Diffusion across a thin film Given this concentration profile, I am to (1) determine at which side substance A will diffuse out of the membrane and (2) chose one of the following profiles below which represents the concentration of a substance A at steady-state.

(1) Diffusion occurs from an area of high concentration to an area of low concentration. Therefore, substance A would diffuse from the left to the right. However, in the question when they ask "at which side will the substance diffuse out of the membrane", should I answer at the right side or that the substance stays within the membrane?

(2) I know that the concentration profile at steady-state should be linear, however I'm wondering what the difference between (A) and (B) is. I assume that (A) is the correct answer to this question since substance A has a higher concentration than the concentrations at the membrane wall and hence the slope should start a little above the concentration at the wall. Is this correct or does steady-state affect the concentration profile of the substance?

(1) determine at which side substance A will diffuse out of the membrane

    should I answer at the right side or that the substance stays within the membrane?

Neither, because the concentration is higher in the membrane than either side, the substance will flow out of both sides initially at least. After a while the material will flow from left to right (high concentration to low concentration) and converge towards steady state.

  (2) chose one of the following profiles below which represents the concentration of a substance A at steady-state.

First thing - When doing diffusion problems immediately refer to Fick's laws of diffusion.

Fick's first law assumes a steady state system and is given as:

$$J = -c D\frac{dx}{dz}\quad \text{or more simply} \quad J = -D\frac{\Delta C(x)}{\Delta z} \quad \text{or more generally}\quad J = -cD\ \nabla x$$

Where $J$ is flux (mols through an cross-sectional area per time), $c$ is total concentration, $C(x)$ is the concentration at point $x$, $D$ is the diffusion constant, $x$ is the mole fraction and $z$ is position.

For steady state to occur, $J$ must by definition be constant through the entire membrane and thus $-D\frac{\Delta C(x)}{\Delta z}$ must be constant through the entire interface. Since $D$ is already a fixed constant, we can assume $\frac{\Delta C(x)}{\Delta z}$ must also be constant. Thus the plot with a constant slope is the steady state profile which is B.

Q.E.D.


Why does equilibrium constant change when the reaction is multiplied by a number and nothing in the real has been changed? As far as I know, a large equilibrium constant shows that the a reaction like the one below  wants to complete:

$$\ce{B <=> A}$$

If A and B are gases, then K is:
$$K_1=\frac{[A]}{[B]}$$

but when I multiply the reaction by 20 , K is
 $$K_{20}=\frac{[A]^{20}}{[B]^{20}}$$ while the same number of moles of B are consumed. If $\small K_1 = 10$ then $\small K_{20} = $$10^{20}$ so the reaction must be approximately complete. Why does this happen while the reaction materials didn't change?

First of all, you switched the numerator and denominator in the definition of $K$. If $\ce{A <=> B}$ , then $K=\frac{[B]}{[A]}$. Products over reactants.

Notice that, if $K$ is large, then mathematically it is necessary for either the numerator to be large and/or the denominator to be small. Therefore, qualitatively, if $K$ is large, then you either have a lot of products, or little left of reagents, both of which imply a reaction which occurs close to completion.

Now here's the problem - we were speaking qualitatively. Your first sentence ("a big equilibrium constant shows that the reaction wants to get complete") is only a rule of thumb. It happens to work well because in reactions where equilibria descriptions are quantitatively reliable/exist at all, the stoichiometric coefficients are relatively small when written with the lowest integers (using the lowest integer coefficients is almost always done). The rule of thumb is borne out of repeated calculations under these implied conditions. They are not the case for your second equilibrium example, so such a rule should not be expected to still be valid.

Indeed, though $10^{20}$ is an impressive number at first, when you actually go do the calculations to find the relative amounts of reactants and products, you will at some point have to take its twentieth root, promptly turning it back into a much smaller number. Another way to think about it is that with such large stoichiometric coefficients, even a small change in concentrations is amplified enormously by the large exponents and has a huge impact on the value of the reaction quotient; a small increase in products will go a long way to reaching the equilibrium condition.


Relating properties and structures of polymers Why would it be more favorable for a polymer designer to design copolymers with irregular chain structures, rather than design homopolymers?

Hint: Discuss intermolecular bonds

I understand that copolymerization allows for greater variation in polymers and there properties, but am unsure how exactly to relate this to intermolecular forces. Would it also depend on the intended use of the polymer?

I think that the best thing for you to start with is to compare the radical formed low density polyethylene (LDPE) and the Ziegler-Natta high density polyethylene (HDPE).

The low density polyethene has plenty of side groups which are randomly arranged through the length of the chain, this can be regarded as a copolymer of a range of different alkenes which are randomly arranged.

The high density polyethene (HDPE) is an almost perfect homopolymer of CH2 units, this polymer has far better mechanical properties, it is a polymer where the chains pack closer to each other than the chains in LDPE.

There is a version of HDPE which has properties closer to that of LDPE, this is formed by using a mixture of alkenes using the organometallic Ziegler-Natta cataylsts.


Creating an efficient synthetic route to this elastomeric block copolymer This was a question my professor gave me in which I have a couple of ideas to create a synthetic route which include: Either an anionic or cationic polymerization? If so, I don't think it would be cationic because I can't seem to see if there would be some sort of initiator system and i don't think there could be carbocation formation either from the looks of it. The only feasible conclusion I have come to is that it would be an anionic polymerization where you create a living polymer and react this with a second monomer to create a block copolymer? I could maybe use an organolithium to start the initation with a polar solvent however I don't actually know which monomer to start off with nor how the reaction mechanism would work out. Can someone help me? Thank you very much!

How about starting with a difunctional isocyanate, adding hydrazine in step 1 and polyethylene glycol in step 2. I would call it an addition polymerization. Your structure does not show the number of repeating units, but that could be largely controlled by varying the monomer proportions.


How to properly model a mechanochemical reaction of barium titanate and copper(II) complexes? I would like to understand, in terms of chemical kinetics, how I could equate the reaction rate between $\ce{BaTiO3}$ and copper(II) complexes under ultrasound effect.

Examples in literature would be great!

Reaction rates can be calculated from the Arrhenius equation $k=Ae^{E_a/RT}$ where $E_a$ is the activation energy found at the transition state.

Please check out the Reference 1 for calculating transition states under force. Following is an excerpt from the paper (Ref.1):

  As a force is delivered to the mechanophore, the energy provided by the work can be quantified by $\Delta E =−F\Delta x$ where $\Delta E$ is the change in energy, $F$ is the amount of force applied, and $\Delta x$ is the change in distance over which the force is applied ($\Delta x$  in the examples of this article will correspond to the distance between the two atoms where the tensile force is applied…

We have recently developed a python version of the code.

Force can also be added along bonds to simulate ultrasound and we have some work in preparation showing this for polymers. Essentially ultrasound creates bubbles which pop and drag the polymer in opposite directions.

References:

Allison G. Roessler, Paul M. Zimmerman, “Examining the Ways to Bend and Break Reaction Pathways Using Mechanochemistry,” J. Phys. Chem. C 2018, 122(12), 6996-7004 (https://doi.org/10.1021/acs.jpcc.8b00467).


IUPAC way of writing prefix of alkyl isomers of an ether Consider this compound:

I am able to identify the base chain as propane. But, I am having trouble naming the other side of the chain. Will it be named as:

2-(isopropoxy)propane,
2-[(propan-2-yl)oxy]propane, or,
2-(2-propoxy)propane

What is the preferred IUPAC way for such a naming of isomeric alkyl groups?

According to Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 (Blue Book) (P-63.2.2.2), the preferred prefix for the substituent $\ce{(CH3)2CH-O\bond{-}}$ is 'propan-2-yloxy'. Both 'isopropoxy' and '1-methylethoxy' remain permissible in general nomenclature.

Correspondingly, the PIN of the symmetrical ether $\ce{[(CH3)2CH]2O}$ is 2-[(propan-2-yl)oxy]propane.

I'm quite sure the double parentheses are required based on other similar examples given in the Blue Book, but am not 100% sure of the rationale for it and would be grateful if somebody could confirm that. I'm thinking one set comes from P-16.5.1.1 as 'propan-2-yloxy' is a compound substituent, and one more comes from P-16.5.1.2 to separate the locant '2' in 'propan-2-yl' from the remainder of the name, but don't quote me on that.


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

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

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

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

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

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

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

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

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

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

Hope it helped!


What Could Formations of Linkage Isomers of [Co(NH₃)₅NO₂]²⁺ be like? I am trying to draw the structure of this complex $\ce{[Co(NH3)5NO2]^{2+}},$ however I am not sure if the linkage isomers of this complex like this:

This Co-ordination compound of were studied by Jorgensen and Werner who observed that were two compounds of the same chemical formula but of different colors. (We can see these two geometries in the image).
A red form of low stability converted readily to a yellow form the $\ce{M-NO2}$ nitro isomer, based on comparison with compounds with similar colors. This conclusion was later confirmed, and kinetic and $\ce{^{18}O}$ labeling experiments showed that conversion of one form to the other is strictly intramolecular, not a result of dissociation of the $\ce{NO2^-}$ ion followed by reattachment.

References:
Miessler, Tarr,Inorg. Chem., 2nd Edition, 1998


1H NMR of Acetylferrocene (source: sigmaaldrich.com)

I'm trying to analyse the proton NMR of this, and there's one singlet for the H on the non-substituted Cp ring and a singlet for the methyl group, as expected.

However, while I was expecting to see a dd signal from $\ce{H_2}$ and $\ce{H_3}$ and a doublet from $\ce{H_1}$ and $\ce{H_2}$, there is only one doublet that accounts for all the protons according to integration.

Can someone help me explain this?

The NMR spectrum of acetylferrocene is very well known and widely published as an undergraduate teaching resource. It would be useful for reference if you were to provide a copy or image of your spectrum. At least you should describe it including chemical shifts, integration and peak appearances. I suspect that your troubles here come from a misunderstanding of your observations of the actual spectrum.

In order to help you with your analysis, you should make sure you complete an accurate integration of all peaks. This will give you some clue as to how many protons are contributing to each of the peaks you are observing. That is, each of your peaks that you think are part of your doublet contribute two protons. Also, when you suggest you are observing a doublet - what is the splitting of this doublet. I doubt it is of a magnitude that conforms to a typical H-H coupling. It is probably closer to 100Hz. So perhaps what you are seeing is in fact two distinct singlets. Or at least two distinct resonances with poorly defined splitting.

Your analysis of the expected splitting pattern for the Me-Cp ring isn't quite correct. From your drawing, H1 and H4 are chemically equivalent, but magnetically non-equivalent. This is because coupling between H1 and H2 is not the same as the coupling between H4 and H2. The same can be said for H2 and H3; H2 and H3 are chemically equivalent, but magnetically non-equivalent.  This means you have an AA'XX' spin system, which is not a simple first order spectrum in appearance. You can read more about these types of spin systems on one of my favourite web resources, Reich. Each signal actually appears as an apparent triplet.

So, why aren't you seeing any splitting in these peaks? There will be a number of reasons why this could be the case, most of which are perfectly reasonable and are no cause for concern. Sample concentration, purity, magnetic field strength, field homogeneity (shimming), temperature and solvent will all contribute to the appearance of this spectrum, and many published spectra for this molecule have splitting which is barely resolved. Without seeing your spectrum, it is difficult to make further interpretation. However, as a suggestion, you should measure the linewidths of your peaks (the width of the peak at half height). You will probably find that the Cp peaks from Me-Cp are considerably broader than the other peaks, especially than for your residual CHCl3.


Particules speed in ideal gasses According to Graham's law, we can calculate de average speed of gasses by using the formula
$$v^2=\frac{3RT}{M}$$
I tried to use this simulation and got these data by keeping same particles amount and increasing the temperature(average speed(m/s) - temperatura(°C)):

T in C
V  in  m/s

27
454

50
480

90
510

130
530

225
590

350
660

425
710

500
740

But, shouldn't be the speed ($v$) a square function of the temperature? ($T=\frac{Mv^2}{3R}$ ) What am I doing wrong?


Effect of Catalysis on Reaction Rate in elementary Reaction I understand that there are many different catalytic mechanisms that suggest different ways of accounting for the effect of a catalyst on reaction rate. My question is:

Given a termolecular elementary (it is known it is an elementary reaction) reaction of the form:
$$\ce{A + B + C <=> D + E + C}$$
Does the rate of reaction simplify down (according to collision theory) to:
$$r = [\ce{C}]\bigr(k_\mathrm{fwd}[\ce{A}][\ce{B}] - k_\mathrm{rev}[\ce{D}][\ce{E}]\bigr)$$

Edit: To clarify my question, I have a gas phase elementary reaction of the form:
$$\ce{A + B + C -> D + E + C}$$
I have another of the form:
$$\ce{D + E + C -> A + B + C}$$
Whether they are exactly the same mechanism is irrelevant. My question relates to how quantitatively does a collision partner's concentration in a gas mixture affect reaction rate. i.e. is forward rate defined by (assuming that I know the reaction acts as an elementary reaction in that there is no intermediate):
$$r_\mathrm{fwd} = k_\mathrm{fwd}[\ce{A}][\ce{B}][\ce{C}]$$
and reverse rate defined by (reverse is also elementary reaction - not concerned if same mechanism just also acts in the rate law as a single collision event):
$$r_\mathrm{rev} = k_\mathrm{rev}[\ce{D}][\ce{E}][\ce{C}]$$
Therefore, the total rate, not assuming they're the same reaction just two elementary reactions that go in opposite directions and we've called on forward and one reverse, the total rate forward, taking into account our other reaction, is:
$$r = r_\mathrm{fwd} - r_\mathrm{rev}$$
$$r = k_\mathrm{fwd}[\ce{A}][\ce{B}][\ce{C}] - k_\mathrm{rev}[\ce{D}][\ce{E}][\ce{C}]$$
$$r = [C]\left(k_\mathrm{fwd}[\ce{A}][\ce{B}] - k_\mathrm{rev}[\ce{D}][\ce{E}]\right)$$
Therefore quantitatively a collision partner does not change the overall rates direction but rather enhances it in both directions. Is this correct?

If it is in equilibrium, you are already looking at it from a macroscopic scale. It can therefore not be an elementary reaction. For example, $$\ce{A + B -> C + D}$$ is an elementary reaction, the reverse $$\ce{C + D -> A + B}$$ is a different elementary reaction. The rates of both reactions
\begin{align}
r&=k_\mathrm{fwd}[\ce{A}][\ce{B}]\\
r&=k_\mathrm{rev}[\ce{C}][\ce{D}]\\
\end{align}
would determine the location of the equilibrium
\begin{align}
\ce{A + B &<=> C + D}& K&=\frac{k_\mathrm{fwd}}{k_\mathrm{rev}}.
\end{align}

For a true termolecular reaction, i.e. three species react in one concerted transition state,
$$\ce{A + B + C -> D + E + F},$$
the rates are always given as
\begin{align}
r&=k_\mathrm{fwd}[\ce{A}][\ce{B}][\ce{C}]\\
r&=k_\mathrm{rev}[\ce{D}][\ce{E}][\ce{F}].\\
\end{align}
You cannot simplify a rate law of an elementary reaction.$%edit$


Does activation energy depend on steric factors? I thought that activation energy depends on steric factors.
By Arrhenius equation it must depend on steric factors since $A$ depends on steric factors. $$k=A\cdot\mathrm e^{-\frac{E_\mathrm a}{RT}}$$ (for bimolecular reaction)

But my book says it doesn’t.

Does it depend on steric factors or not and why?

The Arrhenius model of a rate rate constant is very basic so has to be modified in many cases. One of these is when steric factors (generally undefined exactly) are used as an empirical  multiplier to A and used as a rationale to fit experimental data.


Which is the densest gas known? Which is the densest gas known to exist under normal "shirt sleeve" conditions: room temperature, one atmosphere, won't explode on contact with air, won't kill everyone in the building in trace amounts, etc.

Perfluorobutane is inert and has almost twice the density of sulfur hexafluoride. It is non-toxic enough that it is used in fire extinguishers and injected as a contrast agent for ultrasound. Boiling point: $-1.7\ \mathrm{^\circ C}$.

Perfluoropentane is similar and rarer but somewhat higher density $(\sim13\ \mathrm{kg/m^3})$ in proportion to its higher molecular mass. Its boiling point is $28\ \mathrm{^\circ C}$ (uncomfortably warm, but your thermostat can go that high). This is the densest gas that strictly meets all the criteria in the OP.

If we relax the criteria a bit:

Perfluorohexane is just over the boiling point limit at $56\ \mathrm{^\circ C}$, but it has a molar mass of $338\ \mathrm{g/mol}$ which makes it slightly denser in gas form than tungsten hexafluoride ($\ce{WF6}$).  It's also inert and non-toxic, unlike $\ce{WF6}$.

The ultimate gas density would be uranium hexachloride using depleted uranium and $\ce{Cl-37}$, with a molar mass of $460\ \mathrm{g/mol}$, which makes it over 50 % denser in gas form than tungsten hexafluoride, and 3 times denser than sulfur hexafluoride, but it has a boiling point of $75\ \mathrm{^\circ C}$, decomposes on contact with air, is toxic and slightly radioactive.


Order of basicity of halides The order of basicity of halides is $\ce{F-  >Cl- > Br- > I-}$

My teacher explained this on the basis that the smaller is the ion, the greater is the charge/size ratio, hence it more unstable and more likely to gain a proton.

However if this is the case, why is the magnitude of electron gain enthalpy in the order $\ce{Cl- > F- > Br- > I-}$ and not $\ce{F- < Cl- < Br- < I-}$ since we are calling $\ce{I-}$ the most stable and so on?

(The reaction is going on in an aqueous solution.)

The electron gain enthalpy you've mentioned is measured in the gas phase, not in aqueous solution like the order of basicity.

When you go down on a group, two important properties are increased: nuclear charge and atomic radius. However, the increase in atomic radius is much more pronounced than of the nuclear charge. Considering this, as you go down on the halogens group the electrons added feel less attracted to the nucleus and, therefore, the electron gain enthalpy become less negative.

The order is only apparently changed for $\ce{Cl-}$ and $\ce{F-}$ due to fluorine small size. The additional electron would be added to a "small" $\ce{2p}$ orbital in fluorine (and there would be more electron-electron repulsion) while in the chlorine atom it would be added to a "big" $\ce{3p}$ orbital that can better accommodate the additional charge, so the enthalpy is less negative for the electron gain in $\ce{F}$ then for $\ce{Cl}$.


Comparison of acidity between angelic and tiglic acid Which among angelic acid and tiglic acid (cis and trans isomers of 2-methylbut-2-enoic acid respectively) more acidic and why?

It is given in a book(A guidebook to mechanism in organic chemistry for the JEE by Peter Sykes and B. H. Kishan 6ed. p.3.29 Q.1)that the cis(Z-) form is more acidic but I think the trans(E-) form will be more acidic due to interaction of oxygen in conjugate base with hydrogen on C3

I have some thoughts of my own. However, I am not certain if they are correct.

The key to the question certainly cannot lie in the electronic factor because the cis- and trans- isomers do not differ in this aspect, except for the difference in the overall dipole moment. Thus, I believe we can only rationalise this with steric considerations. Let's first take a look at the chemical structures.

Fig. 1: Structure of tiglic acid

Fig. 2: Structure of angelic acid

The steric interaction I would like to focus on is the repulsion between the methyl groups and the negatively-charged oxygen atom of the carboxylate ion. Due to resonance, the negative charge would be delocalised over the two oxygen atoms of the carboxylate ion conjugate base. In the trans-isomer (angelic acid), the methyl group is located closer to one of the oxygen atoms (i.e. the oxygen atom of the carbonyl group). Thus, there would be a stronger repulsive interaction between the two. In the cis-isomer (tiglic acid), the methyl group is located further away from that oxygen atom, thus the intramolecular repulsion would be lessened, thus it is more stable.

Since a more stable conjugate base implies that the acid is stronger, the cis-isomer (tiglic acid) is more acidic than the trans-isomer (angelic acid). This argument is in full agreement with the information presented in your reference book.


What is the function of PhLi in this reaction? Can someone please explain the mechanism of the reaction when we add 2 moles of $  \ce{Ph3P}$ and $\ce{PhLi}$ to 1,2-dichloroethane.

I know this is a type of Wittig reaction but I am not able to figure out how the reaction will work out.

I was wrong, this is not a Wittig reaction. Just posting the answer to let anybody who comes in hunt for this in the near future.

On googling for about an hour, I discovered that this is indeed a preparation of phosphonium ylide which is the witting reagent.

On adding $ \ce{Ph_3P} $ to 1,2-dichloroethane , $\ce{PhLi}$ acts as an catalyst and eliminates $\ce{HCl}$ from both the carbons resulting in a phosphonium di ylide.


Existence of orbitals Do orbitals exist even when they are not occupied?

For example: $\ce{Cr^{+3}}$ has the configuration $\ce{[Ar]}\mathrm{3d^3}$ with the other two $\mathrm{3d}$ orbitals empty. We know the  other two orbitals exist, since the metal "uses its empty $\mathrm{d}$ orbitals" to form complexes.

But, for $\ce{Na+}$ with configuration $\ce{[Ne]}$, do the $\mathrm{3s,3p}$ and $\mathrm{3d}$ orbitals exist  too?

Do orbitals exist even when they are occupied? Orbitals are mathematical constructs and as such they exist in a mathematical sense regardless of whether they are occupied or not. Whether you can actually say that orbitals exist in reality is a bit dubious. They certainly provide a good description of the observed electron density and behaviour of atoms but they are still only a model. Like everything in science, we develop a model to describe what we observe but that model is not perfect and does not necessarily represent what actually exists. Particularly on a quantum scale, the notion of well defined objects that exist in space starts to become a bit fuzzy.


Nomenclature of organic compounds When I name an organic compound, I follow a preferential order like so: http://www.masterorganicchemistry.com/2011/02/14/table-of-functional-group-priorities-for-nomenclature/ But I don't understand why this order exists. And intuitively (to me at least), a triple bond has a higher preference than a double bond... then why is the order like in the link, i.e., double bond preferred over triple bond? Is there any chemistry behind this?

The naming of organic compounds are governed by IUPAC guidelines, specifically the way the names are formed are by some general principles, which are given on the IUPAC website at their page Guide to Name Construction: General Principles, with naming examples here.
Another rather nice summary is found on the  webpage Nomenclature (Reusch, 1999), states that a organic compound name will have the following features, based on several common chemical characteristics of organic compounds:

•  A root or base indicating a major chain or ring of carbon atoms found in the molecular structure.
•  A suffix or other element(s) designating functional groups that may be present in the compound.
•  Names of substituent groups, other than hydrogen, that complete the molecular structure.

(Specific IUPAC guidelines for different hydrocarbons are included in the link).
Another summary is found on the handout Short Summary of IUPAC Nomenclature of Organic Compounds (Dr. Jan Simek, California Polytechnic State University at San Luis Obispo), which state that the general nomenclature principle is:

IUPAC nomenclature is based on naming a molecule’s longest chain of carbons connected by
single bonds, whether in a continuous chain or in a ring. All deviations, either multiple bonds or
atoms other than carbon and hydrogen, are indicated by prefixes or suffixes according to a
specific set of priorities.


Liquid with low specific heat and high thermal conductivity (relative to water)? As the title suggests – does there exist a liquid with a lower specific heat then water but with a higher thermal conductivity?

I thought of this in the context of sous vide cooking: Ideally, the liquid involved would transfer heat to the food quickly (hence high thermal conductivity), but would also respond to quick changes in temperature (hence the low specific heat), reducing the risk / time of heater element overshoot.

So, are there such liquids? (preferable ones that stay liquid between 20 and 85 degrees Celsius).

Most heat transfer through fluids is actually convection rather than conduction, so you need something that expands a lot with heat and has a low viscosity. Mineral oils or silicone are used as heat transfer fluids when temperatures above 100 °C are needed, but they are much more viscous than water.
However, for your system I guess that heat capacity is actually the most important number, as convection will be sufficient to ensure an equal temperature throughout the system (unless you actively impede it) even with a viscosity slightly higher than that of water.

Eyeballing some of the tables at http://www.engineeringtoolbox.com I’d suggest trying ethanol – its heat capacity is around half that of water, its viscosity is not much higher and it stays liquid over the desired range. Plus of course, unlike some of the more exotic solvents, it’s not a disaster if some of it does leak in to your food. If the food isn’t actually going to be eaten (just for demonstration purposes) then you could perhaps try toluene or petrol (gasoline).


Why are most hydrocarbons less dense than water? Why are most hydrocarbons, like alkanes and alkenes, less dense than water?

There is no truly simple explanation for bulk properties of liquids. But the balance of two factors probably goes a long way to explaining the difference between water and simple hydrocarbons.

The density of a liquid is determined by the mass of the molecules and the efficiency of how they pack in the liquid.

Molecules made of heavier atoms will tend to be denser (e.g. chloroform where the heavy chlorines give the liquid a density about 50% higher than water). Water is made of moderately heavier atoms than a hydrocarbon (oxygen is about 30% heavier than carbon).

The other factor is how well the molecules pack in the liquid. This depends on their shape and the molecular interactions among them. Water molecules have very strong hydrogen-bond interactions causing them to, on average, stay closer than molecules of equivalent size that don't hydrogen bond. Hydrocarbons have only weak van der waals interactions so don't, on average, stay so close.

The interaction of the molecular mass and the forces determine the overall density. Water wins on both compared to hydrocarbons.


H-NMR spectroscopy of [18]annulene Why does [18]annulene show only one peak in its proton NMR at $\pu{100^\circ C}$? Does any structural change happen such that all the hydrogens are alike?

The aromaticity of [18]-annulene1 was confirmed by room temperature 1H NMR (60 MHz) in 1964.2 [τ values have been converted to δ values; δ = 10 - τ] When the spectrum was recorded at room temperature the 12 deshielded, exterior hydrogens appeared at δ 8.9 while the 6-interior hydrogens were highly shielded (δ -1.8). Evidence for flexiblity in the ring at room temperature was confirmed by observing a further downfield shift of the 12 hydrogens (δ 9.28) and an upfield shift of the 6 hydrogen signal (δ -2.99) when the spectrum was recorded at -60oC. Recording the spectrum at 110oC caused the two signals to coalesce to a singlet peak at δ 5.45. This value is close to the weighted average of the resolved chemical shifts. Ostensibly, the coalesence is due to the interchange of external and internal hydrogens. When the sample is cooled, the resolved spectra are restored.

1) Sondheimer, F., Wolovsky, R., Amiel, Y. J. Am. Chem. Soc., 1962, 84, 274.
2) Gaoni, Y., Melera, A., Sondheimer, F., Wolovsky, R. Proc. Chem. Soc., 1964, 397.


Stability of molecules by resonance I am pretty confused by some principles of resonance. Well by my study of electrostatics I know that potential energy of a system of  charges of same nature decreases as we increase the separation between them but same doesn't goes with a system of charges with different nature because their potential energy decreases as we decrease their separation as internal conservative forces do positive work.

I am a newcomer in chemistry so I have a doubt that my book says that energy of a molecule is minimum when they are separated by large distance (under the heading of application of resonance) without caring for nature of charges.This fact goes well for charges of same nature but contradicts the fact for system having different nature of charges.

Please tell me  that why this contradiction exists.

In chemical resonance, the same principle holds as in electrostatics:

"In general, the boundary structures of a conjugated system with the lowest energy are those
...
- in which the distance between charges of opposite sign is as small as possible and the distance between charges of the same sign is as large as possible
... ."

Wikipedia: Resonance (chemistry) - Major and minor contributors:
"Energetically unfavourable and therefore less favorable structures are minor contributors. With rules listed in rough order of diminishing importance, major contributors are generally structures that
...
3. carry a minimum of formally charged atoms, with the separation for unlike and like charges minimized and maximized, respectively
... ."


Dosing leaves with Zinc Nitrate for pollution experiment - Need specific concentration I am trying to assess the pollution tolerance of an insect using Zn as a proxy. I have a supply of Zinc Nitrate Hexahydrate that I want to use to contaminate dried leaves (the insect's food). The LD50 for my insect is around 2000 micrograms g^-1.

I was thinking I could achieve this by using 0.1 g squares of dried leaf that I would contaminate by topically applying 1 ml of Zn solution. To get my desired concentration, I am thinking that I would dissolve 200 micrograms of Zn into 1 ml of distilled water, so that when I apply 1 ml of this solution to a 0.1 g square of leaf I would have a final concentration of 2000 micrograms/gram of food. But if I wanted to apply only 0.5 ml of Zn solution to 0.1 g of leaf but still have the same final concentration of 2000 microg/g food, I would create a solution of 400 micrograms/ml. And if I wanted to use 1.0 gram of leaf and have a concentration of 2000 micrograms/g, I would make a solution of 2000 micrograms/ml. Is all of this correct?

I am also open to suggestions of how to better contaminate the food in a more consistent way. I thought about soaking leaves in a defined concentration, but then I am worried that some leaves will absorb more or less liquid than others, making the final Zn/g food concentrations unequal across different leaves.

Thank you for any solutions (pun intended) anyone might provide.

It seems to me that it will be difficult to get a uniform concentration of Zn in a non-uniform material like dried leaves, unless you measure only in bulk (this pile of dried leaves weighs X grams and contains 2000X micrograms of Zn). Things get more confusing if the insects don't eat all the dried-leaf structure (leaving stems, for example).

I'm also not sure how you'll get all the zinc from 1 ml of liquid onto a 0.1g leaf sample. (And, of course, the leaf won't be dry for a while after that!) I think a more concentrated solution would make your task easier. Your math for adjusting the concentration to match the amount of leaf matter appears correct.

As an aside, is your question confusing LD50 (micrograms per gram of insect) with leaf concentration (micrograms per gram of leaf)? I'm not sure why the concentration in the leaves would need to correspond to the LD50, unless there's something significant about the insects eating their own body weight in dried leaves...?


How is polarization of light measured? Chiral molecules are classified as dextrorotatory (+) or levorotatory (-) depending on whether they rotate plane-polarized light (ppl) to the right or left.

My question is, how is such a measurement made? What instrumentation is used and how would you determine that ppl has been rotated right or left?

The instrument used is called a polarimeter. You basically shine monochromatic linear polarized light through a solution of known concentration and you measure the rotation of the light on the other end of a tube with defined length. You then normalize the rotation according to concentration and length of your tube to get the specific rotation which is then independent from concentration and length and only depends on temperature, solvent and wavelength of the light.

You can determine if it's rotated left or right by changing the concentration since the observed rotation is directly proportional to concentration. You also need to do this to determine if your rotation is more than a full 360°, since a measurement of 10° could also be 370° or 730° or 1090° or also −350°, −710°, and so on.

Let's look at this using a simple example and let us assume our tube has a length of 10 cm (which is the length used for the specific rotation, so we don't need to worry about this). Specific rotation is also normalized to a concentration of 1 g/100 mL.

To determine our specific rotation we could do 3 measurements at 1 g/100 mL, 2 g/100 mL and 0.5 g/100 mL.

1 g/100 mL : +160°
2 g/100 mL : −40°
0.5 g/100 mL : −100°

Now what does this tell us? If the specific rotation would be +160° then double the concentration should be 320° and half should be +80°, so that's clearly not the case. If it would be −200° instead of +160°, then double the concentration would be −400° which would show up as −40° and half the concentration would be −100°. So our specific rotation in this case is −200°.


Pyrotechnic colorants- reactions & safety concerns regarding making charcoal discs I want to create my own charcoal disks/tablets which will be mainly used for incense and I want each of the charcoal disks to glow different colours (eg one disc glows blue, another glows red etc.). The product will be using charcoal and other ingredients including Arabic gum. I will need to add certain chemicals such as metallic salts and so forth to create the desired aesthetic effects such as the colours.

My main concern is the safety of the product, I would like to experiment to see if any of this will work but most of all, I want to know if the end product will not become a dangerous fire hazard and that it will be non-toxic.
I’ve looked up some of the chemicals online but I want to be sure.
I know that you can add metal salts to flames in bonfires to add a similar effect.

(Example here: http://tiny.cc/hqjv9y and in fireworks displays. Can I have that similar desired effect with the charcoal disks as well as the safety with it?

Below are some of the chemicals that I’ve found and would like to know if any of them are safe to add to creating my product:

For a blue flame, you want Copper Chloride or Calcium Chloride
For a light green flame, you want Borax
For a dark green flame, you want Copper Sulfate or Alum
For a strong, red flame, you want Strontium Chloride
For a purple flame, you want Potassium Chloride
For a pink flame, you want Lithium Chloride
For an orange flame, you want Sodium Chloride
For a white flame, you want Magnesium Sulfate

Some chemicals also make coloured sparks, instead of colouring the flame:

For gold sparks, you want Iron (fillings)
For silver sparks, you want (powdered) Aluminum

Source here: https://www.quora.com/What-chemicals-can-change-the-color-of-fire

Other sources: https://www.thoughtco.com/how-to-make-colored-fire-p2-602196

Pyrotechnic colorant: https://en.wikipedia.org/wiki/Pyrotechnic_colorant

https://www.thoughtco.com/how-to-make-colored-sparklers-3980637

https://www.buysparklers.com/product/14-inch-neon-party-sparklers/

I've read that magnesium can be explosive although I also I want to achieve a white hue for one of my discs. Are there any other alternatives for this?

I also would like to add that I may or may not be adding salt peter (potassium nitrate) to the mix. If I do decide to add potassium nitrate, how will each of the selected pyrotechnic colourants behave with it?

I am trying to create a product with as little additives as possible, that burns cleanly, with little to no scent, with hardly any smoke and as mentioned before, is safe and non-toxic.

The best way for you to find out if a chemical is safe for your use case would be to look at the safety data sheets for your chemicals. For your convience, I have linked to relevant ones below.

Copper (II) Chloride
Calcium Chloride
Borax
Copper (II) Sulfate
Alum
Strontium Chloride
Potassium Chloride
Lithium Chloride
Sodium Chloride
Magnesium Chloride
Aluminium Powder

In general, you will find that most metal salts are acutely toxic. Take for instance, the Copper (II) Chloride, which is noted as having the following hazards, as noted in Section 2.1 Classification of the substance or mixture:

Acute toxicity, Oral (Category 4), H302
Acute toxicity, Dermal (Category 4), H312
Skin irritation (Category 2), H315
Serious eye damage (Category 1), H318
Acute aquatic toxicity (Category 1), H400
Chronic aquatic toxicity (Category 2), H411

From the same section in the SDS for Aluminum Powder, you will find that it has the following hazards:

Flammable solids (Category 2), H228
Acute aquatic toxicity (Category 1), H400
Chronic aquatic toxicity (Category 1), H410

You can find out more about these hazards from the EPA GHS Criteria Summary, and the Wikipedia GHS Hazard Statement List.

Overall, you should look over each SDS and make sure that the chemical you are proposing to use is safe for your purposes.

Cheers


By what mechanisms can molecules with the same empirical formula be so different? I understand the difference between empirical and chemical formula. But I always thought that compounds and molecules took their specific shape because it's the lowest energy configuration. How then, can fructose ($\ce{C6H12O6}$) and formaldehyde ($\ce{CH2O}$) both exist?
If for example formaldehyde had overall the smaller energy, why then wouldn't fructose just decompose in 6 $\ce{CH2O}$ molecules? What makes both of those molecules so stable?

I guess this is a very general question of what makes certain elements combine the way they do to form compounds.

What makes both of those molecules so stable?

This is a fundamental question. Let's start by looking at the following diagram; a chemist would call it a "potential energy diagram." The y-axis shows the energy of a molecule as it travels along the reaction coordinate (x-axis; transforms, reacts) from the starting material on the left to the product on the right.

Notice that there are 3 "potential energy wells" (valleys) in the diagram, these reflect molecules that are stable enough to have a finite lifetime.  Notice too that to move from one well to another, an energy barrier must be surmounted.

So to return to your question, a molecule would be "kinetically stable" if it was protected on all sides by high energy barriers.  A molecule would be "thermodynamically stable" if its potential energy well was located at an energy that had a very low value on the y-axis.  A molecule that is very stable in a thermodynamic sense, could still be kinetically unstable (and therefore very reactive) if it was surrounded by low energy barriers.  By the same token, a molecule that is very unstable in a thermodynamic sense (exists at a high energy value on the y-axis) could be kinetically stable if it were surrounded by high energy barriers to reduce the likelihood of further reaction.

Millions of molecules are stable at room temperature, but if you supply enough energy to the system (heat them for example) they can all transform into other compounds.  At room temperature there are high enough energy barriers separating fructose and formaldehyde and they will both be stable in a jar, but under more energetic conditions such transformations might be feasible.


Out of the given reagents, find the ones to be used to convert benzene to p-aminophenol $\ce{Br2/FeBr3}$
$\ce{CH3Cl/AlCl3}$
$\ce{KMnO4}$
$\ce{HNO3/H2SO4}$
$\ce{NBS}$/hv
$\ce{NaOH}$
$\ce{Sn/HCl}$
$\ce{NH4SH}$
$\ce{NaBH4}$

I'm able to get the amino group onto the ring by using 4 and 7.
But I have no idea how to get $\ce{-OH}$ in para position (also specify the order).

Steps:

Use $\ce{Br2/FeBr3}$ to make bromobenzene.
Nitrate with $\ce{HNO3/H2SO4}$ to make 4-bromo-nitrobenzene.
Treat with $\ce{NaOH}$ to displace the $\ce{Br}$ making 4-nitrophenol.
Reduce with $\ce{Sn/HCl}$ to make the required p-aminophenol.


Voltage needed for electrolysis of water Reading the wiki for electrolysis it reads that a potential of 1.23V is needed to separate hydrogen from the oxygen. With that being said, how was that calculated? Does it change with the water volume and conductivity?

I'm assuming you mean this Wikipedia article on the hydrolysis of water in which the method of determining that potential difference is described farther down the page.

I'll address your other questions - what might cause this value to change.

  Volume

No, except see below on concentration. The reaction potential for this electrolysis is independent of the amount of water since it is actually a thermodynamic function. Electrolysis may occur at different rates and take varying time depending on the amount of water, but the potential needed remains the same.

  Conductivity

Conductivity is a measure of the ease of transmitting electrical current (electrons) through a substance. So, increased conductivity will increase the rate of electrolysis, but since it does not change the original or final substances in the chemical equation, does not change any thermodynamic values like the reaction potential.

However, there is a danger here. We typically increase conductivity by increasing electrolyte content. Some of these electrolytes could be electrolyzed at a reduced potential compared to water. For example, if you use sodium iodide as your electrolyte, the iodide ion is oxidized to the triiodide ion at -0.53 V:

$$\ce{3I- -> I3- + 2e-}\ \ E^\circ = -0.53\ \pu{V}$$

  Concentration

The reaction potential -1.23 V is the standard reaction potential, which is determined when all applicable species are at 1 molar concentration and 1 bar pressure. The two half reactions in this case are:

$$\ce{2H2O -> O2 + 4H+ + 4e-}\\
\ce{4H+ + 4e- -> 2H2}$$

Thus, this potential is only valid at $[\ce{H+}] = 1\ \pu{M}$ or $\pu{pH}=1$. As the reaction proceeds, water is consumed and the volume decreases. The amount of protons remains constant, so the pH increases. As the pH increases, so too does the potential required and the electrolysis eventually reaches equilibrium and progresses no further unless you are providing an overpotential.

Note that -1.23 V is equally valid at $[\ce{OH-}] = 1\ \pu{M}$ or $\pu{pH}=14$ thanks to Hess's Law.


Desilylation mechanism with fluoride I want to understand the mechanism of desilylation of trimethylsilyl under fluoride conditions in methanol, because I can't find any useful info that makes me understand the mechanism that takes place.

I did find this:

But my case is a little bit different because a terminal alkyne is bonded to the silyl group instead of oxygen:

Why would that bond be broken and not any other carbon-hydrogen bond?

Thanks in advance!

The mechanism is the same with the generation of the alkynyl anion as the abstract of this paper  ref 1 makes clear:

  Tetrabutylammonium triphenyldifluorosilicate (TBAT) can be employed as a fluoride source to cleave silicon−carbon bonds thus generating in situ carbanions that coupled with a variety of electrophiles, including aldehydes and ketones, in moderate to high yields. Among the examples reported is the first instance of fluoride-induced intermolecular coupling between allyltrimethylsilane and imine derivatives. Also, of particular note is the TBAT-initiated coupling of primary alkyl halides with allyltrimethylsilane. TBAT is an easily handled crystalline solid that has several advantages over tetrabutylammonium fluoride (TBAF) as a fluoride source; it is anhydrous, nonhygroscopic, soluble in most commonly used organic solvents, and less basic than TBAF.

Bear in mind that the $\mathrm{p}K_\mathrm{a}$ of an alkynyl proton is approximately 25 (ref 2) so this is a far more stable anion than any possible alternative. Clearly with $\ce{MeOH}$ as the solvent it will be quenched immediately.


Reason for endothermic nature of NaHCO3 + C2H4O2 reaction Why is the classic baking soda and vinegar reaction endothermic? Here are some possible answers I've come up with:

The dissolution of the sodium acetate absorbs energy,
The release of the carbon dioxide from solution absorbs energy, or
Some combination of the above.

Your "all of the above" answer is correct, with emphasis on number 2.  The phase change from a condensed phase (solid or liquid) to gas is almost always endothermic.  It's the same thing as wetting your hand then blowing on it.  The liquid water undergoes and endothermic phase transition which takes in heat from its surroundings, i.e. it evaporates and your hand gets cold.


How does the room gets heated by leaving a fridge open? Question:

  If the door of the refrigerator is kept open, the room in which the refrigerator is kept:

  gets cooled
  gets heated
  neither gets cooled nor gets heated
  gets cooled or heated depending on the initial temperature of the room.

The answer mentioned is option b) gets heated
It doesn't seem convincing that on opening refrigerator's door the room gets heated. Also the radiator emits out the heat in the surrounding. Thus the heat absorbed from the room is released in it and the must neither get cooled nor heated.

Just before posting this question I have tried to experimentally verify this and observed no change in temperature of the room.

Also the radiator emits out the heat in the surrounding. Thus the heat absorbed from the room is released in it and the must neither get cooled nor heated.

You have forgotten a few parts to your system in how a refrigerator pumps heat. You have assumed a spontaneous steady state cycle of:

$$\text{radiator}\underset{\text{conduction}}{\overset{\text{heat pump}}\leftrightharpoons} \text{fridge}$$

But that is not true to move heat from a low-temperature body body to a high-temperature body requires an input of work which is added to the heat dumped into the high-temperature body.

In this case work is done by the compressor which takes in electrical energy and transforms it into work and heat. Since electrical energy is 100% efficient at heating any electrical energy consumed by the compressor is eventually converted into heat either by ohmic heat, efficiency losses or work being converted into heat at the high-temperature body. Since your high-temperature body and low-temperature body are allowed to conduct/convect/radiate/transfer heat between the two sinks, heat overall is added to the room.

  Just before posting this question I have tried to experimentally verify this and observed no change in temperature of the room.

There are a number of problems with this, and why you may not have observed anything.

Rooms are not closed nor isolated systems thus are subject to many thermal variations from light (radiation), drafts (convection), and conduction of heat through walls to/from the outside which can make your results unreliable.
A the heat capacity of a room is very large and it will take a while for your fridge to impart a change that a thermostat/thermometer can detect.
The energy consumed by a refrigerator is relatively small* and it will again take a while for your fridge to impart a change that a thermostat/thermometer can detect.

All and all, such an experiment would require better control than your average room/kitchen. If you did do such a controlled experiment you would observe that the temperature of the room would rise with time due to the heat losses of the compressor.

 *I understand some parents/utility customers (possible yours) may find this statement controversial. Kids do not quote me at home.


Reactivity of KOH with PVC I Would like to know if its possible to get a reaction between KOH at ~50% (w/w) and an usual PVC pipe? If yes what kind?

If you're just curious for practical purposes (i.e., "Can I put this KOH solution in this PVC container or will it ruin it?"), there are many material compatibility tables for this type of thing. This one suggests that your solution will be OK at low temperatures for a few days at least. It may react after more time or at higher temperatures.

If you want to know specifically about the chemistry, KOH is a strong base. Strong bases are known to crack polymer chains through solvolysis. PVC in particular loses some chlorine atoms to the solvent which weakens and ultimately shortens the polymer chains, as explained in more detail in the link.


Bonding metal pieces using electroplating - what's the process called? So I think the title all most says it all - given say two pieces of copper can we make solid 'weld' using electroplating chemistry? If so whats the accepted terminology for this process so I can do some literature research - also am I correct in my assumption that the join would be as strong as the raw metal?

There's a 1974 Sandia labs paper on "Joining by Electroplating", abstract and citation available here.  They recommend tapering the joint and plating a triangular void closed, which brings up the important point that you need to let fresh electrolyte in to reach the joint as it builds up material.  For example, trying to electroplate two flat disks face-to-face would likely result in most of the deposit forming a ring around the disk edges, sealing a layer of corrosive electrolyte between the disks rather than securely welding them together.

The terminology for this sort of technology has rather permeable boundaries, but includes:

Electroplating, normally aiming to deposit a mostly uniform surface coating over an existing substrate.  Common in a variety of consumer goods, often for corrosion protection.
Electroforming, depositing material over a removable mandrel.  Common in micromanufacturing.
Electrowinning, extracting or purifying material.  Common in mining.


Dissociation of HCl in aqueous solution My question is regarding $\ce{HCl}$. In my book it states that:

$$\ce{HCl (aq) -> H+ (aq) + Cl- (aq)}$$

I understand that when added to water the $\ce{H}$ leaves its electron to the $\ce{Cl}$ atom forming the $\ce{Cl-}$ and the $\ce{H}$ attaches to water to form the hydronium ion ($\ce{H3O+}$). Alltogether:

$$\ce{HCl + H2O -> H3O+ + Cl-}$$

Why does this dissociation actually occur scientifically? What causes the $\ce{H}$ to leave its electron behind to the $\ce{Cl}$ and form the hydronium ion? Why does water ($\ce{H2O}$) take the additional $\ce{H}$ ion? What is the general concept here in dissociation? Electronegativity?

Water is a polar species in which the oxygen atom in $\ce{H2O}$ has a partial negative charge and the two hydrogen atoms have partial positive charges. As you suggested, this is because oxygen is highly electronegative (i.e. it has a high affinity for electrons) and so the negative electrons tend to have a higher density around the oxygen atom than the hydrogen atoms, creating this separation of charges due to water's molecular structure. Take a look at the molecular structure of water, which can be shown below. It has an asymmetry bent geometry and therefore the charge distribution in water is also not symmetric.

The hydrogen in $\ce{HCl}$ does not actually give its electron to the chlorine atom. Rather, the more electronegative chlorine ion is attracted to the partially positive hydrogen atoms in water, and the hydrogen ion in $\ce{HCl}$ is attracted to the partially negative oxygen atom in water. This separates the $\ce{H+}$ from the $\ce{Cl-}$ in $\ce{HCl}$ so that the acid dissociates into its respective ions. You therefore get the equation $\ce{HCl -> H+ + Cl-}$ when in water. If you include water on the reactant side, it can be equivalently written as $\ce{HCl + H2O -> H3O+ + Cl-}$. Here, $\ce{H3O+}$ is simply indicating that the $\ce{H+}$ ion is in an aqueous solution.


Preference for basic conditions in ester hydrolysis It has occurred to me that it is very common see the use of basic conditions for ester hydrolysis in various syntheses. One such sequence commonly seen in the case of the malonic ester synthesis is:

$\ce { OH^-,H2O}$
$\ce {H^+, heat}$

The first step is for the hydrolysis while the second step is for the decarboxylation. It is strange that there is a first step with basic conditions since ester hydrolysis and decarboxylation can both take place in acidic solution. Are there practical reasons for why ester hydrolysis is done with aqueous base?

Precedent - why change conditions that others have shown work well?
Control - it is much easier to react 1 eq of base/nucleophile under mild conditions, whereas acid hydroysis generally requires excess reagent and higher temperatures.
Ease of operation - base hydrolysis gives a carboxylic salt as product which frequently extracts into aq, giving an easy workup and purification. Acid hydrolysis requires neutralisation of excess reagent as a first step.


A problem in solubility of salts with multistep hydrolysis of an ion possible Calculate the solubility of $\ce{MnS}$ in pure water, $K_\mathrm{sp}(\ce{MnS}) = \pu{2.5e-10}$. Assume hydrolysis of $\ce{S^2-}$ ion. $K_1$ and $K_2$ for $\ce{H2S}$ are $10^{-7}$ and $10^{-14}$, respectively.

Let unknown amount of $\ce{MnS}$ be dissolved in $\pu{1 L}$ of water, so $\ce{Mn^2+} = s~\mathrm{moles}$ and $\ce{S^2-} = s~\mathrm{moles}$ where $s$ is the solubility of $\ce{MnS}$ in $\pu{mol L-1}$.

Now let $\ce{S^2-}$ participate in hydrolysis by dissociating its $x$ moles and combining with $\ce{H2O}$ to form $x$ moles of $\ce{HS-}$ and $x$ moles of $\ce{OH-}$, so final moles of $\ce{S^2-}$ is $(s-x)$.

I considered only $K_\mathrm{h1}$ as it is $10^7$ times larger than $K_\mathrm{h2}$, $K_\mathrm{h1}$ is $K_\mathrm{w}/K_\mathrm{a2}$ which is equal to $1$. So I got $s^2 = (s-x)$ and we know that $K_\mathrm{sp} = s(s-x)$. On substituting I got the following equation:

$$K_\mathrm{sp} = s^2 - s^{1/2} \cdot K_\mathrm{sp}^{1/2}$$

I am stuck here. I know that $K_\mathrm{sp} = \pu{2.5e-10}$, so I tried solving for $s$, but could not find the correct answer.

Can you please point out my conceptual mistake?

Sorry, if this question is of homework type, I really tried solving this before posting.

Let's denote all processes of interest which are occuring or might occur in the system:

\begin{align}
\ce{MnS &<=>[$K_\mathrm{sp}$] Mn^2+ + S^2-} && K_\mathrm{sp} = [\ce{Mn^2+}][\ce{S^2-}] \tag{R1} \\
\ce{MnS + H2O &<=>[$K_\mathrm{h1}$] Mn^2+ + HS- + OH-} && K_\mathrm{h1} = [\ce{Mn^2+}] [\ce{HS-}] [\ce{OH-}] \tag{R2} \\
\ce{MnS + 2 H2O &<=>[$K_\mathrm{h2}$] Mn^2+ + H2S + 2 OH-} && K_\mathrm{h2} = [\ce{Mn^2+}][\ce{H2S}][\ce{OH-}]^2 \tag{R3} \\
\ce{H2S &<=>[$K_\mathrm{a1}$] HS- + H+} && K_\mathrm{a1} = \frac{[\ce{HS-}][\ce{H+}]}{[\ce{H2S}]} \tag {R4} \\
\ce{HS- &<=>[$K_\mathrm{a2}$] S^2- + H+} && K_\mathrm{a2} = \frac{[\ce{S^2-}][\ce{H+}]}{[\ce{HS-}]} \tag {R5} \\
\ce{H2S &<=>[K_\mathrm{d}] 2 H+ + S^2-} && K_\mathrm{d} = K_\mathrm{a1} K_\mathrm{a2} \tag{R6}
\end{align}

and also agree on labeling $C_\mathrm{A}$ total concentration of sulfide anions in solution in form of $\ce{S^2-}$, $\ce{HS-}$ and $\ce{H2S}$; $\alpha_2$ – fraction of $\ce{S^2-}$ among those; $s$ – solubility. Then from (R1)

$$K_\mathrm{sp} = [\ce{Mn^2+}] C_\mathrm{A} \alpha_2 \implies s = \sqrt{\frac{K_\mathrm{sp}}{\alpha_2}} \tag{1}$$

$\alpha_2$ can be expressed via known equilibrium constants and $\mathrm{pH}$:

\begin{align}
\alpha_2 &= \frac{[\ce{S^2-}]}{[\ce{S^2-}] + [\ce{HS-}] + [\ce{H2S}]} &&\Bigg| \cdot \frac{[\ce{H+}][\ce{HS-}]}{[\ce{H+}][\ce{HS-}]}\\
&= \frac{K_\mathrm{a2}}{K_\mathrm{a2} + [\ce{H+}] + \frac{[\ce{H2A}][\ce{H+}]}{[\ce{HA-}]}} &&\Bigg| \cdot \frac{K_\mathrm{a1}}{K_\mathrm{a1}} \\
&= \frac{K_\mathrm{a1}K_\mathrm{a2}}{K_\mathrm{a1}K_\mathrm{a2} + K_\mathrm{a1}[\ce{H+}] + [\ce{H+}]^2} \tag{2}
\end{align}

Since hydrolysis might occur partially (R2) or completely (R3), in order to find$[\ce{H+}]$ from

$$[\ce{H+}] = \frac{K_\mathrm{w}}{[\ce{OH-}]} \tag{3}$$

let's estimate whether the equilibrium in (R3) is going to be shifted left or right. Solubility of sulfide anions in the absence of hydrolysis is

$$[\ce{S^2-}] = \sqrt{K_\mathrm{sp}} = \sqrt{\pu{2.5e-10}} = \pu{1.6e-5 M} \tag{4}$$

whereas solubility when $\ce{H2S}$ is forming (from (R6)) would be

$$[\ce{S^2-}] = \sqrt[3]{\frac{K_\mathrm{a1} K_\mathrm{a1}}{4}} = \sqrt[3]{\frac{10^{-7} \cdot 10^{-10}}{4}} = \pu{1.4e-6 M} \tag{5}$$

so that formation of less soluble product would indeed favor (R3). Hence, in order to estimate the increase in $\mathrm{pH}$, it's better to use $K_\mathrm{h2}$:

\begin{align}
K_\mathrm{h2} &= [\ce{Mn^2+}][\ce{OH-}]^2[\ce{H2S}] && \Big| \cdot \frac{[\ce{H+}]^2[\ce{S^2-}]}{[\ce{H+}]^2[\ce{S^2-}]} \\
&= \frac{K_\mathrm{sp}K_\mathrm{w}^2}{K_\mathrm{a1}K_\mathrm{a2}} \tag{6}
\end{align}

On the other hand,

$$K_\mathrm{h2} = [\ce{Mn^2+}][\ce{OH-}]^2[\ce{H2S}] = 4[\ce{OH-}]^4 \tag{7}$$

Equating (4) and (5):

$$[\ce{OH-}] = \sqrt[4]{\frac{K_\mathrm{sp}K_\mathrm{w}^2}{4 K_\mathrm{a1}K_\mathrm{a2}}} = \sqrt[4]{\frac{\pu{2.5e-10} \cdot (10^{-14})^2}{4 \cdot 10^{-7} \cdot 10^{-10}}} = \pu{5e-6 M} \tag{8}$$

Now back to (3):

$$[\ce{H+}] = \frac{10^{-14}}{\pu{5e-6}} = \pu{2e-9 M} \tag{9}$$

Knowing how the $\mathrm{pH}$ of the system has been changed, let's find what fraction of sulfide anions are in solution:

$$\alpha_2 = \frac{10^{-7} \cdot 10^{-10}}{10^{-7} \cdot 10^{-10} + 10^{-7} \cdot \pu{2e-9} + (\pu{2e-9})^2} = 0.0467 \tag{8}$$

Finally, solubility can be determined:

$$s = \sqrt{\frac{\pu{2.5e-10}}{0.0467}} = \pu{7.3e-5 M} \tag{10}$$

e.g. taking hydrolysis into account, solubility of $\ce{MnS}$ is approx. $4.5$ times higher (see equations (4) and (10)).


What is the role of a monochromator in AAS? What would happen if we didn't use a monochromator; i.e. what is its role in Atomic Absorption Spectroscopy?

This question has lingered on the site for a long time, which is a shame because it is a good question to ask.  Atomic absorption spectroscopy (AAS) uses a hollow cathode lamp (HCL) as a source, and HCLs are line sources; therefore, if a line source is being used, why would an instrument need to be made more complex by adding a wavelength selector?

The first point to consider is what the source really looks like.  The emission spectrum from HCLs have multiple lines.  Here, for example, is the emission from a nickel HCL, taken from Spectrochemical Analysis by Atomic Absorption and Emission by L. Lajunen.

A wavelength selector is needed in order to ensure that only one of the emission lines is being used for the analysis.

Additionally, light can be scattered in the flame, which can introduce errors in the analysis if not eliminated.  If you compare the block diagram of an AAS vs a typical visible spectrometer, you'll note that the wavelength selector is typically placed after the sample compartment for AAS to address these concerns about scattered light; however the issue is not that significant in visible spectroscopy so the wavelength selector can be placed before the sample holder.

This representative block diagram for AAS is taken from here.

A representative UV/Vis spectrometer block diagram is from this source

Bottom Line AAS needs a wavelength selector to ensure that the analytical line of interest is the only radiation hitting the detector and it minimizes the negative effects of radiation scattered in the flame.  A good summary can be found here.


why Ammonia is proton acceptor I understand that this is an elementary question, but I tried to find answers which I could not get.

I am trying to understand why Nitrogen in Ammonia is a proton acceptor.

Now as I can see from the structure the octet is complete hence should be stable as it is.

But couple of points here:

1) Nitrogen is more electronegative than hydrogen hence $N$ will pull electrons towards itself. Hence not a good candidate to donate electrons! In addition, the lone pair is in 2s orbital.

2) The other point is the presence of lone pair, it is claimed that this lone pair is more stable if its part of some bond, hence it accepts proton. But if this is the case then lets take the case of Oxygen $O=O$, here each Oxygen atom has 2 lone pairs, and going by the above logic these lone pairs should also participate in bond formation and create an Oxygen polymer(!!), but they do not and Oxygen is quite stable.

Hence with the above points I still could not see why Nitrogen has to accept proton.

The only explanation that I can think of is the lone pair is in 2s orbital and the spread of this electron cloud is larger than electrons participating in bonds. Hence to be stabilized further it accepts a proton $(H_+)$ to form a bond!!

Nitrogen is more electronegative than hydrogen hence N will pull electrons towards itself. Hence not a good candidate to donate electrons!

Yet it makes bonds with hydrogen, as do other electronegative elements (think of hydrogen chloride and water). So polar bonds form in all kind of molecules.

  lets take the case of Oxygen O=O, here each Oxygen atom has 2 lone pairs, and going by the above logic these lone pairs should also participate in bond formation and create an Oxygen polymer(!!)

The concept of stability makes sense when comparing two states with the same number of atoms (e.g. ammonia and hydrogen ion vs. ammonium ion). For elemental oxygen to polymerize, each $\ce{O2}$ molecule would bring in electrons. For a molecule to act as base, a lone pair turns into a bond with hydrogen (where hydrogen provides zero electrons and the base provides two). This does not "break the octet" of the base, and give the hydrogen access to two electrons (completing its duet).

  Hence with the above points I still could not see why Nitrogen has to accept proton

It does not have to. It is a weak base. If you add ammonia to neutral water, most of it remains undissociated (the pH goes up, though, because some of it does dissociate).

  [devils advocate question:] How can I use electronegativity to predict something about acid/base chemistry?

The trick is to compare green apples with red apples, not oranges with bananas. The easiest comparison is isoelectronic species, such as:

$$\ce{HF vs OH-}$$
Hydroxide is a better base than hydrogen fluoride.

$$\ce{H2O vs NH2-}$$
Azanide (had to look up the name) is a better base than water.

$$\ce{H3O+ vs NH3 vs CH3-}$$
The methyl anion is a better base than ammmonia, which is a better base than hydronium.

Or you can compare compounds by exchanging O for NH or CH2.

$$\ce{H2O vs NH3 vs CH4}$$

$$\ce{CH3-OH vs CH3-NH2 vs CH3-CH3}$$

$$\ce{CH2=O vs CH2=NH vs CH2=CH2}$$

This is more complicated because the carbon atoms lack a lone pair while nitrogen and oxygen have one. So you can compare nitrogen and oxygen using an electronegativity argument, but have to exclude carbon because it can't act as a base without the lone pair (well, $\ce{CH5+}$ does sort of exist but is called a super-acid, so $\ce{CH4}$ is hardly a base).

Hydrides

Here is a set of examples from OpenStax Chemistry 2e showing some of the trends across the periodic table. Again, $\ce{CH4 and SiH4}$ don't have lone pairs, so they can't be bases, even though the trend suggests they might be the strongest base in the series. The reason that water is described as neutral is not some special property of water but because the pKa is typically defined in aqueous solution, and a neutral pH references the hydroxide and hydronium concentration in pure water. If liquid ammonia were our solvent of choice, that would be the "neutral" case.


Boiling temperature of salt water at low pressure With a hydro aspirator with re-circulating water (see here), the maximum vacuum you can get is set by the water vapor pressure given the temperature.

Can you lower the vapor pressure measurably by adding salt to the water (I get −700 mmHg, at 25 °C, with regular tap water already)?

According to this 1927 paper, "THE VAPOR PRESSURE OF THE SATURATED AQUEOUS SOLUTIONS OF CERTAIN SALTS" at 25 °C, 6 mol per kg salt solution gives a pressure of 18.5 mmHg instead of 24.5 mmHg for pure water.
So yes there should be a measurable difference (unless salt destroys the diaphragm pump).

Edit: This recent paper "Water vapour pressure above saturated salt solutions at low temperatures", gives the same results


Relation Between Degree of Hydration and Electropositivity Why are salts of strongly electropositive elements less heavily hydrated in aqueous solution than the those of less electropositive elements? Shouldn't it be the other way round as the salts of electropositive elements are mostly ionic and hence strongly attracted by water molecules?

The degree of hydration in solution does not directly depend on electropositivity. It depends on the ratio of charge to volume. The higher charged and the smaller a fragment is, the more water molecules hydrate it.

Note that almost every salt in solution can be considered as separated ionic pairs. For those that do not seperate, i.e. have a significantly large covalent bonding proportion, you would be looking at the solvation of the non-dissociated ion pair meaning that you can’t properly compare that hydration to simple salts’ hydration.

Metals get more electropositive the closer they are to the bottom left of the periodic table with cesium being the (non-radioactive) corner. Electropositivity tends to increase with an atom’s size. But the larger the size, the less the degree of hydration is, because the charge per volume of the metal ion will be smaller!

Also, the most electropositive metals are alkaline metals, meaning the only create $\ce{M+}$ ions. However, the stronger the charge, the smaller an ion gets: $\ce{Ga^3+}$ has an ionic radius of $76\,\mathrm{pm}$ — smaller than $\ce{Li+}$ whose ionic radius is $90\,\mathrm{pm}$ although they are two periods apart! So $\ce{Ga^3+}$ will be a lot stronger hydrated than $\ce{Li+}$ because it contains much more charge per volume.


Why is BCl3 a monomer whereas AlCl3 exists as a dimer? What makes dimerization possible in $\ce{AlCl3}$? Are there 3c-2e bonds in $\ce{Al2Cl6}$ as there are in $\ce{B2H6}$?

Introduction

The bonding situation in $\ce{(AlCl3)2}$ and $\ce{(BCl3)2}$ is nothing trivial and the reason why aluminium chloride forms dimers, while boron trichloride does not, cannot only be attributed to size.

In order to understand this phenomenon we need to look at both, the monomers and the dimers, and compare them to each other.
Understanding the respective bonding situation of the monomers, is key to understand which deficiencies lead to dimerisations.

Computational details

Since I was unable to find any compelling literature on the subject, I ran some calculations of my own. I used the DF-M06L/def2-TZVPP for geometry optimisations. Each structure has been optimised to a local minimum in their respective symmetry restrictions, i.e. $D_\mathrm{3h}$ for the monomers and $C_\mathrm{2v}$ for the dimers.
Analyses with the Natural Bond Orbital model (NBO6 program) and the Quantum Theory of Atoms in Molecules (QTAIM, MultiWFN) have been run on single point energy calculations at the M06/def2-QZVPP//DF-M06-L/def2-TZVPP level of theory.
A rudimentary energy decomposition analysis has been done on that level, too.

Energy decomposition analysis

The dissociation energy of the dimers $\ce{(XY3)2}$ to the monomers $\ce{XY3}$ is defined as the difference of the energy of the dimer $E_\mathrm{opt}[\ce{(XY3)2}]$ and double the energy of the monomer $E_\mathrm{opt}[\ce{XY3}]$ at their optimised (relaxed) geometries $\eqref{e-diss-def}$.
The interaction energy is defined as the difference of energy of the relaxed dimer and double the energy of the monomers in the geometry of the dimer $E_\mathrm{frag}[\ce{(XY3)^{\neq}}]$ $\eqref{e-int-def}$. That basically means breaking the molecule in two parts, but keeping these fragments in the same geometry.
The deformation energy (or preparation energy) is defined as the difference of the energy of the optimised and the non-optimised monomer $\eqref{e-def-def}$. This is the energy required to distort the monomer (in its ground state) to the configuration it will have in the dimer.
$$\begin{align}
E_\mathrm{diss} &=
  E_\mathrm{opt}[\ce{(XY3)2}] - 2E_\mathrm{opt}[\ce{XY3}]
  \tag1\label{e-diss-def}\\
E_\mathrm{int} &=
  E_\mathrm{opt}[\ce{(XY3)2}] - 2E_\mathrm{frag}[\ce{(XY3)^{\neq}}]
  %\ddag not implemented
  \tag2\label{e-int-def}\\
E_\mathrm{def} &=
  E_\mathrm{frag}[\ce{(XY3)^{\neq}}] - E_\mathrm{opt}[\ce{XY3}]
  \tag3\label{e-def-def}\\
E_\mathrm{diss} &=
  E_\mathrm{int} + 2E_\mathrm{def}\tag{1'}
\end{align}$$

Results & Discussion

The Monomers $\ce{XCl3; X{=}\{B,Al\}}$.

Let's just get the obvious out of the way: Boron is (vdW-radius 205 pm) smaller than aluminium (vdW-radius 240 pm). For comparison chlorine has a vdW-radius of 205 pm, too. That is pretty much reflected in the bond lengths and the chlorine-chlorine distance.
\begin{array}{llrrr}\hline
&\ce{X{=}}& \ce{Al} &\ce{B} &\ce{Cl}\\\hline
\mathbf{d}(\ce{X-Cl})&/\pu{pm} & 206.0 &173.6&--\\
\mathbf{d}(\ce{Cl\bond{~}Cl'})&/\pu{pm} & 356.8 & 300.6 & --\\\hline
\mathbf{r}_\mathrm{vdW}&/\pu{pm} & 240   & 205   & 205\\
\mathbf{r}_\mathrm{sing}&/\pu{pm} & 126   & 85   & 99\\
\mathbf{r}_\mathrm{doub}&/\pu{pm} & 113   & 78   & 95\\\hline
\end{array}

From this data we can draw certain conclusions without further looking. The boron monomer is much more compact than the aluminium monomer. When we compare the bond lengths to the covalent radii (Pyykkö and Atsumi) we find that the boron chloride bond is about the length that we would expect from a double bond ($\mathbf{r}_\mathrm{doub}(\ce{B}) + \mathbf{r}_\mathrm{doub}(\ce{Cl}) = 173~\pu{pm}$). While the aluminium chloride bond is still significantly shorter than a single bond ($\mathbf{r}_\mathrm{sing}(\ce{Al}) + \mathbf{r}_\mathrm{sing}(\ce{Cl}) = 225~\pu{pm}$), it is still also much longer than a double bond ($\mathbf{r}_\mathrm{doub}(\ce{Al}) + \mathbf{r}_\mathrm{doub}(\ce{Cl}) = 191~\pu{pm}$).

This itself offers compelling evidence, that there is more π-backbonding in $\ce{BCl3}$ than in $\ce{AlCl3}$. Molecular orbital theory offers more evidence for this. In both compounds is a doubly occupied π orbital. The following pictures are for a contour value of 0.05; aluminium (left/top) and boron (right/bottom)

In numbers, the main contributions are as follows (this is just a representation, not the actual formula):
$$\begin{align}
\pi(\ce{BCl3}) &=
   21\%~\ce{p_{$z$}-B} + \sum_{i=1}^3 26\%~\ce{p_{$z$}-Cl^{$(i)$}}\\
\pi(\ce{AlCl3}) &=
   13\%~\ce{p_{$z$}-Al} + \sum_{i=1}^3 29\%~\ce{p_{$z$}-Cl^{$(i)$}}
\end{align}$$

There is still some more evidence. The natural atomic charges (NPA of NBO6) fairly well agree with that assesment; aluminium is far more positive than boron.
$$\begin{array}{lrr}
                    & \ce{AlCl3} & \ce{BCl3}\\\hline
\mathbf{q}(\ce{X})~\text{[NPA]}  & +1.4       & +0.3     \\
\mathbf{q}(\ce{Cl})~\text{[NPA]} & -0.5       & -0.1     \\\hline
%\mathbf{q}(\ce{X})~\text{[QTAIM]}  & +2.4       & +2.0    \\
%\mathbf{q}(\ce{Cl})~\text{[QTAIM]} & -0.8       & -0.7     \\\hline
\end{array}$$

The analysis in terms of QTAIM also shows that the bonds in $\ce{AlCl3}$ they are predominantly ionic (left/top) while in $\ce{BCl3}$ are predominantly covalent (right/bottom).

One final thought on the bonding can be supplied with a natural resonance theory analysis (NBO6). I have chosen the following starting configurations and let the program calculate their contribution.

The overall structures in terms of resonance are the same for both cases, that is if you force resonance treatment of the aluminium monomer. Structure A does not contribute, while the others contribute to about 31%. However, when not forced into resonance, structure A is the best approximation of the bonding situation for $\ce{AlCl3}$. In the case of $\ce{BCl3}$ the algorithm finds a hyperbond between the chlorine atoms, a strongly delocalised bond between multiple centres. In this case these are 3-centre-4-electron bonds between the chlorine atoms, resulting from the higher lying degenerated π orbitals.

This all is quite good evidence that the monomer of boron chloride should be more stable towards dimerisation than the monomer of aluminium.

The Dimers $\ce{(XCl3)2; X{=}\{B,Al\}}$.

The obvious change is that the co-ordination of the central elements goes from trigonal planar to distorted tertrahedral. A look at the geometries will give us something to talk about.

\begin{array}{llrrr}\hline
&\ce{X{=}}& \ce{Al} &\ce{B} &\ce{Cl}\\\hline
\mathbf{d}(\ce{X-Cl})&/\pu{pm} & 206.7 &175.9&--\\
\mathbf{d}(\ce{X-{\mu}Cl})&/\pu{pm} & 226.1 &198.7&--\\
\mathbf{d}(\ce{Cl\bond{~}{\mu}Cl})&/\pu{pm} & 354.1 & 308.0 & --\\
\mathbf{d}(\ce{{\mu}Cl\bond{~}{\mu}Cl'})&/\pu{pm} & 323.6 & 287.3 & --\\
\mathbf{d}(\ce{B\bond{~}B'})&/\pu{pm} & 315.7 & 274.7 & --\\\hline
\mathbf{r}_\mathrm{vdW}&/\pu{pm} & 240   & 205   & 205\\
\mathbf{r}_\mathrm{sing}&/\pu{pm} & 126   & 85   & 99\\
\mathbf{r}_\mathrm{doub}&/\pu{pm} & 113   & 78   & 95\\\hline
\end{array}

In principle nothing much changes other than the expected elongation of the bonds that are now bridging. In case of aluminium the stretch is just below 10% and for boron it is slightly above 14%, having a bit more impact. In the boron dimer also the terminal bonds are slightly (> +1%) affected, while for aluminium there is almost no change.

The charges are not really a reliable tool, especially when they are that close to zero as they are for boron. In both cases one can see that charge density is transferred from the bridging chlorine to the central $\ce{X}$.

$$\begin{array}{lrr}
                    & \ce{(AlCl3)2} & \ce{(BCl3)2}\\\hline
\mathbf{q}(\ce{X})~\text{[NPA]}  & +1.3       & +0.2     \\
\mathbf{q}(\ce{Cl})~\text{[NPA]} & -0.5       & -0.1     \\\hline
\mathbf{q}(\ce{{\mu}Cl})~\text{[NPA]} & -0.4       & +0.1     \\\hline
\end{array}$$

A look at the central four-membered-ring of in terms of QTAIM offers that the overall bonding does not change. In aluminium they get a little more ionic, while in boron they stay largely covalent.

The NBO analysis offers a maybe quite surprising result. There are no hyperbonds in any of the dimers. While a description in these terms is certainly possible, after all it is just an interpretation tool, it is completely unnecessary. So after all we have two kinds of bonds in the dimers four terminal $\ce{X-Cl}$ and four bridging $\ce{X-{\mu}Cl}$ bonds. Therefore the most accurate description is with formal charges (also the simplest). The notation with the arrows is not wrong, but it does not represent the fact that the bonds are equal for symmetry reasons alone.

To make this straight: There are no hyperbonds in $\ce{(XCl3)2; X{=}\{B,Al\}}$; this includes three-centre-two-electron bonds, and three-centre-four-electron bonds. And deeper insight to those will be offered on another day.

The differentiation between a dative bond and some other for of bond does not make sense, as the bonds are equal and only introduced by a deficiency of the used description model.

A natural resonance theory for $\ce{(BCl3)2}$ gives us a overall contribution of the main (all single bonds) structure of 46%; while all other structure do contribute, there are too many and their contribution is too little (> 5%). I did not run this analysis for the aluminium case as I did not expect any more insight and I did not want to waste calculation time.

Dimerisation - yes or no

The energies offer us a clear trend. Aluminium likes to dimerise, boron not. However, there are still some things to discuss. I am going to argue for the reaction
$$\begin{align}
\ce{2XCl3 &-> (XCl3)2}&
\Delta E-\mathrm{diss}/E_\mathrm{o}/H/G&.
\end{align};$$
therefore if reaction energies are negative the dimerisation is favoured.

The following table includes all calculated energies, including the energy decomposition analysis mentioned at the beginning. All energies are given in $\pu{kJ mol^-1}$.
\begin{array}{lrcrcrcrr}
 \Delta & E_\mathrm{diss} &(& E_\mathrm{int} &+2\times& E_\mathrm{def}&)& E_\mathrm{o} &H &G\\\hline
\ce{Al} & -113.5 &(& -224.2 &+2\times& 55.4&)& -114.7 & -60.4 & -230.4\\
\ce{B}  &   76.4 &(& -111.2 &+2\times& 93.8&)&   82.6 & -47.1 &  152.5\\\hline
\end{array}

The result is fairly obvious at first. The association for aluminium is strongly exergonic, while for boron it is strongly endergonic. While both reactions should be exothermic, stronger for aluminium, the trend for the observed electronic energies ($E_\mathrm{o}$ including the zero-point energy correction) and the (electronic) dissociation energies reflect the overall trend for the Gibbs enthalpies.

While it is fairly surprising how strongly entropy favours association of $\ce{AlCl3}$, it is also surprising how it strongly disfavours it for $\ce{BCl3}$.
A look at the decomposed electronic energy offers great insight into the reasons why one dimer is stable and the other not (at room temperature).
The interaction energy of the fragments is double for aluminium than it is for boron. This can be traced back to the very large difference in the atomic partial charges. One could expect that the electrostatic energy is a lot more attractive for aluminium than it is for boron.
The deformation energy on the other hand clearly reflects the changes in the geometry discussed above. For aluminium there is a smaller penalty resulting from the elongation of the $\ce{Al-Cl}$ bond and pyramidalisation. For boron on the other hand this has a 1.5 times larger effect. The distortion also weakens the π-backbonding, which the additional bonding would need to compensate.
The four-membered-ring is certainly not an ideal geometry and the bridging chlorine atoms come dangerously close.

Conclusion, Summary and TL;DR:

The distortion of the geometry of the monomer $\ce{BCl3}$ cannot be compensated by the additional bonding between the two fragments. Therefore the monomers are more stable than the dimer. Additionally entropy considerations at room temperature favour the monomer, too.

On the other hand, the distortion of the molecular geometry in $\ce{AlCl3}$ is less severe. The gain in interaction energy of the two fragments well overcompensates for the change. Entropy also favours the dimerisation.

While size of the central atom is certainly the distinguishing factor, its impact is only severe on  the electronic structure. Steric crowding would not be a problem when the interaction energy would compensate for that. This is quite evident because $\ce{BCl3}$ is still a very good Lewis acid and forms stable compounds with much larger moieties than itself.

References

The used van der Waals radii were taken from S. S. Batsanov Inorg. Mat. 2001, 37 (9), 871-885. And the covalent radii have been taken from P. Pyykkö and M. Atsumi Chem. Eur. J. 2009, 15, 12770-12779.

Computations have been carried out using Gaussian 09 rev D.01 with NBO 6.0. Additional analyses have been performed with MultiWFN 3.3.8. Orbital pictures were generated with the incredible ChemCraft.


Propose a synthesis of indane from benzene The question is to propose a synthesis of indane from benzene. I started by the following. The black box is what I am not sure about. I know that the stable carbocation should be the middle one, this is why I am not sure. I need help here.

To prepare indane (6) from benzene, make the procedure as simple as possible and amenable to scale-up. Propiophenone (2) is readily available by Friedel-Crafts acylation of benzene with propionyl chloride (1). Removal of the carbonyl group of propiophenone and the creation of a carboxyl moiety can be achieved via the Willgerodt reaction to produce amide 3. Vigorous  acid or base cleavage of the amide is possible but nitrous acid affords a mild route to carboxylic acid 4. One can use the AlCl3 F-C route again but hot polyphosphoric acid is a clean method for the intramolecular acylation without having to form the acid chloride. Finally, Clemmensen reduction of indanone (5) gives indane (6). Wolff-Kishner reduction is also a good option for the final step.


Chemistry underlying blacks spots in banana bread I've noticed that when I bake banana bread with certain brands of self-raising flour, the finished product looks more or less like any other cake.

However, when I use flour with sodium bicarbonate as the leavening agent, black spots appear peppered throughout the cooked cake.

What is the chemistry behind this phenomenon?

There are, at least, two reactions taking place and causing the brown colour of your bread. One is the oxidation of monophenols by an enzyme called PPO and the other is Maillard reaction, which is the same reaction that makes your steak looks deliciously brown.

Reaction 1. Oxidation of monophenols.

Bananas have a lot of polyphenol oxidase (also known as phenolase or PPO), an enzyme that catalyzes the oxidation of polyphenols naturally present in the banana fruit. The product is a colored ortho-quinone, which is brown or dark reddish. When you smash bananas to make a banana bread you end up releasing a lot of PPO which will catalyze the polymerization of the phenols and, therefore, make your bred looks darker then others. Below is a representation of the oxidation of monophenols into the colored ortho-quinone.

Reaction 2. Maillard reaction

The Maillard reaction is one of the most important reactions in food chemistry. Its products can add sensory features to many preparations, from steaks to coffee. Basically, sugars from the ingredients react with an aminoacid's amine group to form a glycosylamine. This compound is then isomerized to a ketosamine, which breaks into smaller substances. These substances are the Maillard products, and one class is called Melanoidins, responsible for the dark spots in your bread.
The reason why it is formed when you add sodium bicarbonate to the mixture is because alkaline media speed up the fragmentation of the ketosamine, yielding more Maillard products. 2

I have included an infographic from CompoundChem with other information and the Maillard reaction in chemical notation.

2Owusu-Apenten R., Introduction to Food Chemistry, CRC Press, 2004


On the unit of standard reduction potential and its independence of elements involved Why is the standard reduction potential in units of volts (Joules/coulomb)? What does this tendency of a material to reduce has to do with the charge in coulombs?

Also, why this number doesn't depend on which elements the reducing material is interacting with? (which material gets oxidized)

A an electric potential (Volt) is equivalent to the unit of J/C. This is because if you move a charge through a potential you gain energy. Volt is kind of a mysterious unit and by the time people reach it the sub-units of J and C are really well understood so it often helps to make sense out of potential. In the same way you can understand speed by thinking of it as distance/time.

The standard reduction potential can be thought of as the potential produced if the reaction happens (if positive) or required to make it happen (if negative). It is determined usually by measuring the total potential (of both reactions) with one half reaction being the one you are interested in and the other which has a zero potential so the standard reduction you are looking for is then just the net potential.

When you have two of those reactions which are both non-zero, the net potential does depend on both.

There is an excellent article on it at CLT :

Standard Reduction Potential


Why might PM2.5 concentration increase or stay the same when SO2 is reduced? The question is in the title. I've made some research, but in every paper I found the correlation between $\ce{SO2}$ and $\mathrm{PM_{2.5}}$ is described as positive. Thus if $\ce{SO2}$ is reduced, $\mathrm{PM_{2.5}}$ should also be reduced.

Your research is correct: increasing $\ce{SO2}$ concentration increases $\mathrm{PM_{2.5}}$ concentration.  Gaseous $\ce{SO2}$ is photochemically oxidized to $\ce{SO3}$ which becomes sulfuric acid on contact with water vapor. This is the process for producing the atmosphere's finest aerosols, as they grow from aggregates of a few molecules.

Your final sentence is also correct: decreasing $\ce{SO2}$ concentration decreases $\mathrm{PM_{2.5}}$ concentration. The logic here is simply the reverse of that for the first statement (that increasing $\ce{SO2}$ concentration increases $\mathrm{PM_{2.5}}$).

Although the overwhelming source of atmospheric aerosols is salt particles from dried ocean spray, and there are other sources like dust storms, these produce much larger particles than those produced by the oxidation of $\ce{SO2}$. So, in a pristine ocean environment far from natural or anthropogenic $\ce{SO2}$ sources for example, you may have more larger aerosols than would be found nearer to $\ce{SO2}$ sources, so that the total particulate matter would be greater in such a region of low $\ce{SO2}$. But, the $\mathrm{PM_{2.5}}$ portion would be very low such that $\mathrm{PM_{2.5}}$ would still be decreased in this region of low $\ce{SO2}$ emission.

For the reasons stated above, the answer to your title question is that there is no significant process by which $\mathrm{PM_{2.5}}$ would increase or stay the same when $\ce{SO2}$ concentrations are reduced.


How many of different types of chemical species are in a certain volume of aqueous solution of an acid The acid in question has $\mathrm{p}K_\mathrm{a}$ about 1.29. Density of 98% w/w solution is $d=\pu{1.323 g/cm^3}$. I would like to find the number of undissociated acid molecules ($\ce{HX}$), water molecules ($\ce{H2O}$), dissociated acid molecules ($\ce{X-}$), and hydronium ions ($\ce{H3O+}$).
The molar mass of the acid ($\ce{HX}$) is $\pu{125.1 g/mol}$. Thus 98% w/w solution corresponds to $\pu{10.57 M (mol/L)}$. If I'm not missing sth, the solution contains 1.5 mole of water and 10.57 mole of acid. What I'm not sure is how these numbers convert into the number of ions/molecules which are, let's say in a volume of $1~\pu{\mu L}$ (one microliter).
Though my attempt would be:

Firstly I have to determine $\mathrm{pH}$ of the 98% (w/w) solution in question; I did some rough calculations and got $\mathrm{pH}$ = 0.509

Then, I need to establish the ratio $\dfrac{[\ce{X-}]}{[\ce{HX}]}$; this can be done using Henderson-Hasselbach equation; I got 0.166

Thus, for let's say 1000 $\ce{HX}$ molecules and 172 $\ce{H2O}$ molecules (this stems from 10.57:1.5 molar ratio), I have at equilibrium: 834 $\ce{HX}$, 166 $\ce{X-}$, 166 $\ce{H3O+}$ and 6 $\ce{H2O}$ molecules.

The solution, seems to be correct. 834 $\ce{HX}$, 166 $\ce{X-}$, 166 $\ce{H3O+}$ and 6 $\ce{H2O}$ molecules should have a total mass of $2.129\times10^{-19} \pu{g}$. Since the density is known, such a group of molecules should occupy a volume of $\pu{160.9 nm^3}$ which is perfectly reasonable.

The following is an approximate calculation, since among other things it makes the approximation that the activities of each substance are equal to their molarities, which is not true in such concentrated solutions. Also I take some shortcuts to make the calculation quicker.

Let's look at a volume of $1\ L$. The mass would be $1323\ g$, 98% of which ($1296.54\ g$) is due to the acid, and 2% ($26.46\ g$) due to water. This initially translates into $10.36\ mol$ of acid, and $1.47\ mol$ of water. However, some of the water and acid are consumed to create hydronium and conjugate base.

To calculate the dissociation of the acid, you may want to apply $K_{a}=\frac{[H_{3}O^{+}] [A^{-}]}{[HA]}$ (the Henderson-Hasselbach equation is derived from this by taking the log on both sides, so they're equivalent), but this is actually a very bad approximation, since you need to take into account the concentration of water in non-dilute solutions. We'll do it anyway and I worked it out to be $[H_{3}O^{+}]=[A^-]=0.704\ mol.L^{-1}$. Since the volume we're working with is $1\ L$, that means $n_{H_{3}O^{+}} = n_{A^-} = 0.704\ mol$. Subtracting those amounts from the initial amounts of water and acid, we reach $$ n_{H_{2}O}=0.77\ mol$$ $$ n_{HA}= 9.66\ mol$$ $$n_{H_{3}O^{+}}=n_{A^-}=0.704\ mol$$
To find the actual number of each entity, just multiply by the Avogadro number $6.022\times 10^{23}$ and you have the amount of each in a litre of solution.


Would it be possible to destroy gold? I'm a writer. I have a scenario in which a sizable amount of gold needs to be rendered unusable, preferably completely destroyed. I know an acid like aqua regia is able to dissolve gold, but would there be a way to make it impossible (or at least not worth the cost) to chemically salvage the gold from the acid, or another way to permanently destroy gold?

Other than a nuclear reactor, the only other chance is to dump it into a volcano. Having a much higher density than magma, it will just fall through until it hits earth's mantle. Then it's really gone. OK, dissolved and diluted in the Atlantic is also quite safe.

P.S. Warning: If the bullion gets stuck in the volcano, which later explodes Krakatau-style, your gold brick might land in someone's front lawn. Making a hole five meters deep, but it may be salvaged easily.


What are some of the canonical texts for polymers? Is materials science more related to polymers than chemical engineering? I am a freshman in college. My major is chemical engineering. I would like some recommendations for a textbook that's an introduction to polymers.

I am wondering whether a major in materials science is more related to polymers than chemical engineering.

I don't know of any undergraduate level texts that would be suitable for a freshman, unless you have already taken physics, statistics, and chemistry up through organic chemistry.

I can tell you that Fundamentals of Polymer Science by Painter and Coleman is a good introductory text, but I used it in the final year of my Chemical Engineering program. There could be better resources as well - that's just the only one I am personally familiar with.

It also depends a lot on which specific aspects of polymers you are interested in. This actually matters a lot for your second question as well. The three main branches of polymers studies are:

Polymer chemistry
Polymer physics
Polymeric materials

There are sub-fields in each of these, and a lot of overlap, but essentially they focus on:

Polymer chemistry - How are polymers made? (this would include engineering and production of polymers)
Polymer physics - How does the behavior on a molecular level affect the macroscopic properties?
Polymeric materials - How can we use the unique material properties of polymers to make useful stuff?

At the undergraduate level, you won't get too specific on any one of these areas, no matter which degree you choose. If you majored in chemistry, you would learn a little more about polymerization reactions. If you majored in physics (and took electives related to polymers) you might learn more about the statistical mechanics of polymers. If you majored in materials science and/or engineering, you would learn more about how we use the polymers, and less about how they are made and why they act the way they do. Chemical engineers learn a little bit of everything, but they don't get too deep into any one particular branch.

The real differences show up if you go on to graduate school or work in a particular role in industry. There, you will definitely focus more, and your degree can make more of a difference. So I would say, for your second question, major in whatever aspect interests you more - if you want to know the details about how polymers are made in a lab, then maybe chemistry is a better option. If you really like math and physics and want to understand how they relate to the behavior of polymers, then physics is good. If you are more interested in how to characterize and use polymers, then materials science/engineering is a good bet. And if you like a little bit of everything, or are interested in industrial-scale production of polymers, then chemical engineering is a good choice.

Keep in mind as well: it is hard to predict now what you will really want to do in four or more years - so if you make a choice based on which aspects of polymer science you are interested in, and then later decide that polymers are actually not what you really like, you will still be studying the "right" stuff for you.


Why does addition polymerisation of high density poly(ethene) use low temperature and pressure? Logically thinking, I thought that it would be high pressure and high temperature that would create high density poly(ethene)... However, that's not the case. Is it because of the catalyst? Why is the condition low pressure/temperature?

In fact, when polymerizing ethylene using a radical initiator at high temperature and high pressure, we favor the formation of low-density polyethylene. This is because the radical at the end of the growing chain can be transferred to any carbon atom of the chain by radical rearrangement (favored at high temperature) and due to close distances among molecules (high pressure) a branch chain can grow on the principal chain, and we get a branched polyethylene.  The chains of a branched polymer cannot align together, and eventually the density of the resultant polymer is low. We have got LDPE.

On the other hand, using a specific catalyst, known as Ziegler-Natta catalyst and under mild conditions of temperature and pressure, we get linear chains of polyethylene that can be aligned together and the density of the resultant polymer is high. We have got HDPE.