Carbocation rearrangement with expansion of five-membered ring? I am confused with this particular problem.

I began the problem with a mechanism as following. In the beginning $\ce{H+} $ protonates the alcohol and leaves as $\ce{H_2O}$. Then everything is about carbocation rearrangement.

But then my professor gave out a different answer saying that ring expansion doesn't happen.

Why doesn't ring expansion take place? Isn't a six-membered ring more stable than a five-membered ring? I am not sure where I am making the mistake.

Your proposed mechanism involves a ring expansion from cyclopentane to cyclohexane. This gives very little driving force in terms of relieving strain in the system. Cyclopentane, in a twisted or "envelope" conformation, has some angle strain ($102-106^o$ vs $109.5^o$ $^{ref~1}$) and it does also has some torsional strain down some of its bonds.

This is mostly alleviated by the move to cyclohexane, if it is in a full chair conformation, but the stabilization of the system is minimal ($25-29kJmol^{-1} $ $^{ref~2}$). Here is a link to a question that deals with ring strain and the calculation of it from $\Delta H$ of combustion.

Along with this, your mechanism proposes this ring opening involving an alkyl shift that moves the carbocation from tertiary to secondary. This involves a destabilizing of the cation due to the reduced hyperconjugation (and inductive effect). Studies have shown that a secondary carbocation is around $67-75kJmol^{-1} $less stable than a tertiary, for simple substituents (and in the gas phase$^{ref~3}$).

In total, the ring opening from a five to six-membered ring in your mechanism does not result in an overall increase in stability. Hence why the second mechanism is suggested to happen instead.


How do Cryocoolers that use Helium as Coolant work? I was unable to find elaborate information on helium based Cryocoolers. I like to know how they use helium to cool samples to cryogenic temperatures and why helium is the best option as a coolant.

A helium based cryocooler works just like an air conditioner, but helium is the gas rather than a halocarbon compound. The boiling point of helium is −268.928 °C, so such a cryocooler could get really cold.

Helium is a rare gas and relatively expensive. So to cool a small experimental apparatus it is much cheaper to buy one helium based cryocooler rather than dewars and dewars of liquid helium.


Kinetic validity; redox reactions What's the kinetic validity of the following way of balancing the above redox reaction?

I've done some searching of the literature; I've never come across oxide anions being mentioned in redox reaction mechanisms. Also, I read something about how two-electron reduction is impossible in a single step - something about spin pairing.

I'm fairly sure the following method is a good thermodynamic method of thinking about the reaction, but what about its kinetic validity?

I'm thinking that the below method isn't kinetically valid, but neither was it meant to be kinetically valid, and neither are the traditional methods of balancing redox reactions kinetically valid. Traditional methods call for balancing "O" and "H." But we aren't dealing with 0 oxidation state oxygens and hydrogens, are we?

Here is the thought process that my professor would take in solving this problem:

1) Skeleton reaction. Note that sulfur is being reduced from a +6 oxidation state to a +4 oxidation state. This suggests that the solid zinc loses two electrons.

$\ce{Zn + HSO4^- ->SO_2 +Zn^2+}$

2) Note that two negative two oxidation state oxygens are "disappearing" from left to right. In other words, what happened to the oxygens in the hydrogen sulfate ion?

$\ce{Zn + HSO4^- ->SO_2 +Zn^2+}$

$\ce{4~ negative ~two~ oxidation~ state ~oxygens -> 2 ~negative ~2~ oxidation~ state~ oxygens }$

3) The special thing about these oxygens is that they're negative two oxidation state oxygens. This suggests that these oxygens are the oxide anion. The oxide anion is highly basic in solution. Therefore, the oxide anions must have reacted with some system component, which caused them to "disappear." This would be the strongest acid in the system.

$\ce{HSO4^- ->SO_2 + 2O^2- (unstable)}$

$\ce{2O^2- + 4H3O^+ ->6H2O}$

4) Also note that a hydrogen proton disappears from left to right in the skeleton equation. Because the proton is highly acidic, it is attacked by the strongest base in the solution - water (since we're in an acidic solution).

$\ce{HSO4^- ->SO_2 + 2O^2- + H^+}$

$\ce{H^+ + H2O -> H3O^+}$

5) Final balanced equation:

$\ce{Zn + HSO4^- + 4H3O+ + H2O ->SO_2 +Zn^2+ +6H2O + H3O+}$

When you balance the equation for a redox reaction, the process of balancing it or the balanced equation itself say nothing about kinetics.  That said, redox reactions, as the name implies, deal with electron transfer.  Unless you are dealing with unusual molecules (bulky ligands for example) or unusual conditions (solids, glasses, very low temperature, etc.) electron transfer reactions are close to diffusion controlled, so kinetics is usually not an issue and is often glossed over in such reactions.


Methods to prepare lactic acid from acetaldehyde I got one method [here](using acetaldehyde and carbon monoxide)1.

But is there any alternative method (using specific catalyst)?

A fairly simple two step synthesis involves nucleophilic addition of $\ce{HCN}$ to the carbonyl group and then aqueous hydrolysis of the nitrile to yield the carboxylic acid product.


Cheap but efficient disinfectant / biocide for home lab EDIT

I am forced to reword my question about a chemical mixing ratio.

I have seen in Wikipedia that peroxide ($\ce{H2O2}$) and Acetic Acid can be mixed to Peracetic acid - this is an equilibrium.

I have also seen in Wikipedia that peracetic acid mixed with acetic acid will be explosive - in other words: If I don't get the mixing ratio just right then the "peracetic acid" will actually be "peracetic acid dissolved in acetic acid". I need to avoid this situation.

I have also read in Wikipedia that "thinning" the peracetic acid will shift the equilibrium and so this might end up in an explosive solution - I need to avoid this.

I have also seen at amazon.de that peressigsäure/alkohol is widely used as a disinfectant and I suppose the ratio of ethanol should be at 69% in order to kill most any microbes.

I need to mix a weak solution of $\ce{H2O2}$ and acetic acid and 69 % ethanol so that the peracetic equilibrium is held to be non explosive. And I also need to mix a more potent solution of $\ce{H2O2}$ and acetic acid and 69 % ethanol so that again, the peracetic equilibrium will fit and not be explosive.

As far as I understood it is not done by having the potent solution in the fridge at 0 degree Celsius and on demand simply thinning the potent solution to a weaker disinfectant - the peracetic equilibrium would then shift and it might become explosive.

Sorry, I am not really fit to calculate concentrations in Vals and Mols - but I could of course deal with any complex mathematical formula (on paper or on the calculator - or I could even code a small mixing calculator in java or c#).

So I would ask you to give me the right formula for my problem:

Component A = 12 % $\ce{H2O2}$

Component B = 20 % Acetic Acid

Component C = 98 % Ethanol

Component D = $\ce{H2O}$

AND I need 1 liter of mixture so that Alcohol appears with 69 VOL % and that peracetic acid appears with $x$ % ($x$ between 2 and 10) and no acetic acid will be in the mixture aside from the equilibrium.

I could then simply vary the $x$ (the conecntraction of peracetic acid) in the formula and mix as needed.

please, show me how to calculate the concentrations.
thanks!

The Amazon website gives the composition of Peressigsaure/Alkohol: Inhaltsstoffe pro 100g:

● 19g 2-Propanol

● 5g Ethanol

● <1% Wasserstoffperoxid

● <1% Peroxyessigsäure

● <1% Essigsäure

Working time 30 seconds, shelf life 12 months.

Don't expect that H2O2 plus acetic acid to give peroxyacetic acid. The full Wikipedia reference is:

Peracetic acid is produced industrially by the autoxidation of acetaldehyde:[2]

O2 + CH3CHO → CH3CO3H

It forms upon treatment of acetic acid with hydrogen peroxide with a strong acid catalyst:[3]

H2O2 + CH3CO2H ⇌ CH3CO3H + H2O

Making the peroxy acid can be done from acetic acid, but requires a strong acid catalyst (to generate CH3CO+).

So, you will have to get peroxyacetic acid from somewhere in some form, in addition to your other chemicals. Duplicating this formula requires more isopropanol than ethanol, but either alcohol should perform equally.

An overview document titled Chemical Disinfectants at https://www.cdc.gov/infectioncontrol/guidelines/disinfection/disinfection-methods/chemical.html mentions two chemical sterilants are available that contain peracetic acid plus hydrogen peroxide (i.e., 0.08% peracetic acid plus 1.0% hydrogen peroxide [no longer marketed]; and 0.23% peracetic acid plus 7.35% hydrogen peroxide. This document also lists the benefits of H2O2, peracetic acid, alcohol(s) and chlorine bleaches. And formaldehyde! About peracetic acid, it says: It is considered unstable, particularly when diluted; for example, a 1% solution loses half its strength through hydrolysis in 6 days, whereas 40% peracetic acid loses 1%–2% of its active ingredients per month.

As regards explosivity, a document of ECETOC at http://www.ecetoc.org/wp-content/uploads/2014/08/JACC-040.pdf gives the formulas of two cleaners as:

1: PAA 15%,H2O2 23%,HOAc 16%,H2O 45% and

2: PAA 15%,H2O2 14%,HOAc 25%,H2O 42%.

I gather from these two formulations that they and anything less in H2O2 or PAA would not even remotely be explosive. No alcohol was listed in these formulations.

Now if you could obtain either of these concentrated sterilants, you would have a reasonable ratio of peracetic acid, H2O2 and acetic acid. The only item missing would be the alcohol. Let's use ethanol, at 70% (or 69%, if you must). The peracetic acid cleaners would have a good shelf life because of their high concentrations. When you need to make up some sterilant, you could take ~800 grams of ethanol and add ~350 grams of the cleaner you choose to get 1 L of final sterilant (density ~1.150 g/cc), which will be 69-70% ethanol, 4.5% PAA, 4.3-7% H2O2, 7.6-4.9% HOAc, remainder H2O.

These levels are 5 or more times higher than the concentrations listed for Peressigsaure?Alkohol Pro, so you could consider adding much less of the peracid cleaner to get in the range of 1% PAA and H2O2.

Overall, it seems that H2O2 by itself is good at killing bacteria, and so is alcohol, by itself, and so is peracetic acid, by itself, but PAA is not stable for long at low concentrations (12 months at 1%??).

And perhaps it would be easier to buy the Peressigsaure/Alkohol Pro, or one of the other sterilants offered by that company, because it already comes in a spray bottle.


Balanced reactions at the anode and cathode A current of 2.25 A is applied to  $\ce{NiCl2}$ solution

A. Write the balanced half reaction that takes place at the anode
B. Write the balanced half reaction that takes place at the cathode

Can someone check if my answers seem logical? I'm not quite sure if I did the problem correctly

Anode: $\ce{O2 + 4H+ + 4e- -> 2H2O}$

Cathode: $\ce{2Ni^{2+} + 4e- -> 4Ni}$

The reaction at cathode seems OK, but at the anode another reaction will occur.

Hint:

At the cathode, 4 e$^-$ are consumed. The electrons have to be "generated" in the system.

Remark:
The value of current applied is irrelevant in this kind of question. More relevant will be the applied potential.


Which nuclei are NMR active? I thought that only way a nucleus can be NMR active is when the atom has an odd mass, which means that there is an odd number of protons or neutrons and an even number of the other particle. I thought that for an atom with even mass, the number of neutrons and protons are equal, and therefore the magnetic moments would cancel.

Now I reading that if both the protons and neutrons are odd numbered (giving even total mass) the nucleus is still NMR active. What is the reason for this?

The spin of atomic nuclei can be predicted based on the knowing the number of protons and neutrons in the nucleus:

Even/Even. Nuclei that contain an even numbers of protons and an even number of neutrons have $I = 0$ and are NMR silent. Examples include $\ce{^{12}C}$, $\ce{^{16}O}$ and $\ce{^{32}S}$.

All other nuclei are NMR active:

Odd/Odd. Nuclei that contain an odd number of protons and an odd number of neutrons have I that are positive integers. Examples include $\ce{^2H}$ $(I=1)$, $\ce{^{14}N}$ $(I=1)$ and $\ce{^{10}B}$ $(I=3)$.
Odd/Even & Even/Odd. All other nuclei (odd/even and even/odd) have spins that are half integral. Examples include $\ce{^1H}$ $(I=1/2)$, $\ce{^{11}B}$ $(I=3/2)$, $\ce{^{13}C}$ $(I=1/2)$, $\ce{^{17}O}$ $(I=5/2)$, $\ce{^{19}F}$ $(I=1/2)$ and $\ce{^{31}P}$ $(I=1/2)$.

A precise formulation is far more complex, and arises from the fact that the key components of the nucleus (proton and neutron) are made up of quarks. The neutron, although neutral in charge, is actually made of 3 quarks (udd), which themselves are charged, and not uniformly distributed about the neutron 'sphere'. Protons also consist of 3 quarks (duu). The nuclear spin comes from the total angular momentum of all these quark components. To delve deeper is probably outside the realm of chemistry, and is probably better suited to a physics forum (at least it is outside my ability to explain).


Amount of substance of a molecule in a solute the same as amount of substance of constituent elements? I’ve done absolutely no chemistry since high school.  Now, 16 years later, I’ve been asked to look at some chemical analyses and I’m having trouble!

Context: I’ve got a series of spectroscopy results giving the concentrations (in mg/l) of chlorine and $\ce{SO4}$ in a sample.  I want to find the molar and mass ratios of $\ce{S}/\ce{Cl}$.

What I’ve done: I convert the mg to g, then (mass) concentration into a molar concentration by dividing by the molar mass (35.5 g/mol for $\ce{Cl}$ and 96 g/mol for $\ce{SO4}$).

Question: am I right in thinking that the molar concentration of $\ce{S}$ will be the same as the molar concentration of $\ce{SO4}$?  My reasoning is that 1 mol $\ce{SO4}$ contains 1 mol $\ce{S}$ and 2 mol $\ce{O2}$.

Furthermore, to get the mass concentration of $\ce{S}$ alone, can I just multiply my concentration of $\ce{SO4}$ by the molar mass ratio, $\frac{M(\ce{S})}{M(\ce{SO4})}$?

You are right thinking that the molar concentration of S will be the same as the molar concentration of SO4, unless there is no other molecule whith S in the sample.

If there is 1 mole of molecules of sulphate (SO4), there will be exactly 1 mole of atoms of sulfur (S), and 4 moles of atoms of oxygen (O).

On the other hand, though "theoretically" it is correct to say there will be 2 moles of molecules of O2, it is really not correct, because there is absolutely no molecule of O2 in your sample. The amount of oxygen atoms is the same in both cases, but how they are organized is not.

If you want to get the mass concentration of S, say in g/l, you would just multiply the concentration of SO4 in g/l by the mass ratium of sulfur within sulphate, (sulfur mass over sulphate mass, 32/96=0.33333). Since for every sulphate gram there is 1/3 sulfur mass and 2/3 oxygen mass, 1/3 of your sulphate mass is due to sulfur.
Another strategy would be to get the concentration of sulphate, which is exactly the same as that of sulfur (1 mol of atoms of S for every mol of molecules of SO4), and multiply that by the molar mas of the sulfur atom (32).

In brief, the molar ratio between S and SO4 is 1, and the mass ratio is 1/3 grams of sulfur for every gram of sulphate


Why does sulfur form S8 instead of S3? P forms $\ce{P4}$ due to its inability to form double/triple bonds. So it makes a sort of tetrahedral structure.
S however makes $\ce{S8}$ cycles w/ $sp^2$ hybridization. Why go all that way to connect the chain? Can't $\ce{S3}$ do the trick? It will still have $sp^2$. What makes $\ce{S8}$ more stable?

Trisulfur does exist, but is not circular, it is ozone-like and unstable at common temperatures.

My guess is that while angle strain does exist in ring form, it is not that harsh. Unlike carbon, having natural bond angle ~109 degree, for bivalent sulfur it is closer to 90 degree (sulfur exhibits little to no hybridization).  However, the S-S bond strength is only 40% of C-C bond and in trisulfur ring  all S-S bonds would have dihedral angle equal to zero, which is unfavorable because of nonbonding pairs interaction. This issue is a lot more severe than in tetraphosphorus, and leads to ring opening.

For further information on sulfur allotropes see the Wikipedia article.


Can a π bond form between the py and d(x2-y2) orbitals along the x-axis? π bond does not form due to overlap between $\ce{d_{x^2-y^2}}$ and $\ce{p_y}$ along x axis? True or false?

Here is a diagram of the orbitals.

This pair of orbitials should be distingished from $\ce{d_{xy}}$ and $\ce{p_y}$, which are shown below.

Perhaps you may (or may not) be familiar with a quantity known as the overlap integral, defined as:
$$ S_{AB}= \int_{\text{all space}} \psi_A(r)^*\psi_B(r) d^3r$$

for 2 atoms, A and B.

In simple terms, it is the amount of "overlap" of 2 orbitals of an atom. Its value ranges between $0$ and $1$, implying no overlap and complete overlap respectively.

Now, to see if two orbitals overlap, in your case $\ce{p_y}$ and $\ce{d_{x^2-y^2}}$, one simply needs to compute the quantity defined above. Now, of course this would be very difficult if you don't know the functional form of the two orbitals, and even if you did know the functional form, evaluating the integral would still be messy.

So we perform some mathematically chicanery. If you are familiar with "even" and "odd" functions. An even function times an odd function is odd, and the product of two odd functions is even.

The orbital $p_y$ is odd as it changes sign if $y \rightarrow -y$. However, $\ce{d_{x^2-y^2}}$ doesn't change signs under the same transformation and thus is even. This is evident if you look at the shape/diagram of the orbital.

Thus their product is an odd function, and when evaluating the integral of an odd function over a symmetric interval (and doing so over all space is necessarily a symmetric interval because 3D space is isometric) is necessarily zero.

Simple proof of this mathematical fact is given below:

Consider an integral of the form (this is the same form as the 3D integral given in the first line, wherein your boudns are $-\infty$ and $+\infty$ for each of the three directions $x,y,z$

$$I=\int_{-a}^a f(x)\,\mathrm{d}x$$
$$I=\int_{-a}^a f(x)\,\mathrm{d}x$$
If $\,f$ is odd or even, then sometimes you can make this simpler. We can rewrite that integral in the following way:
\begin{align*}
I=\int_{-a}^a f(x)\,\mathrm{d}x
&= \int_{-a}^0 f(x)\,\mathrm{d}x + \int_0^a f(x)\,\mathrm{d}x \\
&= \int_0^a f(-x)\,\mathrm{d}x + \int_0^a f(x)\,\mathrm{d}x
\end{align*}

For an odd function, we have $f(-x)=-f(x)$, whence
$$I = \int_0^a (-f(x)+f(x))\,\mathrm{d}x = 0$$


Can Liquid CO₂ dissolve plastic I know that subcritical and supercritical $\ce{CO2}$ are excellent solvents. I've seen some examples of plastics being damaged by the $\ce{CO2}$ being impregnated into the plastic at very high pressures, but in general would liquid $\ce{CO2}$ be able to dissolve plastic?

From the abstract of Solubility of Polymers and Copolymers in Supercritical $\ce{CO2}$

[…] data to temperatures of 270 °C and 3000 bar […]
Over the same range of conditions, $\ce{CO2}$ cannot dissolve polyethylene, poly(acrylic acid), poly(methyl methacrylate), poly(ethyl methacrylate), polystyrene, poly(vinyl fluoride), or poly(vinylidene fluoride) […]


What does the signals in a XPS spectrum mean? I was trying to figure out the different peaks in a XPS spectrum for Molybdenum, but I could not understand them. About XPS, I know that this technique is based on the binding energy of the electrons of different elements

you are quite right, XPS does measure the 'binding energy' of electrons. More specifically, XPS bombards a surface with X-rays and when those X-rays have sufficient energy they are able to sufficiently excite electrons to be emitted from the sample. Each peak in an XPS spectrum corresponds to the energy necessary to emit a given electron.

So what would you expect to see for Mo?

Outer lying electrons (for Mo, we are talking about electrons in the
4p and 4s orbitals) will emit at a lower energy. You should observe
low-energy peaks that correspond to these emissions (approx. 30 - 60
eV).
you should see a cluster of peaks at approx. 230 eV that correspond to the d-orbital electrons.
Finally, you might see discrete peaks at very high energy, approx. 2600 - 2800 eV that    correspond to emission from the 2s and 2p orbitals

In general, outer orbitals emit at lower energy, and inner orbitals emit at higher energy.

Hope this helps!


Derivation of relation between acid and base equilibrium constants While deriving the relation between $K_a$ and $K_b$ in ionic equilibrium topic, say I take an example of

$$
\ce{NH4+ + H2O -> NH3 + H3O+} \tag{1}\label{1}
$$

The $K_a$ of this reaction is:

$$
\frac{\ce{[NH3][H3O+]}}{\ce{[NH4+][H2O]}}
$$

Now $\ce{NH3}$ is a conjugate base of $\ce{NH4+}$. Its $K_b$ can be written as:

$$
\ce{NH3 + H2O -> NH4+ + OH-} \tag{2}\label{2}
$$

$K_b$ of the reaction:

$$
\frac{\ce{[NH4+][OH-]}}{\ce{[NH3][H2O]}}
$$

In my textbooks while deriving relation between $K_a$ and $K_b$, it added the two equations. In the addition of two equations we get:

$$
\ce{NH4+ + 2H2O + NH3 -> NH3 + NH4+ + H3O+ + OH-} \tag{1 + 2}\label{1p2}
$$

My textbook cancelled out both the $\ce{[NH4+]}$ and $\ce{[NH3]}$ concentration terms from the LHS and RHS. $\ce{[NH4+]}$ on the LHS is from equation $\eqref{1}$ and $\ce{[NH4+]}$ on the RHS is from equation $\eqref{2}$.

My question is, how can we cancel them out? How can you say the $\ce{[NH4+]}$ formed from hydrolysis of $\ce{NH3}$ in reaction $\eqref{2}$ is equal to that of $\ce{[NH4+]}$ initially taken in reaction $\eqref{1}$? Even if you initially take same concentration of $\ce{[NH3]}$ for reaction $\eqref{2}$ as formed from reaction $\eqref{1}$, that concentration of $\ce{NH3}$ on hydrolysis can never give same concentration of $\ce{[NH4+]}$ as initially taken in reaction $\eqref{1}$. Then, how can you cancel these different $\ce{[NH4+]}$ concentrations when you are adding reaction $\eqref{1}$ and $\eqref{2}$?

$K_\mathrm{a}$ and $K_\mathrm{b}$ should not have a term for the concentration of water inside.
When you dissolve $x$ moles of $\ce{NH3}$ in water (doesn't matter how much $x$ is), both relations

$$K_\mathrm{a} = \frac{[\ce{NH3}][\ce{H3O+}]}{[\ce{NH4+}]} \tag{1}$$

as well as

$$K_\mathrm{b} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]} \tag{2}$$

must simultaneously hold true. You agree that both of these relations must independently hold no matter what concentration of $\ce{NH3}$ (or $\ce{NH4+}$, or $\ce{H+}$, or $\ce{OH-}$) there is, right? That is why they are called equilibrium constants, after all.

So, in any given single solution with one well-defined value of $[\ce{NH3}]$, one well-defined value of $[\ce{NH4+}]$, ... both equations must simultaneously hold. You can then cancel out the terms as necessary to obtain

$$K_\mathrm{a}K_\mathrm{b} = K_\mathrm{w} \tag{3}$$

If you use two different solutions, with different concentrations of all the relevant species, the relation still holds true. However, the maths will be much harder because for example if you have solution 1 with pH = 1, and solution 2 with pH = 10, then $[\ce{H+}]_\mathrm{sol,1}[\ce{OH-}]_\mathrm{sol,2}$ is no longer equal to $K_\mathrm{w}$. The relation $[\ce{H+}][\ce{OH-}] = K_\mathrm{w}$ is only necessarily true if we are talking about one single solution.


Polycondensation of PMS This is the reaction that I'm interested in. I reckon this is what occurs:

But how does the reaction proceed from there, there is a lone pair on the oxygen in right side of my picture, but how do you get rid of the CH3 group on that oxygen?

Although I do see a problem here that the structure of my final product was due to my aforementioned reaction scheme would still have RO-(PMS)-OR, where R is methyl group, instead of RO-(PMS)-R. I don't know what to do...

"Cationic polymerization can occur with the use of strong or Lewis protic acids. The mechanism of this reaction is yet unknown." https://www.ukessays.com/essays/chemistry/polysiloxanes-synthesis-structure-2201.php

Thereby, we are freed to speculate!

It is more likely that silicon goes pentacoordinate rather than tricoordinate. If a proton attaches to an oxygen on the siloxane, the silicon goes positive, which will attract an oxygen from: 1) a water, 2) a methanol, or 3) a methyltrimethoxy silane. The second and third options can only reform the original reactants. If water attaches after protonation, forming $\ce{H2O-Si-OHCH3}$, methanol could leave, followed by a proton from the water, to give $\ce{CH3Si(OCH3)2OH}$. This becomes a fourth attacking option. If this attacks the silicon of a protonated siloxane thru the $\ce{Si-OH}$ to give $\ce{CH3O(H+)-Si-OHSiCH3(OCH3)2}$, the pentacoordinated silicon can lose methanol, then lose the proton from the oxygen between the silicons, to form $\ce{CH3(OCH3)2Si-O-SiCH3(OCH3)2}$, and that's the first step, a dimer.

Because there are two more alkoxy groups on the silicons which can leave, more reactions can occur, and likely will produce a three-dimensional gel, or even a solid, if you cook it.


Having trouble understanding activity series of metals & electronegativity I want to know how do metals replace hydrogen in a substance such as water. Isn't water happy with its configuration? It has shared enough electrons so that both hydrogen and oxygen complete their orbitals.

I know that alkali metals are very active and they have very low electronegativity. And elements such as gold have much higher electronegativity while being considered very inactive. Maybe I have a wrong understanding of what electronegativity is but isn't the element with higher electronegativity supposed to be more active and be able to replace those with lower ones?

There's electronegativity and then there's electropositivity. Electronegativity is the tendency of a substance to gain electrons and electropositivity on similar grounds is the tendency to donate electrons. What actually is a redox reaction? Exchange of electrons?

A substance with high electronegativity would like to snatch any electron it finds, while a substance with high electropositivity would like to force-fully hand over its electrons, however unwilling the substance on the receiving end may be. Metals are electropositive.

  how do metals replace hydrogen in a substance such as water. Isn't water happy with its configuration?

There was water, it was living happily. Now, water had two components. Oxygen and hydrogen. Let's pull up their electronegativities. Oxygen is at $3.44$ and hydrogen is at $2.2$. This means that the bond that oxygen and hydrogen shared in water was a polar bond. Oxygen being powerful (higher electronegativity) kept the bond in it's favor. The electrons of the bond were more towards oxygen. This gave oxygen a partial negative charge ($\delta -$) and hydrogen a partial positive charge ($\delta+$).

In came a metal which was highly electropositive (think alkali metals, as an example you can take up sodium whose electronegativity is at $0.93$). He gave oxygen a tempting offer:

"Oxygen, you seem to be caught up in a tussle, look at hydrogen, he doesn't want a positive charge because he's quite electronegative. You're constantly fighting with him to keep the electrons of the bond pair pulled towards yourself. Why don't you pair up with me? I'm low at electronegativity. This means I'm electropositive. As much as electropositive species like bearing a negative charge, we electropositive species like bearing a positive charge!"

And oxygen was greedy. A greater difference in electronegativities of elements forming the bond meant a bond which would be richer in ionic character, which appealed to oxygen.


How to draw the 2D representation of ammonia molecule In the 2D representation of the ammonia molecule, we see one solid line, one solid triangle and one dashed triangle

A plane can be made to pass through any three given points. In other words, we can always find a plane which will pass through any three given points in space. So the central N atom and two H atoms will lie in a plane. Only one H will be out of plane. So I think, in the 2D representation, we need to show two solid lines and one solid triangle. Please give your opinion about this. Thanks

According to the rules, yes, Ammonia can be drawn with three of it's atoms in one plane and the other outside the plane. Here's an example, along with a 3D projection:

However, what do you comprehend from this structure? This structure does not convey the trigonal pyramidal geometry of the molecule very convincingly. It might be mistaken for trigonal planar at a glance, and this defeats the purpose of using the wedge/dash notation. Ammonia (and several other tetrahedral molecules) are represented in that manner since it is easiest to interpret their 3d structure from a 2d image.


Does Friedel-Crafts reaction fail with benzoic acid? I know that Friedel-Crafts reaction  (either alkylation or acylation) fails with phenol and aniline, due to formation of a coordinate bond between Lewis acid catalyst and the lone pair of the respective functional group.

But I have read that even benzoic acid fails to undergo this reaction. I couldn't think of a reason for this. Can anyone explain why is it so?

It could be that $\ce{-COOH}$ group is electron withdrawing hence there is an increased electron density in $\ce{-COOH}$ group due to -R resonating effect of electron from benzene ring.

Since friedel craft catalyst is a Lewis acid (electron deficient) the catalyst (anhydrous $\ce{AlCl3}$) reacts with electron rich COOH group of benzoic acid and gets bonded.


What's the difference between the two equations for work? I'm studying thermochemistry and I don't get this. What's the difference between  $w = Fd$ and $w = -p \Delta V$?

Aside from a sign convention artifact, they're the same thing.

Pressure is force per unit area:

$$
 P = {F\over A}
$$

If a surface with a given area is moved through a displacement $\Delta x$, then the volume of displacement is the product of $A$ and $\Delta x$:

$$
\Delta V = A \Delta x
$$

So, the $P\Delta V$ work formula can be rewritten:

$$
w = -P\Delta V = -P\left(A \Delta x\right) =
  - \left({F \over A}\right)\left(A \Delta x\right) =
  - F \Delta x
$$

There are a lot of assumptions embedded in the above, and you have to be a lot more careful than I was about defining positive and negative quantities correctly, but on a high level that's all there is to it.


Are Group 2 metal fluoride salts less soluble in organic solvents than Group 1 fluorides? It is clear that Group 1 metal fluoride salts DO NOT readily form solutions in organic medium. The amount of solvation that can occur leading to the eventual release of a fluoride ion depends on many factors, one is the polarity of solvent (described by dielectric constant). Most examples seen in the literature seem to evolve around using $\ce{CsF}$ or $\ce{KF}$ in the vicinity of 18-crown-6 ether as common examples for the release of fluoride ions (not considering TBAF). However, I cant seem to find any description for $\ce{MgF2}$ or $\ce{CaF2}$ salts and how their solubility in organic medium affects the release of fluoride ions.

Don't get your hopes up.  Wikipedia identifies magnesium fluoride as insoluble in ethanol.  Calcium fluoride fares no differently in acetone.  While these data are hardly exhaustive, they suggest strongly that Group 2 fluorides are not a good way to get fluoride ions into solution.  Of course, we would not expect Group 2 ionic compounds with their doubly charged ions to get vary far in organic solvents generally.
Getting fluoride ions into solution with organic solvents is typically done with bulky 1+ cations, such as N,N,N-trimethyl-(2,2-dimethyl-prop-1-yl)ammonium ion, $\ce{C_{8}H_{20}N^+}$.  See Reference [1] as an example.  The use of cryptand-complex alkali-metal complexes fits this paradigm because such ions are also bulky.
Reference
1.
Victoria K. Davis, Stephen Munoz, Jeongmin Kim, Christopher M. Bates, Nebojslš a Momčilović, Keith J. Billings, Thomas F. Miller III, Robert H. Grubbs and Simon C. Jones, "Fluoride-ion solvation in non-aqueous electrolyte solutions", Mater. Chem. Front.,
2019, 3, 2721


What is the chemical name of the compound H2S4O? I have been studying nomenclature and was asked to find the name of a strange compound: $\ce{H2S4O}$.

In my opinion, it should be dihydrogen-tetrasulfur oxide which seems very weird.

Can somebody advise me on how to name this compound (and whether this compound really exists?, I was able to come up with a structure similar to benzene...)

Section IR-5.4 of IUPAC’s Nomenclature of Inorganic Compounds (Red Book) of 2005 lists the rules that shall apply to naming inorganic compounds by a generalised stoichiometric name which does not carry any information about the compound’s structure. These rules dictate that:

The constituents of the compound to be named are divided into formally electropositive and formally electronegative constituents. There must be at least one electropositive and one electronegative constituent. Cations are electropositive and anions electronegative, by definition. Electropositive elements occur later in Table VI than electronegative elements by convention.
In principle, the division into electropositive and electronegative constituents is arbitrary if the compound contains more than two elements. In practice, however, there is often no problem in deciding where the division lies.
Section IR-5.4.1

In our case, oxygen must be considered electronegative and hydrogen must be considered electropositive since they have the highest and lowest electronegativity, respectively.
Within the classes of electronegative or electropositive substituents, ordering is strictly alphabetical. Electropositives come first, electronegatives last.
Section IR-5.4.2.1 allows the use of multiplicative prefixes as is the case with binary compounds. Further information such as charge, oxidation numbers or structure is not at our disposal.
Therefore, according to the principles laid out, we have two choices to name $\ce{H2S4O}$, depending on whether we wish to consider sulphur electropositive or electronegative:

Dihydrogen tetrasulphur oxide

Dihydrogen oxide tetrasulphide

To the best of my knowledge, this compound does not exist. But the principles of nomenclature are designed in a way to allow the naming of all compounds whether synthesised, hypothesised or fancily drawn on paper.


Is this calculation of the amount of required hydrogen gas correct? I was a bit hesitant to post this question for it may appear like a "homework help" type of question. But upon rethinking it felt like a genuine question. I was studying stoichiometry with my son (from the OpenStax Chemistry 2e book) and suddenly I had an idea to use these calculations in a real chemistry experiment that I am going to perform at home (of course, I truly believe in a "safety first" approach). I wrote it all up here on GitHub, but reposting here to keep it all at one place:
A practical problem: It was decided to inflate a common latex balloon to 5cm radius by the hydrogen gas produced during a single-displacement reaction:
$$\ce{2HCl(aq) + Zn(s) -> ZnCl2(s) + H2(g)}$$ that uses an N/10 i.e. 0.1M HCl solution (available in many drugstores, typically in $\pu{200ml}$ bottles). How much reactants are needed?
Solution: We need to inflate the balloon to about $\pu{5cm}$ radius. Assuming the balloon to be a sphere gives us the volume of balloon, $V_b$, as:
$V_b = \frac{4}{3}\cdot \pi r^3 = \pu{523.599cm^3}$.
Assuming the density of hydrogen at STP as 0.0000899 $\pu{g/cm^3}$, the mass of hydrogen is $\pu{523.599 \times \space 0.0000899 g} = \pu{0.0471 g}$. Since 1 mol $\ce{H_2(g)}$ has a mass of $\pu{2 \times \space 1.008 g} = \pu{2.016 g}$, we need
$\pu{\frac{0.0471}{2.016} mol} = \pu{0.0233 mol}$ $\ce{H_2(g)}$.
Balancing the reaction, we see that to produce 1mol $\ce{H_2(g)}$ we need 2 mol HCl(aq) and 1 mol Zn(s).
This means that to produce 0.0233 mol $\ce{H_2(g)}$ we need $2 \times 0.0233 = \pu{0.0466 mol}\ \ce{HCl(aq)}$. We have a 0.1M HCl (aqueous) solution which means that $\frac{\text{mol of solute}}{\text{litres of solution}} = 0.1$. Clearly, we need $\frac{0.0466}{0.1} = \pu{0.466 L}$ or $\pu{466 ml}$ HCl.
The amount of zinc needed = $0.0233 \times \text{molar mass of zinc} = 0.0233 \times 65.38 = \pu{1.523 g}$.

Kedar's calculation may be right. But it must be realized that the reaction will be rather slow. Even with a $1 M$ solution, the reaction is slow. The reaction of $1.5$ $\pu{g}$ metallic zinc with such a diluted $\ce{HCl}$ solution ($0.1 M$) will last more than one hour. Anyway, in order to inflate such a balloon a higher pressure than the atmospheric pressure must be applied. If not, the balloon will keep its original dimension. How to increase the pressure of $\ce{H2}$ in such a balloon for inflating it, if the hydrogen is produced by the reaction of zinc on $\ce{HCl}$ ?


Why do bubbles form on the side of a plastic bottle? I have a plastic bottle filled with drinking water where only one side of the bottle has air bubbles. I know that water is not pure and there are dissolved gases, and my theory is that there are small microscopic holes on one side of the plastic bottle that make bubble formation possible. Or may be its a temperature difference between the two sides (which shouldn't be the case - the bottle was kept in room temperature, but it was closed for a day or more). What could be the cause?

Your reasoning seems good. A number of things can affect nucleation of bubbles, such as surface defects, which you state, as well as temperature (was one side of the bottle facing the sun, perhaps?), and even sound.

A fun demonstration is to put a cup of carbonated beverage in an ultrasonic bath (e.g. parts cleaner or humidifier). Here, cavitation serves to nucleate bubbles.


What is the mechanism for reduction of alkylmercuric compounds by hydrides? What is the mechanism of the reductive elimination of mercury metal using hydrides that is carried out after oxymercuration of the alkene? I have come across a free radical mechanism, but that suggests loss of the stereospecificity of the product, while the reaction is stereospecific in reality. Also, why isn't an acid sufficient to remove the mercury as it is with alkynes? Why is a reducing agent necessary at all?

The reducing agent NaBH4 is used for this reaction because it is highly selective. If H2SO4 were used in place of NaBH4 in this reaction the molecule would be dehydrated and we would probably see some HgSO4 form as well. The main thing to remember here is that a reducing agent has much more selectivity than an acid solution.


Distillation of aniline/water mixture I read from a source that aniline and water can be separated by steam distillation. But isn't this senseless as steam distillation will simply add more water and make separation more difficult?

Can't we just distill the aniline off?

The boiling point of aniline is around 180 °C. Remember that water boils at 100 °C.

Do water and aniline mix?

Yes, but the solubility of aniline in water is rather low (around $\pu{30 g/l}$).

What's the advantage of the steam distillation?

Aniline and water form an azeotrope with a boiling point around 98 °C.


How to construct a QST2 input file for Gaussian16 for multiple reactants and products? I am attempting to determine the transition state structure for the following reaction using the QST2 method on Gaussian 16:

$$\ce{CH3 + N -> HCN + H2 }$$

However, I am struggling to put together my input file. How do I format my z-matrices for multiple reactants and products? Here is what I have so far (though it is not running successfully):

%chk=sample2.chk
%mem=1500MB

#T UB3LYP/6-31G** opt=(qst2, redundant)

CH3 + N -> HCN + H2 reactants

0 3
C
X 1 1.0
H 1 r 2 90.0
H 1 r 2 90.0 3 120.0
H 1 r 2 90.0 3 -120.0

r 1.0828

CH3 + N -> HCN + H2 reactants

0 3
N

CH3 + N -> HCN + H2  products

0 1
C
X 1 1.0
H 1 rh 2 90.0
N 1 rn 2 90.0 3 180.0

rh 1.0587
rn 1.1326

0 1
H
H 1 r

r 0.741

I'm pretty new to this type of calculation on Gaussian so any help would be much appreciated!

TL;DR: QST2 will not work.

For QST2 calculations you need one reactant structure and one product structure, the program will then attempt a transition state guess. The essential point is, that the supplied structures have to have identical ordering (in Cartesian coordinates), or identical z-matrices, although I am not sure that this works.
For more on how to run these types of calculations, see Finding transition states in Gaussian – focus: Electrophilic Addition Reaction.

QST2 (and QST3 by extension) only make sense, if you know that you will have a single transition state - they are especially difficult to use for dissociation or association reactions because of the implied supermolecular approach. The minima $(\ce{CH3 + N})$, $(\ce{HCN + H2})$ will be very shallow and not well converged. From these structures it would be very difficult to guess the transition state, so you would at least have to supply a guess for it (QST3).

Additionally you are going from a "quintet" state (doublet + quartet) or "triplet" state (doublet + doublet, or doublet - quartet) to a singlet state (singlet + singlet). You could probably force your reactants into a broken singlet state (doublet - doublet), then it's a problem of how much sense this makes. In the other cases it is doubtful that the program will get a good guess for the electronic state of the TS.

DFT is maybe not the best methodology to investigate this kind of reaction in the first place.

In any case, here is an input file (tested with Gaussian16 rev. A.03) that will at least start running. SCF will not converge. Also reactants and products have not been optimised beforehand (which you must do). This is only for illustrative purposes.

%chk=bp86svp.chk
#P BP86/def2SVP
EmpiricalDispersion=GD3BJ
DenFit
opt(QST2,CalcFC,maxcycle=5)
gfinput gfoldprint iop(6/7=3)
nosym
pop=full

N + CH3

0 1
C        0.000000000      0.000000000      1.891780658
N        0.000000000      0.000000000     -2.108219342
H        1.029433990      0.000000000      2.255740536
H       -0.514716995     -0.891515987      2.255740536
H       -0.514716995      0.891515987      2.255740536

HCN + H2

0 1
C        0.649013613      0.189100642      0.000000000
N       -0.817143840      0.083131316      0.000000000
H        0.964811208     -1.929636661     -0.499461795
H        0.964811208     -1.929636661      0.499461795
H        1.696363276      0.452628980      0.000000000


How are Uranium and such materials made I was wondering, how are materials like uranium, plutonium and other radioctive materials made. Do they all come from the old exploded stars and just been in the mountains for millions of years. Is there getting made more of those materials, or are we gonna run out in the future?

Some radioactive materials are made by man, others have been parts of earth's rocks since they were formed

Radioactive materials are not all the same. some are man-made and some have been with us since the creation of the planet (so are indeed the product of old exploded stars).

The best way to tell which is which is to look at the half-life of the atoms involved. Half-life is a measure of how fast an isotope decays. It is the length of time it takes for half of the amount of substance to have decayed to something else. So half-lives with the same sort of size as the age of the earth are quite likely to be primordial (which basically means they were present when the earth was first formed). Other compounds with shorter half-lives would have all disappeared by now even if they were present then. They are synthesised by man.

238U (by far the most common isotope of uranium) has a half-life of ~4.5 billion years (roughly the same age as the Earth) so quite a lot of the uranium present at our planet's birth is still around. 239Pu (the isotope of plutonium used in nukes) has a half-life of ~24 thousand years so even if there was some around when the earth was formed there wouldn't be any of that left. Plutonium is created in nuclear reactors which can be designed to generate enough to make an indecent supply for the military. Having said that, a very small amount may be generated naturally in uranium containing rocks as part of the normal decay pathways of uranium or as part of natural nuclear reactors (occasionally in geological history uranium rich rocks have created the conditions for significant nuclear reactions which produce more plutonium and other materials than simple decay of isolated uranium atoms).

Other, more common, elements sometimes have radioactive isotopes. A small proportion of potassium, for example is 40K which has a half-life of ~1.2bn years (so quite a lot of the earth's original stock is still around and is the largest contributor to radiation in living things which all contains a lot of potassium).

There is little danger of running out of primordial radioactive elements as we are not using them fast enough. The man-made ones can be made in nuclear reactors if they are set up the right way (and we have way more plutonium than we know what to do with; but many rare isotopes used in medical research are made in specialty reactors where the risk is that a reactor shutdown might limit the supply).

If you want to know where all the elements originally came from, there is a good Wikipedia article with a periodic table that explains the detailed origins.


Comparing acidic strength in oxan-2-one and oxan-4-one I need to compare the acidic strength in these molecules:

My Approach:

In compound I: Removing a proton would lead to conjugation between -ve charge, pi bond and lone pair of O. Moreover the inductive effect due to 'O' will stabilize the charge.

In compound II: Removing a proton(from the carbon adjacent to the pi bond) would lead to conjugation, but relatively less than one. Moreover 'O' is far away, so inductive effect does not play a role here.

From this I concluded that compound 'I' must be more acidic.
But the answer says that the second compound is more acidic.

Where am I going wrong?

This has a simpler answer than you expect. First of all, your reasoning is wrong because inductive effect is much less dominant than conjugation of electron pairs. So, if the latter is unequal in both cases, we should first prioritize it over inductive effects.

  Removing a proton(from the carbon adjacent to the pi bond) would lead to conjugation

It seems you already know that the beta hydrogen of carbonyls is most acidic. Both molecules have carbonyl groups, and both have beta acidic hydrogens. So, you're on the right track!

Whichever molecule is more acidic, it will have a more stable conjugate base.

Note that in the second anion, the excess negative charge is completely delocalised over to the carbonyl group. Compare that with the first anion, where both the excess negative charge and the oxygen's lone pair are delocalising with the same carbonyl group. Thus, the excess negative charge is unable to completely delocalise with the carbonyl group.

Can you now identify why the first molecule is less acidic?


SMILES vs. Graph representation in deep learning I have been reading papers on machine learning and deep learning methods for learning molecular space and generating molecules. These methods use different representations of the molecules. The most popular ones in the field include SMILES and graphs [e.g. this and this]. I have seen a shift of interest from SMILES representations to Graph molecular representations in the past couple years. I was wondering what are possible benefits of graph representations over SMILES?

I can think of two reasons:

The uniqueness issue with SMILES: Two SMILE strings might correspond to the same molecule.
SMILES are abstract representations while graphs are more natural representations (although, I can't really see this. I mean, I see what it
means, but doesn't SMILES also contain the same information?)

EDIT:

some paragraph that I found online and I need interpretation for:

it is focused on molecules whose bonds fit the 2-electron valence
model
it handles a limited array of stereochemistry types
there is no standard for handling aromaticity
there is no standard way to generate a canonical representation.

are any of these corrected in graphs? From what I know:

For 2: In graphs, it takes more than just defining atoms and bonds to account for stereochemistry. For example, I should also define chirality for each node.

For 3: While in some graph representations, aromatic bonds have their specific type, I found it better to explicitly use the order of the bonds.

For 1 and 4 I don't know if graphs are any better.

I agree that there seems to be a shift towards using graph representations over SMILES strings. I personally think this is a good thing, and I'll try to explain why, but even if there's nothing inherently better about graph representations of molecules, there is a very practical reason people are moving towards graph representations.

So, first of all, as you point out, both representations suffer from an issue in that they are necessarily a reduction from the complete amount of information needed to specify a molecular structure. For instance, neither a graph nor a SMILES string could distinguish between two isomers which differ only in the direction in which a free $\ce{O-H}$ group points. In fact, a bare graph cannot tell the difference between the boat, chair, or planar hexane molecules as they all have identical connectivity. Nonetheless, a graph is a much more flexible representation than a SMILES string because it is quite common to add weights to edges, which could be bond distances, or add parameters to nodes, which could describe angles to other nodes. So, you can include some information which is in the Cartesian coordinates, without having the problems associated with deriving your molecular representation from Cartesian coordinates.

This hints at the very practical reason why people are using graph representations. Basically, graphs are used in machine learning models in many fields other than chemistry. So, it's very nice to not have to reinvent the wheel, especially when people have so much success with things like transfer learning, where you just take a pre-trained model and re-train it for your own purposes.

Also, very often the first step in training a neural network is making some transformation to your data. For instance, graph convolutional neural networks have been successful in many tasks, so why not just use a convolutional filter on your graph representation of a molecule? You could do this with a SMILES string, but you would probably first just transform the string into something resembling a graph.

As to your specific points about chirality and aromaticity, etc. All of this information can be attached to a graph via parameters belonging to each node, although I would personally avoid giving information that isn't strictly necessary. That is, there is nothing special about a bond which is in an aromatic ring. You need to provide enough data that the model can learn about this on its own. If you tell it that bonds in aromatic rings are special enough to get another parameter, this is probably going to bias the model in some unforeseen way. Chirality is easily handled by a simple parameter attached to each node.

Ultimately, though, the representation depends a lot on the problem you are trying to solve. For instance, if you are trying to learn a representation of the potential energy surface, then graphs can work quite well. What is more common perhaps, is to use so-called atom-centered symmetry functions. In this case, the actual features are abstract vectors which are guaranteed to have the relevant symmetries and smoothness needed in the potential energy surface.

If you're doing something more like a classification problem, then using a representation like a SMILES string might be perfectly suitable.

TL;DR

Graphs are a more flexible representation which are commonly used in fields outside of chemistry. Hence, being able to draw on the knowledge of other disciplines is a huge plus, especially when you're collaborating with computer scientists who know a lot about machine learning and nothing about chemistry.


Redox potential of a lead–acid battery In the German Wikipedia there are two reactions on the poles of the battery shown with the following potentials:

  $$
\begin{align}
\ce{Pb + SO4^2- &-> PbSO4 + 2 e-} &|\pu{-0.36 V}\\
\ce{PbO2 + SO4^2- + 4 H+ + 2 e- &-> PbSO4 + 2 H2O} &|\pu{+1.68 V}
\end{align}
$$

  $$E_\mathrm{Ges}^0 = \pu{1.68 V} - (\pu{-0.36 V}) = \pu{2.04 V}$$

I do understand the potential for the second $\pu{1.68 V}$ since for the second reaction the underlying redox pair is $\ce{Pb^4+ + 2 e- -> Pb^2+} $. For this redox pair the electrochemical standard potential is $\pu{1.69 V}$.

But for the first reaction I think that the underlying redox pair has to be $\ce{Pb^2+ + 2 e- -> Pb}$. This redox pair has standard potential of $\pu{-0.1263 V}$.

This result in a voltage of $\approx\pu{1.55 V}$. But Wikipedia and a book of mine tell the the voltage of this battery type is $\pu{2.04 V}$.

What the reason for the $\pu{-0.36 V}$?

Source: This is from the German Wikipedia article on lead-acid batteries. Unfortunately the English version doesn't contain the calculation of the voltage. I took the standard potentials from the book Elektrochemie by Hamann.

The potentials depend on the form of the compounds.
It is true that in solution

$\ce{Pb^{2+}  (aq) + 2e^- -> Pb (s)} $

is -0.126 V.

But in the case of a battery we have:

$\ce{PbSO4 (s) + 2e^- -> Pb (s) + SO4^{2-} (aq)}$

And in this case the $\ce{Pb^{2+}}$ is in solid form and the potential is -0.356 V.

In a battery the sulphate is insoluble and it is required that it sticks to the electrode, otherwise the reverse reaction can not occur.

A table of potentials can be found here


Can a strong nucleophile carry out an SN1 reaction? For example, if I take a tertiary halide in water + ethanol solution and add I- ions to it (which is a good nucleophile), then shouldn't the reaction be SN1; because SN2 is not possible as the substrate is tertiary. Please explain.

You've pretty much said it all already. SN2 is ruled out at the tertiary centre. Since the attack of the nucleophile in the SN1 mechanism is not rate-determining, it doesn't matter whether the nucleophile is weak or strong; iodide works fine.

In this case, it's more likely to get intercepted by the solvent itself, which is present in much larger quantities and is also nucleophilic, but that's another story.


How thick is the colored oxide layer on heat treated stainless steel? When stainless steel is welded, or just heat treated, it will develop a thicker oxide layer, which in some cases is is rainbow colored, but can become so thick as to appear almost black. From the colors (diffraction effects, I assume), the nicer-looking layers must be on the order of a wavelength of light. Is there any more numerical information on their thickness?
Follow-up question: The colored oxide layers can be removed with a pickling paste (based on HCl) or a passivating cleaner (based on HNO$_3$ plus HF), but the resulting surface is much rougher than the original. Well, it is observably rougher if the original surface before heat treating was mirror-smooth. We call the new surface satinized. Is there any simple way to estimate surface roughness/surface area of the satinized surface?
Further follow-up: would the surface chemistry be expected to change due to the surface area increase (perhaps like shiny Pt vs platinized Pt (~340 mV): https://en.wikipedia.org/wiki/Platinum_black

Not very thick. Carbon steel develops the same colors at much lower temperature/ times . Gold to pink to blue to purple , stainless shows the colors better because the base metal is brighter . It is caused by the thickness of oxides , a matter of a few layers of oxide molecules ; unfortunately I cannot give you a number in microns and can't remember which book it is in.


How hazardous are peroxide-able ethers if stored properly? This question is directed towards laboratory scientists and technicians who actually deal with ethers that can develop peroxides upon concentration (e.g. diethyl ether, THF, dioxane). I am already very familiar with the literature, the normal precautions (dispose of after 1 year, test for peroxides, etc.), how to remove peroxides (shake with ferrous salt or metabisulfate).

My question is how hazardous are peroxides in reality if stored properly. If a peroxide-able ether upon concentration is stored in a cool dark cabinet, is rarely opened, has an inhibitor (BHT or ethanol), and argon is used above the headspace, it is likely that the ether can be used for longer than a year?

Again, I need the input of an actual practitioner. Simply quoting written material will not help me, as I have consulted LOTS of written material and I believe I understand the theoretical issues. I am looking for practical experience.  I have heard informally that some ethers are available in European countries as OTC solvents, and so it leads me to think that the standard warnings may be too severe. Thanks!

Peroxides are no fun.  There have been explosions when ether solutions were concentrated to dryness.  What you really want to do is store your ether over potassium hydroxide - hydroperoxides are slightly acidic, and the base forms hydroperoxide salts and pulls them out of solution.

Also, if you keep solvents for more than a year you clearly aren't turning them over fast enough.  Keeping flammable material that you don't need just increases the danger in case of fire.


Differences between chemical physics and physical chemistry? What makes these two seemingly identical topics separate? What does each field more focus on, like do chemical physics researchers study more the atomic/molecular interactions while the physical chememists study more macroscopic properties? I assume they have quite a bit of overlap.

You are of course right, there is a lot of overlap. There are also the scientific publication series ChemPhysChem and Physical Chemistry Chemical Physics amongst others, that deal mainly with subjects from the both areas.

In principle they are most likely dealing with almost the same subjects. And you often find physicists working in chemistry and vice versa. However, there is a little difference in the general approach to problems.
Physical chemistry usually approaches the problems at hand from a chemist's point of view, focussing (initially) mainly on macroscopic properties of an ensemble and use physical laws to determine them. It is often considered the parent field for many other subjects, like physical organic chemistry, chemical kinetics, and spectroscopy, among others. Most common publication titles are Zeitschrift für physikalische Chemie and The Journal of Physical Chemistry A, B, C, and letters. And many others...
Chemical physics on the other hand studies chemical ensembles and reactions from a physicist's point of view. They focus more or less, like you assumed, on ions, atoms, clusters, surfaces, free radicals, and molecules interacting with each other. This does include, but is not limited to, the study of solvation effects and single entities like quantum dots. Here are fewer specialist titles, like The Journal of Chemical Physics and Chemical Physics Letters. Of course publications are often widely accepted also in other journals of neighbouring areas. (Sometimes it is also treated as a subfield of physical chemistry.)

In modern times the various fields are not separated by a sharp line, they most likely blur into each other. You could probably ask the same question for Biochemistry and Chemical Biology, or Computational Chemistry and Quantum Chemistry, or Polymer Chemistry and Material Sciences. And there are more. Interdisciplinary cooperation is encouraged by many researchers and institutions, it is important not to care too much about the labels anymore.


What does (heated) mean in a chemical equation? I'm supposed to balance the following equation:
$$\text{Potassium chlorate (heated)} \rightarrow \text{Potassium chloride + Oxygen}$$
But I don't understand what is meant by the label "heated" in front of the first element. Do I ignore it, or am I supposed to add another element to the reactants of the equation?

In terms of balancing the equation, the fact that heat must be supplied can be ignored.  It is just providing information about the conditions necessary to run the reaction.


N-alkylation of amides with alkyl halides? Can amides be N-alkylated with alkyl halides? Say I was trying to make N-propyl acetamide. Would iodopropane and acetamide react to form it? Would they react to form N,N-dipropyl acetamide? If so, what catalysts would I need, and if not, why not and how would it be possible?

Sure, there are a number of ways to do this with copper, iridium, or palladium catalysts.  This link should hopefully be helpful.

http://www.organic-chemistry.org/synthesis/C1N/amides2.shtm


Fehling's test arrow pushing mechanism I am currently attempting to generate an arrow pushing mechanism for the oxidation of glucose to gluconic acid using Fehling's solution. My original thought was to go the route of alcohol oxidation using acidified dichromate where the alcohol attacks in and loses the H using a E2 mechanism.

In the aldehyde oxidation, I would have a hydroxide ion attack prior to the H bond collapse leaving an extra electron on the Cu from the $\ce{C-\mathbf{O-Cu}}$ bond resulting in reduction of $\ce{Cu^2+ -> Cu+}$. Which would satisfy the observed empirical data. After an attempt I googled the mechanism finding very little real information.

A post on this very site suggested the mechanism proceeds via enolate formation followed by single electron transfer with $\ce{Cu^2+}$. Unfortunately, there is no elaboration on this or a source, I also can't find a source online which corroborates this mechanism. Does anyone have a good source for this?

The exact nature of the mechanism is unknown

Be aware that some organometallic chemists consider arrow-pushing mechanisms inadequate/inappropriate for a variety of reasons (including the depiction of single electron transfer). For example, the arrow-pushing mechanisms we draw with alkyllithiums are incomplete in that they do not account for the oligomeric nature of the reagent.[2]

The elements which are known:

The aldehyde or α-hydroxyketone must be enolizable. The enolate tautomerization of reducing sugars is discussed in McMurray[3]. This is common to the Benedict’s test mechanism as well.
The rate of enolization determines the rate of oxidation J. Am. Chem. Soc. 1970, 92, 537
Cu(II) has long been known to mediate oxidative coupling of enolates by a radical mechanism J. Am. Chem. Soc. 1971, 93, 4605. Mechanistic studies point to a metal-chelated single-electron-transfer process.5
The active species is the metastable [Cu(tartrate)2]6- ( Eur. J. Inorg. Chem. 2016, 1798).

Drawing one plausible mechanism

Convince yourself that the Cu(II) complex  is a d9 17-electon complex.  That suggests you could plausibly draw an inner-sphere or outer-sphere mechanism.
For the inner-sphere mechanism: coordinate the aldehyde, and generate the enolate (considering C-vs-O coordination and necessity of ligand-slip).
Form the hydrate of the C-bound enolate, then homolyze the bond to form an organic radical and Cu(I).
A second molecule of Cu(II) oxidizes the radical to the cation, which gets quenched the elimination of H from the hydrate to generate the enol form of the carboxylic acid.

For the basics of an outer-sphere single electron transfer see Mark Vander Wal’s Basics of Electron Transfer

[2]: Advanced Organic Chemistry: Part A: Structure and Mechanisms, Carey & Sundberg, p. 413.
[3]: Organic Chemistry: With Biological Applications,  McMurry, p. 758


Why is fluoroantimonic acid so strong? I learned that all strong acids have a 100% degree of ionization and that there are no degrees of strength in acids; however, I also read that fluoroantimonic acid is 20 quintillion times stronger than 100% sulfuric acid !!

Can you please explain the chemistry behind this?

For many common purposes we say there's weak acids and strong acids, and define strong acids to be ones that dissociate completely in water. This is because if the acid dissociates completely in water, then every proton dissociated will just protonate water, and turn it into $\ce{H_3O^+}$ (hydronium ion). The hydronium ion is the strongest acid that can exist in water as far as I understand, and hence why strong acids are simply strong acids in some contexts.

What you're describing is a super acid, that is not used in aqueous solution Take HCl for example. This is a classic strong acid, that will protonate a wide range of structures. But if we take something that is a VERY, VERY poor base (something that really does not want a proton), such as an alkane, HCl can't protonate it. The proton will rather stay with chlorine than protonate the alkane. That is one of the things you can do with the acid you mentioned.

The strength of an acid depends on many different factors, but it all comes down to how strongly the $\ce{H^+}$ is bound to the acid structure. If it is very loosely bound, the structure is high in energy, and the proton would like to find a more stable structure to partake in. In this case protonating an alkane to form a carbocation and molecular hydrogen is low enough in energy, that the acid can protonate it.

But remember that protonation is an equilibrium. If we achieve a vastly lower energy state by having the acid protonate another structure, then the protonation will be complete. If it is the same energy, half of the acid will have protonated (assuming equal concentrations of acid and base).


Mechanism of ring opening of amine combined with formylation What is the mechanism for this reaction? The tertiary amine is converted into an amide and the ring gains a double bond. I can't figure out how the C-O double bond would be formed.

This reaction is very similar to Reimer-Tiemann reaction which also consists of the reagent $CHCl_{3}$ + $NaOH$. The mechanism of that reaction is believed to have dichlorocarbene intermediate too. A good point to remember is that the dichlorocarbene intermediate is often used to form cyclopropane or carbonyl groups.

This is the mechanism for the reimer tiemann reaction. We can see the similarity in the this reaction and the one you have mentioned previously.


What is the (conceptual) difference between binding energy and adsorption energy? I know the equations, but I have some difficulty getting the difference between the two concepts of binding energy and adsorption energy.

For solids, the binding energy is what makes the crystal (if you will) hold together. It is what makes the individual constituents of the compound bind together.

Adsorption however is an entirely different process: Here, some molecules interact with the surface of another compound. Since (strictly speaking) this is no chemistry but mere physics (there is no formation or breaking of bonds) there are other interactions at play, mainly van der Waals-type interactions.

As these are weaker (it's easier to wipe off the water from your windshield than rip out a silicon atom) and aren't based on the same principles as the binding energy, they get a different name.

I hope that cleared things up for you.


Why do transition metals tend to form positive ions instead of negative ions? I have learnt that elements on the left-hand side of the periodic table such as sodium and magnesium prefer to lose electrons to form a cation because this requires less energy to obtain a stable octet, and vice-versa for the right-hand side of the periodic table e.g. fluorine. However, using this reasoning I am not sure why all transition metals tend to lose electrons rather than gain them.

See the outer configuration of some transition elements:

Sc:      $\mathrm{[Ar]~3d^1 4s^2}$
Cr:  $\mathrm{[Ar]~3d^5 4s^\color{red}{1}}$
Co:  $\mathrm{[Ar]~3d^7 4s^2}$
Cu:  $\mathrm{[Ar]~3d^{10} 4s^\color{red}{1}}$
Zn:  $\mathrm{[Ar]~3d^{10} 4s^2}$

If you are wondering about configuration of Cu and Cr read why this happens!

As you may notice, they can form ions by either losing or gaining electron in 4s orbital. In the chemistry of the transition elements, the 4s orbital behaves as the outermost, highest energy orbital. When these metals form ions, the 4s electrons are always lost first, leading to a positive charge on ion. Gaining any electron in 4s orbital would decrease the stability of anion thus formed. So, they form only cations (positive ions).

In general, the outer electronic configuration of transition elements is $n~\mathrm{s}^2~(n-1)\mathrm{d}^{1-10}$. Because of reasons above, the electron removes from  the $\ce{ns}$  orbital, where n = principal quantum number of atom.

See $\ce{Zn^{2+}}$ configuration:  $\mathrm{[Ar] (3d)^{10}}$ as an example. The electron exits from 4s orbital, leading to the formation of a positively charged ion.


Reaction of benzene diazonium salts with different compounds 1. $ \ce{ArN2+Cl-}$ gives addition product with $\alpha,\beta$ unsaturated acids, with decarboxylation being integral part of reaction. For example,
  $$\ce{ArN2+Cl- + Ph-CH=CH-COOH -> Ph-CH=CH-Ar +CO2 + HCl}$$

  2. With $\ce{SnCl2/HCl, ArN2+Cl-}$ given phenyl hydrazine $\ce{ArNH-NH2.HCl}$

I can not find mechanism of any of these reactions, first reaction seems  similar to Meerwein reaction (without $\ce{Cu^2+}$), but certainly can't follow similar mechanism. Second reaction is completely different from either. What are the reaction mechanisms of each?

As to your first reaction that does not utilize copper, a reference to the source would be useful. Nonetheless, the Meerwein Arylation Reaction has been reviewed in Organic Reactions. The reaction is conducted in the presence of catalytic cuprous ion. Meerwein proposed an aryl cation but a radical mechanism appears plausible. Reduction of the aryldiazonium salt 1 produces aryl radical 2 and cupric ion. Addition of the aryl radical to cinnamic acid 3 to the double bond to give the more stable of two possible radicals, ---namely--- the benzylic radical 4. This radical is oxidized by cupric ion to produce benzylic cation 5 and the catalyst cuprous ion. Loss of CO2 forms the stilbene 6.


emulsion - why does it make a differnce if you have have more/less water/oil So I have been wondering why you can see the following:

  Using 10 ml oil and 1 ml water with and emulgator leads to a way better emulsion than if you use 10 ml water with 1 ml oil (if you use the same amount of emulgator)

btw: as an emulgator we used the yellow part of an egg

The two combinations (10/1 and 1/10) of oil and water will give different emulsions: the first will be water-in-oil and the second will be oil-in-water. The difference lies in the continuous phase; the minor phase will be tiny droplets, with a size depending on how much energy you put into shearing it. Neither emulsion will be particularly stable; that's why we need emulsifiers: to stabilize the emulsion, and also to make it easier in the first place.

The yolk of an egg is primarily fatty (that's why it tastes so good). It's so fatty that it does not absorb the egg white, which is 90% water and about 10% protein. The yolk does contain some hydrophilic groups - but not enough to emulsify the albumen. But the yolk can emulsify/stabilize a small amount of water, and the yolk will disperse easily in (some) oils. So the yolk, spread out in oil, can stabilize a smaller amount of water. But you can't get the yolk well dispersed in water: it agglomerates quickly. And extra oil just makes the destabilization quicker and more obvious.


Jahn-Teller Metal : The new state of matter Recently it came in the news that scientists have discovered a new state of matter called Jahn-Teller metal. I tried reading up on it but couldn't understand it clearly.

Please explain in simple terms why this is a new state of matter and/or what's so special about it.

The Jahn-Teller metal is not one of the 3 (or 4 including plasma) states that occur naturally on Earth; this state allows a metal to have the properties of an insulator and a superconductor.

It is an intermediate state of matter confirming the Jahn-Teller distortion; by interacting with certain electrons (a more detailed description can be found in this Science Advances magazine article, a distorted crystalline structure allows a solid depicting characteristics of an insulator to also becomes a conductor.

The main breakthrough this brings is having a superconductors. Superconductors are extremely useful because they possess zero resistance, except the conditions require them to be cooled to extreme temperatures.
This state nonetheless suggests reproducing superconductors at a less extreme temperature.

Making it an economic way to make efficient electronics.


Any example of colloidal solution where liquid Helium is solvent? I would like to see any information, if it exists, about colloidal solutions where Helium is "solvent".

Background of that question is, I'm wondering giving surface energy effects, how potentially small can be water ice particles at low temperatures.

A solution implies that there are solvent-solute interactions. Helium is about the most chemically inert atom there is. In the case of water ice, the hydrogen bonding will be orders of magnitude stronger than the van der Waals forces, so colloids will not be stable at 'normal' pressures. I guess if I were looking for a colloid in He(liq), I'd look at N2 or perhaps H2. We've made a lot of progress since I had colloidal (surface) chemistry class, but I doubt that we could predict size of particles in any such system. But I don't know.


If the polarity of the solvent increases, will the Rf value increase or decrease? True or False: A sample of aspirin has an $R_f$ value of 0.20 when measured on a silica gel TLC plate with a certain solvent mixture. If the polarity of the solvent is increased, the $R_f$ value of the aspirin will become larger.

Would this be false? Because aspirin is polar (I believe would be considered polar because it dissolves in water despite some other sources saying that it is actually non-polar) and the silica in the TLC plates are also polar, aspirin would not be able to travel very far because it would interact with the silica. Thus, shorter distance over longer distance = lower $R_f$ value, with $R_f$ = distance spot moves/ distance solvent moves.

I wish we would stop teaching chromatography in terms of "polar" and "nonpolar."

The aspirin will interact fairly strongly with the silica due to hydrogen bonding/electrostatic interactions of the carboxylic acid and the ester with the silica. If you increased the "polar" component of the mobile phase, it would travel further due to the mobile phase competing more for binding spots of the silica gel.

So increasing the polarity of the mobile phase does raise $\mathrm{R_f}$ of aspirin (or anything) on regular silica gel.


Basicities of meta-nitrophenoxide and para-nitrophenoxide The negative oxygen is the site of protonation. Size and EN are the same for both oxygens. We are mainly looking at resonance (and possibly inductive) effects here. We want the least resonance withdrawal of electron density from the oxygen. And if possible, the least inductive withdrawal of electron density from the oxygen (although this is not a major issue; resonance trumps induction). Recall that the nitro group cannot withdraw electron density via resonance effects at the meta position. Therefore meta-nitrophenoxide should be more basic than para¬-nitrophenoxide.

Is my reasoning correct?

Yes, it is. These resonance structures describe basicity in para-nitrophenoxide.


H and H- atomic and ionic radius A hydrogen anion has a bigger radius than neutral H . Why is this happening ? I mean both electrons exist at the same shell so this is not it. Both H- and H have 1 proton so this is not it as well . Both electrons of the hydrogen ion have the same energy level and they don't shield the charge of the core from one another. What is it then? Pauli repulsion?

Electrostatic repulsion between electrons is the key to understanding $\ce{H-}$. The electrons might be delocalized, but they are also correlated - they avoid each other.

Both electrons of the hydrogen ion have the same energy level and they don't shield the charge of the core from one another.

The Aufbau and Pauli exclusion principles (and the assumption of hydrogenic orbitals) are insufficient to explain the structure and stability of $\ce{H-}$. In particular, it helps to drop the assumption that the electrons occupy the same spatial orbital and differ only according to their spin quantum number [1]:

As  pointed  out  in  the  introduction  Bethe  [5]  and Hylleraas [6] were the first authors to prove the stability of the negative hydrogen.  But it was Chandrasekhar [1]who first introduced a clever wave function to describe the H− system which leads to a beautiful physical picture. The key concept introduced by Chandrasekhar was to break the symmetry between the two electrons, which is a way to introduce implicitly the electron correlation.  [...]
We conclude that the negative hydrogen ion is a special  atomic  system  whose  stability  depends  completely on the electron correlation. [...] mean field theories and other approximations do not take into account correlation, preventing the possibility of predicting stability for the negative hydrogen ion.

Reference
[1]. Ignacio  Lopez  de  Arbina  de  Frutos. Stability of the Negative Hydrogen Ion:  Variational Approach with Electron Correlation.


How can I lower the flash point of stearic acid (Palm Based Fatty Acids) I am trying to build a environmentally neutral firelighter that does not use petroleum based accelerant. I have found that the flash point is way to high to be viable in Stearic Acid (Hexadecanoic acid / Octadecanoic acid). Is it possible to have a "safe", "environmentally friendly" and biodegradable additive to lower the flash point? I am open to suggestions and alternatives. Any assistance in this is appreciated.

could you have a small amount of a more volatile liquid with the hexadodecanoic acid - such as white spirit or kerosene? This would support a flame and raise the temperature to the point at which the hexadodecanoic acid acid will burn.


Artificial nose sensors I'm am doing research in creating an artificial nose. I need to know how this process works by someone with domain expertise; essentially, what technologies are available currently to detect a broader range of particles in a closed system. My understanding so far goes like this on a high-level overview: particle comes into contact with the olfactory cells and particles from the environment have a custom signature which we associate as smell of a banana or orange, now how to do this artificially or how is this done artificially?

"What technologies are available currently to detect a broader range of particles in a closed system" a LOT of technologies allow this: spectrophotometry, NMR, GC-MS, etc. These technologies give a lot of information about the molecular structure (but not about the molecular  function).

The hard part in your idea is trying to replicate the sensivity of a nose, which is, at the moment, impossible as far as I know.

Trying to deduce via a robotic nose the smell of a novel compound is a task that could, theorically, be made if that "nose" had exactly the same receptors and "circuits" of our nose, since receptors interact in a unique way with molecules, and that is the source of all those unique smells that we sense.

A more approximate approach, doable with current technologies, could be a database-based approach: a lot of "smelly" molecules could be categorized and described in terms of smell.
The analyzer (one of the machines mentioned earlier, for instance) could identify the molecule(s) involved, and the database could provide information about the smell in that environment.

The problem? A sistem like this doesn't work like a nose at all! Any unknown molecule couldn't be "smelled".

Maybe in the future, with the improvement of technology, a molecular docking study will be performed in seconds, and that, when all the proteins associated with smell will be known, could allow us to know the exact way a molecule affects our receptors, our brain and our perception, hence creating a working "digital nose".

And finally, let me add a note: you said "particles from the environment have a custom signature which we associate as smell of a banana or orange". One might argue wether that "signature" is in the molecule or in the nose itself. While many compounds are sensed in a similar way within a population, there are genetic mutations between species and even individuals. Puppies are known to be appealed by cat feces, for instance: is that just a behavioural thing, or their perception of the smell of cat's feces is different from ours?
Some people like the smell of truffles, others hate it: it's (at least partially) because of the individual ability of detecting molecules.

The point is not that of detecting a "signature", but in detecting molecules and interpreting them via our specific "signature".
If olfactory proteins didn't exist, smell wouldn't exist: it's not a physical property of the molecule, but the result of an interaction with who "smells".

Don't be fooled: that's not a philosophical sentence: light wouldn't cease to exist if we had no visual receptors. But for sure colours wouldn't exist, since also colours exist only because we are evolved to detect light as a mixture of three different chemical responses (each associated with one of the RGB colours).


Why are yields of some chemical reactions so dependent on the stoichiometry of the reagents? Back in chemistry class in high school I was working with a partner doing a chemical reaction. We measured the amounts of reagents well, but not perfect, and our yield suffered greatly. On the other hand my friend and his partner came up with quite a high yield. He told me it was because they measured each amount of reagent exactly.

So I wonder... Why are these reactions so dependent on perfect measuring of reagents? Why can't you just use more than enough of one reagent to completely react with the other and remove the excess reagent?

Also does this apply to all chemical reactions, or just a select few?

The proper stoichiometry may not be the only reason for the friend's success, but also the overall preciseness of their preparation procedure.

Chemical reactions are not generally that much critical, but there are lot of  cases, that are.

There are few aspects to consider:

Excessive reagents may affect the effectiveness of the product isolation.
Excessive reagents may alter chemical environment in final mixture, affecting formation of products.
Excessive reagent may not remain, but may form another product, instead of the desired one

Typical case for 3. is e.g reaction of $\ce{NaOH}$ with $\ce{H3PO4}$. Phosphoric acid forms 3 salts.

$$\ce{NaH2PO4 + Na2HPO4 + Na3PO4}$$

Let suppose you want to produce
$\ce{NaH2PO4 }$

$$\ce{NaOH + H3PO4 -> NaH2PO4 + H2O}$$

Extra  $\ce{NaOH}$ will create from the wanted salt  other salt...

$$\ce{NaOH + NaH2PO4 -> Na2HPO4 + H2O}$$

... and the amount of the final product decreases.

More detailed answer depends on providing your particular scenario.


How would a pH curve look like for titration of diluted weak base compared to concentrated one? Let's say I take $\pu{80 g}$ of a weak base, dilute it with  $\pu{50 ml}$ of water and titrate it with a strong acid. I get a titration curve. Now I take again $\pu{80 g}$ of the same weak base but this time I dilute it with $\pu{200 ml}$ of water and titrate it with the strong acid. What will be the new titration curve compared to the first one?

And here is what confuses me:

On one hand, the diluted base should have lower $\mathrm{pH}$. But on the other hand, at half-equivalence point, $K_\mathrm{b}=\mathrm{pOH}$, and since it's the same base, $K_\mathrm{b}$ is the same, and thus $\mathrm{pOH}$ and $\mathrm{pH}$ at half equivalence point should be the same. I am also not sure about the endpoint.

I am first going to do the ‘flip side’ of the OP’s scenario and then basically flip it back. So we start with a $\pu{0.1 M}$ aqueous solution of benzoic acid and the initial solution volume is $\pu{25 mL}$. Benzoic acid is a monoprotic weak acid with $\mathrm{p}K_\mathrm{a} = 4.20$. The initial $\mathrm{pH}$ is $2.606$. The strong base used for the titration is $\pu{0.1 M} \ \ce{NaOH}$. At the equivalence point, the $\mathrm{pH = 8.454}$ and the solution volume is $\pu{50 mL}$ ($=\pu{25 mL}$ benzoic acid solution $+ \pu{25 mL} \ \ce{NaOH}$ solution). Thus, the $\mathrm{pH}$ is simply the $\mathrm{pH}$ of $\pu{0.05 M}$ sodium benzoate solution. After all $\pu{50 mL}$ of $\ce{NaOH}$ has been added, the $\mathrm{pH} = 12.523$ and the final solution volume is $\pu{75 mL}$. At the half-equivalence point, $\pu{12.5 mL}$ of $\ce{NaOH}$ has been added and $\mathrm{pH}= 4.185$.

Now assume the initial benzoic acid solution is only $\pu{0.025 M}$, but the initial volume is $\pu{100 mL}$. The total number of moles of benzoic acid is the same as before. The initial $\mathrm{pH}$ is $2.912$: the benzoic acid solution is only 25% as concentrated. At the equivalence point, the $\mathrm{pH} = 8.254$ and the solution volume is $\pu{125 mL}$ ($=\pu{100 mL}$ benzoic acid solution $+ \pu{25 mL} \ \ce{NaOH}$ solution). The equivalence point $\mathrm{pH}$ is closer to $\mathrm{pH}=7$ because the solution is simply a $\pu{0.02 M}$ sodium benzoate solution, i.e., the number of moles of sodium benzoate is now in a volume that is 2.5 times larger, so the concentration is 40% of $\pu{0.05 M}$. After all $\pu{50 mL}$ of $\ce{NaOH}$ has been added, the $\mathrm{pH}= 12.222$ and the final solution volume is $\pu{150 mL}$. At the half-equivalence point, $\pu{12.5 mL}$ of $\ce{NaOH}$ has been added and $\mathrm{pH} = 4.188$, moving slightly up toward $\mathrm{pH}=7$. See the titration curves below:

Now flipping back: the OP’s scenario involving a weak base titrated with strong acid. For reasons that will become clear, assume the hypothetical weak base has $\mathrm{p}K_\mathrm{b} = 9.80$, and that the strong acid is $\pu{0.1 M} \ \ce{HCl}$ solution. Note that the $\mathrm{p}K_\mathrm{b}$ is simply $14$ (treated as exact throughout) $- 4.20$.

As before, start by assuming $\pu{25 mL}$ of $\pu{0.1 M}$ weak base. The initial $\mathrm{pH} = 14 – 2.606 = 11.394$. At the equivalence point, the $\mathrm{pH} = 5.546$ and the solution volume is $\pu{50 mL}$ ($=\pu{25 mL}$ weak base solution $+ \pu{25 mL} \ \ce{HCl}$ solution). Thus, the $\mathrm{pH}$ is simply the $\mathrm{pH}$ of a $\pu{0.05 M}$ weak base salt solution. After all $\pu{50 mL}$ of $\ce{HCl}$ has been added, the $\mathrm{pH} = 1.477$ and the final solution volume is $\pu{75 mL}$. At the half-equivalence point, $\pu{12.5 mL}$ of $\ce{HCl}$ has been added and $\mathrm{pH} = 9.815$.

Now assume the initial weak base solution is only $\pu{0.025 M}$, but the initial volume is $\pu{100 mL}$. The total number of moles of weak base is the same as before. The initial $\mathrm{pH}$ is $11.088$: the weak base solution is only 25% as concentrated. At the equivalence point, the $\mathrm{pH} = 5.746$ and the solution volume is $\pu{125 mL}$ ($=\pu{100 mL}$ weak base solution $+ \pu{25 mL} \ \ce{HCl}$ solution). The equivalence point $\mathrm{pH}$ is closer to $\mathrm{pH}= 7$ because the solution is simply $\pu{0.02 M}$ weak base salt solution, i.e., the number of moles of weak base salt is now in a volume that is 2.5 times larger, so the concentration is 40% of $\pu{0.05 M}$. After all $\pu{50 mL}$ of HCl has been added, the $\mathrm{pH} = 1.778$ and the final solution volume is $\pu{150 mL}$. At the half-equivalence point, $\pu{12.5 mL}$ of $\ce{HCl}$ has been added and $\mathrm{pH} = 9.812$, moving slightly down toward $\mathrm{pH}= 7$. For the titration curves, simply imagine vertically flipping the titration curves, in the figure above, about a horizontal line at $\mathrm{pH} = 7$. Or just look at the plot:

One last thing: The $\mathrm{pH}$ values at half-equivalences were close to the respective $\mathrm{p}K$ values, as expected, but any buffer can be broken by dumping it into swimming pool full of water. So dilution may matter a great deal, or relatively little, depending on the specifics.


What do we learn from an isotope peak (or M+2 peak) in a mass spectrum? Mass spectrum of a molecule showed molecular ion peak and its isotopic
 peak in 1:1 ratio. What is the possibility of the atom present in the
 molecule?

Since it is mentioned that it has a molecular ion peak it gives the fact that it must be aromatics, olefins and not alcohols or branched hydrocarbons since the latter have less probability to show molecular ion peaks . Since they contain isotopic peaks it must mean they have chloro or bromo compounds. But I do not understand how to use ratios to identify the atom present in the molecule

Each element has different natural abundances for its isotopes. Two peaks with approximately 1:1 ratio indicates the presence of bromine. $\ce{^{79}Br}$ is the most common isotope, corresponding to $0.5069$ of naturally occurring bromine.  The isotopic composition for $\ce{^{81}Br}$ is $0.4931$.

What this means is that, for a molecule contaning a single bromine atom, approximately, half of the molecular ions will have $\ce{^{79}Br}$ and the other half will have $\ce{^{81}Br}$, resulting in two peaks with roughly the same intensity with $m/Z$ differing by two units.

References:

[1] https://www.nist.gov/pml/atomic-weights-and-isotopic-compositions-relative-atomic-masses

[2]https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumental_Analysis/Mass_Spectrometry/Organic_Compounds_Containing_Halogen_Atoms


Partial hydrolysis of XeF6 What are the products of partial hydrolysis of $\ce{XeF_6}$?
Is $\ce{XeOF_4}$ formed or $\ce{XeO_2F_2}$ is formed?

Both can be formed. It depends on the amount of water added:
\begin{align}
&&\ce{XeF6 + H2O  &->  XeOF4 + 2HF}\\
&\text{or}&
\ce{XeF6 + 2H2O &-> XeO2F2 + 4HF}\\
&\text{or even}&
\ce{XeF6 + 3H2O &-> XeO3 + 6HF}
\end{align}


Changing pressure on equilibrium Recently I heard my teacher say that when increasing the total pressure of the system, the reaction which is "more dependent on collisions" is favoured.
This seems to make sense, although I've looked at many resources online and on SE, and people have either given a mathematical explanation or simply talked about Le Chatelier's principle (increasing pressure favours reaction to produce less moles)
My question is: is it correct to say that increasing the pressure favours the reaction which involves more molecules colliding?
Take for example the forward reaction here which is favoured upon pressure increase. (4 molecules need to collide for forward and 2 molecules need to collide for reverse reaction)
$$\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}$$
I'd appreciate an intuitive explanation about collisions and interactions between molecules rather than something mathematical. Thanks in advance!

Yes, I agree with your teacher's comment that upon increasing the total pressure of a system, there are select reactions "more dependent on collisions", and such reactions have been a matter of study.
For example, per this 2008 work reported in 'The Journal of Physical Chemistry',  The Temperature and Pressure Dependence of the Reactions H + O2 (+M) → HO2 (+M) and H + OH (+M) → H2O (+M), where '+M' here denotes the catalytic presence of a surface, as can be provided by another gas molecule.
To quote from the abstract:

The reactions $\ce{H + O2 (+M) → HO2 (+M)}$ and $\ce{H + OH (+M) → H2O (+M)}$ have been studied using high-level quantum chemistry methods. On the basis of potential energy hypersurfaces obtained at the CASPT2/aug-cc-pVTZ level of theory, high-pressure limiting rate coefficients have been calculated using variable reaction coordinate transition state theory. Over the temperature range 300−3000 K, the following expressions were obtained...Available experimental data on the pressure dependence of the reactions were analyzed using a two-dimensional master equation.

Note, reactions are being studied via surface collisions involving the addition of  Ar and N2 gases. The selection of the gas relates to chemical inertness and the total collisional cross-section of the particular gas particles, all of which are components of more complex theories surrounding collision frequency and collision theory (see, for example, discussion here).


Maximum Concentration of Benzene possible A question for my PhysOrg course states:

  Assume you are performing a reaction in which benzene is a reagent. What is the maximum concentration of benzene possible? Hint: what facts do you need to know to determine concentration?

I understand concentration is amount per volume in mol/L, but I'm not sure how to proceed from there.

Sorry, I figured it out- just substitute mols for mass/MW and density  for volume and rearrange until you get density/MW


Need a pair of compounds that is similar in formula but different in structure I would like to explain chemical formulas is not representable in terms of the molecular structure of the compound.

Any good suggestions of examples?

The classical example is dimethyl ester $\ce{CH3-O-CH3}$ and etanol $\ce{CH3-CH2-OH}$ pair. They have exactly same brutto-formula, but different structure and very different properties. The first is a gas, the second is a liquid easily reacting with sodium metal.

Carbon is also interesting example. It has two common forms with exactly same chemical formula $\ce{C}$, but very different properties: a very hard and transparent diamond and soft black graphite.


Lines on inside of hexagonal shapes of structure diagrams just a simple question from me today...

I was looking at the structure of Norepinephrine for a school project and was curious about the diagrams on the Wikipedia page for it.

I was wondering why hexagons are used in this diagram and what the purpose of the three lines on the inside of the hexagon mean.

Noradrenaline/norepinephrine is an aromatic organic compound containing a benzene ring (the hexagon). Essentially, the vertices in the diagram are carbon atoms and the number of lines represents the covalent bond order; 1 line (eg connecting an OH to the benzene ring) means a single covalent bond. Any 'spare' bonds are where a C-H bond is. So it can be rewritten like this:

HOWEVER, when it comes to a benzene ring (the hexagon), you do not simply have alternating single and double bonds. In fact you have what is known as a 'resonance structure', where the electrons/bond is shared around the entire ring. One way of thinking of it is that there are 1.5 bonds between each carbon atom in the ring. Another way of thinking of it is that pi-orbitals from each carbon atom merge to form a ring combined orbital structure, so instead of existing in that figure-8 type area around one carbon nucleus, those 6 electrons can be in a donut shaped area above and below the carbon nuclei:

This is why you often see a hexagon with a circle inside it to represent a benzene ring as it is less misleading than the alternate single and double bonds. In that case, you would redraw the noradrenaline diagram like this:

The advantage, though, of having the alternate double and single bonds is in drawing diagrams when working out reactions.