Derivation of formula for Hydrolysis constant While studying acid-base equilibrium I thought I would derive the formula for hydrolysis constant myself. But I ended up getting a weird result and I am not able to understand where I went wrong.

The equilibrium constants for reactions that tend to favour products are large. As the concentration of the reactants decrease the equilibrium constant slowly should move toward infinity. For example the ionisation of strong acids is taken to be a reaction that goes to completion. According to my understanding the equilibrium constant $K_\mathrm a=\infty$. The equilibrium constant for its reverse reaction would be zero. Similarly the equilibrium constant for the ionisation of salts should be $K_\mathrm s=\infty$.
Let us consider the reaction $\ce{HA + BOH <=> AB + H2O}$. This can be carried out in a number of steps:
$\ce{HA <=> A- + H+}$  with $K_\mathrm a$
$\ce{BOH <=> AB + H2}$ with $K_\mathrm b$
$\ce{H+ +OH- <=>H2O}$ with  $1/K_\mathrm w$
$\ce{A- +OH1+<=> AB}$ with $1/K_\mathrm s=0$
So the hydrolysis constant should be
$1/K_\mathrm h=K_\mathrm aK_\mathrm b/(K_\mathrm wK_\mathrm s)=0\cdot K_\mathrm aK_\mathrm b/K_\mathrm w=0$
hence $K_\mathrm h=\infty$
I am sure there is something I havent understood well and hence I've ended up on this result. Can somebody explain where I am wrong?
Note: In most proofs the equilibrium constant of ionisation of salts is not considered but I see no reason to not consider it.

Sorry, I didn't get that. first of all, which hydrolysis are you talking about? I will talk about weak acid and strong base equilibrium. You have three concentrations [HA], [H+], [A-]. For instance, if they took 1L of 1 mM of HOAc (acetic acid) and added 0.3 mmoles of NaOH (concentrated, so no dilution) you get [A-]+[HA] = 1mM Then you say HA <-> H+ + A- has Ka. Then

[H+][A-]/[HA] = Ka. Then you say  [HA] = x, so [A-] = concentration_total- x (in this case 1mM -x). Then you get an equation
[H+]*[concentration_total-x]/x=Ka. Then you solve this square equation.

Regarding your question. I think you are mixing up to things: protonation of weak acid (stuff, relevant to analytical chemistry with a lot of drama, pH swings etc) and the behavior of other salts (they just dissociate completely and we forget about them). In stories like "to a solution of NaCl and K-acetat a drop of acid was added. What happen?" you should first forget about NaCl. It has no impact on the question. You should also forget about K+. Now you have "to a solution of Acetate- a drop of acid was added. What happen to pH?" Now you can easily calculate it using a square equation.

Run away from chemistry. Carcinogens, stinky chemicals, no  jobs... It teaches you how to make poisons and drugs, but how useful is that?


How much does the quality of glassware matter, really? Discussing about the pros and cons of glassware from different manufacturers for chemistry experiments easily becomes a heated debate. Well, that's how it seems if you read the chemistry related blogs, discussions, questions, etc. over the internet. How much does it really matter what kind or type of glass one is using? How big the differences are across different manufacturers?

It would be nice if the answers also elaborated on the differences between laboratory and home chemistry.

In addition to differences in the absorption spectrum, different types of glass are more or less able to deal with temperature or mechanical shocks. But that is less about manufacturer: you'll find different types of glassware from the same manufacturer for different purposes (e.g. Fiolax and Duran are both by Schott).

As @LDC already commented, some glassware is thin walled and more mechanically fragile. OTOH, thick walled glass is more likely to break with a temperature shock. So e.g. for test tubes you have to decide whether you want to have them

suitable for local heating => thin wall borosilicate glass
suitable for centrifuge => thick wall borosilicate glass
cheap => soda glass
UV transmittant => quartz
suitable for huge temp shocks => quartz

If this question is about whether your borosilicate glass ware is called Duran or  Pyrex (or Jenaer or Ilmabor or Rasotherm) I don't see that much of a difference.

The simple glass is not recomended to use in laboratory IF you are using substances thay may emit Ultra Violet light, because the glass absorbs it and we know that that kind of light has a really high energy. Therefore quartz ( silicon dioxide ) is more suitable. And of course, radioactive elements will need special conditions.


What are the products of the reaction between fluorine and ammonia? What are the products of the reaction between fluorine and ammonia? Is it the same as with chlorine and ammonia, which depend upon which reactant is taken in excess?

Fluorine reacts with ammonia to give dinitrogen and hydrogen fluoride as the major product.

The Formation of Dinitrogen Tetrafluoride in the Reaction of Fluorine and Ammonia - J. Am. Chem. Soc. 1959, 81 (23), 6338-6339 - states that dinitrogentetrafluoride accounts for 6% of the yield while the major reaction is:

  $\ce{3F2 + 2NH3 → N2 + 6HF}$

The same article quotes in its first paragraph that the result remained the same irrespective of the concentration of the reactants. (In the case of chlorine it matters which reactant is taken in excess)

When fluorine reacts with an excess of ammonia it forms $\ce{NH4F}$.
When fluorine reacts with a limited supply of ammonia it forms $\ce{HF}$.
Both these reactions yield nitrogen ($\ce{N2}$) gas.


Why can wood burn all the way to charcoal instead of just creating a passivating charcoal layer? I know burning to charcoal is actually incomplete combustion because charcoal can also burn. When a strike anywhere match burns, it burns all the way to charcoal but the charcoal doesn't further burn. How can all of it burn to charcoal? According to https://en.wikipedia.org/wiki/Pyrolysis, the heat causes pyrolysis which undergoes flaming combustion generating more heat. Once some charcoal has already formed, why doesn't it just passivating the wood blocking further endothermic pyrolysis. I know wood is porous but at the small scale, should't it's surface become passivated by charcoal blocking further pyrolysis? Does it occur for a similar reason that iron can continue rusting after it already formed a rust layer by producing a porous rust layer? Could nonporous single crystal cellulose also burn to charcoal? If so, can you give me a detailed mathematical description of how it would continue burning to create a thicker charcoal layer and how iron can just keep on rusting?

Charcoal doesn't passivate and the results you see depend on the physical structure of the fire

The difference between a match and a larger wood fire has little to do with the formation of charcoal and more to do with the circumstances of the reaction. And the charcoal isn't really passivating in the reaction on a match.

Wood consists of a complex mix of chemicals as well as the core cellulose, which is a carbohydrate. Some of the ingredients, like the sap and other volatile components, burn easily. Cellulose burns in multiple steps one of which involves pyrolysis where some components are driven off quickly leaving mostly charcoal. Charcoal itself will burn but it takes more energy to ignite and therefore more time to get going. It isn't so much that charcoal passivates, it is just that it is the last component to burn.

The difference between a match and a larger wood fire isn't about passivation but about the scale of the fire. There isn't enough energy in a match head to drive combustion all the way so only the easy-to-burn components ignite (unless you hold the match upside down so the flame spreads up the matchstick when the heat from the tip is enough to cause much more extensive combustion). In a larger wood fire there is enough energy in the flame to cause more complete combustion. So the volatile components burn quickly but leave enough heat for the charcoal to burn as well. Since heat penetrates charcoal very well, the volatile components under the charcoal will tend to ignite before the charcoal itself (it isn't doing much to passivate the wood underneath). In a badly organised fire the overall heat may never be enough to set off the charcoal.

So the issue isn't passivation just a reflection of the order the components of wood burn and the physical structure of the fire. Charcoal isn't passivating, just the last component to burn and, if the fire isn't well constructed, there may never be enough heat to burn it. Much more of a match will burn if you drive the flame along the stalk by holding it upside down and a well organised fire will burn the whole way to ash.

Also controlled pyrolysis (which is used to create charcoal deliberately) relies on controlling the amount of air available to the work being pyrolysed. If there were enough air, the charcoal would burn but the conditions for pyrolysis involve restricting the air so the end-product cannot burn. Charcoal doesn't passivate anything when there is enough air (which is why you can cook on a barbecue).

The quandary of why burning wood doesn't burn completely is the heat at which the wood is burning. Wood isn't a pure chemical, but rather has all sorts of chemicals within it. So some of the more volatile chemicals burn off at lower temperatures. In fact it is the volatile chemicals burning in the air that give the "flame"  above the match.

If you put the match residue in a crucible and heated the crucible red hot with a Bunsen burner, then you could burn the match residue completely to a powdery ash.


Is CO2 produced when KHCO3 reacts with H2O? We have been doing experiments on the rate of photosynthesis to find out the importance of $\ce{CO2}$ and we used $\ce{KHCO3}$ in one beaker and found that the plant kept in that beaker produced greater amount of oxygen that those without it. Sources say that $\ce{KHCO3}$ acts as a source of $\ce{CO2}$ but I'm not sure of it. What is the reaction taking place to produce $\ce{CO2}$?

Compiling the comments by Max and orthocresol:

  What is the reaction taking place to produce CO2?

$$\ce{HCO3− (aq) + H2O (l)
<=> H2CO3 (aq) + OH− (aq)
<=> H2O (l) + CO2 (aq) + OH−(aq)}$$

or
\begin{align}
\ce{HCO3− (aq) + H2O (l) &<=> H2CO3 (aq) + OH−}\\
\ce{H2CO3 (aq) &<=> H2O (l) + CO2 (aq)}
\end{align}

  $\ce{KHCO3}$ does not readily decompose into $\ce{CO2}$ when dissolved in water.
  It is just that the plant is using the $\ce{CO2}$, so that pulls the equilibrium over, Le Chatelier's principle.

I am not sure about the previous answers but when KHCO3 reacts with water it forms a complex [K(H20)6]+ and bicarbonate ion. And hence no Carbon dioxide is formed.

But if Potassium carbonate decomposes between 373K and 393K it forms CO2.


Why is o-toluic acid (2-methylbenzoic acid) more acidic than benzoic acid? Why does the methyl group at the ortho-position stabilize the carboxylate ion? I recognize that there will be a steric clash between the carboxyl group and the methyl group but I don't understand why this would help increase acidity. If there was any resonance occurring then this steric clash would DESTABILIZE the anion and make it LESS acidic given that the carboxylate group would not be in the same plane as the ring and thus delocalization, which would stabilize negative charge, would be impossible. Clearly there is no conjugation with the aromatic ring in "benzoic acid based compounds" from the carboxylate group. So why does having a methyl group at the ortho-position increase acidity?

Having one methyl group ortho to the carboxylate group will not prevent the carboxylate group from becoming reasonably planar with the aromatic ring.  If you had a second methyl group in the other ortho position, then planarity would be a problem.

A methyl group attached to a benzene ring is electron releasing in both an inductive and a resonance sense.

Inductively, the electrons in an $\ce{sp^3}$ hybridized methyl group prefer to shift towards the lower energy (more s-character in the orbital) $\ce{sp^2}$ aromatic carbon.
The methyl group is also electron releasing in a resonance sense due to carbon-hydrogen hyperconjugation (see figure below).

The resonance structure pictured below on the left is one of many that we can draw for the o-toluic carboxylate anion.  Note that it places a positive charge on the methyl bearing ortho carbon. Both inductively and through resonance (see the hyperconjugated resonance structure below on the right) the methyl group will stabilize this resonance structure.  It is this stabilization provided by the methyl group that makes o-toluic acid a stronger acid (pKa=3.91) than benzoic acid (pKa=4.19).

Note: As pointed out by both @Mithoron and @GaurangTandon, this answer does not explain why p-toluic acid is a weaker acid than o-toluic acid or benzoic acid. The answers in this earlier question better answer this question.

o-toulic acid is stronger than benzoic acid because the conjugate base of o-toulic acid is stabilized by hydrogen bonding (C–H···O interaction).


Orbital overlap and bond strength In Peter Sykes' A Guidebook to Mechanism in Organic Chemistry, it says that sp3 orbitals have greater relative overlap compared to sp bonds.
We know that the $\ce{C-H}$  bond is shorter in ethyne compared to ethane. So the intuitive idea that more overlap would lead to stronger bond formation is obviously incorrect.
So what is the relation between relative overlap of two orbitals and their bond strength? Is there any use that we can get from this overlap?

I suppose the author was talking about the internal overlap which takes place while the formation of sp and sp3 hybrid orbitals.
What I mean is when one s and one p orbital overlap, the extent of overlapping is less than the overlap that happens when one s and three p orbitals overlap.
To understand it more easily, suppose a ball represent a s orbital while a pencil represent p orbital. Think in mind the extent overlap of three pencils and one ball will be more than the extent of overlap of one pencil and one ball.
But the situation changes, when you overlap two sp3 orbitals or two sp orbitals. sp having more s character is relatively stronger then sp3, thus c-c bond is stronger in ethyne than ethane.


Is it possible to separate/remove sugar from common drinks, such as juice? Is it possible to remove sugar from a drink (for example, juice) using common household equipment?

Why not add yeast to your juice and allow it to ferment?  This will depend on the juice having pH and sugar concentrations; look at commercial wine making kits which provide the proper equipment and recipes. If you conduct the fermentation properly without introduction of air, the sugar will be converted to alcohol, which later can be boiled off (see above) to give the sugar-free version of your juice. Otherwise, with air present, the fermentation will go on to form acetic acid and give the juice a vinegary taste.

It is possible to remove sugar partially by decantation.
Sugar solution is nothing more that a dispersion of the sugar molecules in water due to hydrogen bonding. However, the solution is not perfectly stable since the molecule of sucrose is quite large, therefore we can let it sit on the table for a few hours where most of the denser sugar sink and rest at the bottom. Slowly we can decant the less populated solution into another vessel to have a less concentrated form of juice.


Is beryllium an alkaline earth metal? For context: I got a question asking, "Which of the following alkaline earth metals do not give flame colour?". I quickly marked $\ce{Be}$ and $\ce{Mg}$ and got negative marks.

The following is a quote from my book (emphasis in the original):

  The elements of Group 2 include beryllium, magnesium, calcium, strontium, barium and radium. These elements with the exception of beryllium are commonly known as the alkaline earth metals. These are so called because their oxides and hydroxides are alkaline in nature and these metal oxides are found in the earth's crust.

As you can see, it's stated that beryllium is not an alkaline earth metal!

Is beryllium an alkaline earth metal or not?

There is some disagreement in usage among authors, but IUPAC standard nomenclature approves calling beryllium an alkaline earth metal, as explained on page 51 of IUPAC's last Red Book.

In fact, all the elements belonging to group 2, $\ce{Be,Mg,Ca,Sr,Ba,Ra}$, are called alkaline earth metals with IUPAC's approval.

Other common traditional names approved by IUPAC are alkali metals for the elements of group 1 except hydrogen, i.e. $\ce{Li,Na,K,Rb,Cs,Fr}$; halogens for $\ce{F,Cl,Br,I,At}$ in group 17,whose only member excluded from such a designation is $\ce{Uus}$ (now $\ce{Ts}$), and noble gases for all the elements of group 18 except $\ce{Uuo}$ (now $\ce{Og}$), i.e. $\ce{He,Ne,Ar,Kr,Xe,Rn}$.

I think Be is not an alkaline earth metal as its oxide is not purely basic in nature, it is amphoteric. Also its oxide is not found in Earth crust so it must not be an alkaline Earth metal.


Relative acidity of p-chlorobenzoic acid and p-flurobenzoic acid I want to compare the relative acidity of p-chlorobenzoic acid and p-flurobenzoic acid. Both of the facts below point to p-chlorobenzoic acid being more acidic:

The carboxylic acid group is too far away from the halogen for inductive effect to have any significance. The +M effect of fluorine is greater than chlorine (due to more effective orbital overlap). +M effect is not distance dependant and hence p-fluoroobenzoic acid  should be less acidic since the negative charge on its conjugate base is destabilized.
There is resonance of the negative charge of the carboxylate ion with the empty d-orbitals of chlorine which would stabilize it.

However, in reality, p-chlorobenzoic acid is less acidic than p-fluorobenzoic acid . Why?

$\mathrm{p}K_\mathrm{a}$ of para chlorobenzoic acid $\ce{-> 4.03}$.Source
$\mathrm{p}K_\mathrm{a}$ of para fluorobenzoic acid $\ce{-> 4.14}$.Source

For acidity of para substituted benzoic acids, -I effect is helpful and +R effect decreases the acidity.

In 4-fluorobenzoic acid, the +R effect is prominent in $\ce{F}$ due to stronger $\ce{2p}\pi - \ce{2p}\pi$ overlap with nearest carbon. Thus, due to stronger +R effect of $\ce{F}$ than -I effect, it creates a negative charge on your starred carbon. This charge delocalises in the carbonyl carbon  and the electrophilicity is thus decreased of that carbon. Releasing hydrogen also doesn't help much as the negative charge created over $\ce{O}$ of $\ce{-OH}$ group can only delocalise a little.

Whereas in the $\ce{Cl}$, there is much weaker $\ce{3p}\pi-\ce{2p}\pi$ overlap. So, +R effect of $\ce{Cl}$ is much weaker and due to a relatively higher -I than its +R, there is more positive character on the starred carbon due to $\ce{Cl}$.

And, thus, in reality, acidity of 4-chlorobenzoic acid is a little higher than that of 4-fluorobenzoic acid.

In the case of halogens(on ortho/para), the -I effect dominates over the +R effect and hence due to this the negative charge of carboxylate anion is better stabilized by the flourine substituted compound then the chlorine substituted compound. Therefore, the p- flouro benzoic acid is more acidic than the p- chloro benzoic acid.


Reaction of cyclohexene with hydrochloric acid What happens when cyclohexene reacts with concentrated hydrochloric acid?
I formed the carbocation intermediate by protonating cyclohexene.
But now I am stuck. I know the end product will be a dimer.
I tried to form a bond between the intermediate and cyclohexene like this.

Then through rearrangement the plus charge will shift to 3 degree carbon.
In the answer there is a double bond between the two rings.
That's where I am stuck . I can not think anything further.

Have a look at the mechanism for the dimerization of cyclohexene in strongly acidic medium:

After the attack of a nucleophile (pi bond) on the electrophilic carbocation, we have:

A: joining of the two rings with a bond, and the consequent carbocation (why did it get formed on that position? why not on the adjacent ones?)
B: rearrangement of the carbocation (why did it happen?)
C: loss of a proton to form the thermodynamically most favorable product (most substituted alkene)

The double bond was formed because of the loss of the $\ce{H+}$ ion in step C, formation of a negative charge at that position, and then the delocalisation of that negative charge into the adjacent empty p-orbital (of the $\ce{C+}$ position).

Then after the said rearrangement, visualize the lose of proton from the second tertiary carbon atom, giving you your desired product.


Sodium Peroxoborate reacting with Sulfuric Acid and Potassium Permanganate This is one of the reactions I had to undertake in a lab. I think the reason for the test in the first place is to test for the peroxoborate anion in the solution.

From my understanding what's happening is that sodium peroxoborate will liberate $\ce{H2O2}$ when reacted with sulfuric acid. Adding the $\ce{KMnO4}$, the hydrogen will react to reduce $\ce{Mn(VII) ->Mn(II)} $ and form $\ce{O}$

I'm having difficulty interpreting this into balanced chemical equations. This is what I've tried but not sure if it's correct

$$\ce{Na2[B2(O2)2(OH)4] + H2SO4 ->Na2SO4 +B2O3 +2H2O2 +H2O}$$

This equation shows the liberated $\ce{H2O2}$ but not sure how to write a reaction now where $\ce{KMnO4}$ is added.

Any help would be much appreciated. I've been strugging with this for a while and there isn't much information online.

Well $\ce{KMnO4} $ and $\ce{H2O2}$ react as follows:

$$\ce{2KMnO4 +3H2O2 -> 2MnO2  + 2KOH + 2H2O + 3O2}$$

Adding the two reactions together I get:

$$\ce{Na2[B2(O2)2(OH)4] + H2SO4 + 2KMnO4 + H2O2 ->Na2SO4 + 2H3BO3 + 3O2 + 2MnO2 + 2KOH}$$


Molar enthalpy vs Enthalpy of formation? (Note I've refocused a previous unanswered question into a more general question. You can find the original much more specific question here)
There seems to be two main ways of looking at the enthalpy of substances: molar enthalpy and heat of formation. What are the differences/advantages and disadvantages of each and are either one or more or less suitable for different purposes?

Molar enthalpy is the energy released when the substance undergoes combustion. It is essentially the energy required to fragment the substance into its component sub-compounds.

Heat of formation is the theoretical difference between the energy each atom typically has at STP and the total energy that the compound actually has. Essentially, it is the energy that goes into formulating the bonds between the atoms.

Molar enthalpy is generally used a lot more in experimental chemistry because it can be measured directly, and it gives useful values on the energy that can be extracted from a substance.

Heat of formation is preferred in theoretical chemistry, as it is the theoretical maximum amount of energy able to be extracted in perfect conditions.


Can rate constant for forward reaction be equal to rate constant for backward reaction? For a particular reaction, can we have
$K_c = 1$?

My confusion arised when I was studying the effect of temperature on $K_c$ . I see that the temperature must affect forward and backward rate constants, differently for a reaction so that it changes $K_c$ (which is ratio of the two rate constants). So I thought for a reaction with $K_c$ equal to one, both the rates must be equal, so both will get equally affected by temperature, and thus no change in $K_c$ , or does it have something to do with the Arrhenius equation?

I know that there is a lot of lacking in my intuition as I have just began studying Le Chatelier's principle and all the stuff. So a detailed explanation would be much welcome.

When forward and backward reaction have the same rate, $K_c$ is equal to $1$ in the case of a reaction where the total number of reactants is equal to the total number of products in the chemical equation describing the change. If the total of these stoichiometric coefficients is not the same before and after the horizontal arrow, $K_c$ may be equal to $1$, but must have a dimension, like mol/L or L/mol, or any combination of these units.

As the forward and backward rections have different activation energies, their rates change differently with increasing the temperature. So the equilibrium constant $K_c$ can be equal $1$ at only one temperature.


Is this rearrangement in carbocation possible? Generally, in the following type of reactions, positive charge on $\alpha$-Carbon causes ring expansion, but in the following example positive charge is formed on $\beta$-Carbon and the cation is cyclo-propyl methyl.

So is the above rearrangement even practically possible?

yes,  the rearrangement in carbocation is possible depending upon the condition of reaction and the stability of the new carbocation formed
 eg- the general order of stabilty for carbocation is 3degree >2degree > 1degree
NOTE- the stability of a carbocation depends on aromaticity resonance hyperconjucation inductive effect and sometime ring expansion also occurs


Why do the lone pairs of electrons first occupy axial positions before occupying equatorial positions for species with octahedral geometry? Why do lone pairs not reside in equatorial positions in species with octahedral geometry?

The main reason is repulsion with the electron pairs. Compare $\ce{SF6}$ &  $\ce{SF4^2-}$ as an example. To get from $\ce{SF6}$ to $\ce{SF4^2-}$, we need to replace two fluorines with lone pairs. The first replacement is arbitrary; in an octahedral geometry, all the positions are equivalent. Once we have replaced that first fluorine however, the bond positions are no longer equivalent. We will have four equatorial bonds (90 degree separation from the lone pair) and one axial bond (180 degree separation from the lone pair). Lone pairs are very repulsive, especially with respect to other lone pairs, so we want the next lone pair to be as separated as possible from the first. The axial bond is further from the lone pair then any of the equatorial bonds, so it makes sense for the lone pair to go there.

The VSEPR theory tells us that the order of repulsion is :
lone pair-lone pair >lone pair-bond pair >bond pair-bond pair.
It is clear that you need to place the lone pairs such that they have minimum neighboring bond pairs or lone pairs. An equatorial position in an octahedral geometry has 5 neighboring bond pairs/lone pairs ( 3 in the equatorial plane and two in the axial plane ) while an axial position has 4 neighboring bond pairs/lone pairs (4 in the equatorial plane, the other axial position is not a neighboring position).
That is the reason why lone pairs first occupy axial positions in an octahedral geometry. Hope this helps.


Difference between precipitation and crystallisation What is the main difference between precipitation and crystallization.
Is precipitation a chemical change and crystallization a physical change.
How can we say that the resulting solute is precipitate or a crystal?
Is it necessary that when we want precipitates to form then its ksp should be less than its Qsp or the precipitates form even after mixing two unsaturated aqueous salt solutions?

The two words are often used interchangeably (perhaps incorrectly), though there is a slight nuance.
IUPAC defines the following:

Crystallisation: The formation of a crystalline solid from a solution, melt vapour or a different solid phase, generally by the lowering of the temperature or by evaporation of a solvent.
Precipitation: The sedimentation of a solid material (a precipitate) from a liquid solution in which the material is present in amounts greater than its solubility in the liquid.

In essence, a crystallisation is the formation of a crystalline solid (i.e. not amorphous), for instance leaving a beaker of $\ce{NaCl}$ in water out for a long time will eventually grow crystals as the water evaporates.
Precipitation can be thought of a something 'crashing out', for instance if you have certain kinds of amines, you can add $\ce{HCl}$ in diethyl ether to precipitate out the $\ce{HCl}$ salt of the amine as they are not soluble.
There are of course in-between cases. Attempts at 'crystallisation' can yield amorphous solids if the conditions are not right, and, equally, precipitation can sometimes give crystalline solids.
The choice of words is therefore always ambiguous, but in general the IUPAC nomenclature should be followed where appropriate (eg. you can never crystallise to give an amorphous solid).
To answer the other part of your question about chemical vs. physical changes, you are correct that crystallisation is always a physical change (different isomorphs of the same material: amorphous, crystalline needles etc). Precipitation can be either, for instance in the $\ce{PPh3}$ case described by Jan, or in the $\ce{HCl}$ amine salt I previously described.

Crystallisation is formed from a super saturated solution while precipitation is formed due to reaction of 2 substances to form sparingly soluble product


Why does the ionization energy decrease anytime the atom size increases? I would like to have a better understanding of the relationship that exists between the Ionization Energy commonly known as IE and the size of an atom.

I think that the Ionization Energy can be defined as the smallest energy that is required to remove an electron from a gas phase atom.

Why does the Ionization Energy decreases when the atom size increases?

The ionization energy is the lowest amount of energy required to remove one electron(the most loosely attached electron) from each atom in one mole of an element in gaseous state to form gaseous (positive) ion.

There are certain factors that have an impact on the ionization energy value.

To understand the answer to your question, you must understand the effect of all three

Atomic radius (Inversely proportional to I.E. Value)
Shielding effect of electrons (Inversely Proportional to I.E. Value)
Nuclear Charge (Proportional to I.E. Value)

The Atomic radius is the distance between the to-be-removed electron and the nucleus. The greater the distance, lesser the force of attraction. Hence as this distance increases the electrons are easier to remove; lesser force is required to remove it from that attractive field of the nucleus. (This answers your question) Even the smallest change in this distance will cause a change in ionization energy for example The first ionization energy of Magnesium and Aluminum. The ionization energy should increase as the nuclear charge increases across the period but the I.E. of Aluminum is lesser than that of magnesium because there is a change in orbital which increases atomic radius, decreasing I.E.

Now going for a bit more information.
The Shielding effect of the other electrons decreases the I.E. because it doesn't let the outer most electron feel the full force of attraction.

IF both the first factors are CONSTANT (atomic radius and shielding effect of electron) only then the last factor comes in play.
As the atomic charge/Nuclear charge increases, there is a general increase in ionization energy because of the increased force of attraction by the extra proton. for example, again across the period, the general trend is increase in ionization energy because there is no (significant) increase in atomic radius or shielding effect so the I.E. increases proportional to the Nuclear charge.
(the decrease in sulfur is because of the pair of electron in the p orbital repelling each other. Both electrons apply a force on each other. The looser of the two, is removed when energy is applied.)

The electric field around an atom with one electron removed is that of a single positive charge.  The work done, to lift an electron away from a small radius to infinity, is greater than the work done to lift an electron from a large radius to infinity.

That work is the ionization energy.   How could it NOT go down for
larger atomic radii?


The naming of benzene Why is benzene not called as a cyclo triene? Wikipedia says benzene is a IUPAC name, but isn't it a common name?

It would be wrong to say cyclotriene, rather cyclohexatriene would be an appropriate name. And it would also be a highly systematic IUPAC name for said compound, following the rules of e.g. the cyclopentadienyl anion, the cycloheptatrienyl cation, cyclooctatetrene and more.‡

Let’s look at benzene’s history. It was discovered as early as 1825 by Michael Faraday who called it bicarburet of hydrogen. In 1833, Eilhard Mitscherlich isolated it and named it benzin. In 1836 it was isolated by Auguste Laurent who called it phène. The name benzene also dates back to the first half of the 19th century and, like benzin, derives from the gum benzoin. Note that this already predates the founding of IUPAC by about 80 years. The first IUPAC nomenclature was not established until 1921.

At the time when it was discovered, chemists had no clue as to what structure it had. It wasn’t until 1865 when Friedrich Kekulé published his ground-breaking paper[1] on the structure of benzene which in the end turned out to be correct (-ish). At that time, nomenclature of chemicals was still the Wild West, since there was hardly any way for structure elucidation and thus most chemicals acquired trivial names — some of which survive up to date (acetic acid being a prime example).

When IUPAC finally derived comprehensive nomenclatures for organic (and inorganic) compounds, and when structure elucidation was finally good enough to prove structures, the names benzene, benzin and possibly also phène had been in use for many decades, much like many other trivial names. Had IUPAC decided to obsolete them all altogether, they could have just as well trashed their own recommendations, because nobody would have followed them — even today, many chemists are too traditional to take up newer nomenclature schemes even if they have been around for two decades.

Probably to help gain acceptance, IUPAC has always allowed a certain set of retained names to persist under official nomenclature. While the list has gotten smaller over the years, it is extremely unlikely that benzene, phenol, phenyl etc. will ever get removed from it, since calling those compounds that way is so prevalent — and short.

Similar things happened to carboxylic acid. Hardly anybody says methanoic acid or butanoic diacid in place of formic acid and succinic acid. And sometimes, trivial names even become part of the official nomenclature, as was the case with fullerene because it provided a much shorter way to label a certain class of compounds.

So the short answer to your question: IUPAC has retained a common name. The reasons are tradition, prevalence and simplicity.

Note:

‡: Yes, as a chemist I know that the structure of benzene is incorrectly represented by drawing three pure double bonds and three pure single bonds, that in reality the molecule has a very high $D_\mathrm{6h}$ symmetry rendering all bonds equal and the electrons in the π system are delocalised all across the molecule. However, that is not what the question is asking about, and not something IUPAC nomenclature cares about.

Reference:

[1]: F. A. Kekulé Bull. Soc. Chim. Paris 1865, 3, 98.

Cyclohexa-1,3,5-triene does not exist.

It is just a hypothetically possible structure for the formula $\ce{C6H6}$.

The whole discussion is intended to explain that if it were to exist it would have such and such properties but it does not exist.

Instead we have benzene.


How does one calculate chemical shift in a gauge independent manner? The literature on ab initio calculations of chemical shift in NMR experiments usually provide Lamb's and Ramsey's formulae as the solution. Yet the expressions explicitly depend on vector potential gauge choise. Some authors claim that the theory is intrinsically gauge invariant, giving no proper proof yet. Can the solution be expressed in an explicitly gauge invariant form? Some literature on this.

NOTE: I know about London orbitals theory, yet it's still not fully understood by me, how it can deliver a type of expression I want. Some suggestion on where this is clarified are very welcome.

The references below describe the answer.

Prog Nucl Magn Reson Spectrosc. Author manuscript; available in PMC 2012 May 1.
Published in final edited form as:
Prog Nucl Magn Reson Spectrosc. 2011 May; 58(3-4): 176–201.
Published online 2010 Dec 15. doi: 10.1016/j.pnmrs.2010.10.003
PMCID: PMC3058154
NIHMSID: NIHMS248338
Chemical shift tensors: Theory and application to molecular structural problems
Julio C. Facelli

Nucleic Acids Res. 2014 Dec 16; 42(22): e173.
Published online 2014 Nov 17. doi: 10.1093/nar/gku1006
PMCID: PMC4267612
Accurate ab initio prediction of NMR chemical shifts of nucleic acids and nucleic acids/protein complexes
Andrea Victora,1 Heiko M. Möller,2 and Thomas E. Exner1,3,*

A comparison of models for calculating nuclear magnetic resonance
shielding tensors
James R. Cheeseman, Gary W. Trucks, Todd A. Keith, and Michael J. Frisch

J. Chem. Phys. 104 (14), 8 April 1996


What to do with (large) imaginary frequencies for constrained minimum structures? I am performing DFT calculations using ORCA 4.0.1 on an enzyme active site model. The model contains 89 atoms including the substrate (see Animation 1), five of which are fixed in space (the spherical atoms in Animation 1). I am using dispersion-corrected B3LYP with 6-31g(d,p) basis set, with the "RIJCOSX" approximation, and with a CPCM model for the solvent (dielectric constant $\epsilon = 4)$. From what I have seen in the literature, this method is often applied to enzyme models with success, and this is the reason for using it. Once I start computing reaction paths, I plan to try different functionals and basis sets to see if this give drastically different results. My geometry optimization tolerances are

Energy change: 5e-6;
Max gradient: 2e-3;
MRS gradient: 5e-4;
Max displacement: 4e-3;
RMS displacement: 2e-3

The two gradient tolerances have been loosened slightly in order to converge the calculations. By using the default gradient tolerances, the calculations simply do not, I have found (especially the maximum gradient). Additionally, it has been absolutely necessary to compute a better guess for the initial Hessian for the optimization, done at the B3LYP/3-21G-rijcosx level.

The fixed atoms are kept fixed to maintain the orientation of the active site residues. Only atoms where the side chain or backbone has been truncated are kept fixed. Without fixing these atoms, the residues will most likely  deviate way too much from the crystal structure, and no longer represent the enzyme environment we try to simulate.

Now, due to the constrained atoms, I expected imaginary frequencies to occur. However, for one of my presumably converged structures, I have quite strong imaginary frequencies. The list of frequencies start as follows: -220.645934, -69.117105, -65.597557, -61.314719, -48.720497, -40.439949, -32.562415, -21.395791, -16.637113, 6.126145. The two strongest negative frequencies are mainly a sigma bond rotation of a methyl group (of which the carbon is fixed in space during the optimization).

It is difficult to find any good literature on what to do with and how to interpret such negative frequencies in constrained optimizations. Is it "safe" to disregard the frequencies that mainly involved fixed atoms? Should I converge my geometry tighter to come closer to the true minimum? That's not so easy, I find, as the molecule is quite flexible. It took me quite some time to come up with a method that converges, which involve computing a good guess for the initial Hessian used in the optimization (computed at the B3lYP/3-21G level). I hope the imaginary frequencies are artefacts of the constraints, and that I can safely disregard them. However, they seem to have quite higher frequencies than what other people report in the literature (many say that their imaginary frequencies are around 20 cm-1).

This has to be a common challenge for people doing electronic structure calculations on large and constrained systems, but very few address this issue specifically in their publications.

I did read through this question, but it addresses not quite the same issue that I face.

Animation 1. The active site model. A total of 89 atoms.

Animation 2. Imaginary mode with frequency -220.645934 cm$^{-1}$.

What I am trying right now

I have three jobs waiting in the queue right now:

I have made a small rotation of the methyl groups strongly involved in the strong imaginary mode
I have started another geometry optimization of the structure given above (exhibiting the imaginary frequencies), but by reading the computed Hessian (the "exact" one, from same level of theory as optimization)
I have restarted an identical optimization, but by computing the "exact" Hessian from the start, giving a much better guess of which direction to go in.

I can add here later what these attempts result in.

The first I noticed is you said:

  The two gradient tolerances have been loosened slightly in order to converge the calculations. By using the default gradient tolerances, the calculations simply do not, I have found (especially the maximum gradient).

At first sight, the structure of your molecule should not leads to serious problems, but the above is a bad signal, and turns the results unreliable (at least doubtful). The constrains must be considered in the calculation of vibrational frequencies for they make sense.

Maybe no one can give you a foolproof steps series to follow. My suggestions are below.

What to do/check

If comments are right, and Orca does not care about fixing atoms for
the calculation of the vibrational frequencies, then you need to
change the software for this task (because in this case results are not meaningful) (This also answer the question "Is it "safe" to disregard the frequencies that mainly involved fixed atoms?"). The simpler way to check this is by looking at the number of freedom degrees reported (or deduced form the output). I was referring directly to vibrational f.d. because low frequency modes may be treated like rotations. I doubt that Orca lacks of this basic functionality. In such case I bet that this is described their User Manual.
The step you are doing are reasonable. My guess is that maybe you are far from a true minimum, in some unstable geometry, and that is why you have problems with accuracy changes. You said: "Should I converge my geometry tighter to come closer to the true minimum? ". Yes, it is always advisable to get good frequencies. So I would use not only tighter geometry convergence criteria but the TIGHT convergence option for the SCF part.
In the same lines, you can try using modern semi-empirical methods and then make the fine tuning with DFT. Even using a simpler DFT functional.
I agree with comments that you are not using the best basis set out there, but I think that this is not a problem here.
If you want to discard weird behaviors you can try with another functional, but I do not think that this is the problem (because it is known that B3LYP performs well in compounds with the elements involved here).

In brief. I would check if Orca compute correctly the frequencies, if not: change the software. If yes: discard those results, try to converge the structure using cheaper methods (if you are lack of computational resources) from different initial conditions, store those results and start new geometry optimizations and vibrational frequencies calculations starting from those previous converged states. The last step using the tight convergence criteria for SCF and normal convergence criteria for geometries, and using the target basis set (for energy and its derivatives).

Good luck

Note:

  Once I start computing reaction paths, I plan to try different
  functionals and basis sets to see if this give drastically different
  results.

Why do you expect that something like this would happen? You can save time by using a little large basis set and a better functional once you get the B3LYP/6-31g(d,p) results. It should not take too much time.

EDIT DUE TO COMMENTS

This answers generated some discussion about B3LYP or not B3LYP?

In that respect, my viewpoint is as follows:

B3LYP is not the most accurate functional (there should be no doubt about that).
It is reasonably good for geometries in systems like yours. Errors in geometries can be larger due to small basis set than due to B3LYP, or at least of the same order.
While there are functional with better results for geometries, you also must take care of how large basis sets those functional needs to be a better option. I mean, maybe using 5$\zeta$ basis sets they are better, but not using a small basis set.  (and maybe not)
The above discussion can be very long, but in any case, you are using approximations that (in my opinion) are much stronger than the errors due to  B3LYP. For example, the permittivity value, or the geometry of the constained parts, etc..
Also, even if you could perform a FullCI/CBS you shoud notice that the simulated system is different to the real system, so extremely exacts results in the simulated system may no improve your knowledge about reality.
So, if the functionals are accurate enough, what it is important here is obtaining the corresponding results for the functional/basis set. The functionales can be more or less problematic. B3LYP should not be problematic for your system. My guess is what I said in the answer.
Because of the above, I think that B3LYP is reasonable and enough for your case.

I do many calculations on the active site model, but using Gaussian instead. Like what you have also mentioned in the question, the constraint on the truncated atoms caused some imaginary frequencies. However, I was greatly surprised by the -200 cm-1. Try to optimize the structure and calculate the freq at the level B3LYP/6-31**. I would expect much much smaller imaginary there.

update: I thought the calculations were done at the level of B3lYP/3-21G. Sorry about that. Have you found the solution now?


Estimating for atomic number for an "unknown element" A questions tells me there's this element with 3 isotopes (270.51 amu, 34.07% abundance; 271.23 amu, 55.12%; and 269.14 amu, 10.81%), and to solve for average atomic mass. I have done so and got 271 (270.758) for its atomic mass. Then it asks me to estimate its atomic number, and its location on the periodic table. How do I exactly "estimate"?

ps. I dont think this element actually exists...

Well I found it out. It is seaborgium atomic number 106, mass number 271.

To be found in period B6, group 7.


Why is Cr²⁺ is a good reducing agent but Mn³⁺ is a good oxidising agent? The question is the same as the title states. In my textbook,[NCERT Chemistry I for Class 12 , pg no 217.] the following is written

$\ce{Cr^{2+}}$ gets converted to $\ce{Cr^{3+}}$ as the +3 oxidation state has half-filled $\mathrm{t_{2g}}$ orbitals; thus it is a good reducing agent.
On the other hand $\ce{Mn^{3+}}$ gets converted to $\ce{Mn^{2+}}$ as its +2 oxidation state has a half-filled d-subshell.

Now my questions are:

Why can't the $\mathrm{t_{2g}}$ argument be used for $\ce{Mn}$?

Why can't the half-filled $\mathrm{d}$-orbital argument be used for $\ce{Cr}$?

$\ce{Mn^3+}$ could be oxidized to $\ce{Mn^4+}$, which would have a half-filled $\mathrm{t_{2g}}$ set. As oxidation and reduction both can take place in aqueous medium, why wouldn't $\mathrm{t_{2g}^3}$ or $\mathrm{d^3}$ configuration be more stable than $\mathrm{d^5}$ configuration in aqueous medium?
A similar observation is made in case of $\ce{Fe^2+}$ and $\ce{Cr^2+}$, where it observed that $\ce{Cr^2+}$ is more powerful reducing agent than $\ce{Fe^2+}$ (due to reasons explained above).
If possible, I would like to see the crystal field splitting calculations that show this to be the case.

Why can't the $\mathrm{t_{2g}}$ argument be used for $\ce{Mn}$ ?

Because the number of $\mathrm{t_{2g}}$ electrons is not changing upon going from $\ce{Mn^3+}$ to $\ce{Mn^2+}$.  In both $\ce{Mn^2+}$ and $\ce{Mn^3+}$ there are 3 $\mathrm{t_{2g}}$ electrons.  I think the point they are trying to make with $\ce{Cr^3+}$ being half filled is that there are 3 $\mathrm{t_{2g}}$ electrons and no $\mathrm{e_g}$ electrons.  $\ce{Mn^3+}$ and $\ce{Mn^2+}$ both have $\mathrm{e_g}$ electrons.

You could make a similar argument for $\ce{Mn^3+}$ being oxidized to $\ce{Mn^4+}$, which would have half filled $\mathrm{t_{2g}}$ and empty $\mathrm{e_g}$.

  Why can't the half-filled d-orbital argument be used for Cr?

Because neither $\ce{Cr^3+}$ nor $\ce{Cr^2+}$ has a half filled d subshell.  $\ce{Cr^3+}$ has 3 electrons while $\ce{Cr^2+}$ has 4 electrons.


How is drift due to static electricity eliminated from an analytical balance? I've recently encountered a problem when using an analytical (0.1 mg accuracy) scale to determine the mass of a sample of ferrocene powder. The reading in the balance seems to drift either upwards or downwards when I drop some of the powder in an erlenmeyer flask. The reading stabilizes when I add water to form a solution.

I've heard that static electricity might cause this kinds of problems when measuring powders so I suspect this is the case here, because the problem only appears when I add the ferrocene powder. I've try reaserching some ways to get rid of the problem, a static gun seems to work in some cases (no idea if it would work with a powder though) and I've heard that air ionisers also work, but don't have access to either of these.

Although this is the first time I've encountered this situation, I've heard its pretty common, so what do people usually do to get rid of the problem in a cheap manner? Is there an easy solution?

I'm using gloves to operate the scale, which I've been told to do in order to keep the flasks clean of oil from my fingers. Can the gloves be worsening the problem? What can I do about that?

Static charge can affect the measurements made by an analytical balance, as can magnetic materials because the mechanism in modern analytical balances involves a feedback loop between the load that is measured and an electromagnet. Insulating weighing vessels (plastic or glass) tend to accumulate static charge, especially when handled with another insulator (gloves) by the triboelectric effect.

You should first make sure that the body of the balance is grounded properly according to the manual for your model.

As a short-term solution to static charged glass, you might try arranging the sample so that it is father away from the balance. One way might be to place the sample on top of an empty sample bottle on the balance. As you mentioned, a more robust solution for discharging static build-up is to have a commercially available air ionizer nearby. The device will either produce an equal amount of positive and negative ions in the air at high voltage AC or DC electrodes, or with an ionizing radioactive material such as is found in anti-static brushes used for cleaning vinyl audio records.

If you are frequently measuring magnetic materials, many analytical balances can be configured in a bottom-loading arrangement. In this setup a platform is hooked beneath the balance, and the sample is supported sufficiently far away from the force-restoring motor so that it does not influence the measurement.

I have been using gloveboxes for several years in different labs across the world and I find that Ar gloveboxes completely eliminate the problem of static electricity commonly experienced with N2 gloveboxes. I have no explanation for this explanation but I am pretty sure N2 is the culprit because we once switched an N2 box to Ar and the problem disappeared. Before that we spent our time fiddling around the balance with static guns, air ionisers and such, without much success (the only exception being the Po ionisers which are pretty efficient). On the other hand, when purging an Ar box with N2, static electricity will usually appear for a couple of days. In my opinion the price difference between N2 and Ar is simply not worth the trouble.


Shape of PEG in water According to the chemical formula $\ce{H−(O−CH2−CH2)_n−OH}$ of polyethylene glycol, or PEG, I assume that this organic compound has a long chain structure when it is dry and in powder form. (This is also what guys in the chemistry field say.)
But when PEG is dissolved in water, would it form some structure like a cluster or would it remain a long chain?
Can we somehow calculate the diameter of a dissolved PEG molecule when $n$ is given? Or the dissolved PEG molecules band randomly so there are no calculations for an average diameter statistically at all?

PEG is the term usually reserved for shorter chains, while PEO is  used to refer to longer chain molecules. When polyethylene glycols are dissolved in water, they increase the viscosity of the solution. While it might be easy to model short-chain polymers in dilute solution as some kind of folded chain approaching a spherical volume, as the chains get longer and the solution becomes more concentrated (>1%), there is significant entanglement.

The picture above is for a long-chain polyacrylate, which can be adjusted to be very hydrophilic by adding alkali to generate anions all along the chain. The images show various shapes in water: some fairly tight, some looser (spread out), some spread out but very much entangled. At some point in increasing chain length or increasing concentration, it becomes difficult to project the image of a single molecule of PEO into useful predictions of solution properties. Measurements of dilute or shorter chain molecules would include viscosity; as the solution becomes more concentrated or the chain longer, the property includes some elasticity, so we have viscoelasticity, and at very high chain lengths, we have pituosity (more commonly known as stringiness or ropiness). A video shows this effect dramatically:
https://www.neatorama.com/2017/10/09/Fun-with-Polyethylene-Glycol/.
Since PEGs and PEOs are so soluble and compatible with water, the first molecules to be added just fill all the volume of water; you could calculate how long a PEO molecule would be if completely stretched out. Stirring can extend the molecule, showing effects due to viscoelasticity and pituosity.


Chemical methods to extract Oxygen from Rust Let's say I have a large quantity of rust and want to extract the oxygen from it in a way that allows me to collect the oxygen in a container.

What chemical process can I use to do this and what equipment does it require?

It's preferable if the process is doable at home so I can demonstrate the process but it's not mandatory.

Good question; unfortunately, no good answers. The problem is that oxygen, pure oxygen, doesn't like to exist as such; it will form a diatom like many gases, but it's very electronegative, and so it will prefer to get the extra electrons from something more willing to donate them, like the iron. This is why "oxidation" has that name; until fluorine was isolated in the mid-1800s, oxygen was the strongest known electron acceptor ("oxidizer"). It's easier to run a reaction that gives the oxygen something else to oxidize, producing the raw metal, than to get the iron to complex with something else and release the oxygen. In at least one circumstance, this is very good for humans and other aerobic organisms; oxygen complexes with the iron in your hemoglobin for transport to your cells. In most other circumstances, we could do without this behavior.

Rust is insoluble in water but will dissolve in strong acids. It will react with hydrochloric acid to form iron chloride and water (sometimes producing hydrogen gas if there isn't enough oxygen to go around). The easiest "household" method to obtain oxygen from iron oxide, therefore, is to dissolve the rust in muriatic acid forming iron chloride and water, neutralize any remaining HCl by adding lye to form sodium (and/or potassium) chloride and water, distill the water from the chloride salts by boiling and then cooling the vapor, then electrolyze the water with a battery to produce oxygen and hydrogen. That's cheating, though. First, water's water, and you're adding far more to the party with the acid and base solutions than you would ever get from the rust. Second, muriatic acid and lye, while available in relatively pure form for pool maintenance and for soapmaking respectively, are not exactly everyday household chemicals in those forms (both are components of other household chemicals, such as drain cleaner and lime scale remover, but are somewhat difficult to isolate from them).

Theoretically, if you heat the rust enough, the components will dissociate and you'll be left with pure iron and oxygen. Unfortunately, even with carbon in the mix (producing elemental iron metal and carbon dioxide, which is the reaction behind iron ore refinement and recycling), the minimum temperature required for this reaction to become favorable is 842K = 569*C ~= 1056.2*F. I can't find data on iron-oxygen bond enthalpy or entropy, which is needed to calculate the spontaneous decomposition temperature of the compound, but logic states it's much higher than with added carbon, because you're not getting any energy release by forming the carbon-oxygen bond, so every Joule of energy needed to break the iron-oxygen bonds has to come from outside the material. Theoretically it would cost you about 2.5kWh/kg of heat added to the iron; in practice the amount of heat you'd have to generate (and lose to various inefficiencies) to get the metal that hot is beyond any household device's capabilities, unless your hobby happens to be glassblowing.

The "thermite reaction" between iron oxides and aluminum under heat is another well-known method of separating the iron and oxygen. The rust and aluminum powder, given some initiating heat, will exchange the oxygen to form alumina (Al2O3) and pure iron, and enough heat to keep the reaction going (with plenty to spare; a thermite reaction is too bright to watch without a welder's mask and in fact is used in some situations to weld iron). Two problems with this; first, it's illegal in the US to make thermite without BATFE approval, and second, the reason it works at all is because aluminum oxide is ridiculously stable, more so than iron oxide, and so it takes even more energy to reduce the oxide back down to the raw metal (about 8.2 kWh/kg for aluminum, theoretical, versus 2.25kWh/kg for iron). This is evidenced by the highly exothermic nature of the thermite reaction itself; if creating aluminum oxide produces far more than enough heat to smelt the iron to get the oxygen, think of how much would be needed to go the other way.

Pretty much the only direct route to releasing oxygen gas from rust without a blast furnace is to introduce an even more powerful oxidizer. There are very few known to modern science, and none of them are "household chemicals" to say the least. The easiest to work with would be elemental fluorine gas, and the experienced chemists in the crowd will know that for the extremely relative statement that it is. Fluorine is the most electronegative element in the known universe, and so pretty much any substance that isn't already highly fluorinated (like some polymers, such as Teflon) will react pretty vigorously with it. That includes the water, proteins, enzymes and other compounds in your own body; a fluorine gas leak is a "drop everything and run" type of emergency. As used on rust, the fluorine will simply barge in and displace the oxygen from the rust to form a white to pale green iron fluoride, liberating the oxygen as a gas. You then have to separate the oxygen from the excess fluorine, and one of the standard ways in industry to catch excess fluorine is to pass the exhaust gas over or through a powdered sodium oxide filter, which will form sodium fluoride and liberate more oxygen in a very similar reaction as you got with the iron. Sodium oxide isn't a pleasant chemical to work with either; add water (such as by breathing in any loose powder) and it forms the strong base sodium hydroxide, which eats flesh about as efficiently as anything else you could think of.

Two variations on the acid dissolution:
a. Dissolve the rust ($Fe_2O_3$) in 100% sulfuric acid, producing ferric sulfate with some enneahydrate $Fe_2(SO_4)_3.9H_2O$. When the reaction is done, evaporate the $H_2O$ and electrolyze it.
b. Use 100% phosphoric acid, get $FePO_4.2H_2O$. When the reaction is done, evaporate the water and electrolyze it.
And a third (pure speculation):
c. This way probably won't work, but  organometallics do all sorts of crazy stuff. Ferrocene can be made by heating iron with cyclopentadiene at 300 C (https://en.wikipedia.org/wiki/Ferrocene). I think it might also work with iron oxide, perhaps even better (or not).
$Fe_2O_3$ + 6 $C_5H_6$ --> 2 $FeCp_2$ + 3 $H_2O$ + $(C_5H_5)_2$
or something like that. Collect the water and electrolyze it to get the oxygen. Quite doable in a lab - if it works.


how to mathematically predict reaction results I've been trying to computationally achieve the following (I have no lab, I'm 14 years old, and I want to make a Minecraft mod to teach real Chemistry):

I want to mathematically predict the range of temperature and pressure (if the reaction is between gases) or temperature and concentration (if the reaction is in a solution) in which a given chemical reaction can happen, and the rate at which it happens, with the variation of those two arguments.

EDIT 1:

Regarding the first part of @sixtytrees answer I suppose I must calculate $K$.

I assume $\Delta G$ would be:

$$\begin{align}\Delta G &= \Delta G_\mathrm f^°{_\text{methanol}} - \Delta G_\mathrm f^°{_\text{carbon monoxide}} - 2 \Delta G_\mathrm f^°{_\text{hydrogen}}\\[6pt]
&= -162.5 - 137.3 - 0\\[6pt]
&= -299.8\ \mathrm{\frac{kJ}{mol}}\end{align}$$

Now to solve for $K$:
(I believe $R$ must be $0.0083144598\ \mathrm{\frac{kJ}{mol\ K}}$, because $\Delta G$ is in $\mathrm{\frac{kJ}{mol}}$
Also, I think we have to take $25\ \mathrm{^\circ C}$ as the temperature which is $298\ \mathrm K$)

$$\begin{align}\ln K &= \frac{\Delta G}{-RT}\\[6pt]
&= \frac{-299.8}{-0.0083144598 \cdot 298}\\[6pt]
&= \frac{-299.8}{-2.4777090204}\\[6pt]
&= 120.99887336713996006405304847878\\[6pt]
K &= \mathrm e^{120.99887336713996006405304847878}\end{align}$$

And now I realize I must have made some mistake because this number will be too big.
☹

EDIT 2:

I just woke up and I noticed I miscalculated $\Delta G$, because I added them together when I should be subtracting. Let's see how it goes now.

$$\begin{align}\Delta G &= \Delta G_\mathrm f^°{_\text{methanol}} - \Delta G_\mathrm f^°{_\text{carbon monoxide}} - 2 \Delta G_\mathrm f^°{_\text{hydrogen}}\\[6pt]
&= -162.5 - (-137.3 - 0)\\[6pt]
&= -25.2\ \mathrm{\frac{kJ}{mol}}\end{align}$$

Now $K$ will be:

$$\begin{align}\ln K &= \frac{\Delta G}{-RT}\\[6pt]
&= \frac{-25.2}{-0.0083144598 \cdot 298}\\[6pt]
&= \frac{-25.2}{-2.4777090204}\\[6pt]
&= 10.170685820053125395644218884808\\[6pt]
K &= \mathrm e^{10.170685820053125395644218884808}\\[6pt]
&= 26126\end{align}$$

Still not as small as 10 that @sixtytrees predicted.

$$26126 = \frac{[\ce{CH3OH}]^1}{[\ce{H}]^2[\ce{CO}]^1}$$

What now?

EDIT 3:

I've been thinking, and decided Chemical Equilibrium is not something I need to bother with, because as the reaction progresses, it is assumed that the final products are drained out as they are formed, and additional pressure is added to account for the reduction of total molarity. This way I don't need to distinguish between reversible reactions, irreversible reactions, and virtually irreversible reactions (when there are trace amounts of the reactants in the final products).

This only works if the reactants are gases, I don't know how to sustain the rate of the reaction if I take away the products from a solution.

The original question remains, now that equilibrium is irrelevant, is it possible to predict the temperature and pressure/concentration range of the reactants in which the reaction can occur?

I downloaded Orca, by the way. No idea how to use it.

EDIT 4:

Ok, so let´s say this is all happening in a sealed container with 1L of volume, so

$\begin{align}
26126 &= \frac{[CH_3OH]}{2^2 \cdot 1^1}\\
[CH_3OH] &= 26126 \cdot 4 = 104,504
\end{align}$

What does this mean? It can't mean I end up with more than one hundred moles in 1 liter. It makes no sense, I don't have enough reactants to get so much product. So, I am clearly doing it all wrong, can someone explain it to me?

There are two parts to quantification of reactions:

(1) How far it goes? (Thermodynamics, equilibrium). This all is dictated by ground state energy.

(2) How fast it goes? (Kinetics, reaction rates). This is dictated by  transition state energy vs ground state energy.

Equilibrium is much easier to understand. Let [Y] be concentration of reagent Y. You say that for 2A + B = C+3D +4E you have:

K = [C]$^1$[D]$^3$[E]$^4$/([A]$^2$* [B]$^1$)

Note how powers are equal to coefficients in the equation.

$\Delta$G= -RT*ln(K)

$\Delta$G is Gibbs energy (measured and tabulated for all of inorganic chemicals from school textbook).
R is a constant (Universal gas constant)
T is temperature.

How to use it? Well, imagine that you have

2*H2+ CO <-> MeOH

Lets say that K=10 under that conditions. I am sure I am off. Correcr way to find it is to use $\Delta$G= -RT*ln(K). You know T, R is constant. $\Delta$G is known for MeOH, H2 and CO. Calculate the difference to find $\Delta$G. You will need to find $\Delta$G for methanol vapor or take it for liquid MeOH and add $\Delta$G of evaporation.

You start with 1M of CO ( atm at 273C) and 1.5M of H2 (67.2 atm at 273C). Total pressure 112 atm. (Mol to pressure conversion using ideal gas law). What will be the equilibrium amount of MeOH if ?  You say that

Let x be the amount of product at equilibrium. H2 = 1.5 M - 2*x and CO = 1-x.

x/(1.5-2*x)$^2$*(1-x)=10=K

Now you have a cubic equation to solve (1.5-2x)$^2$*(1-x)*10 - x = 0;

With more reagents you have algebraic equation of higher power, but there are methods to solve those numerically and the idea is simple enough to explain.

This gets a bit more complicated when you have multiple reactions going. But This is the basis. When system gets complex you start ignoring some small members to simplify it. (1.0001)$^5$ ~ 1.0005 (reasonably accurate). YOu do the same in chemistry.

Kinetics is trickier. For A+ 2B -> C the first approximation is that

rate = k*[A]*[B]$^2$.

Also "k often increases by a factor of 2-4 per 10C of increase in temperature".

Now to treat it accurately you need to solve a system of equations. each equation is either

r$_i$ = $\delta$[A$_i$]/$\delta$t

or

r$_i$ = ($\delta$[A$_i$]*$\delta$[B$_i$])/$\delta$t

Rate if i$^{th}$ reaction = change of concentration of i$^{th}$ reagent per time period as the period goes to zero).

You end up with a system of ordinary differential equations. But don't get tricked by the word "ordinary". Most of them don't even have solutions in "elementary functions" (functions that you study in school or university as a chemist.

To understand what is a differential equation you would need to know what is an integral. We study differential equations in second year of university and that probably the most difficult subject in university.

Judging by your attempts it looks like you haven't learnt about thermodynamics  yet nor calculus nor advanced probability. The fundamental  problem in your attempts is that you don't take in account the energy transfers, activation energy, etc. These energy changes that occur are scientifically represented in curves, you are trying to solve a certain part of that curve. Therefore, you would use a combination of differentiation  and integration to do that. Secondly, you see if reactions  were mathematically deterministic then there would be no use for a different language for chemistry (I assume you know basic chem). Finally, chemistry is not structured as maths, in other words it is just a hodgepodge of substance that react under the right circumstances.


Comparing the bond strength between a polar bond and non-polar bond According to NCERT class 12 part 2, $\ce{ICl}$ has lesser bond enthalpy than $\ce{I2}$. I have read that non polar covalent bonds are weaker than polar covalent bonds as in polar covalent bonds partial opposite charges develop on the bonding elements which leads to extra electrostatic interaction between them which is absent in non polar bonds. So according to what I read the answer should have been opposite. Please let me know where this logic is wrong. Also using this logic I understood that ionic bond is stronger than covalent bond.

In polar covalent bonds, say $A–X$, assume electronegativity of $X$ is greater than that of $A$ then their is polarisation of bond $A–X$, $X$ gains partial negative charge $\delta–$ and $A$ gains partial positive charge $\delta+$ but this charge is very less (as electrons of sigma bond are held tightly) to account for any significant increase in bond strength of $A–X$ bond but the $–X$ group in $A–X$ molecule tends to attract $A–$ group of another $A–X$ molecule in this way there is electrostatic interaction between two different $A–X$ molecules giving rise to intermolecular forces (Dipole-Dipole interaction, hydrogen bonding etc).

DO NOT CONFUSE
FORCES BETWEEN TWO DIFFERENT MOLECULES AND COVALENT BONDS WHICH EXIST IN A SINGLE MOLECULE. THEY ARE TWO COMPLETELY DIFFERENT THINGS

However, attraction or repulsion may exist between two parts of a single molecule (generally happens for organic  molecules) but doesn't significantly affects bond strength between them rather they tends to change orientation of those groups or may rotate about the sigma bond.

And for $ICl$ and $I_2$, bond of iodine and chlorine is weak due to polarisation. And bond between iodine and iodine is strong covalent bond. I hope you know that polarisation of electron cloud of sigma bond decrease overlapping between orbitals.


Methods of preparation of propane My book States part C as the answer of the following question.

But according to me, part D, that is soda lime decarboxylation can also be a perfect answer besides the Corey House synthesis( that is the part C )the answer given by my book.
Please help in clearing my misconceptions.

Simple carboxylic acid or carboxylate ions (like CH3CH2CH2COO–) do not undergo decarboxylation to yield corresponding alkanes because in this case the leaving group would be a carbanion (in this case CH3CH2CH2–) which is very strong base and hence a very poor leaving group (the reason behind this is that strong base wants to give electrons, but leaving group has to take electrons and leave the compound, strong base is poor leaving group and weak base is good leaving group)

For decarboxylation to happen a keto group must exist to stabilise the negative charge, which is not present in salt of butanoic acid CH3CH2CH2COO–Na+

Yield of propane is very less if you chose decarboxylation, that is why Corey-House synthesis is best for preparation of propane.

I want to know why do you think that Option D is correct, mention in comment.


Are acid chlorides really acidic? If I test a pure solution of an acid chloride with a pH-meter, will I get a reading of pH 7?

When I add water to it, I know $\ce{HCl}$ gas will form, but this gas will dissolve in water to give an acidic solution.

My main confusion is that we call ethanoyl chloride an "acid" chloride, but is its pure solution acidic?

An acid chloride is a covalent molecule.  It has no pH.  If you put a pH electrode into a reactive acid chloride you will dehydrate and acylate its hydrated glass surface ($\ce{#Si-OH}$) and damage its future response when properly used.  pH electrodes are stored sitting in buffer or saline solution.

An acid chloride is an oxo-acid whose acidic $\ce{-OH}$ has been replaced with $\ce{-Cl}$.  Hydrolysis restores the oxo-acid, displacement makes derivatives. Hydrochloric acid is not an acid chloride.

Is there such a thing as a completely pure solution ? Would there not always be some water contamination and hence would you not always have some protons floating around ?

However, as a thought experiment I imagine that in a solution of 'pure' ethanoyl chloride you would have a pH of 7 as there are no free protons moving around in solution.


Availabilty of magnesium ions in phosphate buffer Phosphate buffer is a common buffer in biological applications, it's especially popular for NMR. Magnesium ions are necessary for many biological systems to work, though the amount of magnesium ions you can add in a phosphate buffer is limited by the relatively low solubility of magnesium phosphate.

What I'm curious about is whether the phosphate buffer can affect the availability of magnesium ions even in concentrations where no visible precipitation occurs? Common phosphate buffer concentrations are around 10-100 mM, and magnesium ions are usually 5 mM or below. At those concentrations I see no visible precipitation, but is the magnesium actually fully available or is the phosphate sequestering it and thereby lowering the effective magnesium ion concentration in the solution?

Even just calculating the solubility of magnesium salts in phosphate buffer is not straightforward, as there are multiple species of magnesium phosphate and possibly even other salts like $\ce{MgKPO4}$.

Does phosphate buffer affect the availability of magnesium ions in solution compared to a non-interacting buffer? If that is so, is there a way to calculate the size of that effect?

If you're still interested in this, fluorescence may be a good method to test the phosphate dependence on magnesium activity. First record a fluorescence spectrum of a solution of magnesium and fluorescent dye, then add small volumes of concentrated phosphate and note any changes in the fluorescence. Of course, you would have to do a control experiment to see if the phosphate is directly perturbing the fluorescence. This write-up describes some magnesium-chelating dyes that you could use, though if you don't want to spend a lot of money on a dye, you may already have some chemicals in your lab that will work. Phenol looks like promising alternative as it fluoresces, is decently water soluble (83 g/L), and forms a complex with magnesium which breaks planarity, which would certainly change the fluorescence spectrum.

It would only take a day to get accurate quantitative results if you have access to a fluorometer.

I don't have a source (at the moment) but it is accepted that high phosphate buffers do chelate/sequester Mg2+ ions. Pyrophosphate is particularly infamous for this. During RNA transcription experiments, the solution will get cloudy which is due exactly to this phenomena. It is also seen in cell-free protein synthesis which requires a steady concentration of Mg2+, that the increased concentration of the phosphate buffer requires an increased amount of Mg2+ to maintain the same effective free Mg2+ concentration [1].


How is exact racemization possible? A racemate consists of 50 % $d$ and 50 % $l$ forms of an optically active compound. But how can someone ensure exact 50% quantity comparing molecule by molecule? There will always be some difference in number of molecules in both forms of optically active compounds. Is this sufficient for the mixture to rotate the plane polarised light in clockwise or anticlockwise direction?

According to wikipedia "Racemization can be achieved by simply mixing equal quantities of two pure enantiomers."

Racemization isn't "exact," but rather very very close to equality. It is just simple probability.

Think of flipping a coin, p=probability for heads, and q=probability of tails. Now for a fair flip p=q=0.5. From binomial theory  the standard deviation is $\sqrt{n\cdot p \cdot q}$ where n is the number of flips. Now let's assume 2 standard deviations difference, which is roughly at the 95% confidence interval.

If you flip 10 pennies then a two standard deviation difference is $2\times \sqrt{0.5^2\times 10} \approx 3$ in the number of heads.

Now flip $6.022\times10^{23}$ dimes then a two standard deviation difference is $2\times \sqrt{0.5^2\times 6.022\times10^{23}} \approx 7.8\times10^{11} $ in the number of heads.

But now think of the % difference.

$3$ heads in 10 tries for the pennies is $30\%$.
$7.8\times10^{11}$ more heads when flipping $6.022\times10^{23}$ dimes is only a difference of $1.3\times 10^{-10}\%$ which is an insignificantly small difference.

Stereoselectivity is a very important factor in organic synthesis. Most of the natural chemistry in the world, at least the world that involves biology and chemistry, is chiral, favoring one enantiomer over another. Speaking from the perspective of pharmaceuticals, chirality is critical to biological activity with respect to therapeutic benefit or toxicity. It is often desirable to produce one enantiomer in large excess to increase the yield of the desired "useful" product and reduce expense, waste, or extra steps required to isolate or purify the product, especially when multiple steps are required in synthesis and even small losses accumulate with each step.

In the case of many reactions, the ratio of stereoisomers produced can be random, leading to a near 50:50 split. In other cases, factors such as the intrinsic chirality or sterics of one or more reactants may influence the enantiomeric ratio. In some cases, one enantiomer is preferred almost exclusively because a reactant cannot approach the reaction site because of steric hindrance or factors including molecular conformations, such as hydrogen bonding, solvent, or temperature.

Books have been written about this topic, as well as numerous individual publications and review articles. A few references that might be helpful are listed below.

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

https://www.researchgate.net/publication/316322015_Principles_of_Asymmetric_Synthesis

https://www.researchgate.net/publication/262607017_Editorial_Thematic_Issue_Enantioselective_Synthesis_in_Organic_and_Medicinal_Chemistry

http://www.carborad.com/Volume%20I/Chapter%2011/Stereoselectivity.pdf


Why are there 14 electrons in the third shell of iron atom? The third shell must contain at least 18 electrons. But in iron there are 14 electrons in third and 2 electrons in 4th. Why are there 14 electrons in the third shell of iron atom?

Your doubt is - Total number of electrons in 3rd shell  (i.e. electrons from 3s + electrons from 3p + electrons from 3d) is not yet 18 but still how we filled up the 4th shell?

You have a general misconception than atomic orbitals always follow the 2,8,18... electron filling rule. According to this 2,8,18.. electron filling rule, 2 electrons should get filled up in 1st shell, then 8 electrons in second shell, then 18 electrons in 3rd shell........

Remember this rule mentioned above is completely elementary! Just remember 2,8,18... electron filling rule is actually a waste. Simply I would recommend forget about that 2,8,18... rule. It is very elementary rule... simply forget it. In 2,8,18... electron filling rule, "lower energy level orbitals gets filled up first"- this statement is not considered in that elementary 2,8,18..rule. But this statement which that elementary rule doesn't consider is very important statement to be followed while filling up of electrons. So remember 2,8,18... electron filling rule is an elementary rule for beginners who  just start to read about atoms and orbitals! Forget about that rule.

Its not necessary that 3rd shell has to have 18 electrons for 4th shell filling up to start.
The electron filling depends upon the energy levels of the orbitals. Lower energy orbitals get filled up first.

Iron's configuration is: 1s2 2s2 2p6 3s2 3p6 4s2 3d6. You can note from the configuration, 4s gets filled up before 3d . The reason behind this is lower energy orbitals are filled up first. Even though 4s belongs to 4th shell, still its energy is less than 3d which can be reasoned out by (n+l) rule. 4s energy is less than 3d, hence the two electrons which you expected to be going to 3d goes to 4s before 3d filling starts. Hence you end up having 14 electrons in 3rd shell and 2 electron in 4s. Remember 4s got filled up before 3d and the reason is 4s energy is less than that of 3d.

According to the Bohr model of atoms, the maximum number of atoms which can  be accommodated in the outermost shell is 8. That's why in the atomic shell diagram the 2 electrons are in the outermost shell.


What is the reaction mechanism of a selective catalytic reaction with urea? What is the reaction mechanism of the following reaction:

$$\ce{4NO + 2(NH2)2CO + O2 -> 4N2 + 4H2O + 2CO2}$$

I expect that the catalyst is vanadium oxide and titanium oxide.
Where can I find a detailed description of the reaction mechanism?

After researching this topic, I've found the mechanism to be:


Sulphonation of benzene Will nitrobenzene and benzene have the same rate towards sulphonation reaction, since the formation of sigma complex is not the RDS of this reaction?

Nitrobenzene and benzene would NOT have the same rate towards any type of electrophilic substitution reaction as strength of nucleophile - (nitro)benzene does matter.

Nitro group as in nitrobenzene is highly deactivating group due its electron withdrawing nature and therefore decreases the nucleophilicity of benzene ring.

IJOART_document


How to create black precipitate in clear liquid I am not a chemistry person at all, so my apologies if my question seems dumb.

I am working on the set for a short film and in one of the scenes we would like the main actor to look at a sample of water, drop something into it and watch a black precipitate form. Clouding the water.

I have looked online to see what reaction would yield the desired result, however, my chemistry knowledge is very basic.

Could somebody here help me out and let me know what I could use? Safely.
I understand that we need to handle all chemicals with care.

I can think of iron(II) sulfide($\ce{FeS}$). There are many ways to make iron(II) sulfide but one reaction procedure will suit your purpose. It uses iron(II) sulfate and lime sulfur(calcium polysulfide) to form a clear black precipitate of iron(II) sulfide with a clear supernatent. See this e-book excerpt.

You can use other sulfides as well. See this picture (full picture here):

From left to right: silver sulfide ($\ce{Ag2S}$, black), lead sulfide ($\ce{PbS}$, black), nickel sulfide ($\ce{NiS}$, black-brown), copper sulfide ($\ce{CuS}$, dark indigo). These are formed by reacting $\ce{Na2S}$ with corresponding metal ions. If you want to fully make your solution black, then your options would be to form silver sulfide and nickel sulfide. You could also make your solution black by forming elemental mercury solution but that would be too toxic for your purpose.

Other ways include adding iodide to excess amount of bismuth(III) nitrate  in dilute nitric acid but be careful not to overdo as excess concentration of iodide will make solution orange-red(See here).

If you do not want any reaction to happen then just dump manganese(IV) dioxide in water. They are black and insoluble.

This is not a chemistry answer, but it is not a comment either?  Please remove if it doesn't fit here.

Since you are not familiar with chemistry, why risk it?

I would recommend just dripping dark food coloring or ink into the water, and using post-processing to either replace the drop with another shot of plain water dripping into the beaker, or modifying the color.  There is likely to be a very clear cutoff (especially if filmed parallel to the surface of the water) and a limited time range.  This could be done without expensive software or special skills.


Stability of formyl chloride Why is formyl chloride unstable? I know it is just a combination of CO and HCl. But why it is unstable whereas other acyl chlorides are stable, like acetyl chloride?

It is unlike other acyl chlorides as formyl chloride dissolves in water. At the same time, it reacts vigorously with it, and no other acyl chlorides produce carbon monoxide and hydrochloric acid.


Why are acid catalysed reactions of carbonyl compounds reversible? I am currently studying aldehydes and ketones and their reactions. I've noticed that acid catalysed reactions are represented with reversible arrows, while base catalysed reactions are shown with regular arrows. Why is this so? Is this because the addition of the nucleophilic base stabilises the electrophilic carbon? Or is there some other reason?

Update: Sorry, "all" would not be appropriate here. In my text-book, it is mentioned only once. Nucleophilic substitution by weak nucleophiles is carried out in a weakly acidic medium, while a basic medium is used for strong nucleophiles. The corresponding mechanism use the convention I've spoken about. I can't see the use of reversible arrows elsewhere. I wanted to know why is the nucleophilic addition reversible (as per my book) in a weakly acidic medium.

My text book (not a standard one): Nootan ISC Chemistry XII

The pictures of the reactions (I'm not able to rotate them for some reason)

This is a picture from wiki (Simple aldol reaction). Both Acid and base catalyzed reactions are reversible up to aldol intermediate. Dehydration is the only irreversible step. So this statement " I've noticed that all acid catalysed reactions are represented with reversible arrows, while base catalysed reactions are shown with regular arrows. " is incorrect.

Single arrow means that the step is irreversible. Reversible arrow means that reaction can go both forward and backward. Usually reversible steps are happening before the irreversible step and a happening faster than irreversible step.

This is easy to demonstrate with a pool table that nave a small obstacle near the hole. Reversible reactions are balls that move over the the surface of the table. They can get from any point to any other point of the table. Sometimes they get on that small obstacle near the hole and roll back to the table (they were not able to overcome activation barrier). Irreversible reactions are represented by balls that were able to overcome the barrier and now ended up in the hole. Now they have no chance to go back to the table on their own.

What makes reaction irreversible vs reversible. From thermodynamics perspective irreversible reactions are reactions with a large change in $\Delta$G. From organic chemists perspective reversible reactions are typically those that include protonation/deprotonation steps. Irreversible reactions are those where a strong bond like C-C or O-P forms.

Reversibility is a subject to conditions. 2H -> H2 (recombination of radicals) is effectively irreversible at room temperature, but is reversible at 3000K.

Why do we need to account for reversible reactions? (1) To understand the mechanism. (2) To know how to increase concentration of the reagents right before the first irreversible step to accelerate the overall reaction.

tl;dr: in reactions you are asking about formation of C-C bond is effectively irreversible step.


What is the logical way to understand Iso, Sec, Tert and Neo prefixes? I'm only familiar with Alkanes at a basic level currently.
What seems logical to me is that an Alkyl molecule with the:

"iso" prefix means its an isomer of that alkyl, the usual way i'd see it would be the alkyl group with a substituent (the way an hexane could be isopentane if  I got it right)

as for the "sec" prefix, the alkyl group is connected to the main molecule through a secondary carbon (meaning the carbon on the alkyl is bonded with two other carbons in the alkyl, no consideration given to the main chain here)

I find the "tert" prefix similar to the latter, it is connected to the maine chain through a tertiary carbon (bonded with three carbons on the alkyl)

I couldn't find a basic logical way to address the definition of the "neo" prefix.

Are the ways I described the rest of the prefixes correct?

The prefixes, iso-, sec-, tert-, neo- are all traditional names when structural organic chemistry was still developing. You have already rationalized all except neo-. Neo- is Greek term for new. I looked up the OED etymological dictionary for older meanings, and the earliest use is traced to 1876. Now you can see in no 2, that neo- was a good option.

Used in the names of chemical compounds, minerals, etc., which are closely related to, or new forms of, known substances.
neocatechu  n. (Obsolete) a kind of catechu.
Used in the names of compounds and radicals in which one carbon atom is linked to four others.
Not part of standard chemical nomenclature although well established in neohexane and neopentane (and related terms).
(a)
Categories »

† neo-acid  n. Obsolete
1876   W. Odling in London, Edinb., & Dublin Philos. Mag. 5th Ser. 1 216   The names which call to mind that the normal, iso-, and neo-acids are mono-, di-, and trimethyl & etc. derivatives would seem to be most advantageous.

Neo- even it is used today represents a terminal tert-butyl group, just like the prefix iso- represents a iso-propyl group at the end.

IUPAC names have superseded all.

According to my chemistry book (O. P. Tandon's Organic chemistry) ,if the prefix Iso is present in any compounds name then one methyl group is attached to second last carbon atom of the compound .

If Neo is present then two methyl group is attached to second last carbon atom of the compound.

Sec, Pri and Tert represents the carbon atom to which the functional group is attached to I.e., secondary, primary and tertiary carbon respectively.

Cis and Trans can also be seen as prefixes, trans isomer has zero dipole moment since the atoms present are symmetrical.

The letter n can also be seen which means normal, the compound is straight.


Gattermann-Koch Reaction with Phenol Substrates The Gattermann-Koch Reaction involves the formation of an aryl aldehyde using carbon monoxide and hydrochloric acid as reactants. The reaction is catalyzed by aluminium trichloride in the presence of trace amounts of cuprous chloride.

What I don't understand about this reaction is that according to Wikipedia along with other online texts I've used, the Gattermann-Koch reaction is inapplicable to phenol substrates. The drawback has been left unexplained by all my resources (online and textbooks) and even my professor was oblivious to it.
I am unable to understand why the reaction doesn't apply to phenol and what exact side reactions might occur.

Additionally, why is the explanation so obscure?

[EDIT] I found the explanation. The Gattermann-Koch Reaction is not applicable to phenol and phenol ethers because they "could not be successfully formylated at atmospheric pressure in benzene as a solvent."
"This failure to react was attributed to the insolubility of the cuprous chloride in the reaction mixture."

Can someone please explain this in simple terms?

You have outlined that the formyl chloride is generated as the reactive, electrophilic intermediate under these conditions.

You are planning to perform an electrophilic aromatic substitution, but in a phenol, there's high electron density on the $\ce{O}$ atom. Isn't it conceivable that your main reaction product is phenylformate here, resulting from $\ce{O}$-acylation?

I suspect that, in the presence of the Lewis Acid catalyst, even if a formyl ester is formed, it will rearrange the the acylbenzene, which is irreversible. This may have consequences for the o,p-product ratio,
And as far as the cuprous chloride, use aluminum trichloride. And you could make the Scylla bromide first, with Formica acid an bromine. Much more reactive will likely form the sculpture ester first, in catalyze. Then catalyst to rearrange it


How are cathode rays produced? Why cathode rays are produced if pressure of the gas discharge tube is lowered to about $10^{-4}$ atm?
How bombardment of glass of gas discharge tube from cathode ray result in a faint glow?

"Cathode rays" are simply beams of electrons. The name exists because cathode ray devices predate the discovery of the electron. The sharp shadows produced by Crookes tubes meant that something was travelling in a straight line from the cathode down the tube, so they were named cathode rays.

The vacuum doesn't have much to do with the production of electrons, per se, but without high vacuum, the electrons collide with gas particles before they travel very far. If there's enough gas, you get a plasma rather than a beam of electrons. The electron beam is produced by thermionic emission in hot cathodes—electrons are liberated from a hot filament when the thermal energy is enough for electrons to exceed the filament's work function. In cold cathodes, electrons are released by field emission, where a large applied electric field allows electrons to tunnel away from a metal electrode.

The glow of the glass envelope is luminescence—either fluorescence or phosphorescence (or both). Basically, the electrons striking the glass excite components of the glass to an elevated electronic state which releases light upon returning to the ground state. Normal glass isn't particularly luminescent so tubes designed to produce light (like a CRT) are coated with phosphors.

Cathode rays are the streams of fast moving negatively charged particles originated from the cathode.

At the pressure of 1 mm Hg in the tube, the positive column breaks into alternate and dark dics, then the cathode leaves the glow, this is cathode rays.


Can the number of intermediates for a reaction be more than the total number of transition states? I'm getting results from a transition state guess software, in which the number of intermediates are two and there's only one transition state for a reaction, is that possible?

Let's assume your reactive complex (reactants interacting) are optimised, i.e. its gradient equals zero, and your potential energy surface is continuous.

Then any direction is either uphill or constant (otherwise the optimisation hypothesis would not apply).
There are three possibilities: (a) going from reactants to intermediate gives a transition state which has higher energy than both reactive complex and intermediate; (b) reactants and intermediate lie on a plateau where energy does not change; or (c) one hypothesis is wrong.

So yes, for barrierless processes.

Completely normal, especially for a bimolecular process forming and adduct like that. However, you shouldn't be applying transition state theory for this kind of process - variational methods (minimising the reaction rate) are more accurate.

These systems can be treated very efficiently using energy grained master equation with RRKM theory, for example software packages look at Mesmer or Paper.


Number of hydrogen bonds of organic compounds? How to count the organic molecules ' hydrogen bonds in average?

What are the numbers of hydrogen bonds formed among alcohols ,alkanoic acids ,aldehyde , ketones , amine ,amide , and ether when they are dissolved in water?

Hint you could write down general formula for eg for alcohols it $C_nH_{2n}O$ and also  due to covalent bonds the hydrocarbons arent dissolved in water.


What happens when hydrogen bonds stretch? I'm researching about the Mpemba effect and I came across an explanation saying that the effect occurs because at higher temperatures, the hydrogen bonds between the water molecules stretch and store energy and when the water starts to cool, the hydrogen bonds go back to normal and release the stored energy on top of the normal cooling process, that's why hot water freezes faster than cold water.
I'm quite new to hydrogen bonds, so I don't really understand how hydrogen bonds can store energy. I thought bonding was something like a magnet attracting something, the further you get, the weaker the attractive force becomes. Or am I understanding it wrong?
The article that I saw this in is Hot water freezes faster than cold water; physicists solve the Mpemba effect

Yeah this question is actually surprisingly difficult to to answer.  But let me explain it this way.   In thermodynamics you can't have negative temperature.  Thus anytime you are getting heat, you are actually gaining electrons.  When you gain electrons these chemical bonds change and rearrange themselves, which is how they can store energy.
We do not think about this normally because we look at hydrogen and we just see a proton and electron.  But in reality, unless hydrogen is in an extreme environment, it likely has absorbed energy in some manner.
For instance,  if you put a glass of water under concentrated sunlight it will absorb photons, these photons will be absorbed by the moving electrons inside the atom. Which we understand as "heat", moreover If there are more electrons than water molecules than it will start to evaporate because they run out of room.