Trifluoroiodomethane in basic medium $$\ce{CF_3I + NaOH(aq) -> Major Product}$$

The Major product, according to me, should be ditrifluoromethyl ether $\ce{CF3-O-CF3}$. The hydrogen on the $\ce{-OH}$ group is quite acidic and hence reacts with $\ce{CF3I}$ to give a stable compound according to the following reaction mechanism:

However, the textbook gives the following reaction mechanism and claims that the product is trifluoromethane $\ce{CHF_3}$

Which one of these is correct?

You didn't give a mechanism for your expected product, so I can only speculate as to how you intended to form the final product.

If you thought to use an Sn2 mechanism, I'd say that the reaction is highly improbable as the halide is attached to a trisubstituted carbon, i.e. a carbon with a lot steric hinderance, and therefore, it would be rather difficult for an Sn2 reaction to take place.

If you chose to use an Sn1 mechanism: with the first step, when iodine leaves, you would create a primary carbocation, which is unstable. Additionally, being that fluorine is an electron withdrawing group, the three fluorines would likely serve to further retract electrons from the already electron-poor carbon, and not serve to stabilize the primary carbocation. Therefore, I doubt that this reaction would occur as well.

In the textbook provided mechanism, the oxygen attacks the accesible iodine, which is next to a very psuedopositive carbon. And although this places a  negative charge  on the carbon atom, the electron-withdrawing fluorines serve to help stabilize the carbanion via their inductive properties. In the second step, the carbons rids itself of its negative charge, by taking a hydrogen atom from a nearby hypoiodous acid (pKa 11).

References:
McMurry, J. (2015). Organic chemistry: with biological applications. Stamford, CT: Cengage Learning.

The second pathway looks to be correct because :
1. The carbanion formed is considerably stable
2. $\ce{S_{N^2}}$ dosen't look favourable in the first pathway(in the second step in the first reaction)
The first pathway on the other hand dosen't look to be energetically favorable neither by $\ce{S_{N^1}}$ nor by $\ce{S_{N^2}}$ as the carbon is way too hindered and the supposed carbocation that would form in the $\ce{S_{N^1}}$ mechanism would not be stable in the slightest due to 3 highly electronegative $\ce{F}$ atoms  attached to it


4-Bromo, 2,4-di-methylhexane and 3-Bromo, 3,5-di-methylhexane are the same thing? Came across the following question. The answer only says that option 3 is incorrect which is I agree to but dont you think option 4 is also incorrect. As per IUPAC rules isnt it the Halide that should get lowest possible number (in this case number 3 instead of 4) or is it both are same ? I am confused please help.

Both names, ‘4-bromo-2,4-dimethylhexane’ and ‘3-bromo-3,5-dimethylhexane’ are unambiguous and describe the same compound. However, only ‘4-bromo-2,4-dimethylhexane’ is in accordance with IUPAC recommendations.
First of all, the seniority order of functional classes, which is used to choose the functional class (here: …hexane), does not apply to numbering of locants within the same compound class.
Concerning numbering of locants within the same compound class, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(…)

Furthermore,

P-14.3.5 Lowest set of locants
The lowest set of locants is defined as the set that, when compared term by term with
other locant sets, each cited in order of increasing value, has the lowest term at the first
point of difference; (…)

Therefore, the example is correctly named as 4-bromo-2,4-dimethylhexane rather than 3-bromo-3,5-dimethylhexane since the locant set ‘2,4,4’ is lower than ‘3,3,5’.

Acc. to IUPAC we have to give lowest locant no. to side chain which in this case is 2-Bromo-2,4-dimethyl.
Another reason of this ans. being correct is that we take halides as a normal side chain and it is not given priority over methyl, we just only see lowest locant in this case.


Acidity of o-halobenzoic acids Rank the following compounds in decreasing order of acidity:

2-Fluorobenzoic acid
2-Chlorobenzoic acid
2-Bromobenzoic acid
2-Iodobenzoic acid

My approach
Since the ortho effect is proportional to the size of the substituent, it will be the greatest in the case of iodine and smallest in the case of fluorine. So, the molecule will be more non-planar in 2-iodobenzoic acid.
So resonance will be inhibited in all the above compounds. But what will happen to the acidity?

Acidity data:

The true order of acidity is $\ce{Br = I > Cl >> F}$. Relative to unsubstituted benzoic acid, all of the ortho-halogen substituted benzoic acid derivatives have a greater acidity, clearly having some stabilising effect on the resulting benzoate anion.

  Data retrieved from Journal of Research of the National Bureau of Standards, 1958, 60, p569 - research paper 271. $\mathrm{p}K_\mathrm{a}$ values measured in water.

Rationalisation:

Polar effects All of the halogens exert a negative inductive (-I) effect, removing electron density from the ring. This makes the benzoic acid derivatives more acidic in all 4 cases since the resulting conjugate base is stabilised to a certain extent by the presence of the electron withdrawing halogen.
Steric effects As the halogen gets bigger, strain may force the carboxylate out of planarity with the aromatic ring. Whilst fluorine is roughly the same size as the hydrogen atom, bromine and iodine are both sterically more demanding.
Internal H-bonding This is suggested by another answer below, whilst this has been studied and shown to operate for protic groups such as alcohols and amines, in halogens it has little impact on stabilising the negative charge.

Overall, steric and polar effects operate in tandem to give an overall acidity for the benzoic acid. It's perhaps not intuitive which one will dominate (e.g. fluorine is far more electronegative, but also far smaller than iodine, so which is more acidic).

The table below, taken from the cited paper, attempts to separate out steric and polar effects through calculations on both the acid and its conjugate base. Sadly, they only look into $\ce{F}$ and $\ce{Cl}$ (no $\ce{Br}$ or $\ce{I}$), but it can be clearly seen that:

Hydrogen bonding is making no significant contribution to the acidity of the substituted benzoic acid
Polar and steric effects are both in operation, with sterics being far more important in the ground state (the carboxylic acid) and polar effects dominating for the conjugate base (the carboxylate).

  Table copied from New. J. Chem., 2004, 28, p67

The intramolecular hydrogen bonding between the halogen substituent and the adjacent carboxyl group (as shown below) would decrease the acidity of the halobenzoic acids:

$\hspace{60 mm}$

Since the strength of the hydrogen bonding decreases in the order $\ce{F}>\ce{Cl}>\ce{Br}>\ce{I}$, we can arrive at the order of acidity $4 > 3 > 2 > 1$.


Why is it vital that atoms in the same sample are all given the same charge? Mass spectrometry When being tested in the spectrometer why is it vital that the atoms in the same sample are given the same electrical charge e.g all are c+1 or all are +2 etc

This is a misconception that during ionization, each ion must get a +1 or +2 charge. It is desirable to have +1 so m/z corresponds to the molecular mass of the ion or its fragment. The state of ionization heavily depends on what you are using to ionize your target analyte. For example if you use electron impact, you will see fragments mostly with +1 charge. However in other techniques called matrix assisted laser desorption (MALDI), the state charge can be as high as +12 and so on. They are detected as such. In electrospray ionization, +2 is also common.

There is no way a mass spectrometer "knows" a priori the state of charge of an ion. All charge ratios are detected. However it can distinguish the sign. It depends on the user whether negative or positively charged ions reach the detector. Some ions are better detected in positive mode and some in negative mode.

In mass spectrometry, the position of each peak you observe depends on the value m/z (i.e. mass over charge). If you have only z=1, then the position will depend only on the mass (m/1 = m).(1)

In a mass spectrometer, you will generate a mixture of charges at first but ions with a charge greater than 1 are filtered out.

(1) Obviously, it also works with a -1 charge.


Explain this boiling point order please I read it from a book but don't know the reason. Please explain...

Boiling point orders are -

Ethanamide > Ethanoic Anhydride > Ethanoic Acid

Ethanamide has a higher boiling point than ethanoic acid since amide has two hydogen atoms it can form more Hydrogen bonds than ethanoic acid and their molar masses are 45gmol-1 and 60gmol-1 which makes the difference brought by the London forces somewhat negligible .
you may suppose Ethanoic anhydride should be having a lower boiling point than ethanoic acid and ethanamide since it only has London forces and dipole dipole attractions.But ehanoic anhydride(molar mass-106gmol-1) has a higher molar mass than Ethanoic acid(molar mass-60gmol-1) making the London forces in ehatnoic anhydride more significant.
Hence we have
ethanamide(222 °C)>Ethanoic anhydride(140°C)>Ethanoic acid(118°C).


Variation of current with time in gel electrophoresis During gel electrophoresis, if the voltage of the power supply is kept constant, how would the current in the circuit change over time?

I was wondering whether the electrical resistance of the gel would change over time. There will be migration of ions present in the buffer and it will resist the change in pH. But how does the electrical resistance of the gel change? Can someone explain?

I think this might help you a bit:
https://www.researchgate.net/post/SDS_PAGE_should_be_run_at_constant_current_or_constant_voltage


cooling/refrigeration essentiality for carbonated softdrinks How does cooling/refrigeration help in maintaining the taste and shelf life of food products, especially CSD (Carbonated Soft Drinks)? For example, chilled drinks have a nicer taste than a normal drink and may have longer shelf life. Why is that?

It has a longer shelf life because the bacteria, (which is the reason it will go bad), cannot reproduce as fast in cold temperatures.

As far as taste goes with carbonated soft-drinks, I think it is mostly just about preference.

Refrigeration condenses the CO2 and allows for a "softer" bubble feel in the mouth.
As for shelf life of soda, it should not effect a "canned" soda one way or another.


Why we need the average atomic mass? I know , Average Atomic Mass of an element is - the sum of the Masses of its isotopes,each multiplied by its Relative Abundance percentage (in decimal notation).

But what confusing me is, what is the concept of Average Atomic Mass. In other words,why we want to take average atomic mass of every element on the periodic table,why not Atomic mass ? Why we need it and when to use it?

The average atomic mass is useful because its numerical value is equal to the molar mass of the element. This in turn is useful to know how much of a solid to take when you wish to react it with some known quantity of another reagent, because weighing is typically the easiest way to quantify a substance.

If you know precisely which isotope of the element you sre dealing with then go right ahead and use the molecular mass but generally we dont know which isotope is available to us and moreover there are mamy impurities a single gram might have a mixture of kany isotopes.
So for reducing the error to the maximum limit possible statistically we multiply the mass of that particular isotope with its relative percentage abundance, you can imagine this as if a particular molecular mass is more available in nature than its more probable that is the isotope we have or that isotope is more in abundance in our mixture and hence the value of the average molecular mass we are using as simplification should tend more to that isotope


Feasibility of Electrophilic attack on the Nitrogen over the Phenyl ring of Aniline I have learnt that the '-NH2' substituent attached to the phenyl ring, owing to the substitutent's highly positive mesomeric effect, activates the phenyl ring quite a lot for Electrophilic Aromatic Substitution, so much so that it easily forms a tri-substituted product when Br+ is used as an electrophile with Aniline, and that a protecting group such as Acetyl (introduced using Acetic Anhydride) has to be introduced on the Nitrogen to deactivate the ring for just monobromination to occur.

This made me wonder that if such is the influence of the '-NH2' on the benzene ring then why would the Nitrosonium ion generated in diazotization prefer attacking the lone pair of Nitrogen instead of the electronically enriched benzene ring on the ortho or para position?

Initially I lazily (and wrongly) attributed it to perhaps some stronger bonds forming between similar atoms, i.e Nitrogen-Nitrogen bonds being stronger than the Carbon-Nitrogen bonds.

But, more interestingly I realized that this line of reasoning wouldn't apply to my above stated acetylation of the Nitrogen in Aniline, for in that case too there isn't an Aromatic Substitution taking place and there are no Nitrogen-Nitrogen bonds.

In short, what is the reason for the feasibility of Diazotization and Acetylation in the way they occur with Aniline? Why aren't electrophilic aromatic substitution reactions taking place? I may have erred in my basics, please help me out.

Electrophilic Aromatic substitution in Aniline will occur when the $\pi$-electron cloud of Benzene ring attacks the electrophile generated (mostly in-situ)  instead of the lone pair on the Nitrogen atom. Thus for occurance of Electrophilic aromatic substitution, there always remains a competition between the $\ce{N}$ atom and the $\pi$-electron rich benzene ring. You may think that the lone pair of nitogen remains fully delocalised through the Benzene ring, but there always remains a high enough electronic charge density on the Nitrogen atom to behave as a nucleophile and in fact in greater extent the electronic density of the ring (not always for Oxygen though, being highly electronegative, its less nucleophilic, but Nitrogen is certainly more nucleophilic).

Thus,you will observe that it is difficult to perform electrophilic aromatic substitution reaction on  unprotected Aniline. Thus, most of the reactions on Aniline are initiated by attacking of the elctrophile by Nitrogen atom instead of the ring. It is its general trend. Thus to do substitution in the ring, we first need to decrease electronic charge density on the Nitrogen atom and that's why we first treat the Aniline with $\ce{CH3COCl}$ in Pyridine medium to form acetanilide, and then perform any other reaction on the ring like Bromination, Chlorination, Alkylation etc.

You will also observe that Aniline doesn't undergo Friedel Crafts alkylation/acylation to a great extent. As at the moment you add Lewis acid, Nitogen atom behave as Lewis Base and forms the adduct, which make the ring highly electron deficient due to induced positive charge on Nitrogen.

Thus, the Diazotization  and Acetylation are the normal/expected reaction of Aniline as a effect of higher nucleophilicity of $\ce{N}$(due to lesser elctronegativity and weaker basicity). Instead if you have to perform Electrophilic aromatic substiution in the ring, you have to think about how to perform it.

Adding to Soumik Das' great answer,
The ring would have to be activated to a large extent to overcome the nucleophilicity of aniline's nitrogen if you would like to add the nitrosonium ion at the benzene, as the aromaticity of the ring would have to broken in the rate determining step, which would have significant activation energy demands.And indeed, it has been observed that for tertiary aryl amines, electrophilic aromatic substitution of the NO+ ion takes place majorly.

As a sidenote, if you consider diazotization in the case of tertiary amines, then according to the mechanism, we need to have atleast one hydrogen present on the nitrogen(of the amine) for the tautomeric step to take place. But due to the lack of H in tertiary amines, the addition of electrophile and it's removal from the amine will probably exist as a parasitic equilibrium


Relationship between thermodynamic reversibility and reactions Often we're concerned with changes in thermodynamic properties during reactions, like $\Delta H_p = Q_{reaction}$ (change in enthalpy for an isobaric process). What seems odd to me is that in the derivation of this relationship,

$\Delta H_p = \Delta U + P\Delta V=Q-W+P\Delta V$

we assume the work is given by $W=P\Delta V$, which requires that the process is reversible in the thermodynamic sense. Then the change in enthalpy simplifies to

$\Delta H_p=Q - P\Delta V + P\Delta V=Q$.

Supposedly, measuring the heat transfer during a reaction at constant pressure will give the enthalpy. Even though enthalpy is a state function in general, our case requires that the process is reversible.

To what extent does the restriction of thermodynamic reversibility apply to the reaction? Do we have to assume the reaction happens slowly and can be undone, leaving only an infinitesimal change in the universe? Or does it not matter?

Please correct me if any of my points are wrong. Classical thermodynamics usually doesn't mention kinds of matter, so I'm a bit confused when chemistry comes into it.

The process does not have to be considered reversible, and, if a reaction is occurring in the system, it is, of course, not reversible.  We are trying to find the change in enthalpy between the following two states:

State 1:  Reactants at temperature $T_0$ and pressure $P_0$

State 2:  Products at temperature $T_0$ and pressure $P_0$

We would like to measure the $\Delta H$ in such a way that, for the process path we select, all we need to do is measure the amount of heat transferred to the system Q.  The process path we select involves keeping the system in contact with a constant temperature bath at $T_0$ and controlling the motion of the portion of the system boundary at which the volume change occurs in such a way that the force per unit area is always constant and equal to $P_0$.  Since, in general, the work done by the system on its surroundings is determined by the integral of $dW=P_{ext}dV$ (where $P_{ext}$ is the force per unit area acting on the portion of the boundary at which the volume change is occurring), we have selected $P_{ext}$ such that $P_{ext}=P_0$; and the total cumulative work done by the system when it reaches its final equilibrium state will be just $W=P_0\Delta V$.

From the first law, it then follows that, for this irreversible path:
$\Delta U=Q-P_0\Delta V$.  So $\Delta U+P_0\Delta V=\Delta H =Q$

In chemistry dU = dQ + dW ( not dQ = dU + dW) and dW = -Pdv.The general expression for enthalpy is H = U + PV . Enthalpy is the amount of heat content used or released in a system at constant pressure.
If there is no non-expansion work on the system and the pressure is still constant, then the change in enthalpy will equal the heat consumed or released by the system (q) i.e. (dH = Q).
But,you can calculate change in enthalpy every case throuugh equation (H = U + PV) i.e. dH = dU + d(PV).


Why 4‐methylidenespiro[2.4]heptane cannot be converted to 1‐chlorobicyclo[3.2.0]heptane? According to the solution, the following reaction is not correct:

But why? I have tried solving in the following steps:

Forming a carbocation by breaking the double bond.
Expanding the ring.
Adding a $\ce{Cl-}$ anion.

This should thereby form the compound that is shown. But the solution is given that this reaction is not correct. Can you please explain to me why?

Here the concept of angular strain comes into the picture. The ideal tetrahedral and is around 109° 29'. As the angle between the C−C bond increases, it is more away from the ideal $\mathrm{sp^3}$ hybridization and hence will become more unstable.

For a cyclohexane molecule, the angle between the C−C bonds is 120°, therefore the deviation from the ideal angle is 10° 31', whereas for a cyclopentane the angle between C−C bonds is 108°, therefore the deviation from the ideal tetrahedral angle is around 1.5°. This is a very low deviation.

Since the transition is from cyclopentane which has a relatively lower deviation from the ideal angle to cyclohexane which has a relatively higher deviation from the ideal angle, the above transition does not take place.

Plus the number of carbons are not balanced in your reaction.


Research-backed articles on cast iron seasoning I've always seasoned my cast iron pans with vegetable oil because that was what I was taught. I'd done some research before, and even came across this StackExchange post on asking the best oil for the job.
To my dismay however, when I tried doing some research myself, all sources pointed to blog posts, particularly the Sheryl Canter one. There is not a single source on her post and she doesn't seem to be very near a materials scientist. The closest paper I could find to the topic was this one. Other than that, it was all anecdotal evidence, many people claiming many different things.
It felt almost absurd that I couldn't find a paper on something that seemed relatively common to me. This will bug me till the end of time unless I figure it out, so I need to know - does anybody have any real sources on cast iron seasoning or what factors play a part in the polymerization of different oils? Thanks in advance.

There is this research on Chinese equivalent of seasoning process called "Kitchen God's Blessing" [1].
They argue that the non-stick property of seasoning comes by iron oxide $(\ce{Fe3O4})$ nanoballs.
If they are correct then the oil or its polymers is not exactly what is causing the non-stick property but the those nanoballs on the surface. Their seasoning method is somewhat different however. It uses temperatures up to 450 °C.
Reference

Gao, C.; Yang, N.; Li, C.; Wang, X.; Yu, X.; Zhang, L.; Wei, Z. Seasoning Chinese Cooking Pans: The Nanoscience behind the Kitchen God’s Blessing. Nano Materials Science 2020. DOI: 10.1016/j.nanoms.2020.06.001.

There is 'seasoning' to protect the pan, but on heating the pan, the chemical breakdown products are 'some' things I would not recommend to anyone for continuing good health for yourself or your family (see, for example, Genotoxic and carcinogenic risks associated with the dietary consumption of repeatedly heated coconut oil).
Unfortunately, many oils do breakdown even at low heat, and some at high temperatures, which includes even my favorite healthy (especially when acquired fresh from its harvest) Olive Oil!
However, a safer alternative path may yet exist to resurrect ourselves (but not likely literally).
Electrochemistry to the rescue!
In particular, try employing a sacrificial anode, like a small piece of zinc metal sitting in distilled water plus sea salt, covering your entire pan. Yes, to be explicit, my recommendation on how to best season your iron pan to guard it from future corrosion is zinc metal and saltwater free of oxygen, which unlike an oil coating, is not easily subject to removal or chemical disruption (from air/light exposure).
On the mechanics, your iron metal will no longer be the most anodic metal present, and the zinc may undergo limited anodic corrosion resulting in white ZnO/Zn(OH)2 product. However, even the latter can be consumed in low doses daily whereas, one must be more mindful of iron intake especially with small children. One has also, at low cost and safely, eschewed the dangers of associated heat generated cariogenics from various organics.
Aesthetically, I would also challenge you to compare the results from two iron pans, one stored in the manner I have suggested, to the other pan that was coated in whatever oil-based seasoning formulation you may believe is suitable, after washing both pans.


Why is the boiling point of alkyl iodide is more than alkyl fluoride? As far as I know, more the polar character of the organic compound, more will be the energy required to break the bond and thus more boiling temperature. Here in this case, $\ce{R-F}$ is more polar in nature than $\ce{R-I}$. Then, how $\ce{R-I}$ has more boiling point than $\ce{R-F}$?

The main factor affecting the boiling point is the molecular weight, as can be pictured from the noble gases boiling point in the periodic table (the noble gases are those which have the weakest interactions due to their complete outer shell). But note that if you want to consider all the effects altogether, you must use Kelvins instead of Celsius. So going from 44°C to 88°C is not twice as much but a mere 14% increase.

The other factor affecting boiling point is the nature and strength of intermolecular interactions, usually in the order ionic > hydrogen bond > dipole-dipole > Van der Waals.

In the comparison of RF vs RI boiling point, molecular weight is definitely in favor of higher boiling points for RI, at least for small molecules. It is true that the C-F bond is much more ionized than C-I so it will create stronger dipole-dipole interactions but not sufficiently to compensate for the effect of the mass increase.

HTH

Polarity consideration is important when dealing with interactions between molecules. A hydrogen bond (H-bond) is a good example. Water has polar $\ce{H-O}$ bonds, as a result partially positive $\ce{H}$ interacts with partially negative $\ce{O}$ from a dofferent molecule as a result water is liquid at $T_\mathrm r$ and solidifies at $0\ \mathrm{^\circ C}$, while methane of about the same molecular weight is a gas up until $-168\ \mathrm{^\circ C}$.

The reason for $\ce{HF}$ to be liquid and $\ce{HI}$ to be gas is different. $\ce{H}$ and $\ce{F}$ have similar sized orbitals and the molecule is polar. As a result intramolecular H-bonds are strong. In $\ce{HI}$ the overlap is very poor (minute $\ce{H}$ orbitals don't overlap well with huge diffuse orbitals of $\ce{I}$. As a result $\ce{HI}$ doesn't form strong intermolecular $\ce{H}$ bonds and the bonding between $\ce{H}$ and $\ce{I}$ in the molecule is much weaker than in $\ce{HF}$. Thus, $\ce{HI}$ is both gas and a very strong acid (much stronger than $\ce{HF}$).

I will assume that by $\ce{R}$ you mean an alkyl group. In this situation neither $\ce{R-F}$, nor $\ce{R-I}$ can form hydrogen bonds. For example, the polarization of $\ce{$\mathbf{H{-}C}$H2F}$ is much closer to that in methane ($\ce{CH4}$) than in water, not to mention $\ce{HF}$. Some polarization exists but its effects are small. Much greater effect is the molecular weight and density that translate into much stronger London forces in $\ce{R-I}$ compared to $\ce{R-F}$. Both weight and density favor $\ce{R-I}$ to be less volatile than $\ce{R-F}$. The effect of polarization is not large enough to compensate weight difference.

So, how large is this polarization effect in $\ce{R-F}$? Well, compare bp of ethane ($\ce{CH3CH3}$) and fluoromethane ($\ce{CH3F}$) and iodomethane ($\ce{CH3I}$). $T_\mathrm{bp}(\text{ethane})=-88\ \mathrm{^\circ C}$ and $T_\mathrm{bp}(\ce{CH3F})=-77\ \mathrm{^\circ C}$, $T_\mathrm{bp}(\ce{CH3I})=+44\ \mathrm{^\circ C}$. The effect is there but it is very small.


Electrochemistry conditions for voltage to increase in a galvanic cell I understand that increasing the concentration of reactants in a galvanic cell generally causes the voltage of the cell to increase. But is that really the case always?

Consider a standard cell of zinc and copper, 1 M and 1 L each solution of copper nitrate and zinc nitrate in each half cell. Another cell, increased the concentration of copper nitrate to 4 M, but added 0.25 L of it and kept the same zinc nitrate 1 M and 1 L.

This way I increased the concentration, but the amount of substance is the same. Does the voltage increase in the new cell? I would really appreciate someone to explain to me the specifics of what causes the voltage of a cell to increase based on the reactants

Yes, your voltage will change in accordance with the Nernst equation

$$\Delta E_\text{cell} = - \frac{RT}{zF}\ln Q_{\text{rxn}}$$

So, in your copper-zinc battery, where

$$\ce{Cu^2+(aq) + Zn(s) -> Cu(s) + Zn^2+(aq)}$$

with a standard $E^\circ_\text{cell} \approx \pu{1.10 V},$ setting the $\left[\ce{Cu^2+}\right] = \pu{4 M}$ should raise your cell potential by

$$\Delta E_\text{cell} = -\frac{RT}{2F}\ln\frac{1}{4} = \pu{+0.0178 V}$$

for a new cell potential (at $\pu{25 °C})$ of

$$E_{\text{cell}} = \pu{1.10 V} + \pu{0.0178 V} \approx \pu{1.12 V}$$

As others have answered, yes the potential of the cell will still change. It sounds like you're looking for a more "physics-y" answer than the ones provided so here goes.

The Nernst equation (and equilibrium equations in general) work based on concentration and not total moles. To understand why, start with a homogeneous reaction, like $\ce{HCl + NaOH <-> NaCl + H2O}$, or more usefully $\ce{H+ + OH- <-> H2O}$. In order for one molecule of $\ce{H2O}$ to be produced, we need a proton and a hydroxide ion to collide. If there are lots of them, this reaction will happen often. However, at the same time, some number of $\ce{H2O}$ molecules will dissociate into protons and hydroxide ions (i.e. the reverse reaction). If the forward reaction happens quickly compared to the reverse reaction, then the equilibrium will favor the products. If you double the volume but keep the concentration the same then your increasing the number of collisions of protons and hydroxide ions but also increasing the number of dissociations of water molecules so overall the equilibrium stays the same. Only if you increase the number of protons and hydroxides relative to the number of water molecules will the equilibrium shift. This is what you do when you increase the concentration.

Now let's switch to your battery example and consider the half-reaction of $\ce{Cu^{2+}_{(aq)} + 2e- <-> Cu_{(s)}}$. This is a heterogeneous reaction so we have to think about the interactions at the surface of the electrode. In order for the forward reaction to occur, a solvated copper ion must shed its solvation layer and adsorb to an active site on the surface of the metal then receive two electrons from the copper. At the same time, some number of atoms on the copper metal are releasing two electrons, solvating and entering the bulk electrolyte. Let's say you have a cell in equilibrium. This means that the rate of the forward reaction and the rate of the backward reaction are the same. Increasing the concentration of the electrolyte means that now every second more ions will be hitting the copper metal and it increases the likelihood of a conversion to copper metal. This increase in the rate of the forward reaction pushes the equilibrium farther towards the products. However, increasing the total volume of the electrolyte without changing the concentration doesn't affect the rates of adsorption of ions to the surface of the electrode, since that's an inherently surface phenomenon.

To use an analogy: the number of cars per minute that get through a toll-booth is only determined by how quickly the very front cars pay and drive off. Adding a bunch of cars in a queue behind the booth doesn't change anything. You could make the queue a mile long and still only the very first row of cars actually paying will set the rate of traffic. Of course, having a long line means it will take longer for all the cars in the queue to get through, just as in your case the additional of volume of electrolyte will increase the capacity of the cell.


Producing ketones from secondary alcohols using hydrogen peroxide Usually, to produce ketones from secondary alcohols (e.g. propanone from 2-propanol), people use a solution of potassium or sodium dichromate with a small amount of sulfuric acid. Can other oxidizers like hydrogen peroxide be used, and will the use of hydrogen peroxide produce organic peroxides?

$$\ce{C3H8O + H2O2 -> C3H6O + 2 H2O}$$

This is not for any particular purpose.

I think it is possible to use hydrogen peroxide. This paper discusses it: http://greenchem.uoregon.edu/PDFs/ResourceID88.pdf

Other alternatives:

Pyridinium Chlorochromate (or PCC for short)
Dess-Martin Reagent
Swern Oxidation: oxalyl chloride in DMSO with triethyl amine

Fétizon Oxidation: silver(I) carbonate absorbed onto the surface of celite (Fétizon's reagent). This can be prepared by adding silver nitrate to aqueous sodium carbonate, in the presence of purified celite.

Oppeneaur Oxidation:  excess ketone reagent (such as acetone) and an aluminium triisopropoxide catalyst

Pfitzner–Moffatt oxidation: dimethyl sulfoxide (DMSO) activated with a carbodiimide, such as dicyclohexylcarbodiimide (DCC)

Corey-Kim Oxidation: Dimethyl sulfide is treated with N-chlorosuccinimide (NCS), and then addition of triethylamine

There are several moments:

(1) Chemists are mostly interested in routes yielding products that are otherwise difficult to obtain. Oxidation of alcohols to ketones is a generic operation. There must be zillions of undiscovered methods to produce ketones, but there is no motivation to discover those, because the traditional methods work just as well.
(2) Peroxides are reactive/corrosive/hard to work with compounds. There are copious synthetic methods that use peroxides, but this is done only if milder methods are not known. For the same reason you don't put sodium perchlorate in your soup to enhance taste, since table salt has very similar effect on taste.
(3) You definitely can turn alcohol to a ketone using a peroxide (based on delta G), but the yield might be low, other reaction groups (such as double bond) might not survive and there is zero incentive to come up with yet another way to turn an alcohol to a ketone.

http://www.organic-chemistry.org/synthesis/C2O/ketones/oxidationsalcohols.shtm This page has 40+ common oxidation reactions to prepare a ketone from an alcohol. Choose less toxic alternative if you want.

Bottom line: you will not make a Nature paper with yet another oxidation of alcohol to a ketone. Thus, academia doesn't care. H2O2 is more dangerous (especially considering explosion hazard) than CrO3. Industry doesn't care.


Drying Clothes Outside Below Freezing It is commonly accepted that you can hang your wet clothes outside to dry even when the winter temperature is below freezing. It is mentioned here
http://grist.org/living/ask-umbra-can-i-line-dry-my-clothes-in-the-winter-too/
that water in the clothes will freeze to ice, then sublimate into water vapor, making the clothes dry.

Question: Why does the ice in the clothing sublimate, while the ice on the roads don't sublimate? Why is there a difference between what happens to ice on the clothing versus ice on the roads?

As long as there is less than 100% relative humidity, the ice will sublime to some extent.

Just in the past week it has been in the -12 °C to -5 °C range in my area, and the snow and ice on the ground did sublime, on the order of 1 cm per day, more in the sun less in the shade.

For quantitative information see "Sublimation from a seasonal snowpack at a continental,
mid-latitude alpine site" Hydrol. Process. 13, 1781-1797 (1999).

On cold days, the wet clothes dry due to the water in the clothes that freezes, then sublimates, to create water vapour that escapes the clothes, therefore drying it.


Mathematical properties of free Gibbs energy I'm wandering, if some mathematical properties of the free Gibbs energy (also called thermodynamic potential) are well known.
For instance in chemistry, if $n_i^\alpha$ denotes the number of moles of species $i$ in phase $\alpha$ and G is the free Gibbs energy.
I know that G must be an homogeneous function of degree one in $n_i^\alpha$.

Do we have more information like continuity or coercivity properties of G ?

n is discontinuous. The least amount you can change it by is ~1 /6.022E23 of 1 mole. You'll have to be more specific about what type of (mathematical) continuity you are referring to, but it doesn't seem useful to ask about its fine grained continuity since it is a macroscopic concept.  Statistical (quantum) Thermodynamics is used both for microscopic systems and as the basis for macroscopic quantities such as Gibbs Free Energy. You need to be careful: GFE is one meaning of Thermodynamic Potential, but not the only one. I don't have any idea what you mean by G's coercivity.


What happen to the Salt in Water Electrolysis I haven't found yet what happen to the Salt Ions in Water during Electrolysis and is there an Electron Flow or Ion Flow.
I have expierienced strong magnetic Forces between the Electrodes but couldnt measure an outside Magnetic Field.

When a battery is connected to an electrolytic cell one electrode becomes positively charged the other negatively charged. The ions in the liquid are attracted to the electrode with the opposite charge and move through solution. Ions with a negative charge move to the positive anode (anions), Ions with a positive charge migrate to the negative cathode (cations).

Ionic solids produce electrolytes. When an ionic solid is dissolved in water the aqueous ions are free to move and they carry the current. The anions of the salt and water give up electrons which flow through the circuit. The anions are oxidized at the anode while the cations of the salt and water recieve the electrons and are reduced at the cathode.

Cations will cluster around the cathode and anions will cluster at the anode. It is usually easier to displace hydrogen atoms from water molecules in the aqua ion than to displace the cation from the system  so electrons from the cathode react with water molecules , and hydrogen is displaced.

The result is decomposition of water molecules:

2H-OH(l) + 2e- → H2(g) + 2OH-(aq)

The solution becomes alkaline around the cathode due to the hydroxide ions being formed. If a metal is above hydrogen in the electrochemical series, hydrogen will always be produced at the cathode during electrolysis of solutions of its salts. If the ionic salt consists of a metal below hydrogen in the electrochemical series, electrolysis of its salt solution will always produce metal at the cathode.

Both ion flow and electron flow are occurring.

The intensity of an electromagnetic field reduces significantly the further away from the source.

You're mixing two different concepts.

First a magnetic field has nothing to do with an electrochemical cell. A magnetic field is created when a direct current flows in coil. Think of wrapping a wire around a nail.

As far as the ions in solution, the cations have a mass flow in one direction and the anions have a mass flow in the other. Think of having a theoretical "force field" that can gate anions and cations. As many cations would flow through as anions while current flows in the system.


Can an ethyl ester of L-proline be synthesised from the amino acid and ethyl alcohol in toluene with azeotropic distillation of water? I was told that the ethyl ester of L-proline may not be synthesized from L-proline and ethyl alcohol in refluxing toluene, with an acid catalyst such as p-toluensulfonic acid for example, because toluene and ethanol and in this case water as well form an azeotrope. Is this true?

I do not have a supply of pure thionyl chloride, and that's why I am trying to use this route. I appreciate if anyone may answer this question.

I think this is unlikely to work under the conditions described as the catalytic PTSA is tied up as its proline salt. If you reflux under D-S conditions you will just remove the EtOH due to the azeotrope with Toluene.


Hot and cold air interaction in a sealed environment What would happen if air heated to 1200 degree's Fahrenheit in a sealed environment was injected with colder air?
No specific pressure defined yet just a general idea of potential interaction between the two.

Edit
My apologies, I meant within a sealed environment like an engine cylinder. I know that the hot air would naturally expand and push a piston but how would the injection of cold air alter this, if at all?
For example purposes we can assume the cylinder is iron while the piston is aluminum which is common design today.

In order to work out this problem, several parameter values are required to define the initial state, preferably

amount $n_1$, volume $V_1$, or mass $m_1$ of the hot air in the cylinder
initial temperature $T_1$ of the hot air
initial pressure $p_1$ of the hot air
amount $n_2$, volume $V_2$, or mass $m_2$ of the injected cold air
initial temperature $T_2$ of the cold air
initial pressure $p_2$ of the cold air

Unfortunately, only the initial temperature of the hot air is defined as $T_1 = 1200\ ^\circ\mathrm F$, which is $T_1 \approx 649\ ^\circ\mathrm C \approx 922\ \mathrm K$ in proper units (the use of the unit degree Fahrenheit is deprecated). This leaves us with five unknown variables, which is certainly too many for a vivid analytical description.

Nevertheless, we may investigate this problem by guessing reasonable parameter values as follows. By way of example, we arbitrarily assume that equal masses of hot and cold air $(m_1 = m_2 = 1\ \mathrm{kg})$ are mixed. The assumed temperature of the cold air sample is equal to room temperature, i.e. $T_2 = 68\ ^\circ\mathrm F = 20\ ^\circ\mathrm C \approx 293\ \mathrm K$, whereas the temperature of the hot air sample is given as $T_1 = 1200\ ^\circ\mathrm F \approx 649\ ^\circ\mathrm C \approx 922\ \mathrm K$. The assumed pressures of the air samples is equal to normal atmospheric pressure, i.e. $p_1 = p_2 = 1.01325\ \mathrm{bar} = 101\,325\ \mathrm{Pa}$. These conditions define the initial state and also allow to determine other interesting properties using tabulated reference values for air, e.g. density $\rho$, Volume $V$, specific internal energy $u$, and internal energy $U$.

1. Hot air sample

temperature $T_1 = 1200\ ^\circ\mathrm F \approx 649\ ^\circ\mathrm C \approx 922\ \mathrm K$
pressure $p_1 = 1.01325\ \mathrm{bar} = 101\,325\ \mathrm{Pa}$
mass $m_1 = 1\ \mathrm{kg}$
density $\rho_1 = 0.3827\ \mathrm{kg/m^3}$
volume $V_1 = \frac{m_1}{\rho_1} = 2.613\ \mathrm{m^3}$
specific internal energy $u_1 = 819.4\ \mathrm{kJ/kg}$
internal energy $U_1 = u_1 \cdot m_1 = 819.4\ \mathrm{kJ}$

2. Cold air sample

temperature $T_2 = 68\ ^\circ\mathrm F = 20\ ^\circ\mathrm C \approx 293\ \mathrm K$
pressure $p_2 = 1.01325\ \mathrm{bar} = 101\,325\ \mathrm{Pa}$
mass $m_2 = 1\ \mathrm{kg}$
density $\rho_2 = 1.2046\ \mathrm{kg/m^3}$
volume $V_2 = \frac{m_2}{\rho_2} = 0.830\ \mathrm{m^3}$
specific internal energy $u_2 = 335.3\ \mathrm{kJ/kg}$
internal energy $U_2 = u_2 \cdot m_2 = 335.3\ \mathrm{kJ}$

3. Mixture at constant volume

The hot air sample is confined by a piston in a cylinder. The cold air sample is added, thus allowing the two air samples to mix. The total mass $m_3$ is given by
$$\begin{align}
m_3 &= m_1 + m_2 \\
&= 1\ \mathrm{kg} + 1\ \mathrm{kg} \\
&= 2\ \mathrm{kg}
\end{align}$$

For simplicity’s sake, we ignore any work $W$ performed by or on the system (e.g. the work that is required to inject the cold air into the cylinder). Furthermore, we assume that the process is adiabatic, i.e. no heat $Q$ is transferred into or out of the system. Therefore, in accordance with the first law of thermodynamics, the total internal energy $U_3$ is given by
$$\begin{align}
U_3 &= U_1 + U_2 \\
&= 819.4\ \mathrm{kJ} + 335.3\ \mathrm{kJ} \\
&= 1154.7\ \mathrm{kJ}
\end{align}$$
and the new specific internal energy $u_3$ is
$$\begin{align}
u_3 &= \frac{U_3}{m_3} \\
&= \frac{1154.7\ \mathrm{kJ}}{2\ \mathrm{kg}} \\
&= 577.3\ \mathrm{kJ/kg}
\end{align}$$

At constant volume $V$, i.e.
$$\begin{align}
V_3 &= V_1 \\
&= 2.613\ \mathrm{m^3}
\end{align}$$
the new density is
$$\begin{align}
\rho_3 &= \frac{m_3}{V_3} \\
&= \frac{2\ \mathrm{kg}}{2.613\ \mathrm{m^3}} \\
&= 0.7654\ \mathrm{kg/m^3}
\end{align}$$

The new temperature $T_3$ and pressure $p_3$ can be determined from the density $\rho_3$ and the specific enthalpy $h_3$ using tabulated reference values:
$$\begin{align}
T_3 &= 658\ ^\circ\mathrm F \approx 348\ ^\circ\mathrm C \approx 621\ \mathrm K \\
p_3 &= 1.37\ \mathrm{bar} = 0.137\ \mathrm{MPa}
\end{align}$$

4. Mixture at constant pressure (ignoring work)

Allowing the system to expand to normal atmospheric pressure by moving the piston, the new pressure $p_4$ is
$$p_4 = 1.01325\ \mathrm{bar} = 101\,325\ \mathrm{Pa}$$

We know that the mass $m$ of the air in the closed system does not change, i.e.
$$\begin{align}
m_4 &= m_3 \\
&= 2\ \mathrm{kg}
\end{align}$$
and we still (incorrectly) assume that no work $W$ is performed by or on the system and that no heat $Q$ is transferred into or out of the system. Therefore,
$$\begin{align}
U_4 &= U_3 \\
&= 1154.7\ \mathrm{kJ}
\end{align}$$
and thus
$$\begin{align}
u_4 &= u_3 \\
&= 577.3\ \mathrm{kJ/kg}
\end{align}$$

The new density $\rho_4$ can be determined from the pressure $p_4$ and the specific internal energy $u_4$ using tabulated reference values:
$$\rho_4 = 0.5681\ \mathrm{kg/m^3}$$

Thus, the new volume $V_4$ is
$$\begin{align}
V_4 &= \frac{m_4}{\rho_4} \\
&= \frac{2\ \mathrm{kg}}{0.5681\ \mathrm{kg/m^3}} \\
&= 3.521\ \mathrm{m^3}
\end{align}$$

The volume change $\Delta V$ during expansion is
$$\begin{align}
\Delta V &= V_4 - V_3 \\
&= 3.521\ \mathrm{m^3} - 2.613\ \mathrm{m^3} \\
&= 0.908\ \mathrm{m^3}
\end{align}$$

The corresponding pressure-volume work $W$ can be calculated assuming an external pressure of $p_\mathrm{ex} = 1.01325\ \mathrm{bar} = 101\,325\ \mathrm{Pa}$.
$$\begin{align}
W &= p_\mathrm{ex} \cdot \Delta V \\
&= 101\,325\ \mathrm{Pa} \times 0.908\ \mathrm{m^3} \\
&= 92.0\ \mathrm{kJ}
\end{align}$$

The amount of energy transferred to the surroundings as work $W$ significantly reduces the internal energy $U$ of the system. We ignored this fact when we assumed that $W_4 = 0$ and $U_4 = U_3$. We can eliminate this error by calculating a new state as follows.

5. Mixture at constant pressure (including work)

If the internal energy is reduced (due to energy transfer to the surroundings as work) to
$$U_5 = 1087.8\ \mathrm{kJ}$$
and thus the specific internal energy is reduced to
$$\begin{align}
u_5 &= \frac{U_5}{m_5} \\
&= \frac{1087.8\ \mathrm{kJ}}{2\ \mathrm{kg}} \\
&= 543.9\ \mathrm{kJ/kg}
\end{align}$$
the new density $\rho_5$ after expansion to normal atmospheric pressure
$$p_5 = 1.01325\ \mathrm{bar} = 101\,325\ \mathrm{Pa}$$
can be determined from the pressure $p_5$ and the specific internal energy $u_4$ using tabulated reference values as
$$\rho_5 = 0.6111\ \mathrm{kg/m^3}$$

Thus, the new volume $V_5$ is
$$\begin{align}
V_5 &= \frac{m_5}{\rho_5} \\
&= \frac{2\ \mathrm{kg}}{0.6111\ \mathrm{kg/m^3}} \\
&= 3.273\ \mathrm{m^3}
\end{align}$$
the volume change $\Delta V$ during expansion is
$$\begin{align}
\Delta V &= V_5 - V_3 \\
&= 3.273\ \mathrm{m^3} - 2.613\ \mathrm{m^3} \\
&= 0.660\ \mathrm{m^3}
\end{align}$$
and the corresponding pressure-volume work $W_5$ is
$$\begin{align}
W_5 &= p_\mathrm{ex} \cdot \Delta V \\
&= 101\,325\ \mathrm{Pa} \times 0.660\ \mathrm{m^3} \\
&= 66.9\ \mathrm{kJ}
\end{align}$$

We find that
$$\begin{align}
U_5 &= U_3 - W_5 \\
1087.8\ \mathrm{kJ} &= 1154.7\ \mathrm{kJ} - 66.9\ \mathrm{kJ} \\
\end{align}$$
This result is in agreement with the first law of thermodynamics, which states that the change in the internal energy $U$ of a closed system is equal to the amount of heat $Q$ supplied to the system (which is $Q = 0$ in this case), minus the amount of work $W$ done by the system on its surroundings.

$$\small
\begin{array}{llllll}
\hline
\text{Quantity} & \text{Symbol} & \text{Unit} & \text{State 1} & \text{State 2} & \text{State 3} & \text{State 4} & \text{State 5} \\
\hline
\text{Temperature} & T & ^\circ\mathrm F & 1200 & 68 & 658 & 658 & 580 \\
& & ^\circ\mathrm C & 649 & 20 & 348 & 348 & 304 \\
& & \mathrm K & 922 & 293 & 621 & 621 & 577 \\
\text{Pressure} & p & \mathrm{bar} & 1.01325 & 1.01325 & 1.37 & 1.01325 & 1.01325 \\
& & \mathrm{MPa} & 0.101325 & 0.101325 & 0.137 & 0.101325 & 0.101325 \\
\text{Mass} & m & \mathrm{kg} & 1 & 1 & 2 & 2 & 2 \\
\text{Density} & \rho & \mathrm{kg/m^3} & 0.3827 & 1.2046 & 0.7654 & 0.5681 & 0.6111 \\
\text{Volume} & V & \mathrm{m^3} & 2.613 & 0.830 & 2.613 & 3.521 & 3.273 \\
\text{Specific internal energy} & u & \mathrm{kJ/kg} & 819.4 & 335.3 & 577.3 & 577.3 & 543.9 \\
\text{Internal energy} & U & \mathrm{kJ} & 819.4 & 335.3 & 1154.7 & 1154.7 & 1087.8 \\
\text{Work} & W & \mathrm{kJ} & \text{n/a} & \text{n/a} & 0 & 0 & 66.9 \\
\hline
\end{array}$$

If the volume is constant (sealed container that is not a rubber ballon, for example), the overall temperature would cool.  And the net pressure would be a weighted average of the two pressures if the volume was constant.  Without actual amounts it cannot be determined if the pressure would be higher or lower than the original.

If the container were a rubber balloon, it is not possible to determine the final volume (even as bigger or smaller) without information about relative amounts and the temperature of the 'colder air'.


Which bonds need to be broken to form a liquid? I'm confused whether it's the intermolecular forces, or the bonds that are being broken when a substance changes state? Does this differ among different types of compounds (eg. covalent, ionic, covalent network solid? What is actually happening when they change state?

It could be either chemical bonds or weaker intermolecular forces that change during a change of state. Water is held together by hydrogen bonds and make its boiling point higher than for analogous compounds.

Sulfur changes both ways with temperature:

  Sulfur forms polyatomic molecules with different chemical formulas, the best-known allotrope being octasulfur, cyclo-S8... At 95.2 °C (203.4 °F), below its melting temperature, cyclo-octasulfur changes from α-octasulfur to the β-polymorph. The structure of the S8 ring is virtually unchanged by this phase change, which affects the intermolecular interactions. Between its melting and boiling temperatures, octasulfur changes its allotrope again, turning from β-octasulfur to γ-sulfur, again accompanied by a lower density but increased viscosity due to the formation of polymers. At higher temperatures, the viscosity decreases as depolymerization occurs. Source

Metallic bonds are delocalized, so in melting, there is no real change in bonding (though on boiling, a metallic vapor is quite different).

One could say, though, that this is simply a matter of definition: chemical bonds act over shorter distances, and intermolecular forces over longer, but they are not intrinsically different, the polar or hydrogen bond being an example of one that bridges these realms.

In reality if a substance is in liquid phase , then not all intermolecular bonds are broken between molecules but some of them and this results in the creation of groups of molecules.In gases all intermolecular bonds are broken between molecules.

The intramolecular bonds are not broken during a phase transition and that is why I posted my own answer instead of comment on DrMoisphe's answer.


Why are the nitrate salts hygroscopic? Any metal is never extracted from its nitrate salt because nitrate salts are highly soluble in water, so flow away along with rainwater, or any other source.
But why are the nitrate salts so hygroscopic in nature? Is there any relationship with hydrogen bonding? Or something else is the cause?

Anions like nitrate forms strong hydrogen bond with water. Here water bridges the nitrate oxygen with hydrogen bonds giving it a very stable arrangement.


What is the signal at around 2090 wavenumbers in the nitroguanidine FTIR spectrum? There is a signal at around 2090 apparent in the FTIR spectrum of nitroguanidine  from Sigma-Aldrich (first image). It's also apparent in FTIR spectra we've taken in the lab (second image), but when assigning peaks to bonds, we are unclear to what it is.

We are unsure of how Sigma-Aldrich prepared their pellets, but we used KBr. We are thinking perhaps KCN, but then again - how could that possible form?

Its common Nujol, a hydrocarbon oil used to prepare paste-samples for IR from powdered materials. Hence the C-H valence around 2090 cm-1


Is there a field of chemistry that doesn't do experiments? In other words, a field of chemistry that is all theoretical? I have heard of theoretical chemistry, but my chemistry professor said you can't really major in theoretical chemistry. He said the closest thing is physical chemistry.

I like predicting in Chemistry, but I hate labs or experiments.

You are correct that it is possible to study theoretical or computational chemistry, however these typically only represent a handful of courses in a normal undergraduate chemistry curriculum. Additionally, most undergraduate chemistry curriculum are designed to instruct students broadly in the core disciplines of chemistry, hence it's usually not possible to specialize.

Non-chemisty degrees may be an alternative. However any science undergraduate program will require laboratory-based coursework. Hands-on experience teaches us how scientific concepts work in the real works, giving us an appreciation of how difficult experimental work can be. It also functions to reinforce content from books and lectures; to teach technical skills (which are usually necessary for future employment); and to develop lab safety habits. So there really is no means to escape it. Moreover, in order to take advanced chemistry electives, prior courses which include lab sessions are generally necessary.

That said, there are other routes to get into chemistry. I've know individuals whom have majored in mathematics, engineering, physics, and biology who either did research in chemistry or who did graduate degrees in chemistry subsequent to obtaining an undergraduate degree in those fields. However in order to obtain undergraduate degrees in those fields and to be able to transition to chemistry programs, lab experience and some senior-level chemistry courses are typically required.

However, once in graduate school, it is possible to avoid experiments altogether: theoretical and computational chemistry graduate degrees typically do not require that the candidate engage in any hands-on experiments. Many however do require that the graduate student work as a teaching assistant in undergraduate chemistry laboratories as a condition of funding. In such cases, laboratory experience is either necessary or a significant advantage.

So while it may be possible to minimize the amount of lab work that you engage in, some will always be required. Additionally I would not recommend avoiding lab work, as it greatly enriches the study of chemistry overall.

Theoretical chemistry is the branch of chemistry that focus on non-experimental side of chemistry

https://www.nature.com/subjects/theoretical-chemistry

simons.hec.utah.edu/TheoryPage/WhatisTheoreticalChem.ppt

Your professor seems to be confounding branches. Physical chemistry can be both experimental or theoretical. The same happens with other disciplines, e.g. theoretical organic chemistry.


Is the strength of hydrogen bonding greater in hydrogen peroxide or water? When we compare water and hydrogen peroxide's hydrogen bond strength, which one should have the greater one? I know that peroxide has more hydrogen bonds, but when we compare the strength of each hydrogen bond, which one would be greater?

Online values of the boiling points indicate a greater bond strength in $\ce{H2O2}$. But peroxide has a larger number of hydrogen bonds.

This was a question for comparing bond strength in a test paper.

Online values of the boiling points indicate a greater bond strength in $\ce{H2O2}$.

The boiling point of water is $\pu{100 ^\circ C}$ while that of hydrogen peroxide is about $\pu{150 ^\circ C}$. Here is a list of factors that might contribute to this difference (adapted from source)

Stronger dispersion forces in $\ce{H2O2}$ (larger molar mass)
Higher dipole moment in $\ce{H2O2}$
More hydrogen bond acceptors in $\ce{H2O2}$

As for the number of hydrogen bonds, if we look at the structure of ice, each hydrogen atom of water is involved in a hydrogen bond (each oxygen is involved in two). If we assume that each hydrogen can make one bond at maximum, we would not expect more hydrogen bonds per molecule for  $\ce{H2O2}$. The crystal structure of solid  $\ce{H2O2}$ confirms that.

  But peroxide has a larger number of hydrogen bonds.

We can't infer that from the molecular structure alone. As stated above, in the solid state, the number of hydrogen bonds per molecule is the same. However, the density of solid $\ce{H2O2}$ (1.7 g/mL) is much higher than that of ice. This is true of the liquid forms as well (1.45 g/mL vs about 1 g/mL). This might allow for more dispersion forces (better contacts).

In liquid water, there are 3.4 hydrogen bonds per molecule on average (source). The higher density of $\ce{H2O2}$ compared to water suggests that atoms are closer in $\ce{H2O2}$, but you have to consider that the composition of the two substances are different. On a per-atom basis, the available volume is comparable.

  Is the strength of hydrogen bonding greater in hydrogen peroxide or water?

The question is a bit vague. If you compare one $\ce{H2O2}$:$\ce{H2O2}$ hydrogen bond to one water:water hydrogen bond, the hydrogen bond length is very similar (at least in the solid phase - source). Because the bond partners are the same atoms and the geometry around the oxygen is not that different in the two molecules, this would suggest that the strength of the bond is similar as well.

You are absolutely right, it has a higher boiling point because it has more hydrogen bonds.
However the bonds are slightly weaker since the oxygen are closer to each other and repel slightly more.

" The hydrogen bonds are slightly weaker (12–15%) than those in water " See https://www.google.com/amp/s/www.researchgate.net/publication/230086767_Hydrogen_bonding_in_hydrogen_peroxide_and_water_A_Raman_study_of_the_liquid_state/amp#ampshare=https://www.researchgate.net/publication/230086767_Hydrogen_bonding_in_hydrogen_peroxide_and_water_A_Raman_study_of_the_liquid_state

Edit my appologies, maxW was earlier with approximate the same answer but I did not see his comment. Should I delete this awnser so you can get the credits you deserve? (Glad to know we are on the same page :)


Are vitamins in tablet are more efficient than the vitamins in their natural source? In medical shop I have seen that there are loads of vitamins tablets available but I was wondering like does vitamins in tablet are more efficient than the vitamins which are available in their natural source like in fruits ?

For some vitamins, such as Vitamin C, which are always found in the form of a unique chemical substance, there is no chemical difference between the vitamin in a tablet--synthetic or otherwise--and the same vitamin as found in a foodstuff. Still, how well you absorb a vitamin might be affected by what else you consume at the time - for example a given amount of Vitamin C might be absorbed more or less efficiently as an ingredient in orange juice than as a pill, even though it is the same substance in both cases. But these differences might not necessarily be significant.

Other vitamins, such as Vitamin A or Vitamin D, are found in multiple closely related chemical forms, any of which the body can use to fulfill its need for the vitamin. In that case, the body may absorb or utilize the vitamin more efficiently when it is in one form relative to another. Fats in the diet can further influence your absorption of a fatty vitamin like Vitamin D.

Minerals, such as iron, are notorious for being made easier or harder to absorb depending on their chemical form and how they are bound to other components of the food. So, for example, iron in greens is known to be slower/more inefficient for the body to extract than iron in liver.

For most vitamins, it's much better to consume them in dissolved form (as they would be in most foods). You can test how much dissolves by putting a vitamin tablet into a cup of warm-to-hot (around body temp) vinegar and seeing how much of the pill dissolves. If it doesn't dissolve, it's exiting your body without any chance to use it.

So, for most vitamins, no.


Getting a cleaner GC Peak Resolution I have a GC with a peak pair I'm interested in separating but are not getting resolved cleanly at the moment. Any tips on what temperature programming tweaks might help me?

The problem pair elutes at about 3.3 mins. The temp program was 120 C for 2 min followed by a 8 C/min ramp to 180 C & a final hold of 5 min.

Changing the column, detector or carrier gas is not an option. I want to just get the best I can by temp. programming optimization.

I've tried isothermal runs at 120 & 100 C but not much help. The peaks still come unseparated.

Since it is just this two peaks that interest me mostly, theoretically a ramp is superfluous isn't it? Just figuring out the right isothermal run ought to give me the separation of interest?

Edit:

The column is an OV-351 packed col. 2 m long and 1/8" in diameter. N2 carrier gas with FID. The molecules are:

I just thought that the rearrangement might be occurring during the analysis I.e. The thermal ramp, implying a lower temperature would be better for you. The bp of each compound is too close to separate purely in that way.


Conversion of an epoxide into an alcohol I need help changing an epoxide into an alcohol, i.e. the second step in the scheme below. I found a reaction that created two hydroxyl groups but cannot find one that only produces one:

Superhydride (lithium triethylborohydride) does this. JACS 95 (1973)  8486 reports complete selectivity for unsubstituted over disubstituted carbon with very high yield. HC Brown JOC 45 (1980) (1) reports complete selectivity for 2-butanol from 1,2 epoxybutane and 97:3 for the reduction of styrene oxide.

As I can say about this, we can firstly use MCPBA to form an epoxides and then by observing the product we can use Lithium Aluminum Hydride which gives hydride ion(which favours SN2 mechanism) to conserve the number of carbons in the skeleton.

And now, after this the hydride ion attaches to the less hindered site, so after that we have to do the protonation to form this product.


Why is tetracyanonickelate square planar? Why is tetracyanonickelate, $\ce{[Ni(CN)4]^{2-}}$ square planar?

$\ce{CN}$, for being a strong field, would not follow Hund's rule.
the central atom $\ce{Ni}$ would have a +2 charge so its configuration would change from $\ce{3d^8 4s^2}$ to $\ce{3d^8 4s^0}$
Using the statements above and using crystal field theory I get its hybridization as $\ce{dsp^3}$, which suggests that its shape must be trigonal bipyramidal while it is not.

This is the square planar splitting diagram

Many d8 metal complexes are usually square planar because a very stable electron configuration is achieved if you put the electrons on this splitting diagram -6 electrons stabilizing the compound and only two slightly destabilizing electrons-

Now you could ask, why then not d6 and d7 adopt this cinfiguration? Because for 6 or 7 electrons, an even more stable state is achieved with octahedral fields.

About hybridization... Forget about it when talking about coordination complexes. It's completely useless. It can't explain or predict anything that CFT/LFT can't

It's tetrahedral; as not 6 lone pairs are coming from $\ce{CN}$,  only $\ce{CN}$ are attached. Therefore, only 4 lone pairs will affect the configuration.

So by now applying CFT you will get its hybridization to be $\ce{dsp^2}$ which will make the shape of tetrahedral.


Does vinegar react with tap water? In an effort to prevent mold in my humidifier, I added a little white vinegar to the tap water in the tank. This creates a white bloom that floats in the water. It is odourless and, despite its appearance, is not slimy. It feels more like paper pulp. After a while it adheres to the sides of the tank as a transparent sort of gel. Is this a known phenomenon? What is the white stuff?

Acidic acid (vinegar) does not react with water, and it is not soluble in water. The gel and white bloom may come from a reaction the acidic acid with minerals such as calcium that may be present in the dehumidifier. the gel could also come from the breakdown of the chitin cell wall of the fungi. If the white bloom appeared within a minute or two of the addition of acidic acid it was most likely a reaction with minerals or other compounds.


Saytzeff and beta elimination I am not able to understand the difference between Saytzeff's rule and beta elimination reaction. Also I am writing two reactions below:

\begin{align}
\ce{CH3-CH2-CH(Br)-CH3 + \underset{\text{(aqueous)}}{KOH} & -> CH3-CH2-CH(OH)-CH3} \tag{1} \\
\ce{CH3-CH2-CH(Br)-CH3 + \underset{\text{(alcoholic)}}{KOH} &-> \underset{\text{(major)}}{CH3-CH=CH-CH3} + \underset{\text{(minor)}}{CH3-CH-CH=CH2}} \tag{2}
\end{align}

Are they correct? If they are, then what are they called?

If the reagent is alcoholic KOH, then the reaction mechanism followed is E2 elimination (if steric/other conditions permit)

However, if the reagent is aqueous KOH, it is simply a substitution (SN) type reaction in which the Br atoms are replaced by -OH groups (not to mention, if conditions permit, i.e. SN is actually possible in the given molecule*)

*example: SN type reactions are usually forbidden in compounds like PhBr (bromobenzene) as the π cloud of the benzene ring repels the incoming nucleophile (SN² not possible) and formation of phenyl carbocation is not thermodynamically feasible due to its instability (SN¹ forbidden)

As far as your main question is concerned, i.e. difference between Saytzeff's rule and Elimination -

They are two completely different things, one is a rule to correctly predict major products of elimination reactions and the other is a reaction mechanism, which defines the pathway for the reaction to proceed through.

In fact, Saytzeff's rule states:
'In an elimination reaction, the most substituted product will be the most stable, and therefore the most favored.'

This is not always true, however. Major products are formed when the reaction intermediate is stabilized (product stability is usually secondary)
In case of eliminations, especially if the reaction is to proceed through an E1 pathway which involves formation of a carbocation, stability of the carbocation becomes a very important deciding factor in the product percentage. If the carbocation so formed (assuming reaction proceeds via predicted pathway. Practically, it is not fruitful to talk about such intermediates as they have a very small lifespan and reaction may proceed through other more favourable pathways) is highly destabilised even though it is more substituted, for example, carbon attached to three nitro groups (or any strong electron withdrawing groups for that matter) then Saytzeff's rule is violated and corresponding Hoffman product is found to be major. Of course, there are several other cases in which Hoffman product is found to be major and corresponding Saytzeff product is minor. I believe that discussion deserves a separate post altogether, so I won't elaborate on it here.

Hope it helped!

We know that beta elimination means elimination of beta hydrogen and hydrogen will give its electron to its carbon so that it can make double bond with another carbon . When there is two beta hydrogen in a chain then at the time of beta elimination the position of double bond will be different ( see the reaction you have given , two product came with different position of double bond ) then now we use satyzeff rule to identify which one produce more i.e main product .
If  starical hindrance is more then that product will remain as a main product as its starical hindrance is more other compound cant attack it( in your reaction you have given under the product major)and that compound will be produced more . In your reaction one product have double bond between 2 and 3 number carbon as double bond is in middle of it will have more starical hindrance than the one having double bond in 1 and 2 carbon since in 1 and 2 carbon double bond it have alkyl group in one side only and alky group is main factor here because it creates starical hindrance .

 Photo shows which one have more starical hindrance ( example)

N.B > My answer doesnot represent what is saytzeff rule , i just showed application and mecanism why product is more .


Acid-Base Titration of ammonium dihydrogenphosphate? How can I determine the concentration of ammonium dihydrogenphosphate solution?
In this compound, the ammonium cation and the dihydrogen anion are both weak acids.
I have a weak solution of this salt, $C_A\approx 0.05\mathbf{M}$, that I need to quantitatively determine.
My first thought was acid-base titration because it's cheap and easy but the presence of two weak acids complicates things.
Would it be possible to determine the concentration of ammonium and dihydrogen in one equivalence point, as $2\times C_A$?

The pH can be measured vs. the amount of $0.1$ M NaOH added to your solution with a pHmeter. Then two titrations would be visible : first the titration of dihydrogenphosphate ($p\ce{K_a = 7.21}$) and then the titration of ammonium ($p\ce{K_a = 9.22}$). If you don't have a pHmeter, you would probably use a colored indicator. But the end of the first titration will not be determined with precision, as the two $p\ce{K_a}$'s are not different enough. You will only be able to determine the end of the titration of the two acids together. To do this, a rather basic indicator is needed, like Alizarine yellow whose $p\ce{K_a = 10.5}$.


How to determine the iron content of a screw? In Analytical Chemistry, I was asked to determine the iron content of a screw.

I thought it could be solved in the following way:

Treat the screw with a solution of sulfuric acid or hydrochloric acid.
As the iron in solution can be found in the form of $\ce{Fe^{II}}$ or $\ce{Fe^{III}}$, I would reduce the possible $\ce{Fe^{III}}$ ions with $\ce{SnCl2}$ or Walden's reducer.
Titrate the solution with potassium permanganate.
From the volume spent, the amount of iron present could be determined.

I don't know if it will be OK, it is the first time I have faced a problem of this type.

There is another problem with this.

Potassium permanganate is a very strong oxidizing agent, which is capable of oxidizing Chloride ions to Chlorine. This will result in additional consumption of $\ce{KMnO4}$ solution, which makes you think there is more iron in your screw than there actually is.

To solve this, you should look up the titration of Iron using the Reinhardt-Zimmermann solution. It contains sulfuric acid, Manganese (II) sulfate and phosphoric acid.

The $\ce{MnSO4}$ changes the redox potential so the acidic $\ce{KMnO4}$ solution can no longer oxidize Chloride ions. Because Iron(III) gives an intense yellow solution, the addition of Phosphoric acid is necessary since it will react with Iron(III) to colorless Iron phosphates. That makes it easier to detect the end point.

So the way I did this back in my first semester was to make sure we have all Iron as Fe(II) by adding an excess of $\ce{SnCl2}$. The excessive $\ce{SnCl2}$ was quenched using $\ce{HgCl2}$. Then the Reinhard-Zimmermann solution was added and titration with $\ce{KMnO4}$ could start.

A simple, safe and effective electrochemical based path (absence strong acids which, acting by themselves, are generally inert on stainless steel at intermediate temperatures) to gather surface metal salts for testing is, per a source, to quote:

  The iodine uptake, and corrosion of the stainless steel surfaces was extensive. The observed time-dependent behaviour suggests that the iodine adsorption on the stainless steel surface occurred in two stages, a slow induction period followed by a fast linear adsorption rate region. Iodine deposited on the stainless steel was desorbed during purging with air, but not with N2. These observations could be explained by the formation of metal-oxy-iodide scale, in which iodine is initially incorporated into but subsequently replaced by oxygen. This paper presents the preliminary results of the study, and their potential implication for "calibration" of iodine transmission through stainless steel sampling lines.

Some hot I2 vapor based corrosion on a stainless steel screw in an atmosphere of oxygen is a path to gathering surface formed oxy-iodides. The latter can be used to assess metal content of the alloy which is mainly Fe, but significant alloyed elements of Ni, Cr, Cu,..., also possible.

Also, in time, even very dilute iodized water is problematic on stainless steel, to quote NASA:

  The effects of stainless steel exposure to iodinated water is a concern in developing the Integrated Water System (IWS) for Space Station Freedom.

Interestingly, the rate of corrosion may be reduced (or, even possibly accelerated) in the presence of chromium and nickel, to quote a source:

  Iodine is capable of reacting with iron in the steel, as well as with nickel and chromium. This serves to complicate the situation. As well, there may exist synergistic relationships between these metals. This means that the presence of one metal may inhibit or enhance the reactions of iodine with one of the other component metals. Hellmann and Funke [19] performed a literature review of relevant reactions on this topic.

[EDIT] I found a good albeit dated (like 1913) reference of on the Zimmermann-Reinhardt (ZM) method here. I must correct, however, my misperception above that, somehow, a combination of H2SO4 and 85% H3PO4 in the presence of ferrous or stannous constitutes a 'strong' acid mix that attacks stainless steel. The less appealing truth, having an advanced knowledge of the reaction mechanics, KMnO4 (a substitute for H2O2), fosters powerful acidic radical formations. This is the actual basis of ZM method (which some may cite as historical, and others, as inefficient, reagent exhausting, and possibly dangerous route to the sulfate radical anion ($\ce{.SO4-}$) and likely the phosphate radical anion as well).

Now while, I have proclaimed elsewhere that the nitrate radical anion (see comments and source here), for example, exceeds the nitrating ability of concentrated Nitric acid, this was not well received. Here, the sulfate radical anion apparently assists in the attack of stainless steel and its creation, for the record, does not require strong H2SO4 or HgCl2 or KMnO4, just a source of hydroxyl radicals via a Fenton/Fenton-like reaction with transition metals in their lower valence states at an appropriate pH, or a photocatalyst creating an electron-hole in the presence of water, introducing the powerful hydroxyl radical ($\ce{.OH}$), which furthers reacts with the nitrate or sulfate or chloride ion as follows:

$\ce{ .OH + SO4(2-) -> OH- + .SO4- }$

$\ce{ .OH + NO3- -> OH- + .NO3 }$

$\ce{ .OH + Cl- -> OH- + .Cl }$

$\ce{ .Cl + .Cl -> Cl2 (g) }$

However, apparently couched in the veil of nostalgia ZM is applauded, congratulations.

For those actually interested in modern chemistry, I quote some revealing passages from the cited ZM source:

  If dilute permanganate is allowed to run into a cold dilute solution of ferrous chloride containing hydrochloric acid,  the permanganate is decolorized and the iron is oxidized, but there is a noticeable evolution of chlorine...If, however, permanganate is run into cold dilute hydrochloric acid in the absence of ferrous chloride, there is no evolution of chlorine.

So, an acidic promoted Fenton/Fenton-like reaction with ferrous (or stannous, also mentioned in this source work):

$\ce{ Fe(2+) + H2O2 (or KMnO4) + H+ -> Fe(3+) + .OH + H2O }$

and the creation of chlorine as detailed above. The sulfate radical itself is likely a stabilized transport for the powerful, but transient, hydroxyl radical, to quote source:

  It has also been suggested that the sulfate radical anion may be converted to the hydroxyl radical in aqueous solution.

So, I speculate that the attack of steel may be due to an aqueous sulfate radical assisted release of the hydroxyl radical on the iron surface:

$\ce{Fe + 2 .OH -> Fe(OH)2}$


Rutherford's gold foil experiment Rutherford projected $\alpha$-rays onto the gold foil. $\alpha$-rays are the same as a $\ce{He^2+}$ nucleus and they can ionize atoms. Therefore, why are gold atoms not ionized when the $\alpha$-rays collide with them?

In Rutherford's experiment to show the existence of nucleus in an atom, the alpha-particles were exposed on the surface of certain metal i.e. gold. He observed that more than 99% of these particles were able to go straight and pass through the gold atoms. Thus he was able to conclude that atoms are mainly empty space.
But why did these particles not collide with the electrons present in the gold atoms? I think that the reason behind it is that the speed of the electrons travelling in orbits did not cause any obstacle in the travelling path of the alpha particles.


Why doesn't LiHCO3 exist in solid state? I have read that $\ce{LiHCO3}$ does not exist in solid state. Is it because of the size of anion-cation concept or is it something else? So why is it only for lithium and not for sodium too?

Claiming that a particular chemical compound "doesn't exist" is difficult to prove. The only convincing proof would be to show that neither LiHCO3 nor any of its hydrates can be isolated from (aqueous) solution. Essentially, you would have to show that using any of an almost infinite number of methods, that it couldn't be prepared at any temperature or pressure - not very likely to have ever been done, imho. So, I'll ask you for your source for such a claim, or better yet, I'll ask you to check and verify that your source - which you accept as being accurate - has made stringent efforts that demonstrate the claim as well as having published that work in peer reviewed literature. As an aside, there are plenty of compounds which require "waters of hydration" in their crystal structure. If it really doesn't exist, then the reason is simple: kinetically the reaction 2LiHCO3 → Li2CO3 + H2O + CO2 is spontaneous and energetically favored. It just seems to me that if I placed Li2CO3 in a pressure cell with water and CO2 at a billion gigapascals of pressure, I could make the bicarbonate salt, but perhaps some other structure is more likely...and anyway, generally what they mean by "doesn't exist" is that it "isn't known to exist at standard temperature and pressure".

due to small size of lithium its hydration enthalpy is very high and moreover hydration enthalpy is higher than lattice enthalpy of lithium bicarbonate

hence lithium bicarbonate  exists in solution  state only


A quick method for separating liquids? I'm looking for a way to take 3 separate liquids (liquids which function as paint -- they will all be different colors), put them together into a container, and then rapidly separate them back to distinct colors.

I can handle all the electrical engineering aspects of my project, but this portion is beyond me. Does anyone have any ideas?

When I say "a quick method", I mean something that can be done easily without an excess of time or equipment. Ideally an way to make the paints intrinsically immiscible, but I need to be able to separate them entirely, so that when they are poured together into a container, I can then separate and reuse them.

A stable 3-phase liquid system can be created, as Spießbürger mentioned with three mutually immiscible liquids. The best one that I know of is the organic-aqueous-fluorous 3-phase system.

Fluorocarbons (hydrocarbons in which every hydrogen atom is replaced by a fluorine atom), are interestingly immiscible in both water and common organic solvents. This property has added a dimension to chemical separations using fluorous phase extractions.

If you have three paints, one oil-based, one water-based, and one fluorous-based, then you could separate them easily using a separatory funnel or similar device. You would have to make sure that the pigments in each paint was only significantly soluble in that solvent, which would take experimentation.

Sorry to say no experience but I would think a polymer base cheap achrylic mixed with a 3/4  Mercury metal heated to bond, a symple oil base paint and liquified water color would do the trick in a more than precise time


Why do some covalent compounds form lattice while others do not? Why do some covalent compounds form lattice while others do not? I'm not only talking about amorphous solids but covalent compounds that form molecular lattice in general. For Example: Compounds like SiO2, SiC and AlN form covalent lattices while some covalent compounds like methane and SiF4 form molecular lattices. The question is - Why do only some covalent compounds do this and also, is there a way to identify which compound forms covalent lattice and which forms a molecular lattice? If so, then how?

NOTE: I've looked this up over Concise Inorganic Chemistry by JD Lee and also on the internet but have found nothing.

See, as far as I know, their is no fixed way to compare catenation properties of elements randomly without any context. Some general pointers however, to keep in mind are-:
Carbon has one of the highest catenation powers, as soap molecules like sodium stearate can contain up to 17 carbon atoms.
After carbon, some other elements having significant catenation tendencies are silicon, sulphur, boron etc. (roughly in that order). I will not be elaborating further here as I'm not very sure myself about this except perhaps a few rough pointers such as medium-ish size of atom and the presence of group 13 and 14(which have the maximum number of covalent sigma bonds formed without expansion of octet)
The thing you really need to understand is that the ability to form multiple bonds is more important here to predict the bonding nature of the crystal lattice. Oxygen and nitrogen can form multiple bonds, which would allow them to kind form cross links while satisfying the valency of the central atom. While on the other hand, fluorine will finish it's covalency in satisfying the needs of the central atom (Si) only, so the tetrahedral unit formed will majorly interact via Van der Waals forces (checking it with the facts, this is a pretty good prediction). The exact crystal structure of SiO2 and ALN are quartz and wurtzite respectively, which would be difficult to predict simply by making the covalent structures. However, a fair idea can be taken about the kind of interactions which will prevail in the crystal lattice, provided it is also covalent.


The effect of pH and solubility on coagulation in water treatment In water treatment, a specific pH range must be met in order for the process of coagulation to occur properly. Sources have stated this is due to the pH affecting the solubility of the coagulant, but why is solubility significant?

To start, I assume you are speaking of metal salt coagulants such as aluminum sulfate (alum) and ferric chloride, as these are the most commonly used. As you have stated, their solubility is affected by pH, the simplest representation of the reaction for alum being:
$$\ce{Al^3+ + 3H2O <=> Al(OH)3,_s + 3H+}$$

Small particles in the water such as clay or bacteria are too small to settle out on their own. The purpose of coagulation is to either cause the smaller particles to aggregate and settle or to trap them during precipitation. Coagulation occurs via three main mechanisms: (1) sweep - where the metal salts precipitate (i.e., solubility) and capture particles/contaminants like a net; (2) adsorption/charge neutralization - where the positive charges on the Al and Fe cations neutralize the negatively charged e.g., clay, bacteria, etc., causing them to aggregate and settle out; (3) adsorption and bridging - where a polymer coagulant can be used to "bridge" together like charged particles, causing them to aggregate and settle out. All are affected by solubility, but sweep is the mechanism that uses precipitation of the metal salts to remove particulates.

The mechanism you will want to choose will depend on the source water and the coagulant. The mechanisms are controlled by pH and solubility. Below are the solubility diagrams for alum and ferric chloride, with solubility represented on the left y-axis, dose on the right y-axis, and pH on the x-axis. As you will see, the shaded areas are the regions for achieving optimal coagulation, depending on the mechanism you desire. For example, if your laboratory tests showed that sweep coagulation was the best for your source water and you chose alum, then you would want to be at a pH between ~7 to 8 and an alum dose of ~15 to 60 mg/L alum (the "optimal sweep" circle). This pH and dose would result in the desired solubility behavior to achieve optimum sweep conditions.

*Figure 9-11 from MWH Water Treatment: Principles and Design, 3rd Ed (Crittenden)

When aluminium sulfate (a common coagulating agent) is used for water treatment, it creates byproducts in the form of aluminium residuals.

In the process of coagulation, it is beneficial to minimise the amount of aluminium residuals produced by aluminium sulfate.

Consequently, before the coagulation process begins, the water is corrected to a certain pH dependent upon the dose of aluminium sulfate and the solubility of aluminium.

By treating the water to have a pH which exceeds the solubility of aluminium, these residuals that the process of coagulation leave behind are reduced.

Therefore, solubility is significant because it is a determining factor in the amount of byproducts that coagulation produces.


Why is pyridine used when making tosyl esters from alcohols? Tosyl chloride is used to make a hydroxyl group into a better leaving group. However, when the reaction of tosyl chloride and an alcohol occurs, a weak base such as pyridine should be used. Why?

The function of pyridine is actually not so simple and not so easy to notice at first glance. There is a fundamental reason why pyridine is used to promote the acylation reaction, which is that it can act as a catalyst.

Despite its basicity and the subsequent formation of its chlorhydrate salt after the tosylation reaction, pyridine is also an excellent nucleophile. Pyridine is, in fact, more nucleophilic than the alcohol, and it attacks the acyl chloride rapidly, forming a highly electrophilic (because of the positive charge) intermediate: N-tosylpyridinium chloride. It is indeed this intermediate the actual tosylating agent which reacts with the alcohol to give the ester. Because pyridine is acting as a nucleophile to speed up the reaction, yet is unchanged by the reaction, it is a "nucleophilic" catalyst.

In my experience with acylations and esterificatio, (both O-acyl and O-propionylations) , a very small catalytic amnt of DMAP (100 mg per mole of product) along with reduced qty of pyridine has allowed the room temp acylation of some normally sterically hindered organics in less than 30 mins. In fact, while working with temp sensitive piperidine based compounds, the acylation was performed at - 10 deg C and had 98% conversion in 35 mins. The undesirable temp sensitive side products were at less than 1% and with some optimization managed to get undesired side products down to less than 0.05%. Of course DMAP is considered more toxic than regular pyridine, but it just with pyridine it can be separated from products by similar methods. We were dealing with esp lipophilic compounds acylated in DCM, and removal of pyridines simply involved a wash with acidic water. Even though the products formed HCL salts. Their products are lipophilic nature kept them in the DCM layer. We never experienced more than 5% product loss during acid wash workup and optimization reduced the workup loss to 2%. Overall, DMAP is impressive and stuff. Even with non temp sens acylations, you can still achieve rapid nearly full acylations in 20 mins at room temp and also the use of a small cat qty of DMAP seems to reduce the pyridine requirements, in our case we reduced pyridine qty by a factor of 5 without detriment to our yield or efficiency In my experience with acylations, using acyl and propionylations, a very small catalytic amnt of DMAP (100 mg per mole of product) along with pyridine has allowed the room temp acylation of some normally sterically hindered organics in less than 30 mins. In fact, while working with temp sensitive piperidine based compounds, the acylation was performed at - 10 deg C and had 98% conversion in 35 mins. The undesirable temp sensitive side products were at less than 1% and with some optimization managed to get undesired side products down to less than 0.05%. Of course DMAP is considered more toxic than regular pyridine, but it just with pyridine it can be separated from products by similar methods. We were dealing with esp lipophilic compounds acylated in DCM, and removal of pyridines simply involved a wash with acidic water. Even though the products formed HCL salts. Their products are lipophilic nature kept them in the DCM layer. We never experienced more than 5% product loss during acid wash workup and optimization reduced the workup loss to 2%. Overall, DMAP is impressive and stuff. Even with non temp sens acylations, you can still achieve rapid nearly full acylations in 20 mins at room temp and also the use of a small cat qty of DMAP seems to reduce the pyridine requirements, in our case we reduced pyridine qty by a factor of 5 without detriment to our yield or efficiency.

The temp sensitive reaction was primary alcohol phenylpipedidine to ester acylation. It is temp and acidic sensitivity was needed to avoid problematic olefination of alcohol forming a undesirable phenylpyridine which at the time was suspected to have possible neurotoxicity mediated thru MAO mediated biotransformation that have been known to have highly specific dopaminergic neuron toxicity and have henceforth been utilized to induce Parkinsonism in animals for the testing of anti Parkinson agents, selegline and others have resulted from this line of research. A related study showing phenylpyridinium toxicity to be highly dependent on N-substitution on the piperidine, any group larger than N-propyl showed no reactivity to MAO B or MAO A enzymes, thereby not undergoing the enzymatic transformation to neurotoxicm. Our piperidine was substituted both on the carbon ring and N-phenylpropyl on piperidine nitrogen so there was no surprise when later testing found the olefin byproducts to have no MAO reactivity. Even taking into acct the lack of toxicity of our undesirable byproducts, the DMAP was an impressive catalyst for acylation and ESTERIFICATION that I still would have chosen to work with it out of the convenience of performing rapid room temp acylations with minimal pyridine and easy workup. DMAP is also highly effective in N-acylations with similar benefits in efficiency and yield. I've read of some studies that also show rapid conversion of N-acylations at room temp with tiny catalytic qty of DMAP.


Why can't Schiff's reagent distinguish between glucose and fructose? I have heard that Schiff's base only gives a positive test for aldehydes. Glucose is an aldehyde. However, I was told that glucose doesn't give a positive Schiff's base test. Is this correct? If so, how?

Schiff test is a colorimetric method for determination of aldehyde groups. Both glucose and fructose are monosaccharides and exhibit oxo-cyclo tautomerism. Glucose is an aldose and has aldehyde group in linear form. Whereas fructose is a ketose and doesn't give a positive Schiff test.


CO2 Catalytic Converter We all know how CO2, as a greenhouse gas affects the climate and causes global warming. I was thinking of a small device, which could be attached to the exhaust pipe of an automobile. It could convert the CO2 into something having negligible greenhouse effect, such as water. Is this feasible? Thanks!

Unfortunately this is not feasible
Current catalytic converters are designed to remove noxious pollutants from exhausts. Things like carbon monoxide (poisonous) and nitrogen oxides (noxious and irritating).
The trouble with doing this for carbon dioxide is fundamental. Carbon dioxide and water are the ultimate products of combustion of hydrocarbons because they are the thermodynamically deepest products of combustion (ie forming them releases the most energy from combustion). We are not worried about water. But the trouble with being at the bottom of the thermodynamic well is that, to turn the compound into something else, you need to add energy.
There are known chemical reactions that convert carbon dioxide into other, more useful and less greenhouse-contributing products. We can go the whole hog back to methane or, more commonly, methanol. But these reactions are very energy intensive and require fairly special catalysts and conditions. To do them as a way of reducing carbon dioxide emissions would end up using most of the energy from hydrocarbon burning to run the reaction not to move the vehicle. And there are always energy losses in such reactions so we might end up greatly increasing fuel consumption overall to get a small reduction in emission. It is probably far better to make engines and vehicles more efficient.
On an industrial scale the preferred alternative to carbon dioxide conversion is carbon capture and storage (CCS). Separating carbon dioxide is much cheaper than converting it to something else. Even so, CCS plants are in their infancy and none are remotely economical even when subsidised (or at any feasible carbon price). and these reactions require large chemical plants and could never be viable in a car.

While not likely easily created, other products with air, $\ce{CO2}$ and water, in the presence of an electrical source, are conceivable in my opinion.
For example, there is some discussion around NAIs per this government source to quote:

One of the examples is the negative ion generator using an ultraviolet source to irradiate electrically conductive material, which was patented as early as 1964 (Patent No. US 3128378 A). In this patent, an ultraviolet lamp was used to irradiate metal materials, which photo-electrically eject electrons. The electrons then collide with air molecules and generate NAIs.

Assuming the chemistry involves superoxide radical anions created as follows:
$\ce{O2 + e- -> .O2-}$
And, in a medium of moist air, the superoxide is present as the hydroperoxide anion (see Wikipedia discussion here):
$\ce{H2O <=> H+ + OH-}$
$\ce{H+ + .O2- -> .HO2}$
In an albeit slow reaction, this is a precursor to hydrogen peroxide:
$\ce{.HO2 + .HO2 -> O2 + H2O2 }$
although this source notes an acceleration of the self-reaction in the presence of $\ce{CO2}$. It is also known that:
$\ce{H2O2 + CO2 -> H2CO4}$
as per a source: The Mechanism of Carbon Dioxide Catalysis in the Hydrogen Peroxide N-Oxidation of Amines, to quote:

"The reactivity of the peroxymonocarbonate ion, HCO4- (an active oxidant derived from the equilibrium reaction of hydrogen peroxide and bicarbonate), has been investigated in the oxidation of aliphatic amines"

Also, see Figure 2 at this source. Possibly, a more direct (and rapid) postulated radical reactions (which is in accord with the above):
$\ce{.HO2 + HCO3- -> .HCO4 + OH-}$
$\ce{.HCO4 + e- -> HCO4-}$
Here is an interesting supporting reference: Chemical effects of large impacts on the Earth's primitive atmosphere providing support for a prior significant presence of H2CO4 in the primitive earth's atmosphere, to quote:

Intense bombardment of the moon and terrestrial planets approximately 3.9-4.0 x 10(9) years ago could have caused the chemical reprocessing of the Earth's primitive atmosphere. In particular, the shock heating and rapid quenching caused by the impact of large bodies into the atmosphere could produce molecules such as HCN and H2CO4 which are important precursors for the abiotic synthesis of complex organic molecules.

Bottom line, some more research, and work may provide a path in the future.
[EDIT] The future is near. See The O2-assisted Al/CO2 electrochemical cell: A system for CO2 capture/conversion and electric power generation. To quote from the abstract:

Economical and efficient carbon capture, utilization, and sequestration technologies are a requirement for successful implementation of global action plans to reduce carbon emissions and to mitigate climate change. These technologies are also essential for longer-term use of fossil fuels while reducing the associated carbon footprint. We demonstrate an O2-assisted Al/CO2 electrochemical cell as a new approach to sequester CO2 emissions and, at the same time, to generate substantial amounts of electrical energy... On this basis, we demonstrate that an electrochemical cell that uses metallic aluminum as anode and a carbon dioxide/oxygen gas mixture as the active material in the cathode provides a path toward electrochemical generation of a valuable (C2) species and electrical energy. Specifically, we show that the cell first reduces O2 at the cathode to form superoxide intermediates. Chemical reaction of the superoxide with CO2 sequesters the CO2 in the form of aluminum oxalate, Al2(C2O4)3, as the dominant product. On the basis of an analysis of the overall CO2 footprint, which considers emissions associated with the production of the aluminum anode and the CO2 captured/abated by the Al/CO2-O2 electrochemical cell, we conclude that the proposed process offers an important strategy for net reduction of CO2 emissions.

As such, my opening sentence is completely accurate. Further, if one re-reads my proposed answer, there are some uncanny similarities.
What is quite amazing is that the cited reaction path also involves the presence of electrons, oxygen, the superoxide radical anion, and CO2 leading to a composite compound.
Assertions of a definitive negative answer here, is patently incorrect!