (as Pb)"

What exactly is comprised in "heavy metals", "heavy metals (as Pb)" and what is excluded from each? What is the difference between the two variants? I tried to find a clear standard describing this to no avail.

You are right heavy metals (as Pb) is a vague term. It is being discouraged and individual analysis should be reported. When someone says "I tried to find a clear standard describing this to no avail", it is a good illustration that not all information is on the webpages. It is worthwhile to search books :-)
I used to think this was an old-school analytical terminology and it turns out to be correct because it is based on classical methods. From the Food Chemicals Codex (ISBN 0-309-08866-6), which is a 1000 paged book, has the following information. Later they describe the tests as well.
Heavy Metals (as Pb)

This test is designed to limit the content of common metallic
impurities colored by sulfide ion (Ag, As, Bi, Cd, Cu, Hg, Pb, Sb, Sn)
by comparing the color with a standard containing lead
(Pb) ion under the specified test conditions. It demonstrates that the
test substance is not grossly contaminated by such heavy metals, and
within the precision of the test, that it does not exceed the Heavy
Metals limit given as determined by concomitant visual comparison with
a control solution. In the specified pH range, the optimum
concentration of lead (Pb) ion for matching purposes by this method is
20 micrograms in 50 mL of solution. The most common limitation of the
Heavy Metals Test is that the color the sulfide ion produces in the
Sample Solution depends on the metals present and may not match the
color in the Lead Solution used for matching purposes. Lead sulfide is
brown, as are Ag, Bi, Cu, Hg, and Sn sulfides. While it is possible
that ions not mentioned here may also yield nonmatching colors, among
the nine common metallic impurities listed above, the sulfides with
different colors are those of As and Cd, which are yellow, and that of
Sb, which is orange.
If a yellow or orange color is observed, the following action is
indicated: Because this monograph does not include an arsenic
requirement, As should be determined. Any As found should not exceed 3
mg/kg. If these criteria are met, Cd may be a contributor to the
yellow color, so the Cd content should be determined. If an orange
color is observed, the Sb content should be determined. These
additional tests are in accord with the section on Trace Impurities in
the General Provisions of this book, as follows: ‘‘if other possible
impurities may be present, additional tests may be required, and
should be applied, as necessary, by the manufacturer, vendor, or user
to demonstrate that the substance is suitable for its intended
application.’’


Shouldn't the pressure of gas above a liquid slightly decrease when some of its molecules dissolves in the liquid? Lets say a liquid and gas above are in equilibrium.

(1) Now if I suddenly increase the pressure of the gas it will take some time for some molecules of the gas to enter the liquid and get dissolved. When its finally done and equilibrium is reached shouldn't the pressure of the gas be a little lesser?

(2) Conversely if I violently shake a carbonated water bottle shouldn't the pressure increase a tiny little bit?

In many sources I have read that the pressure doesn't change at all when shaking. It does increase a little bit because of the energy we are giving while shaking.

What I want to know is a detailed explanation of the above two main questions.

1) If you induce a sudden chance in pressure you will indeed take the system out of equilibrium. Lets consider a closed recipient containing some liquid and some gas above. If you suddenly inject more gas into the recipient, the gas pressure is initially going to raise fast and then going to decrease slowly while part of the gas is dissolving, until the system reaches equilibrium again. However, this new equilibrium will have a higher gas pressure AND greater amount of gas dissolved in the liquid when compared to the initial equilibrium. Of the amount of gas you injected part will be dissolved and part will stay above the liquid, contributing to the higher gas pressure. It makes no sense to have pressure under the initial value, this would only be possible if the system hadn't been at equilibrium at the start.

2) Shaking doesn't increase the pressure, unless you shake the bottle long enough to cause a considerable increase in the temperature inside, but that would be quite an impressive amount of shaking. If you could take the system out of equilibrium by releasing some of the dissolved gas from the liquid phase, then that would increase the pressure, but shaking doesn't accomplish that, and I believe it wouldn't be possible to shift a system out of equilibrium like that without transference of matter or heat and without a change in volume or temperature.

There is a very interesting video by Veritasium on youtube about this bottle issue, I recommend you to watch it:
https://www.youtube.com/watch?v=K-Fc08X56R0&t=3s


Viability of proposed carbon-carbon bond-forming reaction Here is my proposed synthetic pathway to make propanoic acid from bromoethane:

  $\ce{CH3-CH2Br ->[\ce{KCN(aq)}][\text{heat}] CH3-CH2-C#N}$
  $\ce{CH3-CH2-C#N ->[\ce{KOH(aq)}][\text{heat under reflux}] CH3-CH2-COOH + NH3\uparrow}$

I have some questions about my proposed pathway:

Is it viable at all?
If so, what might be the disadvantages compared to other carbon-carbon bond-forming reactions such as the Grignard reaction?
If it is viable, what is the estimated yield?
If it is viable, how can I improve the yield? Can I use silver ions to increase the yield of the first step?
If it is viable, how might I isolate the product?
If it is not viable, can I fix it? If so, how?

With some some minor modifications, your two stage reaction

Kolbe nitrile synthesis
alkaline hydrolysis of the nitrile to corresponding carboxylic acid

will work in principle.
In the Organikum, 21. edition, 2001, p. 256ff, a general procedure for the reaction of alkyl bromides with sodium cyanide is provided.
The procedure distinguishes between reactive and less reactive halides.
Reactive halides are benzyl halides (unsubstituted, methoxy- or methyl-substituted.
Less reactive halides are halogen-substituted benzyl halides and plain alkyl halides.
For the latter, the following procedure is described:

In einem 1-l-Dreihalskolben mit Rührer, Rückflußkühler und Innenthermometer werden 250 ml Triethylenglycol, 1,25 mol gut gepulvertes, trockenes Natriumcyanid und l mol Alkylbromid oder Alkylchlorid unter gutem Rühren vorsichtig erhitzt. Der Beginn der stark exothermen Reaktion ist bei den niederen Alkylhalogeniden daran zu erkennen, daß die Lösung aufsiedet. Man steigert die Temperatur langsam auf 14O °C, bei der Umsetzung von Benzylhalogeniden nur auf 100 °C, und rührt bei dieser Temperatur noch 30 Minuten.

(Translation: In a 1-liter three-necked flask equipped with stirrer, reflux condenser and internal thermometer, 250 ml of triethylene glycol, 1.25 mol of well-powdered and dry sodium cyanide and 1 mol of alkyl bromide or alkyl chloride are carefully heated. The beginning of the strongly exothermic reaction in the lower alkyl halides is evident from the fact that the solution settles. The temperature is slowly raised to 140 °C, while the reaction of benzyl halides is only to 100 °C, and stirring is continued at this temperature for 30 minutes.)
The solvent is triethylene glycol with a boiling point of 291 °C, and the reaction is performed under exclusion of water using dry sodium cyanide.
As I have mentioned in my comment, the low boiling point of bromoethane is a disadvantage!
In fact, the Organikum recommends the use of diethyl sulfate (bp 209 °C) for the synthesis of propionitrile (bp 118 °C) and suggests to isolate the product by distilling it off the reaction mixture.
For further purification of the more volatile nitriles with an alkyl chain shorter than $\ce{C5}$, it is mentioned to

wash then with brine
dry over calcium chloride
re-distill over a 30 cm Vigreux column

For this reaction (to propionitrile), the yield is given as 90%. (For valeronitrile, hexanenitrile and heptanenitrile, which all can be prepared from the corresponding bromides, the given yields are around 80%).
According to the Organikum (p. 500) the hydrolysis of primary nitriles is achieved by heating 1 mol of the nitrile with 2 mol of 25% aqueous sodium hydroxide solution under reflux until no more ammonia evaporates (2 to 1o hours).
Whether you really want to use this two stage procedure of perform a Grignard reaction of the bromides with $\ce{CO2}$ is up to you. As mentioned in the comments, the advantage of the Grignard pathway is that it's pretty much a one-pot reaction all the way down to the carboxylic acid. The downside is the requirement of a dry solvent.

Update
In a comment, you asked for a recommendation on which carboxylic acid to synthesize.
I would synthesize a compound if

I need it (for further studies)
it is not commercially available
it costs a fortune. (Make sure that the halide that you need as a starting material is not more expensive than the final carboxylic acid.)

If it is just for practise, I'd go for phenylacetic acid (or a donor-substituted derivative), which has the following advantages:

benzyl bromide is a (weak) lacrymator, but not a CMR compound
the reaction with sodium cyanide can be carried out in refluxing acetone (instead of triethylene glycol)
phenylacetic acid, the final product, has a melting point of 77 °C and a rather low solubility in water (< 2 g/L). The product can be purified by crystallization.


Neutralization reaction between ammonia and nitric acid Complete and balance the following molecular equation (in aqueous solution); include phase labels. Then, write the net ionic equation.
$~\ce{NH_3 + HNO_3 -> ?}$

I thought that the acid $\ce{HNO3}$ would just give its hydrogen to $\ce{NH3}$ and make the resulting reaction:
$$\ce{NH_3 + HNO_3 -> HNH_3 + NO_3}$$
However the correct answer is $\ce{NH_3 + HNO_3 -> NH_4NO_3}$. Why is this?

Your answer is very close to the answer given, except for the following two tidbits (the first being more significant).

An acid-base reaction is not the exchange of a hydrogen atom $\ce{H}$. It is the exchange of a hydrogen ion (or proton) $\ce{H+}$. Thus your answer should be: $$\ce{NH3(aq) +HNO3(aq) -> NH4+(aq) + NO3-(aq)}$$
The given answer combines the two ions produced into a single compound. $$\ce{NH4+(aq) + NO3-(aq) ->NH4NO3(aq)}$$

The result of #2 would be reasonable if you had not been told that the reaction was occurring in aqueous solution. In aqueous solution, the products should have been those of #1, since ammonium nitrate is freely soluble in water.


Probability density "AT" the nucleus? Just had a conceptual question from radial wave functions.

I wanted to know if probability density AT the nucleus is maximum, or zero.

From the attached images, it is clear that the function is at zero for p and d orbitals, but TENDS to infinity as approaching r = 0 for the s-orbitals. Is the value of the function 0 at r = 0?

Intutively, as the distance from the nucleus is the least, there should be maximum probability density.

Additionally, can there be some other interpretation for the phrase "at the nucleus"? It is quite confusing.

Thanks a lot.

EDIT: Apologies, don't know what was going through my head while posting - I initially posted something about PROBABILITY of finding an electron at the nucleus (I know the answer to that, oops).

I have changed it to what I meant to ask.

Hydrogenic Wavefunctions:
A hydrogenic wavefunction, $\psi_{nlm}(\boldsymbol{r})$, can be written as the product
$$
\psi_{nlm}(r,\theta,\phi) = R_{nl}(r) Y_{lm}(\theta,\phi)
$$
where $R_{nl}$ is a radial wavefunction and $ Y_{lm}$ is a spherical harmonic.

$R_{nl}$ is a radial wavefunction
$ Y_{lm}$ is a spherical harmonic
$n$ is the principal quantum number, $n = 1,2,3\dots$
$l$ is the angular quantum number, $l = 0=s, 1=p, 2=d,\dots,n-1$
$m$ is the magnetic quantum number $m = -l, -l+1,\dots,l-1,l$

Probability/Electron Density
The probability density, $\rho$, is equal to the square modulus of the wavefunction:
$$
\begin{align}
\rho_{nlm}(\boldsymbol{r}) &= |\psi_{nlm}(\boldsymbol{r})|^2 \\
\rho_{nlm}(r,\theta,\phi)&= |R_{nl}(r)|^2 \cdot |Y_{lm}(\theta,\phi)|^2
\end{align}
$$
The spherical harmonics $Y_{lm}$ have angular nodes through the nuclei for all values of $l$ and $m$ except $l =m =0$, so we can say that for all $l>0$ orbitals, i.e. all orbitals except s-orbitals, the nuclei will have $0$ probability density.
S-orbitals
Consider s-orbitals, in atomic units and substituting in the solution of the radial wavefunctions in generalized Laguerre polynomials $L^\alpha_k$:
$$
\begin{align}
\rho_{n00}(\boldsymbol{r})&= |R_{n0}(r)|^2 \cdot |Y_{00}(\theta,\phi)|^2\\
&= |R_{n0}(r)|^2 \frac{1}{4\pi} \\
\rho_{n00}(r)&= \frac{Z^3}{\pi n^5} \exp\left(-Zr/n\right) L^1_{n-1}\left(2Zr/n\right) \\
&= A(n) \exp\left(-Zr/n\right) L^1_{n-1}\left(2Zr/n\right)
\end{align}
$$
For all values of $n$, $L^1_{n-1}\left(0\right) = n$, so we know that the nucleus never has zero probability density for s-orbitals, and has probability density
$$
\rho_{n00}(r=0) =  \frac{Z^3}{\pi n^4}
$$
Stationary Points
But we can differentiate with respect to $r$ to find stationary points in the radial probability density.
$$
\begin{align}
\frac{d}{dr}\rho_{n00}(r)&= A(n) \frac{d}{dr} \exp\left(-Zr/n\right) L^1_{n-1}\left(2Zr/n\right) \\
&= A(n) \exp\left(-Zr/n\right)\left[ -\frac{Z}{n} L^1_{n-1}\left(2Zr/n\right) +  \frac{d}{dr}L^1_{n-1}\left(2Zr/n\right) \right] \\
\text{if}\;n=1: \frac{d}{dr}\rho_{000}(r)&= -\frac{Z^4}{\pi n^6} \exp\left(-Zr/n\right) \\
&> 0 \;\forall\;r
\end{align}
$$
Hence the probability density is constantly decreasing away from the nucleus.  So as you previously assumed the maximum for probability density for the 1s orbital is at the nucleus.
$n>1$:
$$
\begin{align}
\text{if}\;n>1 : \frac{d}{dr}\rho_{n00}(r)&= A(n) \exp\left(-Zr/n\right)\left[ -\frac{Z}{n} L^1_{n-1}\left(2Zr/n\right) +  \frac{d}{dr}L^1_{n-1}\left(2Zr/n\right) \right] \\
&= A(n) \exp\left(-Zr/n\right)\left[ -\frac{Z}{n} L^1_{n-1}\left(2Zr/n\right) - L^2_{n-2}\left(2Zr/n\right) \right]
\end{align}
$$
Setting the expression above equal to zero and solving for $r$ will give the positions for the radial nodes and maxima inbetween for a given value of $n$.  I am not aware of a general method/expression in n.
As you asked about nucleus, we can also consider the point $r=0$:
$$
\begin{align}
\text{if}\;n>1 : \frac{d}{dr}\rho_{n00}(r)&= -  \frac{Z^3}{\pi n^5} \left[\frac{Z}{n} L^1_{n-1}\left(0\right) + L^2_{n-2}\left(0\right) \right] \\
&= -  \frac{Z^3}{\pi n^5} \left[\frac{Z}{n} n +\frac{n^2-n}{2}\right] \\
&= - \frac{Z^4}{\pi n^5} - \frac{Z^3 \left(n-1\right)}{2\pi n^4} < 0
\end{align}
$$
Conclusion/TL;DR
Hence the nuclei is always a local maximum in the probability density for an s-orbital.  You would expect the exponential decay in the radial wavefunction to dominate over the Laguerre polynomial term, so would expect local maxima further out to be smaller that the maxima at the nuclei and hence that the point of highest probability density is at the nucleus.
This makes physical sense, as classically the negatively charged electron wants to be as close as possible to the positively charged nucleus to minimise its potential energy, but the quantum mechanical nature of the electron prevents it from tumbling directly into the nucleus.


Which alkene on heating with alkaline KMnO4 solution gives acetone and CO2? Which alkene on heating with alkaline $\ce{KMnO4}$ solution gives acetone and a gas that turns lime water milky ?

  2-methyl-2-butene
  isobutylene
  1-butene
  2-butene

I know that $\ce{CO2}$ turns lime water milky but, I'm not able to find the alkene in the question. Also, I want the mechanism involved.

The best approach, and good practice, is to draw out the reaction products for each case.  You should know that alkaline potassium permanganate replaces a carbon-carbon double bond with two carbon-oxygen bonds, one to each of the original double-bonded carbons.  For instance, with 2-butene we have:

$\ce{CH3-CH=CH-CH3 -> CH3-CH=O + CH3-CH=O}$

Here the product molecules are the same because the alkene is, of course, symmetric.

But wait, there's more.  The carbonyl compounds pictured in the above example are stable in the alkaline permanganate medium if they are ketones, with only carbon atoms attached to the carbonyl function.  If there are aldehydes, the carbon-hydrogen bond(s) to the carbonyl group will be oxidized further.  Thus the "acetaldehyde" identified above for 2-butene is further oxidized to break the carbon-hydrigen bond at the carbonyl group:

$\ce{CH3-CH=CH-CH3 -> 2 CH3-CH=O -> 2 CH3-C(O)-OH}$

where in the alkaline medium the acetic acid will appear as acetate ion.

As you work through your choices you find that there is one aldehyde where the carbonyl carbon is bonded only to hydrogen atoms allowing that function to be oxidized all the way to carbon dioxide (or, again accounting for the alkali, carbonate or bicarbonate ion).  Find it, then see which choice gives that along with the other given product acetone.  Bonus question:  Would the acetone hold up against further oxidation?  How would you tell?


Internal Cannizzaro reaction of phenylglyoxal Phenylglyoxal $(\ce{PhCOCHO})$ can undergo both internal Cannizzaro reaction and between two molecules.

The internal Cannizzaro reaction is like this:

Which reaction will be preferred? Internal Cannizzaro or Cannizzaro reaction between two molecules?

I agree with the other answers that in general intramolecular reactions are faster than intermolecular reactions. Let's look at why this intramolecular reaction might be fast.

We can describe the Cannizzaro reaction as the delivery of a hydride from an anionic hydrate to a carbonyl group. The reaction shown certainly meets those criteria. I see this reaction as a carbocation rearrangement or a (semi-)pinacol rearrangement. In the latter class of reactions, a group undergoes a 1,2-migration to quench a carbocation and generate a carbonyl.

In the phenylglyoxyal rearrangement, if the ketone is considered as its ionic resonance structure (2), then the ketone carbon is a carbocation. We know that hydrogens can and will undergo 1,2-migration if the resulting carbocation is more stable. In this case (3->4), the resulting carbocation is very stable, because it's really a carboxylic acid (5). Furthermore, the carboxylic acid will be deprotonated rapidly under basic conditions (5->6).


What is a simple formula to approximate the change in relative humidity as temperature rises/falls? The moisture-carrying ability of air goes up as the temperature rises. So if I take a volume of air and heat it or cool it (allowing it to expand/contract so the pressure remains constant) how can I calculate the (approximate) relative humidity of the warmer/cooler air that contains the same amount of water vapor?

Say I blow 30°c air at 20% relative humidity over a cool object (but not cool enough to condense water out of the air.)  The air is cooled to 20°. What would the relative humidity of the cooled air be if it contained the same amount of water vapor? How do I calculate that?

I am building an Arduino-based control for a warm air dryer for ski boots and bike gear. I will measure the temperature/humidity of the heated air as it comes out of the dryer, and then measure it again as it escapes the gear that I am drying. While the gear is still damp, the waste air should pick up additional moisture, and so its relative humidity should be higher that it would be if I simply cooled it down. When my gear is completely dry, the cooler waste air should contain the same amount of water vapor that it started with, so the relative humidity should be the amount calculated for the measured change in temperature.

I want a "good enough" solution, not one that's mathematically pure. If it gets me answers that are within a percent but involves less computation, that's better than an ideal solution that involves lots of time-consuming calculations. The Arduino microcontroller is fairly slow and cannot do high speed floating point calculations.

(Would this question be better on an engineering forum? I could see arguments either way.)

If you want to convert the %RH from one temperature to another you can use one of various parameterizations, for instance the Antoine equation:

  range: 273. - 303 K,
  Bridgeman and Aldrich, 1964,
  https://webbook.nist.gov/cgi/inchi?ID=C7732185&Mask=4#Thermo-Phase

  A = 5.40221;
  B = -1838.675;
  C =   -31.737;

  log(P) = A + B/(C+T);

Then obtaining the $RH_2$ at $T_2$ from that at $T_1$ is as simple as
$RH_2 = RH_1 P_1/P_2$, where $P_1$ and $P_2$ are the saturation vapor pressures of water computed with the equation above, assuming the ambient pressure (and therefore the partial pressure of water) is constant.


Why lactols being tetrahedral complexes do not break? The answer given is (D) but I think (B) should be the answer as the concerned benzylic carbon in option (D) is an unstable tetrahedral complex and it should break. Besides, the medium is acidic so it should catalyse this process.
In generalising, my question is: are lactols stable?
Edit:
I found online that there is an equilibrium between lactols and corresponding hydroxyaldehyde. So, why lactol is favoured more in this case?

You may notice that B is an intermediate to eventually reach D.  Indeed, it is quite possible, performing the reaction in the lab, B would spontaneously yield D with only little chance to isolate this intermediate as such.
$\ce{NaBH4}$ reduces the aldehyde to an alcohol.  Then, mediated by the acid, the actetal is converted into an aldehyde to yield B.

The aldehyde in B already is more reactive than a ketone would be, but the presence of acid (still present from converting the acetal) may increase the reactivity of the carbonyl group even further.  The subsequent cyclization is favored

because the chance that alcohol and aldehyde are close enough to form a new bond is much higher for an intramolecular reaction, than for an intermolecular one.  This is a general rule worth to retain. In addition, the formation of five- and six-membered rings equally is energetically favorable.


Does the addition of base to 3-pentanone give the cis or trans enolate? I've spent several hours searching the WWW with entries like "stereoselective enolate formation" and similar ones, but I find almost only entries for Li-amide bases (LDA, LMTP) and no other bases like t-BuO, NMe3, NaH, n-BuLi concerning the stereoselectivity.

I'd like to know if the addition of these four bases to 3-pentanone all results with preference in the cis or trans enolate (and if possible with an explanation why).

Related: Stereoselective enolate formation with different bases: Does the addition of a base NR3 to a ketone affords the cis or the trans enolate?

As a prenote: My answer is somehow speculative because I have no experimental data or reference values of similar reactions used. In other words: The argumentation pathway below might be wrong.

Addition of LDA mostly gives the trans-enolate if a primary alkyl substituent next to the carbonyl group has been used.

Carbonyl(O) is smaller than the methyl group [ref.], therefore the allylic A13 interaction of Me//Me or even Me//i-Pr is bigger than the allylic A12 interaction.

An allylic A13 Me//Me interaction is about 3,7 kcal/mol, an allylic A12 Me//Me is may be around 3,2 kcal/mol (this is just a guess).

If HMPA is added Li might coordinate to four HMPA molecules forming a cluster and inhibiting the cyclic transition state of LDA and hence no longer providing the trans-enolate, meaning an open transition state must be lower in activation energy for the cis-enolate formation.

So I tried to draw the conformations of 3-pentanone and then treated the base as "point bulk", meaning the base is like a sphere with different sizes.

Hard bases have a bigger repulsion to the carbonyl oxygen.

I have used arbitrarily numbers for the stereoselectivity.

Here is my answer which base gives more cis-enolate than the other:


What chemical could dissolve low-density polyethylene? What chemical could dissolve plastic aside from acetone? Preferably, low-cost and environmentally friendly. Any suggestions would be helpful.

I'm targeting for thin plastic (low-density polyethylene). Environmentally friendly for me is that it follows the EPA standards from the USA. I'm targeting for around 150 USD. I'm okay with solvents that make the plastic swell, but a chemical that could dissolve it is preferable.

The safest solvent might be supercritical carbon dioxide, $\ce{CO2}$. Supercritical carbon dioxide is also used in "green" dry-cleaning. Use for dissolving polymers requires a pressure vessel for ~5 to 50 MPa, able to work at ~40 to 100°C. There are (paywalled) articles at ACS and NCBI, if you want to investigate further.
According to Agilent, 1,2,4-Trichlorobenzene (TCB) is effective.
Some other solvents mentioned at Research Gate include toluene and xylene.

N.B. The aromatic solvents all are toxic, some quite so, and are volatile and flammable. All of them require temperatures near the melting point of polyethylene, ~100°C, to dissolve much of the polymer, so a reflux or pressurized system is required.

Supercritical $\ce{CO2}$, while not very toxic, can cause asphyxiation above ~5%, and the container is, effectively, a bomb containing a warm or hot pressurized fluid.


Can G, H, and S be negative? Consider Gibb's free energy of a reaction in a closed system $G = H - TS$. Can $G$ ever be negative? (note, I am not talking about $\Delta G$)

I would suspect no because there is always some energy on every system.

What about $H$ and $S$? Are they nonnegative?

I would suspect yes because if there are bonds then they have internal PE and KE. I wouldn't know how to interpret negative disorder.

Is $H \ge TS$ always?

Variation on 1)

G and H are like altitude.  They are not absolute; rather their values are always relative to whatever you set as a reference. So they can be either positive or negative.  There is no general constraint on the sign of their values.  For this reason, we usually talk about $\Delta G$ or $\Delta H$, since it's the changes that are meaningful; giving them arbitrary values isn't typically useful.
By contrast, there is a non-arbitrary zero for entropy.  In classical thermodynamics, it's the entropy of a pure substance at absolute zero. Since the temperature can never be lower than absolute zero in a system at thermodynamic equilibrium, $S$ (as opposed to $\Delta S$) can never be negative.
In statistical mechanics, there are various equivalent ways of defining the entropy, depending on the ensemble you are using.  The simplest definition is that for the microcanonical ensemble:
$$S = k_B \ln \Omega,$$
where $\Omega$ is the number of possible microstates of the system, and $k_B$ is the Boltzmann constant.  Since the minimum number of microstates is 1 (which happens at absolute zero), the minimum value for $S$ is zero.


Why do we get slightly more inversion product than retention product in SN1 reaction? Theoretically, in $\mathrm{S_N1}$ reaction we should get a racemic mixture as the product because the nucleophile can attack from either side of the formed carbocation.

However, many textbooks claim that there's slightly more inversion product than retention product.

Is there any specific reason for this to happen.

It is well known that SN1 reactions often give incomplete racemisation:

Although many first-order substitutions do give complete racemization, many others do not. Typically there is 5–20% inversion, although in a few cases, a small amount of retention of configuration has been found. These and other results have led to the conclusion that in many SN1 reactions at least some of the products are not formed from free carbocations but rather from ion pairs.

We can explain this by realising that the SN1 and SN2 mechanisms are not really discrete, as often taught, but rather two extremes of a 'mechanistic continuum'. This has implications for both the stereochemical outcome of many substitutions (i.e. proposed SN2 are not always as stereospecific as one might expect), and also the kinetics (i.e. measuring the rates of substitutions won't always give data consistent with purely SN2 or purely SN1 mechanisms).

At the SN1 extreme, there is no covalent interaction between the reactant and the incoming nucleophile during the transition state (TS) for the cleavage of the leaving group.
At the SN2 extreme, the TS has covalent interactions in which there is concerted formation of a new bond between the incoming nucleophile and the reactant , with cleavage of the bond to the leaving group.

This mechanistic continuum was first explained by Winstein, who studied both the kinetics and resulting stereochemistry of nucleophilic substitutions at saturated carbons.[1] This model is widely accepted, though rarely taught at the undergraduate level.
Winstein's theory proposes the presence of several additional intermediates along the reaction coordinate- these are known as 'ion pairs' and represent varying levels of association between the reagent and the leaving group. The energy levels between these ion pairs are small, and nucleophilic attack can occur at any stage (hence the continuum).

 Fig1: Winstein model for nucleophilic substitution. Taken from Carey's Advanced Organic Chemistry. Part A. /2
The Winstein model fits with what is commonly taught, namely that SN1 gives racemisation and SN2 gives inversion:

SN2  reactions occur when the nucleophile intercepts the contact ion pair and give stereospecific inversion of configuration since the leaving group is still associated with the carbocation, essentially shielding one face from attack
SN1 reactions occur when the nucleophile intercepts the fully dissociated free carbocation and gives complete racemisation as both faces of the planar carbocation are equally accessible

Which mechanism operates is largely determined by how stable the carbocation is. Primary carbocations are highly unstable and as such there is little (if no) dissociation. Tertiary cations are stable and as such complete dissociation can occur.
The interesting case is where nucleophilic attack occurs on a solvent separated ion pair - we see neither complete inversion nor complete racemisation, as some association still exists at this stage (partially, but not completely blocking one face of the cation).
Modern Physical Organic Chemistry[3] gives an overview of all of the possibilities, though the figure is a little hard to follow:

References

Winstein, S.; Clippinger, E.; Fainberg, A. H.; Heck, R.; Robinson, G. C. Salt Effects and Ion Pairs in Solvolysis and Related Reactions. III.1Common Ion Rate Depression and Exchange of Anions during Acetolysis2,3. J. Am. Chem. Soc. 1956, 78 (2), 328–335. DOI: 10.1021/ja01583a022.

Carey, F. A.; Sundberg, R. J. Advanced Organic Chemistry, Part A: Structure and Mechanisms, 5th ed.; Springer: New York, 2007.

Anslyn, E. V.; Dougherty, D. A. Modern Physical Organic Chemistry; University Science Books: Sausalito, CA, 2006.


What is the maximum pressure that JB Weld can withstand? JB Weld, a brand name of steel-reinforced epoxy, makes claims of being the "world's strongest bond".

What is the maximum pressure it can withstand, and what is it about the interaction between the steel and the epoxy that makes it so strong?

This is only a partial answer, addressing the maximum pressure a JB weld can withstand. I am still thinking about the reason for the strength so I will leave it to the 'real' chemists among you for now.

According to Repair Products the JB weld in fully hardened state has the following properties (in psi):

Tensile Strength:   3960
Adhesion:           1800
Flex Strength:      7320
Tensile Lap Shear:  1040

Tensile strength is simply the pressure needed to break the material by pulling it apart. Flex strength is the pressure needed to bend the material which, in case of a perfectly homogeneous sample would equal the tensile strength.

The adhesion is the pressure needed to pull the bond from the surface when pulling perpendicular to the surface (parallel to the bond). The tensile lap shear is also an 'adhesion', but measured when pulling parallel to the surface (perpendicular to the bond).

Since the latter is the lowest it means that this would be the maximum pressure the JB weld could withstand, but only if it is applied exactly perpendicular to the bond. If the direction is in a different direction than you will get some mixture between the adhesion and the tensile lap shear pressures.

What is clear from these numbers is that the bond itself will not break but rather it will detach from the materials it is keeping together.


What are the states of HX when reacting with alcohol and alkenes respectively, and Br2 when reacting with benzene? I couldn't find this in any textbooks, so I guess I will ask this here. What is the physical state of $\ce{HX}$ when reacting with alcohol (nucleophilic substitution) and alkenes (electrophilic addition) respectively?

For electrophilic substitution of benzene with bromine, what is the state of bromine?  (e.g. aqueous, liquid, solid)

If possible, can you explain why they have the particular state?

The first reaction ($\ce{HX}$+Ethanol) happens in solution.  High concentration of HX mixed with ethanol will produce the product. However, for $\ce{X}$ other than $\ce{Cl}$, more often one put concentrated phosphoric or sulfuric acid with alcohol and a salt of X (for example, $\ce{KBr}$ or $\ce{KI}$).  Concentrated phosphoric/sulfuric acid are easy to obtain and store in high concentration, as opposed to $\ce{HX}$. It is, therefore, a liquid phase reaction.

For ethene + $\ce{HX}$, it can be done in gas-phase, in an aqueous environment, or you can also add $\ce{HX}$ gas to an ethene solution in a solvent such as $\ce{CCl4}$. So, it can be either gas phase or liquid phase.  It is almost always more preferable to do a reaction in solution in a laboratory set up. (So much that we call the experimental part of the chemistry wet chemistry.)  This constraint is, however, lifted in the industrial world, with the availability of expansive specialized reactors.

And for benzene + $\ce{Br2}$, both reactants are liquid and $\ce{Br2}$ is also soluble in benzene, so the reaction is in also the liquid phase.


Aromatic Apparent Substituion Reaction Mechanism The first molecule reacts with $\ce{NaOH}$ to form the second molecule. I am told that it does't undergo via a substitution mechanism. I am asked to draw the mechanism for it.

I am thinking that the double bond will attack the OH group (addition reaction) so that it will add onto the meta carbon.  Then the bromine group will leave (elimination reaction) which then the OH group attacks the para carbon.

I am not really sure if that is correct. Could someone please help me out. Thanks.

The question is slightly misleading. It is usually referred to as a substitution; nucleophilic aromatic substitution to be precise; but it is actually an addition-elimination. What the question is trying to say is that it is not a regular $S_N1$/$S_N2$ substitution (which I think you correctly realised anyway).

The mechanism involves attack at the carbon attached to the bromine to form an intermediate anion which is stabilised by the electron withdrawing acetyl group, para to the site of attack.

I'll let you draw the arrows and come up with a complete mechanism.


Which compound or compounds are responsible for color change in blue butterfly pea flower tea? Blue Butterfly Pea flower tea is blue by default. Add citrus juice and it turns purple. What compound or compounds are responsible for this reaction to change in pH?

Wikipedia doesn't say, nor can I find any source on the internet that does.

OK, that was simple. This

https://www.google.de/search?q=Clitoria+Ternatea+blue+color

gave this

https://www.ncbi.nlm.nih.gov/pubmed/14568080

which lead me to this

https://en.wikipedia.org/wiki/Anthocyanidin

and I guess that's the culprit. I judge from that abstract linked above that the actual plant can contain a wide mixture of those anthocyanidines.


What is the logic behind the division into groups of anions and cations in qualitative inorganic analysis? I find it very confusing and hard to memorize.
Is there a way to deduce it, like understanding how it is derived, or is it just one of a lot of ways you can do it but is the standard one?
Do people use to test every possible reagent or combination of reagents and see if something happens or is it deducible?

What is the logic behind the division into groups of anions and cations in qualitative inorganic analysis?

There is a somewhat established classic procedure of qualitative and quantitative chemical analysis (also known as "systematic analysis") based on treatment of the solution by specific set of reagents in specific order. Cations are divided into groups based on the stage they are detected in the procedure. There is NO other logic behind this division, nor it has any meaningful chemical significance.

  Also, how does one get at the various tests, like, did people use to test every possible reagent or combination of reagents and see if something happens or is it deducible?

Somewhat deducible. Solubility product constant can be used to check if the solution is prone to precipitation of particular salt barring cases of redox reaction. For redox reactions standard electrode potentials and Nernst equation might be used for a quick check. Still, knowledge of these particular constants is mostly (if not completely) empirical.


Are there any common and identifiable hazards to being a Chemistry Teacher? I'm a High School Chemistry teacher. I have only been teaching for a few years and know very few other chemistry teachers. Of the ones I do, One has died of cancer, one is a cancer survivor and one is being tested and currently in cancer therapy while an equal number remain healthy (but have taught fewer years). That's scarry.

Could it be coincidence or some other more localized cause? Yes. Although cases were up to 100 miles apart. I realize the sampling size is small and this is all very anecdotal but I'm looking for answers nonetheless.  It seems to me that a possible culprit might be the haphazard way that stock handling and chemical lab supply making is done in High Schools as compared to higher education institutions or research facilities. Many schools don't vent their stock rooms even though it's required (there are no consequences) and teachers take the stock put it out on the counter and measure out what they need and make solutions. They don't bother with gloves, lab coats or aprons either. In a professional laboratory this would be done in a fume hood to minimize long term exposure to many substances with proper protective clothing.

Are there any readily identifiable health hazards for a High School Chemistry teacher based on the above information?

I have been teaching chemistry at the college level (if you include my years as a TA in graduate school) since 1999.  While there are certainly dangers involved in dealing with hazardous chemicals, I would not say that the job is unusually dangerous if you follow safe practices and use decent equipment.  I have known many chemistry teachers and professors who lived well into their 80's and beyond.  You're probably going to find certain workplaces which, for whatever reason, are more dangerous than most.  I have heard of (but not verified) an incident at a local college where five female biology professors were diagnosed with uterine cancer, an event which is extremely improbable to occur by chance.  Obviously, there is a substantial need to investigate such a situation and see if there is a common carcinogen present which is causing the cancer.  Such a situation could conceivably occur in the vicinity of a chemistry lab, although it seems unlikely if all standard precautions are followed.

The greatest source of danger is likely to be the students themselves.  When you have many young people together who are (1) easily distracted, (2) not particularly aware of their surroundings, and (3) lacking in manual dexterity (e.g. clumsy), they are far more likely to be a source of injury than the chemicals on their own.  So long as you are willing to invest the time in enforcing safety in your labs on a consistent basis, I think you're going to be just fine.


Can a meso compund be rotated in ANY sort of way in order to show its achirality? Are there any limits to rotating a compound in order to prove that its chirality or achirality? (assuming that you're rotating about a single bond)

I think what you may be asking is: could there could be a nominally "meso" compound that has restricted rotation about its central single bond, causing it to exist in the form of two non-interconvertible stereoisomers. Example- start with meso-tartaric acid, and make the di-tert butyl ether from the OH groups and di-tert-butyl ester from the COOH groups. Maybe these groups would be bulky enough to restrict rotation. You could mentally rotate around the central C-C bond until the molecule has a plane of symmetry, so it is still a nominally "meso" compound, but physically you could separate the two stereoisomers.


What is the relationship between ortho, meta and para-dimethyl benzene? What is the stereochemical relationship between ortho, meta, and para-dimethyl benzene, given that interconversion between the three structures requires you to break and form new bonds?

I know that ortho relates to the 1,2 position, the meta position relates to the 1,3 position and the para position relates to the 1,4 position.

However, what does it mean by 'stereochemical relationship'? Does it just mean that if the methyl is placed at the different carbon numbers, they're different molecules?

These compounds do not have a stereochemical relationship. Stereoisomers have the same connectivity of atoms (constitution) with different arrangement in space. The (1,2), (1,3), and (1,4) in the names of these compounds indicate that this is not the case. These are not stereoisomers of any kind.

Not that the qualifying text about having to break bonds to interconvert is unhelpful and irrelevant. All isomers require bond-breaking to interconvert.

So, what then are the relationships between these three compounds? Let's look at a decision flowchart for relationship between structures:

  Do they have the same formula?

Yes.

  Do they have the same connectivity?

No. Thus, they are constitutional isomers. Constitutional isomerism is not a type of stereoisomerism. The two main types of stereoisomerism are enatiomerism and diastereomerism.

  It would be worth contacting your instructor and asking for clarification, noting that these are not examples of any kind of stereoisomerism. It could be an honest mistake in the wording of the question. As someone in the teaching profession, my experience is that most instructors are happy when a student understands the content at a level high enough to catch mistakes.


What is the difference between absolute and relative configuration in stereochemistry? Can you help me understand the difference between absolute and relative configuration?

Let's say that we have synthesized a new compound, C(1)(2)(3)(4), where (1)-(4) represent 4 different groups attached to the central carbon, so the molecule is chiral.
Let's say that we find a method to separate the two enantiomers present in the new compound and that we placed the enantiomers into separate test tubes.
Next we might take each enantiomer by itself and place it in a polarimeter.  We would find that the material in one test tube rotates light in the (+) direction and the material in the other test tube rotates light in an equal, but opposite direction (-).
At this point, all we know is that we have two enantiomers that, as expected, rotate light in equal and opposite directions.  We don't know which test tube contains the "R" enantiomer or which contains the "S" enantiomer.  We just know their relative configuration with respect to each other, they are mirror images of each other.
To find out which is "R" and which is "S" we would either need to 1) put a crystal of one of enantiomers in an X-ray diffractometer and get a three-dimensional picture of the structure of the molecule from which we could see whether it is the "R" or "S" enantiomer or 2) carry out a series of chemical transformations where the stereochemical consequences of each step are known (e.g. does each step proceed with inversion or retention of configuration at the chiral center) and convert it to a compound of known absolute configuration
Let's say we pick one of our test tubes and are able to convert that enantiomer to a chiral compound C(A)(B)(C)(D) where we know that R-C(A)(B)(C)(D) rotates light in the (+) direction.
We find that the C(A)(B)(C)(D) we created from our C(1)(2)(3)(4) enantiomer rotates light in the (-) direction; therefore we have produced the "S" enantiomer of C(A)(B)(C)(D).
For simplicity, let's say that all of the reactions we used in our conversion of C(1)(2)(3)(4) into C(A)(B)(C)(D) are known to proceed with retention of configuration.  Then we know that the test tube of C(1)(2)(3)(4) used in our reaction contains the S-C(1)(2)(3)(4) enantiomer. Now we know the absolute configuration of our sample.

EDIT: d, l and D,L

The lower case "d" and "l" can be used interchangeably with "+" and "-" respectively.  The letters stand for dextrorotatory and levorotatory.  They simply indicate how the plane of polarized light will be rotated when a sample of the material is placed in a polarimeter.  So l-2-butanol is the same as (-)-2-butanol.  Use of "d" and "l" is discouraged because it can create confusion with "D" and "L".

Upper case "D" and "L" are somewhat archaic.  Way back when, the two enantiomers of glyceraldehyde were separated, purified and placed in a polarimeter.  One enantiomer rotated the light in a dextrorotatory manner; this enantiomer was called D-glyceraldehyde.  The other enantiomer rotated light in the levorotatory direction and was referred to as L-glyceraldehyde.  If you take what ever chiral molecule you are studying and convert the chiral carbon to glyceraldehyde, and it rotates light in the "+" direction, then you've made D-glyceraldehyde, and if all of your reaction steps involved retention of configuration (see above) then you would have the "D" enantiomer of your material. The terms "D" and "L" relate whatever chiral center you are studying to the chiral carbon in glyceraldehyde.


How to carry out basis set fitting and extrapolation I am playing around with basis set extrapolations, and i would like to perform some simple tests by doing some parameter fitting and checking the errors.
The functional form of the fitted curve is this
$$
E(X) = E_\infty + Ae^{-\alpha X}
$$
where $X$ is the cardinal number of the basis set used, and $E$ is the energy. This function makes sense because the rate of decay empirically fits the convergence (of correlation concistent basis sets) and the exponential term vanishes as $x\rightarrow\infty$.
We want to use energies from two low-cardinal numbers, and extrapolate to the $E_\infty$ as given by the function and the fitted parameters.
Assume that for the cardinal numbers $X$ and $Y$ we have the corresponding energies $E_X$ and $E_Y$. We then have the relationship
$$
E_X = E_\infty + Ae^{-\alpha X} \\
E_Y = E_\infty + Ae^{-\alpha Y}
$$
By using that $A = \frac{E_X - E_\infty}{e^{-\alpha X}}$, we can solve for $E_\infty$ to obtain
$$
E_\infty = \frac{E_Ye^{-\alpha X} - E_Xe^{-\alpha Y}}{e^{-\alpha X} - e^{-\alpha Y}}
$$
and solve for $A$ to obtain
$$
A = \frac{E_Y - E_X}{e^{-\alpha Y} - e^{-\alpha X}}
$$
From what I can see, if all I want is to obtain the extrapolated energy, then I do not need take $A$ into any consideration. I just need the cardinal numbers, the energies, and $\alpha$. In Python syntax this would be something like this
def extrapolate(X, Y, E_X, E_Y, alpha):
    gamma = np.exp(-alpha*X)
    omega = np.exp(-alpha*Y)
    return (E_Y * gamma - E_X * omega) / (gamma - omega)

Is the above approach the normal one for curve fitting and extrapolation?
In some tests I find that the extrapolated energy is virtually identical to the energy with the highest cardinal number used in the extrapolation. So either there is no benefit of my extrapolation, or I am doing something wrong.
(Note that I am not using Gaussian basis sets, but a MultiWavelet basis, and that the cardinal numbers are not really cardinal numbers, but a defined precision relative to the CBS. Still the approach should be the same regardless - even if the function I use for the fitting perhaps not is the perfect choice.)

If I well understand your question you have a set of experimental points $(E_1,X_1), (E_2,X_2), ... , (E_n,X_n)$. You want to fit to this set of points the function :
$$
E(X) = E_\infty + Ae^{-\alpha X}
$$
So, you want to approximately determine the parameters $E_\infty$ , $A$ and $\alpha$.
This requires a non-linear regression calculus. For example see https://mathworld.wolfram.com/NonlinearLeastSquaresFitting.html
This calculus is iterative and requires to set some "guessed" values of the parameters to start the iterative process.
Another method (much simpler but for approximative result only) is not iterative and doesn't require "guessed" values is explained in : https://fr.scribd.com/doc/14674814/Regressions-et-equations-integrales
You could try it in folowing the numerical example below with $\quad y(x)=E(X)$ , $a=E_\infty$ , $b=A$ , $c=-\alpha$

Note : This simplified method is not reliable if the number of points is too low.


How are bubbles formed when heating water? My thinking is that bubbles are formed due to the dissolved gases in water. Consequently, upon heating, bubbles are formed and do not remain intact with water molecules, and thus the bubbles are formed. And since the bubbles are less dense than water, they goes up and pop due to air pressure. Water vapor is then formed in the above air. Is this all correct?

I have researched this topic and it's very difficult for me to understand. Please explain it as if you're explaining it to someone who knows very little about chemistry.

Saying that the liquid is converted to bubbles is not correct. The liquid simply becomes a gas (so inside the bubble there is the gas phase of your liquid) this is a phase transition and depends on the pressure and temperature of the portion of liquid close to the bottom of your container.

The points where the bubbles appear depends on the local variations of temperature and pressure.


Steric Hindrance and Brønsted Basicity Does steric hindrance affect Brønsted basicity?

I understand that basicity is a thermodynamic quality. Not a kinetic factor. So it doesn't matter if it takes a million years for the proton to reach the site of basicity; time is nothing in thermodynamics. We only care about the initial and final state. So this suggests to me that steric hindrance does NOT affect Brønsted basicity

In addition, the hydrogen proton is quite small. So I wouldn't expect it to be particularly sterically hindered by anything (unless we're in solution and the proton is heavily solvated). Still, once the proton reaches the site of basicity, it's reached it, and if I understand thermo correctly, thermo doesn't deal with anything in the middle of the energy graph. Only the points on the two ends.

Also it would be nice if one could comment on the difficulty of this problem in general. Just curious. Personally I think it's a fascinating problem. (On the other hand I'm not the one actually doing the problem).

Anything that stabilize one of the species in an acid-base equilibrium influences the free energy difference between the reactants and products, which alters the equilibrium constant:

$$\ce{HA <=> H+ + A-}$$

$$\Delta_r G^\circ = -RT\ln K_{eq}$$

You forget the solvent. Acidity and basicity depend on the interaction between the acid/base and the solvent. While steric hindrance will decrease the rate solvent access, it more importantly also decreases solvation of the acidic or basic site. More steric hindrance prevents additional solvent molecules from reaching the site to stabilize an ionic conjugate acid or base.

Consider the difference in $\mathrm{p}K_a$ between methanol, ethanol, 1-butanol, and tert-butanol. These are equilibrium constants, so they most certainly derived from thermodynamic phenomena!

Methanol $\ce{CH3OH}$ has a $\mathrm{p}K_a = 15.5$ (in water)

Ethanol $\ce{CH3CH2OH}$ has $\mathrm{p}K_a = 15.9$ (in water)

1-Butanol $\ce{CH3CH2CH2CH2OH}$ has $\mathrm{p}K_a = 16.1$ (in water)

t-Butanol $\ce{(CH3)3COH}$ has $\mathrm{p}K_a = 17$ (in water) (I've also seen 18 as its $\mathrm{p}K_a$)

The difference between methanol, ethanol, and 1-butanol (all primary alcohols) is small, while t-butanol has a larger pKa. If sterics had no effect, then both butanol isomers should have similar acidities since both isomers have similar inductive destabilization of the conjugate base by electron donating alkyl groups.

However in methanol, ethanol, and 1-butanol, more water molecules can solvate the alkoxide anion conjugate base, stabilizing that conjugate base more than the conjugate base of tert-butanol, which due to sterics cannot have as much solvation of the anion.

Now, gas phase acidities, in which there is no solvent, should have no solvent and solvation effects, and thus no (or at least reduced) steric effects.


Why is the principle of separation in GFC opposite to electrophoresis despite using same material for the gel? In GFC, larger proteins get eluted out first while in electrophoresis the larger proteins move slower. Why is this?

Gel electrophoresis and gel filtration chromatography both use gel materials (though often polyacrylamide gels are used for GE and cross-linked dextran gels for GFC), but their operating principles are different.

In GE, the gel is a solid block of gel which contains only small pores for proteins to pass through.
The proteins generally have to be denatured (and given a charge) to pass through the pores as native proteins are too bulky.
The denatured proteins are like long strands of noodles being drawn through the gel and shorter proteins pass through more quickly because they don't have to contort as much to fit through all the pores—the larger the protein, the more interactions with the gel to slow it down.

However, GFC is a subset of what is known as Size Exclusion Chromatography.
SEC techniques generally also use porous gels, but instead of using a solid block of gel, the gel is made into small beads.
A similar situation to GE exists where small things can enter the pores of the gel and larger molecules suffer more resistance to pass through the gel, only now, the larger molecules have an alternative route with lower resistance—going completely around the beads.
The effect of this is that small molecules pass into the pores of the beads and slow down, while larger molecules bypass the beads entirely and travel faster (what is considered small and large depends on the gel's pores).

GE has the benefit of much better size separation, but requires denatured proteins.
SEC can handle native proteins (that are still functional on the other side of your column), but the separation is more akin to dialysis, where small and large molecules are easily separated from each other across a cutoff molecular mass, but there is otherwise poor resolution between species.


Should I plate cast iron with copper first before nickel plating? I have some antique handplane lever caps I would like to retore back to their former glory. They are made of nickel plated cast iron. The plating  pin holed and flaked in areas, creating rust spots.

I am planning on stripping the nickel using hydrochloric acid and electrolysis. I've heard that nickel has a much better affinity for copper than iron. Would it be a good idea to first plate the lever caps with copper and then plate them with nickel for a more durable finish?  Any other suggestions for improving the quality if the finish?

The standard technique is to copper plate then nickel ,then chromium, if you want chrome. The copper protects the steel then the nickel provides a silver surface color ,normally significantly thinner than the copper.


What is 'Retention period' in chromatography and what is its importance? I heard people quite often speak about retention time/retention period in Chromatography and Mass Spectrometry. I see somewhere that it is the time duration between the injection time and the target compound peak. I have the following questions:

1)What is the purpose of defining that time?

2)If the compound peak is absent, how one can speak about retention time (I mean if the compound completely fragments)?

3)Is it has importance in mass spectrometry as well?

To answer your three questions in series:

1) The time is a measure of how strong the interaction of a compound is with the column used. The stronger the interaction the longer the compound will stay on the column and this is therefore a way to separate two compounds. For example one molecule with a weak interaction spending a minute on the column, the other spending 5. Collecting fractions of solvent from a liquid chromatography column will yield pure compounds. Note that the time is also a measure for the amount of 'solute' flushed over the column. Many chromatograms also report retention in column volumes of solute. I will not go into detail about that here. It might be worth a separate question.

2) If no peak is found, no retention time can be reported. In chromatography it never (actually almost always) good when a compound fragments/degrades. That means the technique is not suited for purification or analysis. In unique cases, the degradation can be used to analyse whether or not a compound will degrade when interacting with certain materials.

3) In mass spectrometry a retention time in the machine is, as far as I am aware a 'Time of Flight'. Compounds are quickly brought into a gaseous and charged state and are pulled through a magnetic field. Depending on the weight of a compound, the time it flies, is shorter or longer. That is a way of calculating the weight of a molecule. In some cases, for example small proteins, the molecules are purposefully fragmented. The fragments can then be used to reconstruct of what amino acids the protein is made of.

Hope this answers your question! If any clarification is needed, please, do ask :)


Transition state - multidimensional surfaces I've been reading an explanation about transition states in Smith's Organic Synthesis:

  A transition state is an energy maximum along the reaction coordinate. However, a chemical
  reaction where the reactant contains N atoms actually involves movement over an energy surface
  with 3N-6 dimensions, and a transition state is actually an energy minimum in each of the remaining
  (3N-7) dimensions. Liken a chemical reaction to a drive across a mountain range. The “goal” is to
  go from one valley to another, just as the goal of a chemical reaction is to go from one minimum
  on the energy surface (the reactants) to another minimum (the product). The road moves about a
  two-dimensional surface whereas the energy surface for a chemical reaction may involve dozens
  or even hundreds of dimensions. There is no need to climb to the top of a mountain (a maximum
  in both dimensions) to go from one valley to another. Rather, is sufficient to go through a pass
  between mountains (a maximum in one dimension but a minimum in the other). Similarly, it is
  only necessary to pass through an energy maximum in one coordinate (the reaction coordinate)
  for a chemical reaction.
  Because a transition state is an energy maximum (albeit only in one of many dimensions) it cannot
  function to “trap” a molecule, let alone collect a population of “trapped” molecules. This means
  that a transition state cannot even be observed experimentally let alone isolated or characterized.
  Quite simply, a transition state does not exist in the sense that a molecule exists.

This multidimensional explanation completely confounds me. Could someone explain this in a more understandable language? What is this surface with 3N-6 dimensions? How did the author use it to conclude that TS can't be observed experimentally?

The picture below shows a simple potential energy surface with a transition state at the maximum point on the reaction path, which is at the 'saddle point'. The reaction path is the solid black line. (Oscillations in this show that the reactant and product are vibrating). The contours are at different energy with large negative being lowest. This picture is based on a calculation using known experimental parameters.

Along the reaction path the energy is as small as it can be, even though it increases as we go along, because by moving to the right or left (perpendicular to the general direction of the path)  the energy is even higher. So the saddle point is just like a pass between high mountain peaks. At the highest point on the reaction path the only way to lower the energy is to go down, either to return on the original path or to continue to products. The time it takes to cross the transition state is definitely sub picosecond, < $10^{-12}$ s and possibly up to a 100 times faster.

It is not possible to visualise a multidimensional reaction path, so we have to make do with 2D plots, as below, of one reaction coordinate vs. the other with contours at different energy. In principle many similar plots could be needed for a complex reaction but very often most of the atoms in the molecule are bystanders and it possible then to only look at the separation and energy of the atoms in bonds being broken and made.

Trajectory of $\ce {Cl + H2}$ in a gas phase reaction. The H atom approaches the chlorine molecule at the right. The yellow circle shows the transition state.


Industrial synthesis of hydrazine Not a duplicate of Hydrazine synthesis - I know what hydrazine is and am not going near it voluntarily!

Looked at this resource on the Internet which cites three major commercial-scale processes:

$\ce{H_2O_2}$ process (in use at Arkema and Mitsubishi Gas Chemical, developed by Produits Chimiques Ugine Kuhlmann and patented in 1976)
Bayer process (at the eponymous company)
Urea process

What processes do other major large-scale producers use and what are the driving forces behind adoption of a particular hydrazine synthesis? I can guess that Chinese producers may use the urea process, but I'm not sure.

Background: am a Space Exploration.SE regular and am interested in learning more about worldwide production of hypergolic propellants.

The Kirk-Othmer Encyclopedia of Chemical Technology reports that: "About 60% is made by the [Bayer] hypochlorite–ketazine process, 25% by the peroxide–ketazine route, and the remainder by the Raschig and urea processes." Here is a link to the original article, if you can access it (e.g. at a university library): Hydrazine and Its Derivatives, by Eugene F. Rothgery. The data cited dates back to 2004, so it may have changed somewhat, but probably not too much.

Consider this also: the above makes hydrazine hydrate. Propellant-grade hydrazine must be anhydrous, so there is an extra step in which water is removed from the hydrate, typically by azeotropic distillation with aniline.


Clarifications about the mechanism of the Wittig reaction Ok so this is the proposed mechanism for the Wittig Reaction as stated by my textbook, LG Wade. I have some doubts regarding this mechanism:

  1) Why is only triphenyl phosphine used here? Why not some stronger
  nucleophile like say trimethyl phosphine?

  2) Why would the lone pair on P attack the C of the alkyl halide in the
  first place? Isn't the lone pair in resonance with the benzene ring?

  3) Why is butyllithium used particularly.Can't any other reagent be
  used to abstract H ?

  4) Why is the four membered ring formed in step 2?

  5) Why does the ring collapse in step 3?

Note: I'm just a beginner in organic chemistry (Org 2) so it would be helpful if you can keep adjust your answer to suit my level of understanding.

1) Why is only triphenyl phosphine used here. Why not some stronger
  nucleophile like say trimethyl phosphine?

Trialkyl phosphines can particpate in the Wittig reaction just fine, but they create some other problems.  Let's say you were going to create the phosphonium salt from ethyl iodide and trimethyl phosphine.  It would form in high yield, but in the next step where you deprotonate it and form the ylide you would have a problem.  You could form the ylide from deprotonation of the desired ethyl group, or you could form an ylide by deprotonating any of the 3 methyl groups.  This would give rise to a mixture of ylides and consequently a mixture of products.  On the other hand, there are no protons on triphenyl phosphine that can be removed once the phosphonium salt is formed.

The following additional reasons also favor usage of triphenyl phosphine:

it is cheap
it is very cheap
it is a solid and easy to handle and weigh out, many of the smaller trialkyl phosphines are liquids
triphenyl phosphine is less toxic and doesn't smell as bad as many of the smaller trialkyl phosphines

  2) Why would the lone pair on P attack the C of the alkyl halide in the
  first place? Isn't the lone pair in resonance with the benzene ring?

Triphenyl phosphine has the aromatic rings arranged in a non-planar, propeller-like geometry which makes delocalization less likely.  Also $\ce{2p-3p}$ overlap would be involved which is not as effective as $\ce{2p-2p}$ overlap.

  3) Why is butyllithium used particularly. Can't any other reagent be
  used to abstract H?

In the phosphonium salt, the protons alpha to phosphorous are not very acidic (pKa ~ 20-35), a strong base is required to remove them.  Other strong bases could be used, but BuLi is commercially available in high purity at a reasonable cost.  When BuLi abstracts the proton butane is formed and escapes as a gas - one less side product to worry about separating out.

  4) Why is the four membered ring formed in step 2?

The 4-membered ring may not be as strained as you imagine.  Remember, phosphorous is in the third row of the Periodic Table.  It has longer bond lengths and can tolerate smaller bond angles.

  5) Why does the ring collapse in step 3?

The bond between phosphorous and oxygen is an extremely strong bond, in other words its formation is very exothermic.  Therefore, this last step is being driven by thermodynamics.  It is the exothermicity of this step that drives the reaction to completion.


Hydrogen covalent bond It's stated: when two hydrogen atoms form a covalent bond, the distance between them corresponds to the lowest energy. The letters A, B and C show the correspondence between shifts in electron density and the distance apart of the atoms

I am having a difficulty understanding why the figure looks the way it does and what is the energy plotted on the y-axis, can anyone explain?

The energy on the y-axis is the total energy of the system. It uses a reference of E = 0 for when 2 H atoms are infinitely far apart.

As to why the plot looks the way it does...

When the H atoms are infinitely far apart, they are not interacting at all. When brought together, you get a covalent bond which forms. This can be treated at various levels - In the simple classical picture, the 2 electrons are "shared" between the two atoms, leading to lots of electron density in between the atoms, which the 2 protons are then attracted to, and this is what holds the 2 H atoms together. Now we have a slight problem - why do they stop? You could say due to the nuclear repulsions, but if you have a simple situation, using  $\ce {H2+}$:

$\ce{H+}$---e---$\ce{H+}$

With the electron just sat halfway between 2 nuclei. Now let's analyse this situation with just classical mechanics, all units set = 1 for convenience. q = Charge on electron and nucleus. d = Distance from nuclei to nuclei:

Force on each $\ce{H+}: $
$\ F = q^2/d^2 - q^2/(d/2)^2 =  -3q^2/d^2  $

We have $\ce{q^2/d^2}$ term for the repulsion between the 2 nuclei and the negative $\ce{-q^2/(d/2)^2}$ term is due to attraction between nucleus and electron. In this situation the covalent bond results in an attraction of both nuclei together for all values of d - IE eventually we get a singularity. This is why the covalent bond can only really be understood in a quantum mechanical framework.

Taking again the situation of $\ce{H2}$ but this time in a quantum mechanical way.

We start off with 2 hydrogen atoms infinitely far apart. They each have an electron in H 1s, with OPPOSITE spins. When they are brought closer together the wavefunctions change. Now what happens is this. The electron in H 1s of one atom is able to tunnel into the potential well of the other hydrogen atom. By doing this the total volume in which the electron resides is larger - Obviously the electron can go anywhere in the universe, as per usual QM, but the H atom is localised less. This leads to a decrease in the electron's kinetic energy - It is spread out more, and so uncertainty in position is larger, so uncertainty in momentum can be less, and so leads to a lower KE. The potential energy which the electron experiences won't change a huge amount - Yes there are 2 potential wells but it doesn't get closer to them. This is the story at first glance.

However there is the virial theorem which states:
$2KE+PE+R(dE/dR)=0$

So the decrease in KE by bringing 2 H atoms closer together must be balanced out somehow - Either an increase in KE by another means, or an increase in potential energy (PE). If we shrink the 1s orbitals contributing to the bond, this in turns increases KE and decreases PE - PE is negative, and so becomes more so. Thus we find the orbitals shrink in bonding. This all leads to a picture like this:

As 2H atoms draw closer together, electron on one can tunnel onto the other. This leads to a decrease in KE, and as at the equilibrium bond distance $2KE+PE = 0$ this decrease in KE must be balanced out by a contraction in the H 1s orbitals contributing to the bond, leading to both a larger KE term and a more negative PE term. In other words a covalent bond forms because it reduces the kinetic energy of the contributing electrons and increases the potential they feel, due to the contraction of orbitals via the Virial Theorem pathway.

Now back to your plot. At far distances, E=0, because no interaction, and this is taken as 0 point - IE referenced relative to infinite separation. As 2 H atoms brought together, get interaction described as above. At around 2 bond lengths, the KE term reaches a minimum. It then rises - the 2H atoms are so close now that more overlap - IE more tunneling of electron from well to the other well - has no more effect. Instead the wavefunction of the electron is found to contract. This raises the KE term - electron now confined to a smaller space = Larger KE term - but the PE term becomes more negative by a larger amount. At the equilibrium bond length, the virial equation applies as written above:

$2KE+PE=0$

now applies. Thus the KE term is positive with value x, and the PE term is negative, with value 2x. Thus if
$E=PE+KE$
we have a negative binding energy, as expected.

Interestingly the involvement of KE in bonding leads to a new class of bonds, called charge shift bonds - the prime example of which is difluorine, $\ce{F2}$. This molecule has an overall bond order of 1 according to MO theory, but has a large antibonding effect due to filling of antibonding pi orbitals. Calculations using VB theory indicate that the covalent contribution to the structure never actually has a minimum - IE antibonding at all distances. In this case adding in ionic structures is entirely responsible for the bond observed.

This stackexchange answer explains it well, if you're interested, and there's a number of interesting papers listed there as well.

What is charge shift bonding?


Why is there a charge on COOH⁻¹? The dissociation of formic acid ($\ce{HCOOH}$) is:
$$
\ce{HCOOH -> H+ + COOH-}
$$

Why is there a charge on $\ce{COOH^{-1}}$
 though?  Is it achieved through:
$$1\times(\ce{C^{4+}}) + 2\times(\ce{O^{2-}}) + 1\times(\ce{H-})?$$

Does carbon exist as a 4+ ion?

Also, as a somewhat unrelated question, can an acid donate multiple $\ce{H+}$ ions (or can a base receive multiple $\ce{H+}$ ions)?

Thanks in advance!

Not all bonds are ionic bonds. While saying that "Carbon is 4+ valency and is a $\ce{C^{4+}}$ ion" works for carbides like $\ce{Be2C}$, it will not work for a covalent compound like $\ce{HCOOH}$.

Instead of ionic bonds, here we have covalent bonds. Whereas in an ionic bond one atom donates an electron to the other, in a covalent bond, two atoms each contribute some electrons and share them.

For example, with $\ce{O2}$, each atom shares two electrons with the other. Neither "give up" any electrons, they are just shared and contribute to the octets of both atoms:

The formate ($\ce{HCOO-}$) ion is like this:

The central carbon shares one electron with a hydrogen, and two electrons with an oxygen. It also shares one electron with a negatively charged oxygen. This completes the octets (or doublets in the case of hydrogen) of all four atoms.

With electrons drawn on:

Note that while the charges are not of the $1\times(\ce{C^{4+}}) + 2\times(\ce{O^{2-}}) + 1\times(\ce{H-})$ form, the oxidation numbers are.