Oxidation of conjugated dienes into carboxylic acids by potassium permanganate In the book -  MS Chouhan; Advanced Problems In Organic Chemistry; 11th ed, Chapter Alkenes Q32 - the following reaction is given:

The given solution says that:

  Conjugated dienes on oxidation by $\ce{KMnO4/\Delta}$ give oxalic acid

However, the mechanism for this reaction is not given. Also, the formation of a dicarboxylic acid on the parent ring is not explained. I am also unsure what would the products be in a simpler case like buta-1,3-diene. What products will it give?

I checked the internet but could not find a mechanism for the same. The related question of oxidation of arene subsituents by $\ce{KMnO4}$ is not a duplicate.

What is the mechanism for this oxidation?

Here is a relevant extract from a PDF (Dr. P.G. Hultin, 28 July 2016) depicting oxidative cleavage.  For the compound in hand (your question) the process will be done twice. $\ce{H2O2}$ will serve as the source of $\ce{H3O+}$.

  When alkenes are treated with $\ce{KMnO4}$ in acidic solutions, the diol is not formed. Instead, the alkene is cleaved. The reaction proceeds by the same mechanism at the start, forming a cyclic manganate ester (although since the reaction is under acidic conditions the structure is protonated). In the schemes below, the alkene carbons are highlighted throughout, so you can see where they end up in the product.

  Now, under these conditions the manganate ester is not very stable and it undergoes a cyclic fragmentation process, which results in breaking the C-C bond between the two oxygens. Notice that in this case, since there was a hydrogen atom attached to each of the alkene carbons in the starting material, there is a hydrogen attached to the carbonyl carbon in the product and therefore the product that is initially formed is an aldehyde

  Aldehydes are very easily oxidized to carboxylic acids, and thus the aldehydes formed in the cleavage reaction do not survive. They are rapidly transformed into carboxylic acid groups, by a complex reaction whose mechanism you need not worry about.

  Now, if the alkene had not had any hydrogens attached, the product in that case would have been a ketone
  rather than an aldehyde. Ketones are not easily oxidized further, and the reaction would have stopped at that
  stage.

  If one of the alkene carbons had a hydrogen substituent, while the other did not, then we would get both acid
  and ketone groups in our product, as shown below.


Notation of beta decay Is it an ion or a neutral atom that is created through beta decay?

For example: $_6^{14}\mathrm C \to ~ _7^{14}\mathrm N + \mathrm e^- + \mathrm{v{_e}}^{-}$

Isn't $_7^{14}N$ an ion since neutron gives us only a proton and an electron, not a proton and 2 electrons? If so, why isn't it stated in the example above (it's from Wikipedia)?

An ion is created. Usually this is not noted down in the nuclear reaction because while talking of nuclear reactions we only concern ourselves about nuclei, not the entire atom. The reaction is to be read as "A 14-$\ce{C}$ nucleus decays into a 14-$\ce{N}$, an electron, and an antineutrino."

Also, note that the neutrino has no charge.


ChemDraw temporary label for chemnum LaTeX Currently I'm searching for a nice way to implement my ChemDraw graphics into $\mathrm{\LaTeX}$.

There is a package called chemnum which will auto-label the compounds in the eps-graphic which sounds nice. The package describes, that I need to put "temporary markers" inside the document where the number should be.

How do I get these markers? I guess it is not a normal text area, isn't it? I'm using ChemDraw 16.0.1.4.

The ChemNum package documentation that can be found on CTAN specifies:

  Create the scheme and place temporary tags like TMP1, TMP2 and so on where you want the compound labels to be.

A question on TeX.SX specifies:

  When you create a text box and type into it, it will store it as a string correctly in the EPS. However, it was (sometimes) automatically putting it into chemical formula mode - so that in TMP1, the 1 was subscript (TMP1). Even if you fix this manually, it now saves the TMP and 1 separately in the EPS, meaning CHemNum fails to find TMP1.

Thus, it seems all you need to do is create textboxes with the appropriate labels.


Why do certain compounds dissolve in oil? Why do some types of compounds dissolve in oil, yet not water? Sorry if this is an obvious answer. I tried rewording this question every way I could, but search engines came up with nothing- they all skirted around this exact question.
Why do certain compounds, such as fatty acid or Vitamin A, dissolve in oil?
Polar substances can dissolve in water because water molecules are polar and literally pick apart the [polar] molecules in the solute by attracting to them. However, oil is nonpolar and has no such ability.
How, then, does it make certain compounds dissolve? Is there some other bond it can form with certain compounds?
On a second question, I read something about London dispersion and Van der Waals forces being the main force of attraction between oil solvents and solutes that can dissolve in oil. If this is the case, that means the electrons in orbit around the solute's molecules shift around and form temporary dipoles with each other.
But what makes it so that some solutes can only dissolve in oil-- and not water [molecules], which are polar, and could also interact with these temporary dipoles?
edit: A comment said that this not energetically preferable. So I think the second question can be considered answered.
Thanks!

I propose a logical approach, to explain the rule "the like dissolve like". Why don't we admit that all liquids should be able to mix, exactly like gases may be mixed in all proportion ? This is the case for non polar molecules, like oils. And if two liquids don't mix, this is due to the existence of attracting forces between molecules A (hydrogen bonds for exemple), and that mixing them with another stuff B may destroy too many of these bonds. If the mixing operation A + B creates about as many new bonds as those broken during the mixing, the liquids A and B may be mixed. This is the case in the mixture water - ethanol. If the mixing operation A+B does not create enough new bonds between molecules A and B, like water and oil, Nature refuses to break the bonds between A molecules without any counterpart, and prevents the A molecules (water) to be separated from one another in order to accept B molecules, and A and B are not miscible.


Synthesis of 3-Methylenecyclopentene from 1-methylCyclobutene How do I make 3-Methylenecyclopentene

from 1-methylcyclobutene?

I thought to do ring opening reaction to add 1 carbon atoms but this won't add double bond in product. How to start please?

Here is my try.

Ozonolysis
Aldol condensation using a hindered base (like LDA)
Wittig reagent to convert $\ce{C=O}$ to $\ce{C=C}$

Step 1:

Step 2:

I used $\ce{LDA}$ which abstracts a proton from the least hindered carbon near to a carbonyl group. (So that the resulting carbanion is stable).
I am not drawing the enolate ion formation etc. You can find the reaction mechanism for aldol condensation here and the reaction using $\ce{LDA}$ here.

After protonation, you can heat the compound to remove $\ce{OH-}$ which is a simple dehydation. There are two possible alkenes. One of them (the one given below) is more stable because of conjugation.

Step 3:

We are almost done. To replace $\ce{C=O}$ with $\ce{C=C}$, use wittig reagent.


Equilibrium constant for a multi-step reaction If a reaction proceeds in steps, is the equilibrium constant of every step taken into account to arrive at the final constant? Or can we just calculate equilibrium constant for the overall reaction with reactants on one side and the products on the other?

To give a specific example, let's take this set of reactions:

$$\tag{1}\ce{A + B <=> I_1}$$

$$\tag{2}\ce{I_1 + C <=> I_2}$$

$$\tag{3}\ce{I_2 <=> I_3 + D}$$

$$\tag{4}\ce{I_3 <=> E + F}$$

To get the net reaction, we add up the equations and cancel the intermediates:

$$\ce{A + B + C <=> D + E + F}$$

When the system is a equilibrium, all the reactions are at equilibrium. You can write the equilibrium constant expression based on the net reaction:

$$K_\text{net} = \frac{[\ce{D}][\ce{E}][\ce{F}]}{[\ce{A}][\ce{B}][\ce{C}]}$$

You would arrive at the same expression if you take the product of the individual equilibrium constants (and cancel):

$$K_\text{net} = K_1 K_2 K_3 K_4 =\frac{[\ce{I_1}]}{[\ce{A}][\ce{B}]} \times \frac{[\ce{I_2}]}{[\ce{I_1}][\ce{C}]} \times \frac{[\ce{I_3}][\ce{D}]}{[\ce{I_2}]} \times \frac{[\ce{E}][\ce{F}]}{[\ce{I_3}]}$$

The concentrations of the intermediates will not be zero because they are in equilibrium with the other species. They are intermediates in the sense that we care a bit less about them since they don't appear in the net equation of interest. However, they are part of the mass balance and may well be among the major species.


Flood Control - electrolysis - limit to rate of conversion? Flood control - is it possible to use electrolysis to remove flood waters by converting water into oxygen and hydrogan gas?   Is there a limit to the amount of water that can be converted at a time?

I know that this is a bit crazy and I respect what you are saying, BUT I am not trying to vaporize the entire Mississippi river, only enough to keep flood waters from ebbing into a house.

Maybe a different appreach - how about freezing the water place (in blocks) and then transporting it elsewhere

if only water was compressible....

Floodwaters are immense in quantity.  Let's estimate how much energy would be required to convert that quantity of water to $\ce{H2}$ and $\ce{O2}$.

The Mississipi river has a (non-flood) flow rate of about 593,000 cubic feet per second, or 17000 cubic meters per second.  That's $1.7 \times 10^7$ liters per second, or $1.7 \times 10^{10}$ grams per second, or roughly $10^9$ moles per second.
Faraday's constant is the conversion factor between moles and Coulombs.  It's value is about 96000 ampere-seconds per mol.  Let's approximate it as $10^5$ ampere-seconds per mole.
It takes four moles of electrons per mole of water to complete the electrolysis.
Multiply the numbers together, and you get that to electrolyze all the waters of the Mississippi river, you'd need $4 \times 10^{14}$ amperes of current.  That's about the amount of current present in a billion bolts of lightning.  Yikes!
Exactly how much voltage would be required depends on your design of electrolytic cell you use for this giga-bolt scale.  Let's say it's 5 volts.  That means the power you need for the electrolysis is $2\times10^{15}$ watts.  That's 2000 terawatts.  But the world's current total energy consumption is about 12 terawatts...


Reduction of carbonyl groups In such reactions as in the pic, how can we know which carbonyl group will undergo a reaction to give an alcohol. Also, can both of the sites undergo the reaction?

Sodium borohydride is a weaker reducing agent than, say, lithium aluminium hydride, so, although it will react with ketones and aldehydes, it is not strong enough to react with carboxylic acids or esters. This generalization is quite broadly applicable, so you already know your borohydride will reduce the ketone carbonyl on that molecule. $\ce{NaBH4}$ is a weaker reducing agent than $\ce{LiAlH4}$ because boron is less electropositive than aluminium. Boron has a Pauling electronegativity of 2.04 while aluminium's pauling electronegativity is only 1.61, therefore $\ce{NaBH4}$'s hydrides are significantly less nucleophilic and they won't be able to react with a weaker electrophile.

To answer your question directly, you have to think about which carbonyl is more electrophilic. Generally, resonance effects are more important than inductive effects when considering something's nucleophilicity/electrophilicity.

The ketone has two resonance forms, while the ester has three, therefore the partial positive charge on the carbon is more delocalized on the ester, making the ketone more electrophilic. The weak $\ce{H-}$ nucleophile provided by $\ce{NaBH4}$ will only be able to react with the stronger electrophile, the ketone.


Existence of Dichlorine Pentoxide I searched the net and a lot of books, but I was unable to find any information about dichlorine pentoxide.

Does dichlorine pentoxide($\ce{Cl2O5}$) exist?
If it doesn't exist, then why so?
If dibromine pentoxide and iodine pentoxide do exit, then why not dichlorine pentoxide?

The crux of this (Russian) article is that $\ce{Cl2O5}$ is not stable enough to be isolated in a pure form, and if it were, it would be strongly acidic oxidizer.

This web site is representative of several others that simply give example reactions of the compound (the second reaction being the most common exampe):

$$\ce{NiCl2 + 3O2 -> NiO + Cl2O5}$$
$$\ce{H2O + Cl2O5 -> 2HClO3}$$

The statement in the first article and the simple fact that there is such a dearth of information about this compound suggest that it really only exists as a transient, unstable, oxidizing acid that apparently has not been isolated as a stable compound.  Furthermore, information can be found in Wikipedia and/or PubChem for $\ce{Cl2O_x}$, where $x \in 1, 2, 3, 4, 6, 7$.


Why is the molecular geometry of a molecule with 5 bonds not uniform? I'll note that I'm a student in a rather introductory course, and today we learned about molecular geometries.

I'm curious as to why certain geometries (say, $\ce{CH4}$, tetrahedral) are uniform (evenly distributed in a sphere) while other geometries, say "$\ce{SbCl5}$," (5 bonds to the central atom) are less-uniformly distributed - two of the atoms are linear and 3 of them of trigonal planar.

Is this just a simplification for teaching introductory students, or is this how the molecule actually behaves?  I would think that to achieve "equilibrium" from the $\ce{Cl}$ atoms repelling each other, the distribution of the $\ce{Cl}$ atoms would be perfectly normal around the sphere.

You may want to look into VSEPR theory. To a good first order approximation you can think of electron pairs and external atoms on a central atom as taking up about the same space.
This explains why water's bond angle (104 degrees) is very close to the 109.5 degrees exact equidistant angle associated with methane, or $\ce{AlCl4+}$.

Your example with $\ce{SbCl5}$ is a good example where the theory gets a bit more complicated than simply maximizing distance between steric centers. The bonds to the chlorines are not equal. And yes this is how the chemical would look: see the Wikipedia article on orbital hybridisation.

Now, as your example has 5 identical groups, inter-conversion can occur, where the axial group "switches" with an equatorial group. But at any moment there will be two distinct chemical environments around the compound.


Does 4-methylbenzaldehyde give haloform and aldol reactions? The following question was asked in JEE Advanced 2017:

The answers given are option A and B. I was able to get B as the correct answer but A look like the wrong option to me.

My approach for option (A) and checking the reactions for Q:

Since (Q) of option (A) is able to give both haloform and aldol reaction, it should be incorrect. The reason I say that aldol reaction makes the option incorrect is that cannizaro reaction is given by aldehydes or ketones which have non enolisable hydrogen atoms. But aldol reaction requires enolisable hydrogen.

Can anyone confirm what should be the correct answer and why is my approach wrong if the answer given is right.

First, Cannizzaro reaction is not given by ketones (your statement says ketones give cannizzaro reaction).

Counter to your analysis, the following reaction will not occur since $\ce{H}$ bonded to methyl group is not sufficiently acidic for hydroxide ion to attack. Refer to $\mathrm{p}K_\mathrm{a}$ values given by Mathew Mahindaratne.

Under the given conditions  P on ozonnolysis gives formaldehyde and 4-methylbenzaldehyde. Both are nonenolizable aldehydes capable of undergoing Cannizzaro reaction. Both compounds do not undergo haloform reaction.

Scheme 1

Under the given conditions R gives acetophenone and formaldehyde. Acetophenone gives haloform test and it also undergoes aldol condensation, but not cannizaro reaction.

Scheme 2

Choice B also meets the conditions given.
Thus choices A & B are correct.


How can the number of moles in the product side be more than reactant? A rigid stainless steel chamber contains $\pu{0.32 bar}$ of methane, $\ce{CH4}$, and excess oxygen $\ce{O2}$, at $\pu{200.0 °C}$. A spark is ignited inside the chamber, completely combusting the methane. What is the change in total pressure within the chamber following the reaction? Assume a constant temperature throughout the process.

I wrote out the following chemical equation:

$$\ce{CH4 + 2O2 -> CO2 + 2H2O}$$

I know I have $\pu{32 kPa}$ of methane, so using up all the moles of methane would mean that pressure decreases all the way to $0$.

The solution writes the following: it multiplies the 1 mol of $\ce{CH4}$ by $1$ to get the number of moles of $\ce{CO2}$ and by $2$ to get the moles of $\ce{H2O}$ produced.

Why can we do that? I would assume that by multiplying in such a way, you're implying that all of the $\ce{CH4}$ is being used up to create $1$ mol of $\ce{CO2}$. This means that you have no more moles to spare for creating $\ce{H2O}$.

I know my conceptual understanding is wrong, but this is what I thought while doing the question and the solution doesn't intuitively make sense to me.

I was wondering if someone could clarify my misunderstanding for me.

A mole is a way of counting things, in the case of chemistry those things are molecules.

But the number of things is a chemical reaction are not necessarily preserved: some reactions create more molecules (electrolysis of water (H2O), for example, starts with two molecules of water but delivers one molecule of oxygen gas (O2) and two molecules of hydrogen gas (H2)).

So moles are not conserved if the structure of the molecules changes because of the reaction. The point of writing equations is to match conservation of matter (atoms don't disappear) to the structures of the molecules involved in the reaction. Sometimes reactions just rearrange the atoms and give the same number of moles of product, sometimes they create more molecules, sometimes they create fewer molecules.

In an ideal gas there is a neat relationship between the number of moles and the volume or pressure (at constant pressure the volume is proportional to the number of moles of gas; at constant volume the pressure is proportional to the number of moles of gas). So your question is asking you to compare the number of moles resulting from the reaction to the number going into the reaction. since they are the same, the pressure won't change.


Molecules having higher enthalpy of vaporization and boiling point than another but with lower entropy of vaporization? I came across the Trouton's rule that predicts the entropy of vaporization of most molecules to be around 85~88kJ/(K mol). It is said to fail when there is hydrogen bond between molecules.

When I checked the entropy of vaporization for linear alkanes, it increases with molecular mass which looks reasonable to me because a liquid with greater volume should expand more once it transits completely into gas than those with lesser volume. Eventually, the increase does slow down as the values get close to ~90kJ/(K mol).

For linear monohydric alcohols, their entropies of vaporization were higher than that predicted by Trouton's rule which I have found to be explained by their lower entropies in liquid phase due to hydrogen bond. However, I found that the entropy of vaporization of alcohols decreases instead with molecular mass. But since the boiling point and enthalpy of vaporization both increase with molecular mass. Shouldn't their entropy of vaporization increase with stability like alkanes?

So my question is: Is there any way to explain the decreasing ratio between the enthalpy of vaporization and the boiling point with increasing molecular mass for linear monohydric alcohols?

The enthalpy of vaporization is greater for hydrogen-bonding molecules than for plain alkanes.  For low-molecular weight alcohols, this effect is pronounced.  The longer the alkane chain becomes, the more the compound behaves like a pure alkane.  For example, using data obtained from the NIST Webbook:

Table 1. Enthalpy of vaporization in kj/mol:
$\ce{Methanol\ \ \ \ \ 38}$
$\ce{Ethanol\ \ \ \ \ \ \  42}$
$\ce{n-propanol\ \  47}$
$\ce{n-butanol\ \ \ \  52}$
$\ce{n-pentanol\ \  57}$
$\ce{n-hexanol\ \ \ \  61}$
$\ce{n-heptanol\ \  67}$
$\ce{n-octanol\ \ \ \  71}$
$\ce{n-nonanol\ \ \  77}$
$\ce{n-decanol\ \ \ \  82}$

Keeping in mind the relative molecular weights of the compounds, you can see there is a decreasing effect of the hydrogen bonding (and other) effects on the n-alcohol series as we move to larger chains and become less alcohol-like and more alkane-like.  This is much more obvious in the next table, which is simply Table 1 normalized to a per-carbon basis:

Table 2. Enthalpy of vaporization on a per-carbon basis in kj/mol:
$\ce{Methanol\ \ \ \ \ 38}$
$\ce{Ethanol\ \ \ \ \ \ \  21}$
$\ce{n-propanol\ \  16}$
$\ce{n-butanol\ \ \ \  13}$
$\ce{n-pentanol\ \  11}$
$\ce{n-hexanol\ \ \ \  10}$
$\ce{n-heptanol\ \  10}$
$\ce{n-octanol\ \ \ \  9}$
$\ce{n-nonanol\ \ \  9}$
$\ce{n-decanol\ \ \ \  8}$

Table 2 even more drastically illustrates the deviation in $\mathrm{\Delta}H$ changes as we move from alcohol-like compounds to alkane-like compounds. So, given that the enthalpy of vaporization is greater for hydrogen-bonding molecules than for plain alkanes, this should explain your findings.


Calculating self-diffusion coefficient from MSD programmatically According to the Einstein relationship, the diffusion coefficient $D$ is
$$\lim _{t\rightarrow \infty} \frac{\langle \left(\mathbf{r}(t)-\mathbf{r}(0) \right) ^2\rangle}{6t} = D$$
I have run a MD simulation with $N$ particles, and I have a file which has the location of each particle at every time step.
If I want to calculate $D$, I would have to do the following:
$$\langle (\mathbf{r}(t)-\mathbf{r}(0))^2\rangle = \frac{1}{N} \sum _{i=1}^N (\mathbf{r}_i(t)-\mathbf{r}_i(0))^2$$
Divide the above by $6t$, then take the limit as $t$ goes to infinity - which in my case is the final time step of my simulation.
My questions:

Is this the right way of calculating MSD?
If it is the right way of doing this, and my system reaches equilibrium and doesn't diverge, as $t\rightarrow \infty$, shouldn't my diffusion coefficient always be $0$, because of that infinity in the denominator?

In 1D diffusion along the $x$ axis by Brownian motion, let the mean jump lengths be $a$ each taking a time $\tau$ and this should be large compared to the time between successive collisions, which are $\approx 10$ fs, i.e. $a$ is the sum of all displacements during time $\tau$. Assuming this means that there is no correlation between collisions at time $t$ and $t+\tau$. From this the diffusion equation (Ficks Law)  can be derived and the diffusion coefficient defined by $D_x=\langle \Delta x^2\rangle/2\tau$ where $\langle \Delta x^2\rangle$ is the variance of the jump lengths $a$ or $\langle\Delta x^2\rangle = ( x-\langle x\rangle)^2$.  In 3D as there is no correlation between the motions on each axis,  $D=\langle \Delta x^2\rangle/6\tau$. This equation is subtly different from your equation and so you now need to connect these terms with what you have calculated. You should find that your diffusion coefficient settles down to a constant value (plus or minus statistical fluctuations) as the calculation proceeds.


How do I lookup a polymer by its structure? This is more of a meta-chemistry question, but, here goes...

Given the structure of a polymer, how can I search online for which polymer it is?

Specifically, I have an image like this:

I'd also like to know which polymer specifically this is, but also a method for figuring this out in the future by myself :)

That looks like bilirubin, from hemoglobin catabolized in the liver  Monomer, dimer..tetramer...oligomer...polymer...high polymer.  It is a tetrapyrrole, not even an oligomer.


Rationalizing the Planarity of Formamide In formamide, the nitrogens appear to be $\ce{sp^3}$ hybridized, implying tetrahedral geometry. However, analysis shows that the molecule is actually very nearly planar with bond angles close to 120 degrees.

EDIT: as suggested by Martin and another poster, hybridization is a rough concept. So perhaps the hybridization of nitrogen upon further analysis should best be described as somewhere inbetween $\ce{sp^3}$ and $\ce{sp^2}$. This however would still necessitate planarity, correct? Pi bonds are formed through the above and below electron-pairing in p-orbitals; effective bonding is achieved when these p-orbitals are parallel with respect to each other.

I'm thinking this has to do with the partial double bond character in the molecule (also appears to be some ionic character to the molecule - likely due to electron-withdrawing effects of the nitrogen and the oxygen).

This is the standard answer. However, would intramolecular hydrogen bonding also play a role? Couldn't there be a hydrogen bond between the peripheral hydrogen on the nitrogen and the oxygen? Couldn't this also help with achieving the 120 degree bond angles?

Most of the amides are planar (due to steric reasons the restriction may be lifted) and so is also formamide.

The carbon is obviously $\ce{sp^2}$ hybridised (as this concept is very well applicable here), hence organising it's ligands in one plane with roughly $120^\circ$ angles. Naturally one would assume the nitrogen to be $\ce{sp^3}$ hybridised, which is the case for most amines. However, the inversion barrier for these molecules is (depending on the substituents) very low. $$\ce{[NH3]^{pyr-top} <=> [NH3]^{TS-plan} <=> [NH3]^{pyr-bot}}$$ For the nitrogen that means going from $\ce{sp^3}$ to $\ce{sp^2}$ and back again. You can now stabilise the intermediated structure with conjugation, and that is exactley the case here. In your chart, that refers to entry 2. This causes the nitrogen to most likely be of $\ce{sp^2}$ hybridisation and the lone pair of entry one would be in a $\ce{p}$ orbital.

The conjugation happens as stated in you chart by the overlap of that orbital with the antibonding $\pi^*~\ce{C-O}$ orbital. This causes the $\ce{N-C}$ bond order to increase, while the $\ce{C-O}$ BO has to decrease.

All those resonance structures are only descriptions of extreme states, the truth lies between them. The following scheme considers the most common ones and adds a third, that might explain delocalisation (in a non-traditional Lewis way) up to a certain visual point.

In molecular orbital theory you can form 3-centre orbitals from all molecules perpendicular
to the molecular plain. If you choose this plane to be $xy$ then the contributing orbitals will be $\ce{p_{$z$}}$. The following scheme may support this claim, the dipicted orbitals were obtained by a BP86/cc-PVTZ calculation. (The HOMO is an in-plane lone pair of oxygen.)

While there is certainly no intra molecular hydrogen bond ($d(\ce{O-H}\approx 2.57$ same level), there will also most certainly be an attraction between the $\ce{C-O}$ and $\ce{N-H}$ bond helping to stabilize the planarity. But I will not go into detail about that, because it would involve breaking away from the very handy hybridisation concept.


Stoichiometry: efficiency My book gives me the following problem:
"A mixture of $125.0 g$ $\ce{N2}$ and $32.0 g$ $\ce{H2}$ reacts to $36.5 g$ $\ce{NH3}$. Calculate the efficiency."

My method to do this is - one I learned from my teacher, to put every given molecule's amount of substance (moles) in a table with 3 steps: start (S), gone (G)/formed (F) and rest (R).

$125g\:N_2 \leftrightarrow n=4.462$ moles

$32g\:\left(3\right)H_2\leftrightarrow n=16.13$ moles

\begin{matrix} &N_2&+&3H_2&\longrightarrow &2NH_3\\S&4.462&&16.13&&\\G&4.462&&13.386&F&8.924\\R&0&&2.74&&8.924\end{matrix}

($\rightarrow$ abundance of $H_2$)

Now, what is here the efficiency? Is it $\frac{36.5\:g}{8.924\:moles\cdot M_{NH_3}}\approx 24\%$?

Yes, that is completely correct. Let me summarize the approach.

If you look at the chemical reaction:
$$\ce{N2 + 3H2 \rightarrow 2 NH3}$$

You can see that you can, at best, make 2 mole of $\ce{NH3}$ per mole of $\ce{N2}$ (for the case here where $\ce{H2}$ is abundant). Given that you have $4.46\; moles$ of $\ce{N2}$ you could at best make $8.92\; moles$ of $\ce{NH3}$ which would be $151.9\; g$. Since you have made only $36.5\;g$ that means that your efficiency is $36.5/151.9=24\%$


Why are ionic compounds bad conductors of electricity in solid state? I understand the fact that ionic compounds are good conductors of electricity in molten state. But why aren't they good conductors in solid state. Cannot ions vibrate about their mean position and transfer electricity in the same way as they transfer heat?

Electric charge is transferred by physically moving charged particles around. In the case of an electric current moving through a wire (for example), the electrons are moving.

In an ionic compound, the ions are locked in place. They can move around a little bit, but there is not much translational motion - the ions stay in their places on the crystal lattice. In addition, the ions are "happy" with the number of electrons that they have. The ions formed in the first place by giving up or accepting electrons in order to minimize the overall potential energy of the system. If an anion were to transfer an electron back to a cation (for example) the energy of the system would increase, and so in general, transfer of electrons after the compound has formed is not favorable.

In solution or in a molten state, the ions themselves can move around - they become the charge carriers. In a solid, the ions can't move, and so electricity cannot be easily transferred.

You mentioned heat transfer - heat is the transfer of the kinetic energy of atoms and molecules. Heat can still be transferred (in some cases quite easily) in an ionic solid because, as you said, ions can vibrate about a mean position. When this happens they bump into their neighbors, which spreads the kinetic energy around.

In summary, ionic compounds don't conduct electricity very well because the charge carriers can't move through the crystal. They can conduct heat because the kinetic energy itself is the "heat carrier" - it can be transferred without moving ions too far from their mean positions.


Mechanism for the intra-molecular exchange of an ester and a nitrile group When the isolated product 4 is treated with sodium ethoxide in ethanol it rearranges to 5 as shown:

My TA says that you need to start with a reverse Claisen at the ketone but I am not quite sure where to go from there. I do not know how to move the nitrile alpha to the carbonyl furthest from the methyl group.

Following is a reasonable suggestion for the rearrangement:


Electrolysis of brine solution gives sodium hydroxide , chlorine and hydrogen gas but catner kellener cell it forms amalgam Sodium hydroxide is generally prepared
commercially by the electrolysis of sodium
chloride in Castner-Kellner cell. A brine
solution is electrolyzed using a mercury
cathode and a carbon anode. Sodium metal
discharged at the cathode combines with
mercury to form sodium amalgam.
My question is, why sodium ion discharge on the cathode and why not hydrogen ion?

I like what Maurice said "When nobody understands a scientific phenomena, we give it a name. Here the name is overpotential. Here the hydrogen is said to have a big "over potential" on mercury cathode. That is a bright and remarkable way to hide our ignorance."
This is little on the extreme side but people have spent their life on studying the hydrogen evolution reaction. It is a kinetic issue. It was one of the favorite topics of electrochemists. As one can see in the figure, each electrode evolves hydrogen at a different potential in water. Using mercury in the Castner-Kellner cells, allows sufficient negative voltage that can reduce sodium ions even in the presence of water. Additionally, mercury can form amalgams with many metals. Sodium is not an exception. The moment sodium is reduced at the mercury surface, it is dissolved in the mercury pool, which is periodically tapped off.
Figure taken from: An Electrochemistry Experiment: Hydrogen Evolution Reaction on Different Electrodes in the Journal of Chemical Education.


Why does acetate aldol with Evans auxiliary give almost no diasteroselectivity? The following acetate aldol reaction gives no diasteroselectivity:

Because I don't know what the basis of above statement is (I heard it in a lecture and read it on some internet web pages), I expect the two products are formed as a racemic mixture.

For the related propionate aldol we can draw two (considering R of aldehyde to always be equatorial) possible transition states:

The question is why we don't get good selectivity if we compare the acetate aldol with the same transition states for the propionate aldol?

As a side-note: in practice chlorine is used instead of Me to obtain good selectivity and the chlorine will be eliminated finally by use of Zn/AcOH.

1. The acetate aldol reaction
Before looking at selective variants, its worth pointing out what an acetate aldol is.
An acetate aldol (below), is the simplest kind of aldol in which the β-hydroxyketone product has no methyl group at the α-position to the carbonyl. This is in contrast to the (more usual?) propionate aldol reaction in which the α-position contains a methyl group (as is common in polyketides).

Source: Modern methods in stereoselective aldol reactions, Wiley.

Early studies into acetate aldols quickly established that it was far harder to set a single stereo centre than to introduce two stereocentres concomitantly. This was due to the cyclic transition states involved (Zimmerman-Traxler model) in which the methyl group bound to the enolate (in the propionate aldol) exhibits a level of control which isn't present in the acetate aldol.
2. The Evans' auxiliary and its applications in the aldol reaction
Whilst many aldol reactions exhibit diastereoselectivity, they show no inherent enantioselectivity. Several approaches were developed to fix this including chiral auxiliaries and chiral reagents. One of the most 'famous' of these was Dave Evans' oxazolidinone auxiliaries.
These are used to gain enantioselectivity, but can also be used to override substrate control to get the desired product.

Source: Dave Evans' CHEM206 Lecture notes, Harvard

With propionate aldols, this works excellently (see above), giving d.r.'s well in excess of 20:1 (about the maximum NMR can detect), with a 300:1 d.r. being quoted by Evans'. With acetate aldols however, the selectivity drops down to 1:1.
Let us consider the TS for the propionate aldol.

Source: Dave Evans' CHEM206 Lecture notes, Harvard

Evans' himself discusses the two diastereomeric TS structures. The favoured on top we now rationalise by opposing the dipoles between the auxiliary and enolate.
An alternate reason for why the disfavoured TS was disfavoured was provided by computational work, in which it appears that the methyl group in the propionate system clashes with the auxiliary.

Source: Evans' and Houk.

By disfavouring one TS (Steric clash) and favouring another (opposing dipoles), we gain good diastereoselectivity.
When the α-methyl group is taken away (as it is in the acetate aldol), we lose the unfavourable interaction (no more steric clash between the enolate and the auxiliary) leaving us only with the favourable dipole-dipole minimisation.
There is perhaps also another argument here which is that without the α-methyl group the enolate is considerably smaller, meaning that the TS doesn't even need to be quite so organised in order for the reaction to proceed.
3. Selective acetate aldols
As a closing point, switching from boron --> tin, and using a slightly modified auxiliary does allow for selectivity to be achieved.

Source: Unknown

This works because the sulfur and tin remain coordinated throughout the reaction rather than with the Evans' chemistry where the oxygen points into space to allow the dipoles to oppose. This results in a much more ordered TS.


Conditions for spontaneity So the two conditions for a reaction to be spontaneous are:

tendency to achieve minimum energy
tendency to achieve maximum randomness.

Both these things are contrasting and the only way I see them working are when they work independent of each other. Do these work simultaneously?

I am really having a hard time in figuring out a reaction where energy decreases and entropy increases. Can this happen?

Luckily for us, a really smart guy named Josiah Willard Gibbs figured out how these two tendencies work together back in the late 1800's. As a result, in chemical thermodynamics we now have a state function called the Gibbs Free Energy that describes what is called the thermodynamic potential of a system at constant pressure, temperature, and number of molecules.

Since most of the systems we work with under lab conditions are at constant pressure and temperature and are typically closed to mass transfer (abbreviated as NPT), Gibbs free energy is a very convenient way of determining what will happen to a given chemical system under normal lab conditions.

The reason it works is that it is a potential energy function - it describes how a system's potential energy will change as different state variables change, and therefore lets us predict where the system will "try" to go. An analogy that works pretty well is to imagine it like the surface of a planet, where gravity is the potential energy. If the derivative or slope is negative in a given direction, then things will tend to "roll downhill" at that point. On the other hand, to make things go uphill, you need to add energy.

The other thing we can get from this analogy is that given the opportunity, things will eventually move from a state of higher potential energy to a state of lower potential energy. So we can compare the energy at two different states (the reactants and products, for example), and decide based on the difference in potential energy whether the reaction would be spontaneous under those conditions.

The equation for Gibbs free energy is:

$G = H - TS$

Where $H$ is enthalpy (a.k.a heat of reaction), $T$ is absolute temperature (usually measured in Kelvin) and $S$ is entropy.

If you take the derivative at constant temperature and pressure, you get:

$\Delta G = \Delta H - T\Delta S$

(This is for large changes. for small changes replace $\Delta$ with $\delta$). What this tells us is that change in Gibbs Free Energy is a function of change in enthalpy, change in entropy, and the absolute temperature.

To illustrate how this works, take a look at the following diagram.

Here the line represents the Gibbs Free energy "surface" - the thermodynamic potential as a function of some variable (a reaction coordinate, for example). The ball represents the system at some point along that coordinate axis. If this were regular potential energy, it would be easy to see what happens - the ball will roll downhill if it gets the chance, and it will stop at the lower energy state. To get back up the hill, someone would have to put in that amount of energy. For thermodynamic potentials, it works the same way - the system will move along the coordinate axis in the direction of decreasing thermodynamic potential energy ($G$ in this case).

This means that under constant NPT conditions, any process that involves an overall decrease in Gibbs Free energy will be spontaneous. It will also tend to move in the "direction" that has the most negative slope in G at any given time. In mathematical terms,

$\Delta G < 0$ - overall process is spontaneous

$dG < 0$ - process will be moving in that direction

$dG = 0$ - process is at equilibrium

So how does this all fit in with what you described about systems trying to reach minimum energy, and maximum entropy? The answer is: those are both thermodynamic potentials for different types of systems. For a system with constant entropy, volume, and number of particles, (NSV) the internal energy (or total energy) is the thermodynamic potential. For a system at constant entropy and pressure (NSP), enthalpy is the thermodynamic potential. And for a system at constant volume and internal energy (NVE), the negative of entropy is the thermodynamic potential.

In other words, all of these thermodynamic variables are interconnected, and a change in one affects all of the others. You can hold three constant at any given time and still allow the system to "move" through phase-space. Which thermodynamic potential you need to describe how the system will move depends on which variables you choose.

Let's look at the equation for $\Delta G$ and see how changes in enthalpy and entropy affect it.

$\downarrow \Delta G = \space \downarrow \Delta H - T \Delta S$

If $\Delta H$ decreases, it will make $\Delta G$ decrease as well. This makes sense, since we know that exothermic processes tend to be spontaneous, because they are releasing energy and therefore the final system energy is lower than the initial.

$\downarrow \Delta G = \Delta H - T \space \uparrow \Delta S$

On the other hand, $\Delta G$ tends to decrease as $\Delta S$ increases - this is because the change in entropy is subtracted in the equation. This also matches up with what you know - an increase in entropy indicates a spontaneous process.

For your last question:

  I am really having a hard time in figuring out a reaction where energy decreases and entropy increases. Can this happen?

Yes, it can! In fact, under these conditions, the process is guaranteed to be spontaneous - if $\Delta H$ is negative, and $\Delta S$ is positive, then $\Delta G$ has to be negative - the reaction would be spontaneous at any temperature under these conditions.

Let's look at the other possibilities:

$\Delta H > 0; \Delta S < 0$

In this case, since $\Delta H$ is always positive, and we are subtracting a negative $\Delta S$, the reaction can never be spontaneous.

$\Delta H > 0; \Delta S > 0$

$\Delta H < 0; \Delta S < 0$

In these two cases, the reaction could be spontaneous or it could be non-spontaneous - it depends on the relative magnitudes of the enthalpy and energy terms as well as the temperature at which it occurs.

To summarize - all thermodynamic variables are related to each other in fairly complicated ways. When you hold three of them constant (two if you don't count number of molecules), you can derive thermodynamic potential energy functions that describe the behavior of the system in terms of a single quantity. For most cases in chemistry, Gibbs Free Energy is the thermodynamic potential that we use. It gives us the relationship between enthalpy, entropy and temperature under constant pressure conditions. Since it's the thermodynamic potential, it also lets us predict how the system will behave - how it will move through phase space.


What does "at%" refer to in sol-gel method? I'm reading on the preparation of ZnO thin films for UV stimulated emission and I found a method called Sol-Gel. During this I found a paragraph saying, " the ration Ag/Zn in the sol is 3 at%......". so I want to know what at% refer to.

Atomic percentage. It can be converted to weight percentage (wt%) too.


Copper layer doesn't stick after Copper electroplating I am playing around with copper plating using Copper Sulfate solution with electricity.
I am using 2-2.5 volts as power source and around 100-500 mA.
It is actually a very small container around 100 ml of the solution.

I have used copper sulfate anhydrous to make the solution and I have mixed around 20 mg with around 200ml of tap water in normal room temp.

I have tried several materials (iron, copper , steel, graphite) but I end up with a layer of copper that can be easily removed, even I leave it for fair amount of time(up to 15 min), the result layer is actually is not thin at all but still can be easily removed!!

What is the thing that I am missing here, I have seen people doing this on the internet for just like 5 min. and the layer that they create seems very strong!!

Roughen the deposition surface, perhaps by buffing with very coarse sandpaper or an abrasive scrub pad or similar.

If the surface isn't rough enough, the deposited coating has nothing to mechanically "grab onto" and it can be peeled off just as you've observed.  With a sufficiently rough surface, the deposition penetrates into the nooks and crannies of the surface texture and adheres very strongly.


How do we experimentally know the strength of bonds? How do we experimentally measure the strengths of covalent versus ionic bonds? Also, for a given molecule, how can we know if two pairs of atoms is covalently or ionically bound?

Related question: Textbooks say that in solution of water, two neighboring water molecules (call them A and B) will be 'attracted' because the O in molecule A will attract the H of the 'nearby' molecule B. It is said that H will alternate states between A and B. How do we know this is true experimentally and how can it be measured?

You measure the energy of any bond by carrying out reactions and seeing how much energy (usually in the form of heat) has to be invested.  For example, if you react one mole of $\ce{C}$ with one mole of $\ce{O2}$, you get $\ce{CO2}$ and heat.  Measure the energy with a calorimeter and divide by how many atoms and you get the energy you invested to produce one bond.  Not coincidentally, bonds are generally measured in units of energy/mol (e.g. kJ/mol).

Whether a bond is ionic or covalent has to do with whether the bonding electron resides on one atom or both.  A perfectly covalent bond finds the electron half on one atom and half on the other.  A perfectly ionic bond has the electron 100% on one atom and not at all on the other.  Real systems are a mix of both.  When we say a bond is ionic (such as $\ce{NaCl}$), we are really referring to the fact that we have measured a circumstance where the electron truly favors one atom.  There are various ways to measure this, one of the most common is to melt the solid and see if it conducts electricity.  In the case of $\ce{NaCl}$, the bond is so ionic that a small voltage will cause the ions to wander from their neighbor.

With a covalent bond such as you find in $\ce{SiO2}$, atoms can't wander without breaking the bond, so a small voltage doesn't cause a current.  A large voltage may break the covalent bonds and cause a current, but it also changes the substance (in this case releasing oxygen gas and producing silicon metal).


Aluminium, heat and biological safety I'm currently working on a really simple project, an ashtray that can stop cigarette combustion.

I planned to put an aluminium block into my ashtray, with some kind of curves or holes in it, in order for the cigarette to fit in, then stop combusting by lack of $\ce{O2}$, and maybe by heat transfer, as the cigarette tip and the aluminium will be in contact.

I choose aluminium, because I've got some, because it seems quite inert to me, and because it's not too hard to work with. But I can't solve the safety estimation for this material.

Here is some information I found around :

A cigarette tip being around $\pu{900^oC}$
Aluminium point of boiling at $\pu{2519^oC}$
Aluminium toxicity seems still debated, but aluminium is more or less eatable, i.e. used as pigment in food (E173)

Of course, aluminium won't burn, but I'm afraid that heating it may produce a layer of aluminium oxide, which may scatter on the cigarette tip, or get combined with ash components.

So, how can we estimate the safety of this material under these condition?

The use of aluminum in the construction of your ashtray as you've described will not result in the ingestion/inhalation of any form of aluminum in any significant way.

All aluminum that has been exposed to air is covered in a chemically and physically stable oxide. This aluminum oxide has an even lower vapor pressure than aluminum metal, such that the tip of the cigarette will actually be more than $\pu{2000^oC}$ below it's boiling point. By physically stable, I mean that aluminum oxide adheres strongly to the bulk aluminum and will not flake off as in the case of many other metal oxides.

Although your question was about potential safety issues, I will make one comment regarding your discussion of the design. You suggested that heat transfer into the aluminum might play a role in extinguishing the cigarette, which is not the case. So long as there is sufficient oxygen available for the combustion, the cigarette will continue to burn.


Photoelectric effect (AP Chemistry question) Which of the following postulates regarding the photoelectric effect is considered to be INCORRECT according to the quantum mechanical model we currently understand?
(A) Increasing the intensity of light does not change the energy per photon.
(B) The energy of a photon which ejects an electron from a metal surface must be equal to the binding energy (a.k.a. work function) of the electron plus the KE of the electron.
(C) Increasing the frequency of a photon can allow that lone photon to eject multiple electrons from a metal surface.
(D) Some electrons in a metal are easier (require less energy) to eject than others.

Please see question above and the teacher also supplied us with answers, so I know that answer is C.
Now, I am trying to reason why C is an answer, and A seems to be reasonable from what I read in my book. D looks somewhat good even though books keep mum about it. But I am puzzled about answer B. I looked at every AP Chemistry source I could get, and I am not sure where I can find formula for energy of photon, which would be equal binding energy + KE of electron. I can imagine this might exist outside of AP Chemistry, but within a subject? I found the equation in Wikipedia, for example:

The energy of a photon causing the photoelectric effect is found through E = hf = KE + w, where h is Planck's constant, 6.626X10^(-34) J*s, f is the frequency of the electromagnetic wave, KE is the kinetic energy of the photoelectron and w is the work function for the metal.

Any help on this matter would be highly appreciated. Desirably, reference to AP Chem book with page number. Thank you so much in advance!

I think the teacher pointed out that (c) is the correct answer because the postulate is generally regarded as incorrect according to the normal convention of photons behaving as individual packets of energy.

However what I think is interesting is that more than one electron can be ejected provided the photon is sufficiently energetic (this information will not be useful during an examination, so I suggest forgeting about this when you take an exam or do your homework, but if interested read on). For more information you can search for "multielectron ejection in auger spectroscopy" which provided references [1] and [2] below. Reference [2] opens up with the following statement:

  Absorption of an energetic photon by a target should in first approximation lead to the ejection of one electron only, as this interaction is purely mono-electronic. However it is well established that sometimes several electrons can be ejected. This can happen in valence multiple ionization or after innershell ionization when secondary electrons are emitted sequentially (cascade Auger decays) or simultaneously (direct double Auger decays). Such processes and especially the ‘one-step’ or direct ones (direct double ionization or direct double Auger decay) are extremely interesting to investigate as they revealthe strongelectron correlation phenomena which are at their origin.

Not sure what Einstein would have thought of this in any case.

An Auger process does not violate the principle that intramolecular electronic transitions are quantized and therefore that the energy required to eject electrons is quantized, irrespective of the number ejected. So the first two postulates are still correct.

References

P. Lablanquie, M.A. Khalal, L. Andric, J. Palaudoux, F. Penent, J-M Bizau, D. Cubaynes, K. Jänkälä, Y. Hikosaka, K. Ito, K. Bučar, M. Žitnik, Multi-electron coincidence spectroscopy: Triple Auger decay of Ar 2p and 2s holes,
Journal of Electron Spectroscopy and Related Phenomena, Volume 220, 2017, Pages 125-132, https://doi.org/10.1016/j.elspec.2017.04.003.
https://pure.mpg.de/rest/items/item_737492/component/file_737491/content


At boiling point, is the total pressure twice the atmospheric pressure? We know that one definition of boiling point is that it's the temperature at which the vapor pressure is same as the atmospheric pressure.

Assuming a closed container, does that mean that at this temperature, the total pressure exerted on the liquid is twice the atmospheric pressure, i.e., the vapor pressure plus the pre-existing atmospheric pressure?

edit: Based on some comments, it appears that closed container assumption is not very helpful. The questions still remains, even without that assumption: Does the total pressure become twice the atmospheric pressure at the surface?

Pressure is not additive

If you mix two gases of different pressure, the resulting gas does not have a pressure equal to the sum of the two pressures. Treating the gases as ideal gases, the pressure will be the average of the two pressures, weighted by the volumes. What is additive is the partial pressure of a gas mixtures, where the sum is the total pressure.

Closed container with constant volume

To simplify things, let's place a closed bottle filled with the liquid of interest into the container, adjust the pressure to 1 atm, and close the container. If we heat the container, the pressure will go up just because the gas temperature rises. If we open the bottle, some of the liquid will evaporate until the partial pressure is equal to the vapor pressure. Again, the container will have a pressure higher than atmospheric pressure. So the liquid will not boil when the temperature reaches the normal boiling point because the pressure will be above atmospheric pressure at that point, for the two reasons mentioned.

Closed container with variable volume

If the experiment is done in a closed, variable volume container that expands into the surrounding kept at 1 atm, the volume will increase as the temperature is raised (because the gas expands, and because the vapor pressure increases with increasing temperature). When the temperature reaches boiling temperature (slightly above, actually), the liquid will boil until it is all gone, and the volume will increase further during this boiling process. The partial pressure of the liquid will never reach 1 atm (because there are other gases present), so the vapor pressure will remain higher than the partial pressure throughout the boiling process.

  Does the total pressure become twice the atmospheric pressure at the surface?

No. Either there is room to expand, and the gas produced by boiling will push away the other gas (and then quickly mix with it), or we keep the volume constant, and then we don't call it boiling (because the total pressure exceeds 1 atm, so the liquid will not boil. However, when you heat a liquid to the boiling point and let the system reach equilibrium, the partial pressure of that substance will be 1 atm, and if any other gases are present, the total pressure will be higher. How much higher depends on the starting conditions. If our starting condition is a completely filled closed bottle, with inert gas (e.g. nitrogen or air) already at the boiling point of the liquid and at 1 atm, when we open the bottle the total pressure will reach 2 atm.


How do I know what the end products of the reaction of glucose and fructose? I am unsure about why the end products of the reaction are what they are.
Let's say I was combining glucose and fructose in a dehydration synthesis reaction to create sucrose. The equation is:

$$\ce{C_6H_{12}O_6 + C_{6}H_{12}O_{6} \rightarrow C_{12}H_{22}O_{11} + H_{2}O}$$

How do I know it is not:

$$\ce{C_6H_{12}O_6 + C_{6}H_{12}O_{6} \rightarrow C_{12}H_{24}O_{12}}$$

There are three possibilities. The first is knowing the reaction. I.e., you just know because you rote memorised it, that one molecule of water will be lost and then you apply that.

The second is knowing the mechanism. If you take a good look at it, a water molecule must be displaced for the mechanism to work out. Anything else would generate something highly unstable.

And finally, if you know the product structure, you can calculate double bond equivalents. Both reactants have a single double bond equivalent: $(2 \times 6 - 12)/2 + 1 = 1$. This corresponds to the ring in the reactants’ structures. The product has two rings, thus it requires two double bond equivalents. The number of carbons must stay the same, though. Thus: $(2 \times 12 - n)/2 + 1 = 2$; rearranging gives $n = 22$. Two hydrogens must be ejected and that biochemically typically happens with water.


Finding Pv in Pv=nRT for molecular hydrogen I'm attempting to write a small program that will calculate the values for the Ideal Gas Law, specifically for molecular Hydrogen (H2) in space.  Eventually, this will grow to be a sort of "simulator", but right now I seem to be having problems with remembering my college chemistry and physics.

Here is a picture of the form that I created:

So in my program (written in C#), I have the following as initial values for those variables:

        P = ?
        v = ?
        n = 2.0158 (weight in grams of H2)
        T = 2.725 (average temp in space)
        R = 1.67E-27 (1.67 * 10^-27)

My problem is that, how does one go about trying to solve for P or v?  Do you assign an arbitrary number for what you suppose the volume of the gas would be in space, and then calculate P based on that?  Or is it the other way around?

Extra, yet entirely unneeded info:
This was brought about because my 7 year old daughter asked me how stars were made.  I gave her the explanation that anyone who's got even a remote interest in it gives to their kids, but thought it would be pretty cool to make a little program that would give her a visual.  Here was my explanation:

  "we think that Hydrogen stuck together to make even larger clouds of
  Hydrogen, which increased the total mass of the gas cloud, increasing
  the gravitational pull on other atoms/molecules of nearby gas, etc.
  This eventually got to the point that the gases started to heat up as
  they were squished together by their own mass, but that increase in
  temperature created an outward push (pressure), which got into a
  tug-of-war with gravity acting on the mass of the gas itself.  And
  later, we got a star.  That's how most of us non-scientists think it
  happens, anyway!"

You can solve for either $P$ or $V$.  Choose one, say $P$, and then divide that into the right hand side $nRT$ to get $V$; i.e. $V=\frac{nRT}{P}$.

More seriously, you need to check your units!  You seem to be using Kelvins for $T$, which is right, and $n$ is just a count in moles.  But I do not recognize your value of $R$.  The SI value for $R$ is $8.3144621(75)$, in units of joules per Kelvin-mole.  Joules, in turn, are Pascal-meters$^3$. Personally, I think most people find Pascals too small a unit for pressure.  Far more practical would be Torr-liters, and then $R=62.36367(11)$.  You can examine lots of different unit combinations for $R$ at the ideal gas constant page on Wiki: Ideal Gas Constant


Can Ethanol be used to carry out SN1 reaction? My book says that the reaction of

$$\ce{Cl-CH2-O-CH2-CH3}$$ with ethanol to form
$$\ce{CH3-CH2-O-CH2-O-CH2-CH3}$$
occurs through $\mathrm{S_N1}$ mechanism, but shouldn't it actually be $\mathrm{S_N2}$?

My understanding of this statement is that the substrate is a $1^\circ$ halide and on top of that ethanol is available to us as $\ce{C2H5-O-}$, which is a considerably strong nucleophile. All possible conditions are pointing towards an $\mathrm{S_N2}$ reaction but then why does the reaction proceed through $\mathrm{S_N1}$ mechanism?

Be careful!  Ethanol is not a strong acid ($\mathrm{p}K_\mathrm{a} \approx 16$). So its conversion to ethoxide anion is not so easy. Rather, it would attack the $\ce{C}$ center by its lone pair and then it would release proton to form the product.

Actually here what I think is, the hetero atom $\ce{O}$ just near to the $\ce{C}$ undergoing substitution will assist by its lone pair to kick out chloride and then being planar ethanol can attack from either side. And finally, you get the product of $\mathrm{S_N1}$ type.

But don't mess up between ethanol and ethoxide ion. They are totally different in nucleophilicity. Treat them as two different nucleophiles.


How do cloves break down plastic? When my family buys powdered cloves in bulk, we put them in plastic until we get home - which takes at very most an hour. As soon as we get home, we transfer them into glass because we have had issues with them eating through the plastic cups until they are soft and have holes in them - this takes maybe a week.

I do not know exactly what type of plastic the cups are made out of; if this helps, they have the recyclable number 6 on them.

So, my question is, what is it in the cloves that is causing them to eat through the plastic at such a high rate? Are they really acidic? Or is it perhaps due to their oils?

Most of the odor of cloves comes from volatile molecules, especially eugenol, found in the cloves.  Eugenol is a water-immiscible, oily liquid at ambient temperatures.

Plastic with resin identifiction code #6 is polystyrene.

Polystyrene is not compatible with most organic solvents, because it tends to be softened, swelled, or even completely dissolved by them.

Thus I suspect that the problem is that oil in the cloves is attacking the plastic that you use.  You could avoid the problem (mostly) by using a plastic with resin code #2 or #5 (i.e. high-density polyethylene or polypropylene) instead of #6.  But, you are probably right that glass would be better than any plastic for storing your cloves.


How is 4H‐1,4‐oxazocine aromatic? I’ve found 4H‐1,4‐oxazocine in my chemistry book and I’m perplex as it states as aromatic:

I can count 6 electrons on p orbitals which could satisfy Hückel’s $n = 1,$ but the oxygen is sp3-hybridized. Even if we’d consider one lone pair as one sp2-bonding, the second lone pair would add 2 electrons to the p system, counting a total of 8 electrons, which doesn’t satisfy Hückel’s rule.

According to Huckel's rule Four Criteria for Aromaticity
The molecule needs to be (1) planar, (2) cyclic, (3) fully conjugated, and has (4) 4n+2 electrons.
Your molecule does not seem to be planar (b/c of N and O atoms) but assuming that it has a nearly planar conformation it satisfies all other criteria.
You only need to count one of Oxygen's lone pair and the only Nitrogen lone pair:
10=4*2+2
So it could be aromatic (if it is nearly planar)


What is CuCl · CO · 2 H₂O? What is the chemical $\ce{CuCl.CO.2H2O},$ formed via the reaction of copper chloride and carbon monoxide?

Does this compound have any use or appear in any noteworthy reaction?

The compound is called copper(I) chloride dihydrate - carbon monoxide adduct. It is formed during the Gatterman-Koch reaction.

Gatterman-Koch reaction is a reaction that produces aromatic aldehyde from aromatic hydrocarbons. The reagent used are carbon monoxide, hydrochloric acid and a Friedel-Craft catalyst like copper(I) chloride ($\ce{CuCl}$) or aluminium chloride ($\ce{AlCl3}$). So, the adduct is formed by the reaction of copper chloride and carbon monoxide during the course of reaction. If aluminium chloride is used, then a complex would be formed i.e aluminum trichloride complex, $\ce{[HCO(AlCl4)]}$.

To know more about the mechanism and role of reagents, see this previous question on chem.SE..

Other References

https://en.wikipedia.org/wiki/Gattermann_reaction
https://www3.aiche.org/proceedings/Abstract.aspx?PaperID=358705


If you can smell something does that mean that this thing loses molecules? When you can smell something there must be some molecules in your nose which signalize the smell in your nose. Does that mean, that e.g. a cheese is losing molecules which fly into your nose?
And does that effect the weight or amount significantly?

Yes, the thing you smell surely does lose some molecules, and consequently some weight. It can be said that it slowly evaporates or sublimates. But the strength of smell varies greatly in different compounds, as does the rate of evaporation (sublimation). Some compounds may be gone completely in a few minutes, and some would stay for years, or even decades.


How to increase volatility? Is it possible to put a substance with low volatility into a highly volatile solvent, thus make the original substance also vaporize quickly?

If not, are there anyway to increase it's volatility?

The important thing is the solvent does not dissolve the substance. Then their vapour pressures sum up and they boil together.

This technique is massively used for obtaining aromatic oily essences from plants, that are mostly insoluble in water.

The destilate usually forms a stable emulsion, that gets extracted by lipophilic volatile solvent, that is later evaporated to got the substance, or a mixture of them.


Assuming ideal gas behavior, how can I tell when the change in enthalpy is equal to the change in the inner energy? If ideal gas behavior is assumed, for which of the following reactions does $\Delta H = \Delta U$?

\begin{align}
\tag{a} \ce{N2O4               &-> 2 NO2(g)}\\
\tag{b} \ce{CH4(g) + 2 O2(g)   &-> CO2(g) + 2 H2O (l)}\\
\tag{c} \ce{SO2(g) + 1/2 O2(g) &-> SO3(g)}\\
\tag{d} \ce{Br2(l) + 3 Cl2(g)  &-> 2 BrCl3(g)}\\
\tag{e} \ce{Cl2(g) + F2(g)     &-> 2 ClF (g)}\\
\end{align}

The correct answer is (e), and I just want to make sure I understand why.  For $\Delta U = \Delta H$, no work can be done as $\Delta U = H - W$ (where $W$ is work done by the gas).

This means that the total number of molecules can't change, otherwise a change in volume will occur, causing work to be done upon molecules or the molecules to do work.  Likewise, phase changes will also require work to be done on the molecules.  Thus, we rule out (a), (b), (c), and (d)?

I'm not sure if this is too simplistic or the correct way of thinking about this problem.

You got it right, and there is really nothing to add. Same number of molecules means no change in volume, hence no work is done, hence $\Delta H=\Delta U$.

On a side note, I find it hardly believable that burning of methane (b) will result in liquid water, but if the problem statement was given to you exactly like this, well, so be it.


How to identify zero order reactions? I have come across many reactions that are zero order reactions but at one glance I'm unable to tell if they are zero order or not.  Is there any criteria that can be used to identify these reactions or are they purely experimentally determined?

Determining the rate of the kinetics of a reaction isn't done with only one measurement, or looking at only one graph.

What zero order kinetics means, is that the rate of the reaction is independent of the concentration of chemical species. To determine the kinetics of a process, you need to take a series of measurements while altering one variable, namely the concentration of one of the chemical species under investigation.

In zero order kinetics, you should observe that as you change the concentration of reactant, for example, the rate of the reaction remains constant.


How to disarm and correctly dispose of tear gas spray? I have a can of spray for self-protection which has expired way back in 2008. It hasn't been used at all, and it doesn't seem a good idea to just throw it away without disabling it somehow first. I'm looking for ways to disarm it with minimum risk to myself and others. Obviously, I can just spray it away, but I'd rather not if I can avoid it.

The active ingredient is 2-chlorobenzalmalononitrile.

It's not water-soluble, but I was thinking of some combination of cooking oil and alkaline agent (baking soda, for example) to neutralize it. I'm still doing research on it, and I won't be doing anything rash just yet :-)

Any particular advice how to neutralize this spray?

If jurisdiction is somehow important, assume European Union.

Do not open, do not try to neutralize it yourself! Bring it to Recycling center (DE: Wertstoffhof). If your city is 100k+ there should be one. That's the place where you bringing your old furniture, the rubbish after renovations, and so on. They have a part for disposal of hazardous waste (DE: Sondermüll). What usually comes in there are paints, old car batteries, old pesticides, etc.... Another possibility - if you have some friend at chemistry school of some local university, give it him. Each Chemistry faculty has some "disposal" facility.


What is incorrect about this depiction of the hydrolysis of an ester? First of all, I'd like to apologize for again asking a question solely featuring the deficiencies of my textbook.

I have a few specific questions, and a general one about the image at the bottom of this question:

Is the product really an acid? It doesn't even have an $H^+.$
It's stated that hydrolysis can happen when esters are treated with an acid as well as a base. But acids only donate $H^+$ so where would the extra $O$ come from to produce an alcohol as a product?
It's stated that a molecule of water is produced. Is this statement incorrect, or is the diagram incorrect?
In general, how would we change the diagram/information below to make it correct?

If you can answer any/all of these questions, I would greatly appreciate it.

Thanks again for this professional community.

There are indeed two mechanisms for ester hydrolysis, one using base catalysis and one using acid catalysis. The mechanisms of both are provided below (as an aside, you mention that your textbook isn't great- you should consider getting a different one, even if your teacher gives assignments out of the other book, it'll give you a second opinion and something to cross check your book against).

Base catalysed ester hydrolysis

Source: Organic Chemistry (Wothers and Clayden)
Acid catalysed ester hydrolysis

Source: Organic Chemistry (Wothers and Clayden)

Is the product really an acid? It doesn't even have an H+.
Yes, it is an acid. The acid is actually initially formed however because of the basic conditions of the base catalysed mechanism it gets deprotonated to give the carboxylate. Upon workup the carboxylate salt (in your example the sodium salt of the carboxylate) would protonate to give the carboxylic acid (COOH).
It's stated that hydrolysis can happen when esters are treated with an acid as well as a base. But acids only donate H+ so where would the extra O come from to produce an alcohol as a product?
The product is a gain a carboxylic acid, and as you can see from the mechanism above, water is involved, which is where the additional oxygen comes from.  Its worth noting that with the acid catalysed mechanism the whole thing is reversible.
It's stated that a molecule of water is produced (in the base catalysed mechanism). Is this statement incorrect, or is the diagram incorrect?
It could be water or an alcohol depending on whether hydroxide or the alkoxide does the deprotonation. Water is most likely however, as shown above. In the acid catalysed mechanism it is the alcohol that is formed.
In general, how would we change the diagram/information below to make it correct?
See the mechanisms above, taken from Clayden and Wothers. In general the diagram in your textbook looks essentially 'fine', though I dont entirely like the way the structures are drawn (its a very american textbook way of doing it.)


How to test for heavy metal in water? Is there a way to test if water has toxic heavy metals. Samples to both the water and people who drink it are available. Preferably, it should be done with at hold materials. I could also borrow materials from a high school chemistry lab potentially.

I have already tried boiling, but that just showed that it is hard water, which luckily isn't dangerous. Any ideas?

For qualitative tests to work properly, you need a concentration sufficiently high to observe the effects, e.i. precipitation of heavy metal salts and/or color changes of solution. If heavy metals are present, they are likely to be in small concentrations (depending on how your water is filtered before it reaches your house).

If you are concerned about your water quality and want to check if heavy elements are present, you should contact a trustworthy laboratory that routinely analyzes water samples. They are likely to use trace-element techniques such as inductively coupled plasma mass spectrometry (ICP-MS) and ICP optical emission spectroscopy (ICP-OES), or maybe atomic absorption (AAS) techniques. This way you get accurate and precise quantitative measurements of almost all elements in the periodic table.


Frequency Factor of the Arrhenius Equation The question came to my mind as to whether the effect of a catalyst on a reaction included the modification of the frequency factor (or pre-exponential factor) of a reaction, shown in Arrhenius' Equation below as "$A$":
$$ k = A\exp{\left(\frac{-E_\mathrm{a}}{RT}\right)}$$
On the one hand, we know that catalysts reduce the activation energy of a reaction, thus enabling more particles of reactants to acquire the minimum energy needed to break bonds and form products, therefore speeding up the reaction. But, does the frequency factor stay the same? My guess is that it does, and the effect of the catalysis is manifested solely on the change in activation energy. My question is if this intuition is correct and, if so, why.
Maybe it would help if someone could give a little more insight into the actual meaning of the frequency factor. Is it constant a constant specific to each reaction? What does it represent?
Thanks!

The short answer is yes, we can imagine effects on the frequency factor. Without going into too much detail, the frequency factor combines two conceptual elements - the frequency of molecular collisions if reactants are present at 1 M or 1 bar (called z in collision theory) and the probability that such a collision occurs with the correct spatial orientation that a reaction is possible (called $\rho$ in collision theory).
Let's start with $\rho$. A catalyst that binds multiple reactants can do so in such a way that they are optimally spatially oriented to react. Examples can be found in enzymes which are often large enough to bind the reactants in very specific conformations. So in addition to stabilizing the transition state of the reaction (thus lowering the activation energy), they can also increase $\rho$.
Now let's consider z. This one is a bit more complicated. In homogeneous catalysis, we could imagine z actually decreasing, since we now need the reactants and the catalyst all coming together via sequential collisions. In heterogeneous catalysis, the situation is even more complicated, since the mathematics of colliding are more complicated when the participants are in different states.


Which salt will best be able to buffer the addition of HCl? From these three salts, which will be best able to buffer the addition of $\ce{HCl}$?

ammonium chloride

sodium carbonate

ammonium acetate.

I think sodium carbonate is the right choice because when it dissociates in water forms a basic solution, so when $\ce{HCl}$ mixes with sodium carbonate it will be the best buffer, but it's a weak base so I'm confused. So can someone explain which is best?

To pick the best salt in this case, we should pick the least acidic cation and the most basic anion.
While sodium cannot dissociate into $\ce{H^+}$ ions, ammonium can, so the ammonium salts will be more acidic than their sodium counterparts.
In addition to being the most basic ($\mathrm{p}K_\mathrm{b}(\ce{CO3^{2-}})=3.67$,  $\mathrm{p}K_\mathrm{b}(\ce{CH3COO^{-}})=9.24$, $\mathrm{p}K_\mathrm{b}(\ce{Cl-})=20.3$), carbonate is also a diacidic base whose conjugate acid is also a stronger base than the other two anions ($\mathrm{p}K_\mathrm{b}(\ce{HCO3-})=7.63$).
So, sodium carbonate is the best choice.


Are there terms for ions of chlorine other than chloride? An ion of chlorine with a charge of −1 is called a chloride ion/anion.
Are there terms for other ions of chlorine, for example, $\ce{Cl+}$ or $\ce{Cl^2-}?$

Assuming that "term" in your question refers to chemical nomenclature, the authoritative source of information would be the current edition of IUPAC Red Book [1]. From [1, p. 70]:

IR-5.3.2.2 Monoatomic cations
The name of a monoatomic cation is that of the element with an appropriate charge number
appended in parentheses.
[…]

$\ce{I+}$ iodine(1+)

Analogously, the name of $\ce{Cl+}$ is chlorine(1+).
Further from [1, pp. 72–73]:

IR-5.3.3.2 Monoatomic anions
The name of a monoatomic anion is the element name (Table I) modiﬁed so as to carry the
anion designator ‘ide’, either formed by replacing the ending of the element name (‘en’,
‘ese’, ‘ic’, ‘ine’, ‘ium’, ‘ogen’, ‘on’, ‘orus’, ‘um’, ‘ur’, ‘y’ or ‘ygen’) by ‘ide’ or by directly adding ‘ide’ as an ending to the element name.
Examples:

chlorine, chloride

[…]
Charge numbers and radical dots may be added as appropriate to specify anions fully.
Examples:
14. $\ce{O^2-}$ oxide(2−), or oxide

So, the name for $\ce{Cl^2-}$ is chloride(2−).
Reference

IUPAC. Nomenclature of Inorganic Chemistry, IUPAC Recommendations 2005 (the “Red Book”), 1st ed.; Connelly, N. G., Damhus, T., Hartshorn, R. M., Hutton, A. T., Eds.; RSC Publishing: Cambridge, UK, 2005. ISBN 978-0-85404-438-2.


Do atoms in a molecule actually keep rotating freely in case they have sigma bond? While reading about cis-trans isomers I found the concept of free rotation of atoms.it says 2 atoms connected by sigma bond do not have trans sis isomer because of free rotation.while in case of pi bond free rotation is not possible. My question is : do we mean actual rotation of atoms in molecule when we say free rotation? Does the rotation happen spontaneously? What things are responsible for the rotation? Or is it just a concept to understand the concept of cis-trans isomers?

You really are referring to molecular conformations:

  Conformation, any one of the infinite number of possible spatial arrangements of atoms in a molecule that result from rotation of its
  constituent groups of atoms about single bonds.

Below are given two possible (out of many!) conformations of the ethane $\ce{C2H6}$ molecule:

The single $\sigma$ bond allows free rotation around the inter-nuclear axis of the two $\ce{C}$ atoms. One consequence is that around the single bond no cis-trans isomers can exist.

But if we add a $\pi$ bond, so we have a double bond and ethene $\ce{C2H4}$, then the methylene groups can no longer freely rotate about the bond.

That's because the electron probability density added by the $\pi$ bond doesn't lie on the inter-nuclear axis, but above and below it. Double bonds, and by extension also triple bonds, are rigid, with respect to rotation of the atoms they bond. This can give rise to cis-trans isomers.

Below: highly schematised rendition of the electron density of a $\pi$ bond:


What is the molecular mechanism of ammonia's toxicity? How does it damage proteins/lipids/DNA? There seems to be surprisingly little literature on its toxicity.

Ammonia is reportedly not toxic in bacteria even at molar conentrations (PMID 16604417).

However, yeast, plants, and animals (presumably, all eukaryotes) are unable to tolerate high intracellular ammonia concentrations.  The best explanation for this phenomena I've seen is a hand-waving assertion that intracellular ammonia taxes overtaxes the cell's ability to maintain membrane gradients (PMID 7764110).  However, that isn't a particularly detailed or satisfying response.

I agree with your assertion that there's a paucity of literature on the subject - we probably don't know the answer yet!


Why does the Förster resonance energy transfer scale as 1/r⁶? The title more or less contains my question - why does the Förster resonance energy transfer (FRET) (photon?)-based energy transfer mechanism, which folks sometimes use as a spectroscopic ruler to measure the inter-particle spacing between dye pairs, exhibit $1/r^6$ scaling for energy transfer efficiency?  Since we're in 3-space when performing a FRET measurement, I could understand some $1/r^3$ contribution due to us having an expanding sphere of radius $r$, but where does the rest of the distance-dependent efficiency dropoff come from?

From this article by Förster himself:

  This coupling is strongest if the corresponding optical transitions in
  both molecules are allowed for electric dipole radiation. Then these
  transitions are coupled not only to the radiation field but also to
  each other. Naturally, the interaction energy is of a dipole-dipole
  nature, depending on an inverse proportionality to the third power of
  the molecular distance. The probability of energy transfer is then
  proportional to the square of this interaction energy and decreases,
  therefore, with the sixth power of the distance.

This paper does not give a detailed derivation though. But in his seminal paper from 1948 Förster goes through the derivation in detail (unfortunately it is not in English but in German and the condition of the text is not very good).


How can we balance one-half of an oxygen molecule, or any other diatomic, in a chemical reaction? How can we balance oxygen or any other diatomic molecule with a fractional coefficient in a reaction?  I mean, if we can balance oxygen atoms in an odd number what happens to other oxygen atom, and can this sort of reaction even happen?

Whenever a chemical reaction is written with fractional coefficients, implicitly it's meant to be interpreted as the amount of substance in moles of each species that is involved.  So, it's not really one half of an oxygen molecule reacting, but 0.5 moles of $\ce{O2}$ reacting.

bon's comment is spot-on -- if you are trying to think about a reaction written this way on a molecular level, you need to multiply the reaction through by whatever constant eliminates all the fractions.


Determine which molecule has stronger intermolecular force? SiH4 and CH4

The only intermolecular force they both have is London Dispersion forces

Strength of LDF is determined by molar mass

molar mass of SiH4 = 32.132

molar mass of CH4 = 48.42

Therefore shouldn't CH4 have a stronger LDF??

The answer says that SiH4 has a greater intermolecular force...why is that?

Look at the boiling points.  Interaction is based on geometry (surface area) and polarizability, among other things - and the phase: solid, liquid, vapor.  Molecular weight is not diagnostic.  SiH4 is "fluffy" versus methane being "hard."  Compare the melting and boiling points of decane and isodecane  versus adamantane, all being $\ce{C_{10}}$.  Compare the melting and boiling points of n-octane versus 2,2,3,3-tetramethylbutane, both $\ce{C8H_{18}}$.  Complicated.

You might check your MW for methane.  Bells and whistles should have gone off, $\ce{CH4}$ versus $\ce{SiH4}$.


Will alcohol or soap damage plastic or rubber? I have read conflicting advice on various sites about the appropriate methods for cleaning plastic and rubber items and would like to ask people who actually know what's going on what they would recommend.

I realize some of my assumptions may be false:

I am (perhaps wrongly) assuming that most plastics would have similar properties as it relates to cleaning with alcohol or soaps/detergents and that you don't need me to give you the actual chemical structure of a specific polymer.
I am (perhaps wrongly) assuming that most soaps and detergents would act similarly on either a plastic or a rubber item.  If this is false please give your understanding of what an "average" soap or detergent would do.

I am not assuming that plastics and rubber will respond the same to either alcohol or soaps and detergents.  If they are different, please note which will have what effect.
I am not assuming that synthetic and real rubber will act the same.  If they are different and you know the answer for both, please mention the difference.

I am hoping that it's okay to ask "applied" questions here.  This seems like the best of all stackexchange.com sites, not just because none of the less theoretical sites are deal with this sort of question, but also because it really boils down to a matter of chemical reactions.

Update: by "alcohol" I mean rubbing alcohol, or isopropanol.

Will soap damage plastic or rubber?

Probably not. Most plastics and rubbers are resistant to aqueous solutions and ionic compounds (including most surfactants and soaps).

  Will alcohol damage plastic or rubber?

Maybe. There are two problems, in that neither alcohol nor plastic refer to specific compounds, but are rather generic terms encompassing hundreds of compounds each. The number of possible combinations is endless. As Leonardo suggests, the best approach is either to look in the manual or experiment in a small inconspicuous part of the product.

However, assuming that by alcohol you either mean ethanol, which is the drinking kind of alcohol, or isopropanol, which is the alcohol in many cleaning products, then I can give you some advice based on common plastic types. Both are similar enough in their solvent properties that we can generalize.

Alcohol will damage some plastics, but not all. I will list my answers by the resin code, or "recycling symbol", found on most plastic items:

Poly(ethylene terephthalate), PET or PETE - PET is not very soluble in ethanol or isopropanol, but prolonged exposure may cause crazing or stiffening due to the dissolution of plasticizers.
High-Density polyethylene, HDPE - HDPE is resistant to most things.
Poly(vinyl chloride), PVC - PVC is not very soluble in ethanol or isopropanol, but prolonged exposure may cause crazing or stiffening.
Low-Density polyethylene, LDPE - LDPE is resistant to most things.
Polypropylene, PP - PP is resistant to most things.
Polystyrene PS - PS is not very soluble in ethanol or isopropanol, but prolonged exposure may cause crazing or stiffening.
This stands for "other", but the most common is polycarbonate, which is not very soluble in ethanol or isopropanol, but prolonged exposure may cause crazing or stiffening.

Rubber - most rubbers are probably resistant to alcohols, but prolonged exposure will cause loss of elasticity


Is there a heatformed vinyl-like plastic that is hard as shellac when cold? I recently started collecting shellac records, and got the idea of releasing my own music on 78rpm shellacs.

But the exact formulas and techniques for making shellac records seems pretty much lost to history, and it seems it was a very hard material to use anyway. You can use vinyl, but then it would not be playable on the old hand-cranked acoustic type 78rpm record players, as those needles have too high pressure and would destroy the softer vinyl.

So this is probably only doable if there is a plastic that is soft and possible to press to a record shape when heated, like vinyl, but that becomes a lot harder than vinyl, like shellac, when cold, and also doesn't distort when cooling. Anyone knows of something like that?

I found a web article, "The Origin & Many Uses of Shellac" by R.J. Wakeman, to be particularly illuminating and well-written source of information on the composition and manufacture of shellac records, for instance it explains:

  Emile Berliner’s first disc records were pressed in celluloid (1894) followed by the use of vulcanized rubber (1895).(36) In 1896 Duranoid shellac records were developed and became the standard medium.(11) The formula used for making shellac records varied from firm to firm and from decade to decade; it was usually a company secret. The shellac component represented approximately five-eighths of the material used for the better quality records.(33) Essential components would include shellac, ground rock, carbon black, and cotton floc.

The article lists a few patents among the references that you could quickly search for additional preparation methods, although I found them to be somewhat evasive with regard to details, preferring to amplify the scope of the application by making general claims. The number of patents describing the manufacture of records with shellac and competing resins with a variety of additives is considerable. There is therefore no universal recipe for which to search. What common properties do the formulations share? The key is that the shellac serves as a binder, and that strength and hardness, to protect against wear of the record by the steel needle, to actively wear the needle into shape, or to prevent shattering, is largely conferred by the ground mineral filler and other additives such as abrasives and plasticizers. Apparently the fineness of the ground rock and the quality of the shellac are important.

  Abrasives were frequently added to the record matrix to help wear the point of the needle to conform to the shape of the record grooves

and

  When a needle enters the groove of a 78 rpm shellac record the needle point should quickly abrade to fit the record groove.

Regarding the preparation of shellac records, Wakeman writes:

  To mold shellac records, powdered shellac, ground stone, carbon black, and cotton floc were measured by a formula and mixed in a revolving drum or Banbury mixer.(16) The shellac needed to be finely ground for thorough mixing with and coverage of the other components. High quality orange shellac with low wax content and few impurities was the standard.(16) One report states the shellac was ground to 80 mesh, the fillers to 200 mesh, and the coloring matter to less than 0.4 micron in size. The mixture was steam heated, causing the shellac to melt and form a thick dough which was then passed onto heated rollers and formed into a long roll. The roll was cut into sections (called “biscuits” in some factories) with adequate material to form a 10 or 12-inch record.(8,31)

So in answer to the question, "what alternatives exist to shellac with similar properties?" you would like to replace shellac with other resins, while retaining the other essential additives, such as abrasives. You would not be the first to attempt to do so. As Wakeman explains:

  After the war shellac was combined with other resins to form softer and less brittle 78 rpm records.(16) Shellac was reduced in the matrix to form 15 to 30% of the mix.(16) Vulcanite (also called ebonite), a hard rubber, was sometimes included in the mix, as were Vinsol (a pine tree resin) and Manila gum (16).

Another important property is time to harden. I looked up Durium, a relatively simple shellac competitor. Durium was patented as a fast-hardening resorcinol-formaldehyde resin. Production of Durium records was faster since, unlike shellac, Durium hardened quickly and records were fabricated by pressing rather than moulding. While it has properties that represent a significant advantage over shellac (price, hardening speed, method of impression), unfortunately it also requires working with a variety of unpleasant chemicals.

For an early vinyl patent, developed during shellac shortages in WWII, see US patent 2335986. For a more recent patent providing a number of formulations alternative to shellac, see US patent 2787602.

Regarding characterization of shellac composites, this is a difficult topic to address, since the properties can be expected to depend strongly on the specific additives. I did find an old reference quoting a value for the modulus of elasticity of processed shellac as on the order of $\pu{10 \times 10^3 kg/cm2}$. You have many choices regarding a resin to match the required strength and softening temperature.  You may want to contact a company such as Ensinger plastics for guidance on selection of a thermoplastic composite.


What does CH4 form when dissolved in aqueous solution? So I understand $\ce{CH4}$ has a low solubility, but when it does dissolve, what does it form? E.g. $\ce{HCl_{(g)} + H2O -> H3O+_{(aq)} + Cl^{-}_{(aq)}}$. So what does $\ce{CH4 + H2O}$ form?

Methane dissolved in water is simply methane. It does not react with the water under normal conditions.


Mechanism for ring contraction from 2-methylenecyclobutanol to 1-methylcyclopropanecarbaldehyde I am stuck between two paths. I was going to protonate the alcohol to give a better leaving group, leaving a 2 degree carbocation and then go on with a hydride shift. But, I'm not sure what that may turn out to look like, and when exactly the ring is supposed to contract. Or, I was going to have the alkene attack the acid, but after that I wouldn't know what to do with the alcohol.

Mechanistically, both of those protonation options (at the alcohol or the alkene) are reasonable. Without any other information, I would probably try both and then see if either gets closer to the product.

However, two things suggest to me protonation of the alkene:

1) As suggested, protonating the alcohol generates a good leaving group (water). However, the product has a C-O double bond, so breaking the C-O single bond appears to be unproductive. That's not to say it's impossible, but it leads me to think about trying something else first.

2) The product has a methyl group, but the starting material does not. Somewhere in the mechanism, that methyl group has to be formed. If protonation of the alkene is the first step, then a methyl is formed, which I see immediately as productive. Again, that's not to say that this protonation is definitely correct, but at least something productive has happened.

One last tip before I show my suggested mechanism. For any mechanism that is difficult to draw, but especially rearrangements, it can be very helpful to draw in the hydrogens. In this case, drawing in the hydrogens just about gives away an exact mapping of the carbons from the starting material to product. With a mapping, it becomes easier to determine all of the bonds broken and formed, which is essentially a checklist of all of the goals of the mechanism.

I'll leave it to you to add the curved arrows, but the intermediates are shown below. Protonation of the alkene in 1 gives carbocation 2, which contains the methyl group seen in the product 4. Next is the ring contraction, sliding a C-C bond over the the carbocation and generating new carbocation 3. Compound 3 might appear to be less stable than 2, but it turns out to be quite stable due to resonance structure 3', which is fully octet. Deprotonation gives 4.


Why are cations ordinarily smaller than anions? I'm reading on ionic crystal structures and am confused by this from my textbook:

  Because the metallic elements give up electrons when ionized, cations are ordinarily smaller than anions

I understand that having a positive charge will pull the electrons closer, but shouldn't the atom from the larger period (row on periodic table), have more electron shells and be the larger atom?

It isn't true that any cation will be smaller than any anion.

You intuition that atoms with larger electron counts from lower rows of the periodic table will be bigger is correct. Size for atoms is determined by the highest occupied electron shell or orbital as the interactions with other things that define size are, crudely, caused by interactions between the electrons in their outer orbitals.

So a Rb+ cation radius is about 166pM but an F- anion is smaller at about only 119pm. There is a good table of such values in the Wikipedia article on ionic radii which also explains some of the trends and some of the difficulties in actually defining size for ions.


Proton NMR chemical shift of water peak in different solvents Why does the water peak appear at different chemical shift values (ppm) in different solvents in proton NMR spectra? For example, the water peak in DMSO-d6 appears at nearly 3.33 ppm, but the same moisture peak in $\ce{CDCl3}$ appears at 1.56 ppm.

What is the reason behind it? Why should the same species ($\ce{H2O}$) give rise to two different chemical shift values?

Consider also water in addition to the solvents you mention:

water-d2 (or, $\ce{D2O}$) is an extensively H-bonding polar solvent (dielectric constant $\epsilon = 79$)
$\ce{DMSO-d6}$) is polar ($\epsilon= 47$) but aprotic (no-H-bond donation ability)
chloroform-d is an apolar aprotic solvent ($\epsilon=4.81$)

Now consider the state of a trace amount of $\ce{H2O}$ in each of these solvents, giving rise to an NMR signal:

in $\ce{D2O}$ the trace $\ce{H}$ can exchange and forms $\ce{HDO}$ which retains an extensive H-bond network. $\ce{H}$ is strongly deshielded by bonded and H-bonded electronegative oxygen and due to the formation of oxonium ion with $\ce{H}$ carrying a significant positive partial charge. The chemical shift is $\approx \pu{4.7 ppm}$
in DMSO-d6, $\ce{H}$ is strongly coordinated (H-bonded) by DMSO oxygen atoms, resulting in substantial shielding. The chemical shift is $\approx \pu{3.33 ppm}$
in chloroform-d interactions with the solvent are comparably weak and mainly dipolar or dispersion interactions. $\ce{H}$ experiences deshielding mainly due to directly bonded oxygen. The chemical shift is $\approx \pu{1.56 ppm}$

The following table shows shifts for residual water $\ce{H}$ in different solvents ([1]; dielectric constants are for non-deuterated solvents):

$\begin{array}{c}
\begin{array}{c|c|c}
\text{solvent} &\text{shift(ppm)}&\epsilon\\\hline
\ce{C6D6 }&0.40 & 2.28\\
\text{toluene-d8} &0.43 &2.38\\
\ce{C6D5Cl} &1.03 &5.69\\
\ce{CD2Cl2 }&1.52 &9.08\\
\ce{CDCl3} &1.56 &4.81\\
\ce{CD3CN}  &2.13 &36.64\\
\text{THF-d8} &2.46 &7.52\\
 \ce{(CD3)2CO} &2.84 &21.01\\
 \ce{(CD3)2SO}  &3.33 &47\\
\text{TFE-d3} &3.66&8.55 \\
\ce{CD3OD} &4.87 &32.6\\
\end{array}
\end{array}$

Note that the most deshielding solvents are the protic H-bonding ones (the bottom two).
Next in effectiveness are those with oxygen atoms with lone electron pairs available for H-bonding.
Note also that the top three have an additional effect on water shifts due to the aromatic ring current.

Reference
[1] Fulmer et al. Organometallics 2010, 29, 2176–2179


Grignard Reagent and Ester Is this answer wrong? I feel like there should be a hydrogen instead of a carbon where I indicated in red. Thank you.

The answer is correct. After the first nucleophilic attack from $\ce{MeMgBr}$, the ethoxy group left and the ester is transformed into a ketone (Step 2 in the link), then another $\ce{MeMgBr}$ attacks and introduce an extra methyl group, resulting in a $3^\circ$ alcohol after hydrolysis. The process is well described in this page.