Comparison of Resonance Energies of some Carbocations We were given the following carbocations and asked to compare their resonance energies.

$\ce{L-CH2+}$, where $\ce{L}$ is:

$\ce{NH2}$
$\ce{OCH3}$
$\ce{Ph}$
$\ce{F}$
$\ce{Cl}$
$\ce{NO2}$

and the order was given as:

$1 > 2 > 3 > 5 > 4 > 6$

I am clear with the other orders but I am unsure about $\ce{F}$ and $\ce{Cl}$. Why does $\ce{Cl-CH2+}$ have a higher Resonance Energy than its $\ce{F-CH2+}$ counterpart even though in resonance, $\ce{Cl}$ forms $\ce{{3p}-{2p}}\ \pi$ bonds compared to $\ce{{2p}-{2p}}$ ones in $\ce{F-CH2+}$, which are less stable due to less efficient overlapping? $\ce{F-CH2+}$ does have a higher electronegativity, but is it enough to overcome the higher stability of the $\ce{{2p}-{2p}}$ bond over the $\ce{{3p}-{2p}}$ one?

It is not like π bond between fluorine and carbon is very stable and that is where your line of thinking is wrong. While it is true that $\ce{{2p}-{2p}}$ π bonds are more stable in general than their $\ce{{2p}-{3p}}$ counterparts, one should not take that alone as a decisive factor.

The question is comparing the electron-donating capabilities of fluorine and chlorine, and that italicised word there alone should ring a bell. Fluorine is much more electronegative than chlorine which means that a carbon bound to fluorine has a much smaller electron density which destabilises cations much more than could be offset by the overall weak π donating capacity of fluorine.

Remember that for electrophilic aromatic substitutions, halide substituents are classified as deactivating — mainly due to their $-I$ effect — and only weakly ortho/para-directing — due to the smaller $+M$ effect. This should serve to show which effect predominates and it only gets more pronounced if instead of a large aromatic system you only consider the atom directly adjacent.


Is there a good sol gel method for carbon nanotube thin film deposition? I was searching for a sol gel method for deposition of carbon nanotube thin films and there aren't too many articles on that.  Does anybody know the reason?  Is it because Carbon nanotubes are difficult to disperse in many solvents or?  Does anyone know if it is possible to effectively use a sol gel method to make a CNT thin film?

I think it may depend on what you mean by "effectively." I imagine that it would be difficult to get a thin film of oriented CNTs using a sol-gel method, since the gel would necessarily have randomly oriented tubes.

The other potential issue is making the sol-gel in the first place. Unless they are functionalized, carbon nanotubes won't gel in hydrophilic solvents. It might be possible to create gels in hydrophobic solvents, but a quick search didn't turn anything like that up.

This link has a brief description of a typical process - you can see that the idea is to make the sol-gel with something that will gel in aqueous solution first (silica, for example), and then add the nanotubes.

So, if you want pure CNT thin films, that would probably be a lot more difficult.


Heat of combustion among substituted cyclohexanes Arrange in ascending order of heat of combustion:

I know heat of combustion is inversely proportional to stability. Also, to solve I did try making chair conformations of the substituted cyclohexane.
In same type of, say, 1,3,5-conformations one can do comparisons by seeing equatorial axial positions, as bulky group tend to be in equatorial positions. But how to compare and arrange when there is 1,2,3 and 1,2,4 and 1,3,5 substitutions?

Ashish: Welcome to SE! You are quite correct that heats of combustion and formation are inversely related. Since all of your compounds are trimethylcyclohexanes (C9H18), their sum will be equal to the heat of combustion of 9 moles each of graphite and hydrogen. To approach the solution of this problem, we can start with methylcyclohexane and evaluate the effect of adding ultimately two more methyl groups. In the somewhat busy diagram below, all numbers are in kcals/mol. Known heats of formation (NIST) are located inside the ring. Calculated heats of formation (ΔHfo) are in red and values over the arrows are the incremental changes in heats of formation (ΔΔHfo). The more stable conformation of all compounds is assumed. Equatorial (e) and axial (a) assignments are made.
There is no methyl/methyl interaction in the formation of 1,4-isomers A (cis) and B (trans) or 1,3-isomers E (cis) and F (trans). The addition of an equatorial methyl (B and F) is worth -7.0 and -7.1 kcal/mol, respectively, while addition of an axial methyl (A and E) has a value of -5.2 kcal/mol. Note that the difference between an axial and equatorial methyl is worth 1.9 kcal mol. The axial methyl necessarily makes the ΔHfo value less negative than addition of an equatorial methyl. However, there are gauche butane interactions in 1,2-isomers. Because C is less stable than D, the addition G --> C (cis) is -4.1 kcal/mol and G --> D equals -6.0 kcal/mol.
Using a value of -7.0 kcal/mol for the addition of an equatorial methyl, the conversion G --> R gives a value of ΔHfo = -51.0 kcal/mol [NIST = -51.48 kcal/mol]. Note that 1,3,5-isomer U's value is what is expected given the difference between a non-interacting equatorial vs. axial methyl group.
The reader should be able to corroborate the values assigned to P, Q, S and T. The stability (heats of formation) of P through T are R>S>T>Q>P. Therefore, the heats of combustion are in the order R<S<T<Q<P.


Preservation and bleaching of sliced mushrooms with hydrogen peroxide According to the article Extending the Shelf Life of Fresh Sliced Mushrooms, it states that 5% (v/v) hydrogen peroxide is an effective means for the preservation of sliced mushrooms.

In regards to the preservation:

  In the USA hydrogen peroxide
  is an approved bactericide for some dairy products and is used for disinfecting fruit
  and vegetables. Its activity is due to its oxidising effects on
  bacteria and it also bleaches mushrooms during the soaking period.

How does oxidising effects cause the bleaching, hence preservation of sliced mushrooms?

Hydrogen peroxide is quite a strong oxidizer. First, let's address the bleaching effect: It bleaches by oxidizing colored compounds, making them colorless. From Wikipedia:

  An oxidizing bleach works by breaking the chemical bonds that make up the chromophore. This changes the molecule into a different substance that either does not contain a chromophore, or contains a chromophore that does not absorb visible light. This is the mechanism of bleaches based on chlorine.

This is also the mechanism of hydrogen peroxide as a bleach, since both chlorine-based bleach and hydrogen peroxide are oxidizers.

As for the antibacterial effect: hydrogen peroxide in this particular concentration (5%) may or may not be effective against bacteria (and viruses, fungi, etc.) due to the presence of the enzyme catalase in some bacteria and fungi. In those organisms which it is effective against, hydrogen peroxide would probably work by causing oxidative stress in the organism.


Why do alpha-hydroxy ketones respond positively to Tollens', Fehling's, Benedict's, Schiff's and HgCl2 tests? I learnt from a certain textbook that aldehydes and formic acid respond positively to Tollens', Fehling's, Benedict's, Schiff's and HgCl2 tests.

However, in another textbook I read that even alpha-hydroxy ketones respond positively to these tests. What is the reason behind this? I thought ketones are not oxidizable (unless under drastic conditions).

Fructose is an alpha-hydroxy ketone, but all alpha-hydroxy ketones are not fructose. In fructose there is a terminal hydroxy group which facilitates Tollens' test, etc. in basic medium. But why do alpha-hydroxy ketones like benzoin give these tests even though there is no terminal hydroxy group? That part is still unclear to me. Could you please answer that part?

Its because of two tautomerizations happening, the ketone becomes en-ol, and then another tautomerization happens with the original hydroxy group (which is terminal), making it an aldehyde group. All three forms exist in equilibrium, so it responds to Tollens' test.


How does expansion of electron shells work? How does it work? I've read that its about the similarity of the energy levels of $3d$ and $3p$, but shouldn't the electron shell $4s$ be filled before $3d$ if filled? So what is done with the $3d$?
I'd also like to know in what cases will atoms of elements expand their valence electrons? Which blocks or groups or periods do this?

You're right! The 4s subshell is filled before 3d. Electrons fill the shell   with least energy first and the 4s shell has a lower energy than the 3d subshell.
Use the Aufbau Principle to find the shell with least energy.

Basically it says that shells with a lower n+l (n and l are the principal and azimuthal quantum numbers) value are filled first.
If two shells have the same n+l value, then the shell with lower n is filled first.
There are a few exceptions to this rule - Copper, Chromium, Palladium, etc.

Reason: n+l gives the number of nodes in an orbital. The more the number of nodes, the less stable the orbital and so the higher the energy.
Valence shells are filled in the s and p blocks. The inner shells are filled in the d and f blocks.


How to calculate concentration of unknown saturated stock solution from diluted solution whose concentration is known? I want to measure the solubility of a compound. I prepared a saturated stock solution of my compound in 100ml of water by dissolving it in excess and filtering out the undissolved sample.

Then I took 1ml of saturated solution, diluted it to 10ml and measured the absorbance on a UV spectrophotometer. After that I calculated the concentration of diluted solution by plotting a calibration curve of various solutions of known concentrations.

Now how can I calculate the concentration of my saturated stock solution from diluted one?

The concentration of diluted solution is 17ppm/10ml.

Unless I misunderstood (please give more info), you've already done the most difficult part of the job by determining the concentration of your diluted solution!
And you just diluted your stock solution ten times, so you'd only have to multiply by ten the value that you found for your diluted concentration...


Auto-catalysis Mechanism I just read about the theory of auto-catalysis, and here's one thing which is sort of unclear to me:

  In this type of catalysis, one of the reaction product catalyses the reaction. For example, in the oxidation of oxalic acid by potassium permanganate, Mn(II) cation is formed which is known to accelerate the reaction. So when potassium permanganate is added to a warm solution of oxalic acid in presence of dilute sulphuric acid (acid medium catalysis), initially there is a time lag before decolourisation occurs. As more permanganate is added, the decolourisation is observed to be nearly instantaneous.

Points I didn't understand:

How does the formation of Mn(II) cation accelerate the reaction? Could someone please help me with understanding the mechanism?
What is the reason for time lag in decolourisation of the solution? Also, what exactly do they mean by decolourisation here? My guess is that it's referring to the bright purple coloured permanganate solution losing its colour after getting reduced to Mn(II) cation.
How does the time lag reduce? Are there ways to estimate or determine the same? Of course, it'd depend on the reaction kinetics; but I'm looking for a mathematical viewpoint on how the catalyst influences reaction kinetics, except for decreasing the activation energy.

Thank you.

(1) There are several web sites giving details of this reaction and so this need not be copied out here.

(2,3) Rather than describe a complex reaction scheme it is easier to understand a generic autocatalytic reaction. The general autocatalytic reaction of species A with catalyst B has the form $\ce{A + B $\to$ P + 2B}$ and the rate equation is deceptively simple and is $da/dt =- kab$ if $a$ and $b$ are the concentration A and B. The important point here is that species B is both reactant and product. As more B is produced than that there initially, the reaction accelerates as it proceeds. This is the reason for the time lag: the reaction rate is initially small but as reaction proceeds the extra B produced speeds it up. Eventually A runs out and so the reaction stops. (The colour change you mention is just a measure of the concentration of permanganate. Note also that from the Beer-Lambert law describing transmission of light the absorbing species concentration is related exponentially to light transmission (and so absorption) and this probably complicates what you observe by eye).

If $x$ is species A concentration, the rate equation is, $\displaystyle \frac{dx}{dt}=-kxb=-kx(a_0+b_0-x)$ or $\displaystyle \int\frac{dx}{(a_0+b_0-x)x}=-kt+c$ which can be integrated using partial fractions;

$$\frac{1}{(a_0+b_0-x)x}=\frac{1}{(a_0+b_0)}\left[\frac{1}{x}+\frac{1}{a_0+b_0-x} \right]$$

which produces two log functions when integrated. The initial conditions are $x = a_0$ when $t$ = 0 then the integration constant is $\ln(a_0/b_0)$ and the rate equation,

$$ \ln \left(\left| \frac{xb_0}{a_0(a_0+b_0-x)}     \right|  \right) =-(a_0+b_0)kt$$

and the absolute value is added because the log cannot be negative. Rearranging produces $\displaystyle x = a_0\frac{a_0+b_0}{a_0+b_0e^{-(a_0+b_0)kt}} $.   This and also the increase of $B =a_0+b_0-x$ are plotted.

Figure Autocatalysis $\ce{A + B $\to$ P + 2B}$. Initially species A decreases slowly as the concentration of B is small. As the reaction proceeds, more B is produced and although A is reduced overall, the rate increases and A is consumed even more rapidly. At longer times, the concentration of A becomes so small that even though that of B is large the reaction rate is slow.


How to calculate the percentage yield for a reaction with excess reagent? A chemist mixed $12~\mathrm{g}$ of phosphorus with $35.5\ \mathrm{g}$ of chlorine gas to synthesize phosphorus(III) chloride (phosphorus trichloride). The yield was $42.4~\mathrm{g}$ of $\ce{PCl3}$. The equation is
  $$\ce{2P + 3Cl2 -> 2PCl3}$$
  Calculate the percentage yield.

I think that the amount of phosphorus the chemist used is in excess. I figured that only 1/3 of the amount of phosphorus the chemist used reacted to produce phosphorus trichloride. But the question didn't mention anything about products in excess that is why I am not so sure of my answer, which is $92.51$.

First I found the amount of substance of phosphorus used which is $12/31$. Then I calculated the amount of substance of chlorine gas used (and it's $0.5$). But $12/31\ \mathrm{mol}$ of $\ce{P}$ needs $1.5 \times (12/31)\ \mathrm{mol}$ of chlorine gas, so $\ce{P}$ is in excess. The amount of substance of $\ce{P}$ that will react will be $$0.5/1.5 = 1/3$$ Now $1/3\ \mathrm{mol}$ of $\ce{P}$ produces $1/3\ \mathrm{mol}$ of $\ce{PCl3}$ and the molecular mass of $\ce{PCl3}$ is $137.5$ so the mass that is supposed to be produced is $137.5 \times 1/3$. Then I divided $42.4$ by $(137.5 \times 1/3)$ and multiplied by $100$.

The yield is typically calculated according to equation $(1)$.

$$\text{yield} = \frac{\text{actual amount of substance obtained}}{\text{theoretical amount of substance to be obtained}}\tag{1}$$

(Whether it is given in percent or as a fractional number is not important, but percentages are the usual value.) So for any reaction, you need to figure out:

the reaction equation
the amount of each reactant substance involved
the amount of desired product theoretically possible by stoichiometric coefficients
the actual amount of desired product.

The first is rather easy, especially since the equation $(2)$ has been given previously.

$$\ce{2 P + 3 Cl2 -> 2 PCl3}\tag{2}$$

Thus, for every mole of phosphorus, we expect one mole of product and for each mole of chlorine gas, we expect two-thirds of a mole of product. The amounts of the reactants are easily calculated using molar masses:

$$\begin{align}M &= \frac mn\tag{3}\\
n (\ce{P}) &= \frac{12~\mathrm{g}}{30.97~\mathrm{g/mol}}\\
&= 0.39~\mathrm{mol}\tag{4}\\[1em]
n(\ce{Cl2}) &= \frac{35.5~\mathrm{g}}{70.9~\mathrm{g/mol}}\\
&= 0.50~\mathrm{mol}\tag{5}\end{align}$$

For each of these two we would need to check the amount of product we would expect using the stoichiometric coefficients.

$$\begin{align}n_\text{theor}(\ce{PCl3}) &= \frac 23 n(\ce{Cl2})\\
&= 0.33~\mathrm{mol}\tag{6}\\[1em]
n_\text{theor}(\ce{PCl3}) &= n(\ce{P})\\
&= 0.39~\mathrm{mol}\tag{7}\end{align}$$

Since $0.33 < 0.39$, chlorine is the limiting reagent and our theoretically possible yield is calculated from the amount of chlorine added to the reaction. Therefore, we now know our denominator which is $0.33~\mathrm{mol}$. What is the numerator? Again, it is easily calculated:

$$\begin{align}n_\text{exp}(\ce{PCl3}) &= \frac{42.4~\mathrm{g}}{137.32~\mathrm{g/mol}}\\
&= 0.31~\mathrm{mol}\end{align}\tag{8}$$

And now all we need to do is calculate the yield:

$$\frac{0.31~\mathrm{mol}}{0.33~\mathrm{mol}} = 0.93 = 93~\%\tag{9}$$

This is consistent with your result.


Mechanism for chloromethylation of benzene with formaldehyde and HCl What is the mechanism of the above reaction? I have thought of one possibility:

Would this work? How exactly is the chlorine installed on the alkyl chain?

This reaction is chloromethylation, similar to Blanc chloromethylation, but using $\ce{AlCl3}$ as co-catalyst instead of $\ce{ZnCl2}$. These are reactions belonging to a group related to Friedel-Crafts reactions, but characterized by usage of protonation instead of coordination with molecular Lewis acids

Mechanism of your reaction (not accounting for influence of co-catalyst) can be written as:

(source)

$\ce{AlCl3}$ coordinates to $\ce{HCl}$ creating adduct ($\ce{HAlCl4}$) which is much stronger acid.


Mechanism of hydrolysis of acyl chlorides The following mechanism for acyl chloride hydrolysis was given by my professor :

However I do not understand why in the second step, oxygen has a partially negative charge, should it not gain a positive charge as it is bonded to three atoms, furthermore shouldn't the $\text{O}-\text{H}$ bond breaking homolytically form a $\text{H}^-$ ion and not a proton?
Should the arrow be reversed in the second step?

This reaction is mostly ignored in basic organic texts and the mechanism you give is straight out of Wikipedia with some modification. I am sure some serious study exists some where but not on the web. Please bear with me because I can't[or just don't know how to] do drawings.  The first step is probably as written and is most likely slow because it involves charge separations. You are right there is a + charge on the O of water. [btw there is no homolysis involved, that generates radicals, and there is no reasonable path to generate hydride ions] The second step seems to require the participation of a base this may be loosely described as an E2 reaction where a proton is removed from the protonated O atom by another water molecule and the developing negative charge on the other oxygen expels the chloride ion. This mechanism[if plausible] requires enough water for active solvation. This mechanism possibly explains the success of the Schotten-Bauman method where base is added to the reaction as a "catalyst" [tho it is consumed]. If water is limited the base may be the attached chlorine and references do state that dry reactions evolve HCl gas. These are my thoughts on the reaction; library research is in order.


Why doesn't KI(aq) react with HCl(aq)? When concentrated sulfuric acid is added to anhydrous potassium chloride and the fumes produced are bubbled into aqueous potassium iodide solution, the observed solution would be colourless solution.

I think the first reaction is:
$$\ce{2KCl(aq) + H2SO4(aq) -> K2SO4 + 2HCl(g)}$$

I assumed the second reaction would be:
$$\ce{HCl + KI -> KCl + \frac{1}{2}I2 + \frac{1}{2}H2 }$$

However the second reaction is wrong according to the answer book as $\ce{HCl}$ won't react with $\ce{KI}$. From what I know, $\ce{Cl}$ is a stronger oxidising agent than $\ce{I}$, so shouldn't $\ce{I-}$ in $\ce{KI}$ be oxidized to $\ce{I2}$? I am an A-levels student so would appreciate a simpler answer.

In your first reaction, you added concentrated sulfuric acid to anhydrous potassium chloride. In your equation, you wrote $\ce{KCl (aq)}$, which is incorrect. Under dilute conditions, that reaction would not take place because all reagents and products would be in aqueous ionic state, meaning there is no reaction.
In your second reaction, the equation is completely wrong. It is true that elemental $\ce{Cl2}$ a stronger oxidizing agent than elemental $\ce{Br2}$ or elemental $\ce{I2}$. But in current situation is ionic $\ce{Cl-}$ versus ionic $\ce{I-}$ where both are aqueous. If all reagents are in ionic form and expected products are also ionic, there is no reaction. See your reaction below:
$$\ce{H+(aq) + Cl-(aq) +K+(aq) + I-(aq) <=> H+(aq) + I-(aq) +K+(aq) + Cl-(aq)}$$
There is no gas, liquid or solid formed, but all ions. Therefore, there is no reaction.


How do I make a 0.5 M citrate buffer at pH 4.5? I need to make a citrate-sodium citrate buffer at 0.5 M concentration and pH 4.5. I calculated the molarities of the citrate and disodium citrate based on the Henderson Hasselbalch equation, using a pₐ of 4.76 . To cross-check my calculations, I tried a couple of online calculators, but the results are way off from mine. I think the formulae should change based on where I use monosodium, disodium or trisodum citrate. What is the right way to go about making this calculation?

There are many ways to make simple pH buffers like this one. As @Poutnik noted, preparation via titration works fine and in very commonly used. To illustrate how this is done, several possibilities are presented. Assume the desired buffer characteristics are pH = 4.5, concentration = 0.5 M, final buffer volume = 1 L, and the following reagent grade chemicals are available: citric acid, sodium dihydrogen citrate, disodium hydrogen citrate and 2 M NaOH solution. Also assume a laboratory quality pH meter is available and that the pH meter has been properly calibrated with buffers of 4.00 and 7.00, bracketing the desired pH of 4.5.  Before proceeding, it is useful to examine citric acid’s $\alpha$ dissociation fractions.

Brief refresher on $\alpha$ dissociation fractions.

Citric acid, denoted $\ce{H_3Cit}$, is a triprotic weak acid. The successively less protonated citrate species are $\ce{H_2Cit^-}$, $\ce{HCit^{2-}}$, and $\ce{Cit^{3-}}$. Since all the citrate species are in the same volume of solution, molar concentrations (denoted by the usual square brackets, e.g., $[\ce{H_3Cit}]$) are proportional to numbers of moles. Define $D$ as the sum of $[\ce{H_3Cit}] + [\ce{H_2Cit^-}] + [\ce{HCit^{2-}}] + [\ce{Cit^{3-}}]$. Then $\alpha_0$ is $[\ce{H_3Cit}]/D$, $\alpha_1$ is $[\ce{H_2Cit^-}]/D$, $\alpha_2$ is $[\ce{HCit^{2-}}]/D$, and $\alpha_3$ is $[\ce{Cit^{3-}}]/D$. The sum of the $\alpha$ values equals unity, as expected: every citrate species must be somewhere in the solution.

At any given solution pH, all four citrate species are present (though possibly at negligible concentration), as determined by the pH, the three acid dissociation equilibria and their associated equilibrium constants: $\mathrm{p}K_\mathrm{a1}$, $\mathrm{p}K_\mathrm{a2}$, and $\mathrm{p}K_\mathrm{a3}$. For citric acid, the $\mathrm{p}K_\mathrm{a}$ values are 2.13, 4.77, and 6.40. At any given solution pH, the $\alpha$ fractions are simply the proportion of total citrate that is in the relevant ionic or molecular species.

The screenshot below shows the $\alpha$ fractions up to pH = 8 (it is boring above pH = 8):

For a desired citrate buffer at pH = 4.5, the figure shows that only $\ce{H_2Cit^-}$ (green $\alpha_1$ curve) and $\ce{HCit^{2-}}$ (blue $\alpha_2$ curve) are present in significant concentrations. Furthermore, $[\ce{H_2Cit^-}]/[\ce{HCit^{2-}}] > 1$.

Method 1

Weigh out an amount of citric acid equal to approximately 0.5 moles, i.e., about 91 g. Dissolve that, with stirring, in about 0.5 L of water in a large beaker. Be careful to not drop a magnetic stir bar into the beaker!  If 375 mL of 2 M NaOH solution were to be added, the pH would be about 4.76, which is a little too high. The reason is that 0.75 moles of $\ce{OH^-}$ would neutralize the first citric acid proton and lead to a roughly equimolar ratio between $\ce{H_2Cit^-}$ and $\ce{HCit^{2-}}$. This is too much added NaOH solution because $[\ce{H_2Cit^-}]/[\ce{HCit^{2-}}] = 1.82$, as is easily determined using the Henderson-Hasselbalch equation or the $\mathrm{p}K_\mathrm{a2}$ expression.

Add 250 mL of 2 M NaOH and then, with constant stirring and monitoring of the pH, slowly add more 2 M NaOH solution until the pH increases to 4.5. Finally, the last step is to add water until the final buffer volume is 1 L.

Method 2

Start by assuming $[\ce{H_2Cit^-}]/[\ce{HCit^{2-}}] = 2$, so the pH will be a little below the target value of 4.5, and weigh out about 71 g of $\ce{NaH_2Cit}$, i.e., about 1/3 mole. Likewise, weigh out 39 g of $\ce{Na_{2}HCit}$, i.e., about 1/6 mole. The total number of moles of citrate species is about 0.5. Dissolve the $\ce{NaH_2Cit} $ and $\ce{Na_{2}HCit} $, with stirring, in about 0.8 L of water in a large beaker. Then, with constant stirring and monitoring of the pH, slowly add 2 M NaOH solution until the pH increases to 4.5. The last step is to add water until the final buffer volume is 1 L.

What happens if a mistake is made and the pH is too high? Simply add a little more of the most acidic species, making sure it dissolves, and then titrate back up with NaOH solution. It can be helpful to have a lower concentration NaOH solution available, e.g., 0.1 M, either for this corrective procedure or for tweaking the pH.

Other possibilities include 1) citric acid plus trisodium citrate or 2) citric acid plus disodium hydrogen citrate or 3) sodium dihydrogen citrate plus trisodium citrate, but these are not considered further.


What is the reason for strong acids/bases dissociating in water? I have read that the actual definition of a strong acid is one which will fully dissociate in water, so please do not turn the problem on its head when answering. It is also my understanding that every acid reaction is in fact reversible (just as with any other reaction) and yet the thing that is different about strong acid forward reactions is that they are shifted so far forward (the equilibrium constant is very high) to the extent that hardly any of the reactants remain after reaction. The acid reactions follow the laws of chemical reaction kinetics and the reason for this is that the Gibbs free energy decrease of the acid reaction is fairly high and so according to a Maxwell-Boltzmann Curve of energy distributions, only a very small fraction of the molecules will have enough energy to actually revert back to their undissociated form.

My real question is what is it about the electronic (or otherwise) structure of these compounds, for example $\ce{HCl}$, as opposed to weak acids, e.g. $\ce{CH3COOH}$, that makes them more easily dissociated in water. It might also be helpful to learn the same about strong and weak bases. I think it might have something to do with the electronegativity of the ions involved.

If you can help answer the above, please could you also explain the weak/strong conjugate rule (that if a weak/strong acid/base reacts it will produce a strong/weak conjugate base/acid respectively).

The easy answer:

It comes down to the strength of the bonds. Or that covalent bond strength is directly related to the bond dissociation energy and thus dissociation.

A little Physical Chemistry:
(The following is how Pauling explained this.)

In physical chemistry, they have something called bond dissociation ($D_o$).

From Atkins (9th Ed 10.4): "bond dissociation energy, $D_o$, the energy required to separate the atoms to infinity or by the well depth $D_e$." - What is meant here by well is a potential well – for now just think of that as the 2D box talked about in physics.

Now lets shift gears. You may be familiar with bond enthalpy (sometimes just called bond energies). A common formula used in general chemistry is $\Delta H_\mathrm{rxn} = n(\text{bonds broken}) - n(\text{bonds formed})$

It turns out that $\text{bonds broken}$ or $\text{bonds formed}$ is actually referring to the bond energy or bond enthalpy.

It actually works out that there is a direct relationship to bond dissociation energy and bond enthalpy; which makes sense. It is elegantly related by:

$\Delta H^o_\mathrm{bond} = D_o + \frac32 RT $

I guess the natural question to ask now is why do some bonds have a greater enthalpy of formation?

This was not well understood for a long time. In reality there is a lot of factors that can come into play. The way that Pauling initially came up with dissociation energy, $D_o$, in The Nature of the Chemical Bond followed this protocol:

Spectroscopy, (I did this with FTIR for example, but there are several choices) to get a molecular potential energy curve.
Extract the well depth, $D_e$
Find the dissociation Energy using $D_o = D_e - \frac12 h\nu$

In modern thermo:

Because of the relationship with enthalpy, we can measure enthalpy with calorimetry and back calculate.

But the best description uses statistical thermodynamics … Which I will not get into.


Converting Lactic Acid to Potassium Lactate I am looking to try an experiment incorporating potassium lactate into soap making. This seems to be new territory for soap makers in that sodium lactate is usually used. I found a bottle of "Lactic Acid 88%" at a beer making supplier and already have 90% potassium hydroxide but do not know how much of one is needed to neutralize the other. Can anyone point me to the right tables or resources to do the calculations?

OP wasn't clear about OP's procedure. Yet, OP is clear that OP needs a table to calculate amounts of lactic acid (MW: $M_\ce{LA} = \pu{90.08 g mol-1}$) and potassium hydroxide (MW: $M_\ce{NaOH} = \pu{50.11 g mol-1}$) need make certain amount of potassium lactate. The chemical equation regarding this calculation is as follows:

$$\ce{CH3CH(OH)COOH + KOH -> CH3CH(OH)COO^-K+ + H2O}$$

This means one mole of lactic acid reacts with one mole of potassium hydroxide to give one mole of potassium lactate.

Suppose OP needs $x$ moles of potassium lactate. Thus, OP needs $x \ \pu{mol}$ of $\ce{KOH}$. However, I believe OP has only $90\%\mathrm{(w/w)} \ \ce{KOH}$, meaning if you weigh $\pu{100 g}$, you get only $\pu{90 g}$ of $\ce{KOH}$.  Thus, mass ($m$) you need to measure to get $x \ \pu{mol}$ of $\ce{KOH}$:

$$m_{\ce{KOH}} = \frac{\pu{56.11 g}}{\pu{1 mol}}\times x \ \pu{mol}\times \frac{100}{90}$$

Similarly, since OP has only $88\%\mathrm{(w/w)}$ lactic acid, meaning if you weigh $\pu{100 g}$, you get only $\pu{88 g}$ of lactic acid.  Thus, weight ($m$) of lactic acid OP needs to measure to get $x \ \pu{mol}$ of lactic acid:

$$m_{\mathrm{LA}} = \frac{\pu{90.08 g}}{\pu{1 mol}}\times x \ \pu{mol}\times \frac{100}{88}$$

Now, using these equations, OP can measure any amount of reagents in order to make desired amount of potassium lactate. For example if OP need $\pu{5 mol}$ of potassium lactate ($x=5$ in both equation), then OP can simply weigh following amounts of $\ce{KOH}$ and lactic acid respectively, and react them together:

\begin{align}
m_{\ce{KOH}}
  &= \frac{\pu{56.11 g}}{\pu{1 mol}}\times \pu{5 mol}\times \frac{100}{90}
 = \pu{311.7 g}\\
m_{\mathrm{LA}}
  &= \frac{\pu{90.08 g}}{\pu{1 mol}}\times \pu{5 mol}\times \frac{100}{88}
 = \pu{511.8 g}
\end{align}

Keep in mind that these calculations are valid only if purity of both reagents is given in $\%\mathrm{(w/w)}$. Otherwise, additional information such as density is needed.


How does SO2 have 2 π bonds? The hybridization of $S$ in $SO2$ is sp2 which means it has only one unhybridized p orbital to form a $\pi$ bond. How does it form two?
Could it be one electron from a $p$ orbital getting promoted to a $d$ orbital leading to one $d-p$ and one $p-p$ $\pi$ bond. Is this even possible or are $\pi$ bonds only between p orbitals? Could the electrons possibly be delocalized in the $d-p$ and $p-p$ $\pi$ bonds?

Yes, you are right. One electron from 3p orbital is unpaired and excited to a 3d orbital. The hybridization of sulfur atom is sp2 hence a lone pair and two bond pairs(due to sigma bonding) reside in these hybrid orbitals.

The unpaired electrons are 3p and 3d hybridized orbitals are used in pi bonding with oxygen's unhybridized 2p orbitals. Hence two pi bonds are formed which are p-p pi and d-p pi bonds. Because of delocalisation these bonds are identical and have properties intermediate to the two different pi bonds.

To understand how the electron is able to make a jump with high energy difference, you will have to understand that d orbitals usually participate in bonding only when central atom is joined to many highly electronegative atoms. Because of polarization, there is a slight positive charge on the sulfur atom due to which orbital contract. This contraction is very very significant in d orbitals. Because of this electrons can be excited to the less energetic and tightly held d orbital. Also, the bonds formed will have a greater extent of overlap because d orbitals are not that diffused anymore. This is the reason that PF5 exists but PH5 does not.

Also, note that we cannot use dz^2 orbital for pi bonds as its lobes have the same phase and hence net overlap during pi bond will become zero.


In acid/base chemistry, are there amphoteric substances that undergo something that resembles disproportionation in redox chemistry Some substances disproportionate. This means a species with an intermediate oxidation state yields two species with higher and lower oxidation state. For example:

$$\ce{Hg2Cl2 -> Hg + HgCl2}$$

My question is if there are substance that have the analogous acid/base behavior, where an intermediate protonation state yields the conjugate acid and the conjugate base. You could write it like this:

$$\ce{2AH -> A- + AH2+}\tag{1}$$

An example would be water:
$$\ce{2H2O <- OH- + H3O+}$$

What I am looking for, however, is a case where the mixture of conjugate acid and the conjugate base (of the same amphoteric starting material) are the major species, i.e. a reaction of the type (1) with an equilibrium constant larger than 1. This would mean that the pKa values are in the opposite order than usual, i.e. it is "easier" to lose the second proton than it is to lose the first. For disproportionation, there is a similar feature of the reduction potentials (it is easier to accept the second electron than it is to accept the first).

Are there any such examples?

Update: To clarify my question, I am looking for substances that skip a protonation state, i.e. pKa2 < pKa1. If you could isolate the intermediate protonation state, it would yield the higher and lower protonation state. This is similar to species that skip an oxidation number (or have the corresponding species at very low concentration) because the reduction potentials favor disproportion of the (hypothetical or barely detectable) species with intermediate oxidation state.

As I mentioned in the comments, I know of one case where a polyprotic acid has a second proton which is more acidic than the first: the aqueous pervanadyl complex, $\ce{[VO_2(H_2O)_4]^+}$. According to Wikipedia,

$$
\begin{array}{rcl}
\ce{[VO_2(H_2O)_4]^+ &<=>& H3VO4 + H+ + 2 H2O} &  \mathrm{pK_{a_1}} = 4.2\\
\ce{H3VO4\ \  &<=>& H2VO4^- + H+} &  \mathrm{pK_{a_2}} = 2.60\\
\ce{H2VO4- &<=>& HVO4^2- + H+} &  \mathrm{pK_{a_3}} = 7.92\\
\ce{HVO4^2- &<=>& VO4^3- + H+} &  \mathrm{pK_{a_4}} = 13.27
\end{array}
$$

Though it certainly feels quite odd that $\mathrm{pK_{a_2}<pK_{a_1}}$, I suspect this feature is not absurdly uncommon in larger/more complex systems, where structural reorganisation upon deprotonation can be severe. Perhaps what makes the pervanadyl complex unusual is its comparative simplicity.

However, Wikipedia provides no source for those acidity constants. I looked around for an article discussing the speciation of vanadium ions in aqueous solution, and discovered this nice reference:

  Huang, J.-H.; Huang, F.; Evans, L.; Glasauer, S. Vanadium: Global
  (Bio)Geochemistry. Chemical Geology 2015, 417, 68–89.
  DOI:10.1016/j.chemgeo.2015.09.019

The relevant portion in this article is section 4.1. (Aqueous speciation chemistry of vanadium). The authors use thermodynamic data obtained elsewhere to establish the following equilibria:

$$
\begin{array}{rcl}
\ce{[VO_2(H_2O)_4]^+ &<=>& H3VO4 + H+ + 2 H2O} &  \mathrm{pK_{a_1}} = 3.67\\
\ce{H3VO4\ \  &<=>& H2VO4^- + H+} &  \mathrm{pK_{a_2}} = 3.40\\
\ce{H2VO4- &<=>& HVO4^2- + H+} &  \mathrm{pK_{a_3}} = 8.06\\
\ce{HVO4^2- &<=>& VO4^3- + H+} &  \mathrm{pK_{a_4}} = 13.28
\end{array}
$$

As you can see, the first two constants are rather different from the values quoted on Wikipedia, and now they've become difficult to distinguish. However, clearly this is still an unusual case. They take these data and then calculate speciation curves to obtain the following graph:

Evidently, the speciation curves don't really look anything out of the ordinary. It's likely this is partly due to the proximity of the values of $\mathrm{pK_{a_1}}$ and $\mathrm{pK_{a_2}}$ used, and also because the solution is highly dilute. It would be nice to know what happens in more concentrated solutions, but unfortunately we get something like this:

Like some other transition metals, vanadium has a tendency to oligomerise in solution, forming complex mixtures of polyoxometalates. Therefore, it's basically impossible to get a good visualisation of the effects of $\mathrm{pK_{a_2}<pK_{a_1}}$.

Fortunately, we can forget the chemistry and just study how the mathematics behaves. For simplicity, let's consider a diprotic acid $\ce{H2A}$, where we choose the values of $\mathrm{pK_{a_1}}$ and $\mathrm{pK_{a_2}}$. Also for simplicity, let us safely ignore the autodissociation of water by working with relatively concentrated solutions. Therefore, the mole fractions of the species $\ce{H2A}$, $\ce{HA-}$ and $\ce{A^2-}$ will be:

$$X_{H_2A} = \frac{[H^+]^2}{[H^+]^2 + K_{a_1}[H^+] + K_{a_1}K_{a_2}}$$

$$X_{HA^-} = \frac{K_{a_1}[H^+]}{[H^+]^2 + K_{a_1}[H^+] + K_{a_1}K_{a_2}}$$

$$X_{A^{2-}} = \frac{K_{a_1}K_{a_2}}{[H^+]^2 + K_{a_1}[H^+] + K_{a_1}K_{a_2}}$$

Let's keep $\mathrm{pK_{a_1}}=5$ and decrease $\mathrm{pK_{a_2}}$. Here are the individual speciation curves for $\mathrm{pK_{a_2}=10}$ to $\mathrm{pK_{a_2}=-1}$, in decrements of 1 unit of $\mathrm{pK}$ (click on graphs for a larger version).

As you can see, there is no sudden catastrophe as the second proton becomes more acidic than the first. As the second ionisation becomes more favourable, it smoothly and continuously suppresses the formation of $\ce{HA-}$ . Something nice does happen at the crossover point, where $\mathrm{pK_{a_1} = pK_{a_2}}$, namely all species are present in the same concentration when $\mathrm{pH = pK}$. However, this not surprising if you look at the equations above.

For good measure, here are all the plots once again, now with a logarithmic y-axis. It's nice to see that even with the suppression of $\ce{HA-}$, it still exists as a trace in the background, and still behaves smoothly.

And to finish things off, now here are graphs showing the mole fraction of each unique species as $\mathrm{pK_{a_2}}$ varies from 10 to -6, in decrements of 1 unit of $\mathrm{pK}$. I rather like the subtle way the curves for the fully protonated and fully ionised species change, as the intermediate ion stops being a significant contributor in the equilibria.


Does benzene's resonance structure allow it to enter DNA? According to this link, benzene is able to insert itself into the human DNA. It isn't an authoritative source and appears to be quite biased, so I'm wondering if there's any truth to this.

  The resonance hybrid structure of the benzene molecule causes it to have electrostatic potential, be very stable, and insert itself into human DNA. Taken together, these molecular and electrical characteristics of benzene make it carcinogenic to humans.

I will start my answer with a preface that the website linked to in the question is a pseudoscience website (and I am glad that it has vanished from the face of this earth, only accessible via the Wayback Machine). These people use scientific-sounding jargon that sounds impressive to the lay reader, but any actual scientist will know that it is simply rubbish. The objective is usually to fear monger, or to promote some product of theirs that has no special properties. Unfortunately, these seem to be becoming more and more prevalent recently.

However, the premise of the question is sound: benzene itself is known to cause cancer in humans. It is listed as a Class 1 carcinogen, meaning that there is evidence for carcinogenicity in humans.

As with many things in molecular biology, we do not actually know the full mechanism by which benzene causes cancer. However, there have been a number of papers written on the topic: some of them are linked at the bottom of this post.

Exactly which metabolite of benzene causes cancer is not known. It is likely to be a combination of them. The first step is oxidation of benzene by the enzyme cytochrome P450 2E1 (CYP2E1) to benzene oxide, which exists in equilibrium with oxepin (via a 6π electrocyclic reaction). The cytochrome P450 enzymes are found in the liver, and their role is generally to insert oxygen atoms into molecules, making them more polar and water-soluble, so that they can be excreted. For more information about this process, see Drug metabolism on Wikipedia.

These undergo further reactions, catalysed by various other enzymes, to give a large range of metabolites:

In particular,

The benzoquinones have been shown to inhibit DNA topoisomerase II, an enzyme that makes temporary cuts in double-stranded DNA in order to "unwind" DNA that has been entangled. It plays a key role in many cellular processes such as DNA replication, DNA repair, and chromosome segregation (during cell division); therefore, inhibition may lead to chromosome breakage or failure to segregate.
The same quinones can undergo a process called redox cycling, where they undergo an enzymatic reaction in which a single electron is added to them to form a radical anion. These species are then released, and react with molecular oxygen $\ce{O2}$ to give the superoxide anion, $\ce{O2^-}$... and then the process repeats itself. The buildup of $\ce{O2^-}$ (and other reactive oxygen species) leads to oxidative stress and DNA damage.
(E,E)-muconaldehyde has been recently shown to form an adduct with two molecules of deoxyguanosine, i.e. the guanine bases in DNA. Here, R represents the rest of the deoxyribose sugar. Intra- or inter-chain links can be formed via this method, which then lead to inaccurate replication or chromosomal aberrations.
A mechanism was proposed in reference 3. It is not reproduced here but it is not difficult to imagine how such a reaction might happen: nitrogen atoms in guanine are nucleophilic, and literally every carbon in muconaldehyde is electrophilic.

The literature on the topic contains much more information than I can write in here. Reference 4 is a relatively recent review on the topic, which would be a decent starting point to find further information.

Regardless of what mechanism it is, one thing is certain: benzene itself and its molecular properties are not likely to be the cause of its carcinogenicity. It is almost certain that multiple metabolites of benzene are the culprits.

References

Chen, H.; Eastmond, D. A. Topoisomerase inhibition by phenolic metabolites: a potential mechanism for benzene's clastogenic effects. Carcinogenesis (Oxford) 1995, 16 (10), 2301–2307. DOI:10.1093/carcin/16.10.2301
Rappaport, S. M.; Kim, S.; Lan, Q.; Li, G.; Vermeulen, R.; Waidyanatha, S.; Zhang, L.; Yin, S.; Smith, M. T.; Rothman, N. Human benzene metabolism following occupational and environmental exposures. Chem.-Biol. Interact. 2010, 184 (1–2), 189–195. DOI:10.1016/j.cbi.2009.12.017.
Harris, C. M.; Stec, D. F.; Christov, P. P.; Kozekov, I. D.; Rizzo, C. J.; Harris, T. M. Deoxyguanosine Forms a Bis-Adduct with E,E-Muconaldehyde, an Oxidative Metabolite of Benzene: Implications for the Carcinogenicity of Benzene. Chem. Res. Toxicol. 2011, 24 (11), 1944–1956. DOI:10.1021/tx2002838.
Hartwig, A. The role of DNA repair in benzene-induced carcinogenesis. Chem.-Biol. Interact. 2010, 184 (1–2), 269–272. DOI:10.1016/j.cbi.2009.12.029.


Real boiling temperature I've been thaught that water at 1 atm water boils at 100 °C (the so called “normal boiling point”).

However, I came across a more rigorous description of the boiling process saying that in order to fulfill thermodynamic equilibrium between liquid and vapour the pressure inside the bubble must be higher than that outside and accordingly the liquid results superheated (to be at the same temperature of the vapour).

I also understand the nucleation process, but here for sake of simplicity I assume that the bubble is already present.

So my question is why we say that water boils at 100 °C (at 1 atm)?

Simply speaking: The rising bubbles in your pot shrink wherever it is too cold (< 100 °C), and grow whereever it is too hot (> 100 °C), equillibrating the temperature, by exchanging some heat of evaporation. If you keep your thermometer at a distance above the heat source, it will show (mostly, see below) exactly 100 °C, or whatever is the boiling point at the given outside pressure.

Very small bubbles have to have a measurably increased inside pressure, but while the water is boiling, the bubbles grow, and so this difference vanishes. You can calculate it with the surface tension of water (0.06 N/m at 100 °C) using the https://en.wikipedia.org/wiki/Kelvin_equation, and it is negligible for bubbles of millimeter size.

Of course with increasing distance to the water surface, the temperature rises  because of the pressure increase of ~ 0.1 bar/m. Slightly less, because water at 100 °C is less dense.


Signs of electrode potentials Why is the oxidation potential given a positive sign and the reduction potential a negative one?

Also, in my book a bunch of formulas are given without any explanation:

$E^{o}_{cell}= $ Oxidation potential of anode + Reduction potential of cathode

$E^o_{cell}=$ Standard reduction potential of cathode - Standard reduction potential of anode

$E^o_{cell} =$ Oxidation potential of anode - Oxidation potential of cathode

I am unable to get any intuition for this and the physics behind these signs. Could someone please provide an intuitive explanation of what's going on?

We use a convention for the sign of potentials. The standard reduction potentials are referenced wrt  hydrogen reaction $\ce{2H+ + 2e^- <=>H2_{(gas)}}$ which is given the value $0.0$ V. Oxidising agents have positive values, for example $\ce{Ag^+ + e^- <=> Ag_{solid}}$ $E^\mathrm{o}=+0.799$ V, or $\ce{Cl2_{(gas)} + 2e^- <=> 2Cl^-}$, $E^\mathrm{o}=+ 1.36$ V, and reducing agents negative values, e.g. $\ce{Cr^{2+} + 2e^- <=>  Cr_{solid}}$, $E^\mathrm{o}=-0.91$ V.

The oxidation potentials of these reactions are just of the same magnitude but of opposite sign. Thus for the same species  the oxidation potential + reduction potential add to zero. Normally we only deal with reduction potential so if you have an oxidation potential just reverse the sign and it becomes a reduction potential.

Note that the same species can be an oxidising agent or a reducing agent depending what it is being reacted with.

A reducing agent is an electron donor, and oxidising agent an electron acceptor. If a species is oxidised it looses electrons but gains electrons if reduced. Thus a reducing agent is itself oxidised;  confusing, yes  but its just how it is.

The cell potential is defined in your second definition as this uses the standard reduction potentials. The others are just variants by messing with reduction vs oxidation and changing signs as appropriate.

( Note that in some papers and books you may come across the redox potential values quoted wrt the saturated calomel electrode, not hydrogen, and this has a value $+0.246 $ V so its always worth a quick check. )


Calculating vacancies in crystal structure I have this question on a practice exam. I'm not quite sure how to find the number of vacancies when replacing atoms.

Question:
The stable oxide for lithium is $\ce{Li2O}$. How many oxygen vacancies in $\ce{SiO2}$ will be created if 100 Si atoms are replaced by 100 Li atoms (remember, O has a -2 charge in an ionic material)?

I think that because Li is +1 charge and Si is +4 that when I replace the Si with Li it leaves 150 unbound O. So 150 vacancies? Am I totally wrong?

That's it, you have the right answer. Conservation of charge (electroneutrality) is what you use to determine the number of vacancies.


When is the activity coefficient greater than 1? I am quoting a sentence from Non ideal solutions:

if the interactions (between solute and solvent) repel each other, the
activity coefficient is greater than 1.

How can interactions between molecules repel each other? Don't they always line up so that the negative end is in contact with positive (and vice versa)?

The classic case of having activity coefficient greater than 1 is the solution of very hydrophobic organic solvents in water. For example, if you have toluene or benzene in water, these molecules do not prefer to stay in water. This type of work was a popular research topic in the 50-70s.
Note that here we are talking about neutral solution so the concept of ordered lining up does not arise.
There are plenty of papers on this topic. Just search activity coefficient of benzene in aqueous solutions.


Can the ions of dissolved salts be physically separated with a strong enough external electric field? Say you had a arbitrary amount of sodium chloride dissolved in water. Could an external electric field (e.g. the electrically charged plates in Millikan's oil drop experiment) physically pull apart the dissolved ions in the water if it was strong enough? What forces would it need to overcome?

If it is possible, how powerful would the electric field have to be? What would be the outcome if the $\ce{Na+}$ and $\ce{Cl-}$ ions were then physically separated into two different containers?

Could an external electric field (e.g. the electrically charged plates in Millikan's oil drop experiment) physically pull apart the dissolved ions in the water if it was strong enough?
What forces would it need to overcome?

Yes, this is the principle behind electrophoresis and electric double layer supercapacitors where charge is separated by an electrostatic field shown below.

Diagram of Double Layer Capacitor charging and Discharging

What would be the outcome if the $\ce{Na+}$ and $\ce{Cl-}$ ions were then physically separated into two different containers?

Assuming you could so this without any interaction with the container, you would have 2 Leyden Jars that were opposite in charged and storing a lot of energy.


Does the largest known cycloalkane contain 288 carbon atoms? If you Google "largest known cycloalkane", you will find this result:

For context, there has been discussions on "largest known alkane". There is a related chem.SE question. There is also a scientific paper on it. The paper says that the longest linear alkane is $\ce{C59H120}$. OTOH, the Wikipedia article for cycloalkane mentions that the larger cycloalkanes, with more than 20 carbon atoms are typically called cycloparaffins. So, there has to be a known "cycloparaffin" with highest number of carbon atoms.
The above result came from a random HW-QnA site. From where does the result come from?
Question: Does the largest known cycloalkane really contain 288 carbon atoms? If not, how many carbon atoms are there in the largest cycloalkane that has been actually observed and isolated?

According to Makromol. Chem., Rapid Commun. 1985, 6 (3), 203–208[1], Kwang Sup Lee and Gerhard Wegner, attempting the synthesis and study of $n>100, \ce{C_nH_{2n}}$ cyclo-alkanes and linear alkanes, were successfully able to synthesis a linear alkane of $\ce{C384H780}$ and cycloalkanes with 288 carbons ($\ce{C288H578}$)

Linear alkanes up to $\ce{C384H770}$, (2k; M = 5378) and cycloalkanes up to $\ce{C288H576}$(3g; M = 4040) were prepared by this strategy.

A brief method of preparation is as follows:

The strategy of the synthesis of long chain linear and cyclic alkanes was developed starting from the work of Schill et al. who had shown that the oligomers obtained by oxidative coupling of 1 following Eq. (1) can be separated by adsorption chromatography.
\begin{align}
\ce{\underset{\textbf{1}}{HC#C-(CH2)20-C#CH} ->[Cu^2+] \\ \underset{\textbf{2}}{HC#C-(CH2)20-C#C-\left[C#C-(CH2)20-C#C\right]_m-H}} \\
\end{align}

A similar strategy was used to synthesize the cycloalkanes as well.

The cyclic oligomers were synthesized by the same reaction according to Eq. (2)
under conditions of high dilution of the terminal diacetylene

Finally, the linear and cyclic oligomers were hydrogenated over a Pd/C-catalyst in a conventional manner to obtain the analytically pure alkanes.

As for the assignment question, I was able to link it to a introductory exercise in Organic Chemistry by Janice Gorzynski Smith.[2].
Reference

Kwang Sup Lee, Gerhard Wegner, "Linear and cyclic alkanes ($\ce{C_nH_{2n+2}, C_nH_{2n}}$) with $n > 100$. Synthesis and evidence for chain-folding," Makromol. Chem., Rapid Commun. 1985, 6 (3), 203–208 (https://doi.org/10.1002/marc.1985.030060316)
Janice Gorzynski Smith, Organic Chemistry, Second Edition, McGraw-Hill book Company Inc.;2008, ISBN: 978-0-07-066720-4


Separation of sulfur dioxide and sulfur trioxide in the contact process When gases leave the catalytic reactor in the contact process of sulfuric acid generation, both $\ce{SO2}$ and $\ce{SO3}$ will be present. According to descriptions I have read, the trioxide reacts with the water spray, but the dioxide does not, and instead rises as a gas and is recirculated into the system.

I thought that sulfur dioxide was highly soluble in water. If so, should it not dissolve into the water spray and thereby contaminate the oleum?

You are true that $\ce{SO2}$ is very soluble in water but it depends on the temperature. Also, $\ce{SO3}$ reacts with water then dissolves in it in a very exothermic way. My understanding is that, if the quantity of water is well adjusted, you will actually get hot concentrated sulfuric acid and sulfur dioxide is not quite soluble in it (ca. 8 g/kg at 120 °C). So $\ce{SO3}$ reacts with water and sulfur dioxide gas can be recirculated.

HTH


Thermochromic, fluorescent fluid I'm a total layman in chemistry and am working on an interactive sculpture project and therefore am researching into fluids that have certain properties - long story short:
we are thinking on having a transparent sculpture where a fluid is flowing through it, a fluid that reacts visually on the heat of peoples hands on their touch - so if people touch the sculpture, the fluid changes colour or even emits light in a fluorescent way.
Sorry, I am conscious this may sound super vague and really naive - I'm just trying to reach out to check out possibilities and may come up with interesting insights.
Is there any known way of achieving such an effect? Where could I continue looking after? Any help or pointer towards a certain direction is highly appreciated. Thanks!

While thermochromicity already was known for me, for example as thermometer for fish tanks like here, a search based on "thermochromic fluorescent fluid" as keyword yielded this video.

Beside safety, yet indeed, it is easier to fill the sculptures with a liquid / a solution of a fluorescent dye excited by black light from the inner of the (glass) sculpture and to wrap the container with thermochromic dyes. These are even commercialized as here and here.  "Black light" is an other name for ultraviolett light, and there are even hand-size battery powered lamps found in the chemistry lab, or in civil life to check banknotes.

Disclaimer: I have no relationships to the said corporation.

Addendum:  Examples of fluorescence triggered solely by the gentle warmth of a hand are not known to me (yet).


How can I find the eutectic point of gallium and aluminum? I've been interested in liquid metals for a while, but I've stayed away from mercury for obvious health reasons. I have a sample of gallium and I've seen a liquid metal mixture of gallium and aluminum. I'd like to make this myself, and I want to achieve the lowest melting point I can, but I can't find much information on the eutectic point of these metals. Is there a way I can calculate it?

The Al-Ga system is reasonably well studied, and there are a number of articles in the Calphad (Calculation of Phase Diagrams) literature. For example, Al-Ga-Zn System: Reassessments of the Three Binary Systems and Discussion on Possible Estimations and on Optimisation of the Ternary System, M. Mathon et al., CALPHAD 24(3) 253-284 (2000) has the thermodynamic analysis yielding the fairly simple model for the Al-Ga system. These parameters may be used in one of the many phase diagram software packages, such as ThermoCalc. Or, one could write a simple simulator in Python, yielding something like:

The eutectic is way over on the Ga side. The paper listed above also references a number of experimental papers with measured eutectic points.

For other systems of interest, one easy place to start is the ASM Alloy Phase Diagram Database, assuming your institution has access. The phase diagram entries have a set of references, often containing both experiment and calculation papers. If you have access to Calphad, Journal of Alloys and Compounds, and Journal of Phase Equilibria and Diffusion, those are good places to search for a given binary or ternary system.


How to change the chirality of amino acids? Is it possible to change the chirality of the amino acids? I understand such a process is called "stereoinversion".

Take a levorotatory amino acid and turn it into it's dextrorotatory counterpart (enantiomer, optical isomer).

Anyone tried to do that? Can a high temperature change the chirality of an amino acid for example?

Racemization is good enough for me. Changing the chirality of just a small amount of the total amino acids in the sample is good enough.

Temperature is likely not going to work. To racemise something that has a chiral center, you need to break a bond: if you want to supply that much energy by heating, you are going to fry your molecule.

The easiest things to do are usually extraction of an acidic proton with a strong base, formation of a double bond involving the formerly chiral carbon, and then you go back to the initial form, but obtaining a racemate because the attack on the two sides of the double bond is equally likely.


Are chiral quasicrystals possible? Quasicrystals are materials that have long-range atomic order but lack the translational symmetry of conventional crystals. All quasicrystalline tilings and packings I have read about thus far, whilst lacking translational symmetry, have rotation axes and reflection planes. I'm wondering if anyone is aware of experimental evidence of quasicrystals which lack reflection symmetry (and thus possess implicit chirality), or can provide a theoretical rationale for why such structures can or cannot exist.

To clarify, a cursory google search turns up a paper interpreting a viral capsid as having chiral quasicrystalline order - Whilst I don't have access to the paper, I specifically exclude this sort of intepretation from consideration as I'm talking about structures with potentially infinite spatial extent, which viruses do not have.

Some recent developments shown in article Transversely quasicrystallographic 3D photonic chiral lattices: Polarization-controllable complex photonic band gap structures, where autors claimed experimantal results on produsing 3d chiral photonic quasicrystals.

  Figure. Computer simulation of the light intensity distribution of
  the interference pattern for transversely quasicrystallographic 3D
  chiral photonic lattices respectively with rotational symmetries
  (a)5-fold (d) 8-fold (g) 12-fold. (b), (e) and (h) are their respective
  x-y planes. (c), (f) and (i) are their respective x-z planes

  "So the fabricated class of 3D PQC chiral lattices integrate both these wonder features on asingle platform in view of their possible applications in new generation integrated optical photonic devices with tunable polarization band gaps as well as frequency band gaps"

I also found fascinating that some researchers, use combination of structures like simple lattice and chiral lattice. This presentation Photonic Crystals and Metamaterials
has a lot of pictures, which could be good "food for thoughts".

Hope this was helpful, I think your area of research is very interesting.


What is the correct structure for sulfur dioxide? A question from the National Chemistry Olympiad (Germany-2008) goes as follows:

  Besides the composition of a substance the structure of its molecule or ion is of interest. To predict the structure you may use the electron repulsion theory
  (VSEPR).
  The following sulphur compounds are choosen as examples
  $\ce{SO2}$, $\ce{SO3(g)}$, $\ce{SOF2}$, $\ce{SO4^2-}$, $\ce{SF4}$, $\ce{SOF4}$, $\ce{SF6}$.

  b) Draw the structure of these species. Insert the free electron pairs of $\ce{S}$.
  Give the name of the geometrical shape of these compounds or ions.

I'm only interested in the structure of $\ce{SO2}$, that is represented in the answer sheet as $\ce{O=S-O}$ instead of the one I believe is the correct one $\ce{O=S=O}$.

What do you think?

See this.
It's because it can be thought as two interchangeable resonant structures. What happens in reality is that the electron is delocalized over the three atoms.


Troubles with cinnemaldehyde extraction - A few general questions I'm trying to do a cinnemaldehyde extraction.  I've tried both a simple and a steam distillation.  I'm using the video: [https://www.youtube.com/watch?v=bDoaPIKT_Xg] - as a guide.  One of the main problems is that, at the end, the receiving flask is full of water and has very few drops of oil whereas the video shows I should have a foggy mixture.

First off, what's up with that?  Second, third and fourth:
What is a good solvent to use other than DCM/ether?
I have read a lot but haven't figured out: What information does one use/how does one know what makes a good solvent given a situation?  Through some research, I decided to use methanol as an alternate solvent - I ended up with a purely clear substance in the separation flask - No layers to separate.  I have isopropyl, methanol, acetone and naphtha to choose from as solvents.

Finally, in the video, I'm having difficulty understanding what the guy is doing after separation.  I've watched the video several times but it still makes little sense to me.  If this last paragraph doesn't make sense, please just focus on the first two - Theoretically, I could figure out the last part through trial and error - that's why I'm doing such a benign experiment as my first.

One of the main problems is that, at the end, the receiving flask is full of water and has very few drops of oil whereas the video shows I should have a foggy mixture. [...] what's up with that?

It's tough to say without seeing your exact setup, but here are some possibilities for you to think about:

Too much water and not enough cinnamaldehyde is coming over in your distillation.  Try heating the still less rapidly.  You might also try using more cinnamon bark and/or less water.
You are using a degraded or very impure cinnamon bark that does not contain very much cinnamaldehyde.  Is there a way you can test a different batch?  What does the distillation smell like?  What does the condensate smell like?
Other possibilities I haven't thought of...

  Second, third and fourth: What is a good solvent to use other than DCM/ether?

The intended application of the solvent is extraction of cinnamaldehyde from an aqueous emulsion of the water-insoluble cinnamaldehyde.  Thus, the main criterion for the solvent is that it be insoluble in water.  If it isn't insoluble, adding it to the aqueous emulsion will just dilute the emulsion.  Nothing will be transferred from one phase to the other, because there will only be one phase.

A secondary feature of DCM in the video is that it is heavier than water.  This means that in the separatory funnel, the DCM/cinnamaldehyde layer will be lower than the aqueous layer.  In this particular extraction, the aqueous layer is repeatedly washed with fresh DCM, and the DCM layer is removed each time.  Doing this is a bit easier if the solvent layer is the bottom layer.  If it were the aqueous layer at the bottom of the funnel, each wash with the 60 mL of solvent would require draining the aqueous layer into one flask, then draining the solvent/cinnamaldehyde layer into another flask, then re-adding the water to the separatory funnel and washing it with fresh DCM, etc.  This requires more flasks and more manipulations than just being able to leave the aqueous phase in the funnel the whole time.

  I have isopropyl, methanol, acetone and naphtha [available].

Now that you know why DCM makes a good solvent, you can probably see why methanol wasn't a great choice.  It is miscible with water, so that explains why you didn't get multiple layers.  Of the solvents you have, only one, naptha, is immiscible with water.  The other two, isopropanol and acetone, are miscible with pure water, but are not miscible with very salty water.  So you might be able to extract the cinnamaldehyde by adding a lot of salt such as calcium chloride, magnesium sulfate, or sodium chloride to the distillate, and then trying to wash the now-salty emulsion of cinnamaldehyde to the solvent for washing and phase separation.  However this is likely to be more difficult than if you could access DCM.  (Getting DCM may be less difficult than you think, by the way.)

  I'm having difficulty understanding what the guy is doing after separation. I've watched the video several times but it still makes little sense to me.

I don't know what part is confusing or exactly what "separation" you mean when you say "after the separation".  But if you mean after the separatory funnel phase separations, then roughly the steps are:

"Drying" the DCM solvent.  Even though DCM and water don't completely mix (they form two phases, which, as explained above, is the reason why it works to extract cinnamalehyde from an aqueous emulsion), some water does leach into the DCM phase during the separation.  "Washing" the DCM with brine helps rid the DCM  of the small amounts of water dissolved in it after the phase separation.  Adding calcium chloride powder to the DCM is an even more effective drying agent that removes the last bit of water from the DCM.
DCM distillation.  After drying the DCM, all the cinnamaldehyde is still dissolved in a (now "dry" or water-free) DCM phase.  To get rid of the DCM and obtain the pure cinnamaldehyde oil, another simple distillation is performed.  Because DCM boils at around 40 °C and cinnamaldehyde doesn't boil until 248 °C, it is fairly easy to remove nearly all the DCM by simple distillation, leaving the oil.
NMR.  The very end shows a proton NMR of the oil.  This is something that requires big fancy equipment and won't be possible for home chemists.  Don't worry about it and enjoy your cinnamaldehyde!


Can we do column chromatography for the purification of metal complexs? In case of organic ligands there are several techniques for purification, but in case of metal complex (using $\ce{Cr,Mn,Fe,Co}$ as central metals), I have not seen any literature for the purification of them except recrystallization.

Can we do column chromatography for the purification of metal complexes?

For some yes, for some no. Typically, the ‘more organic’ a complex gets, the easier it would be to purify it by column chromatography.

Example: I synthesised zincporphyrin complexes in my master’s thesis and used silica gel columns. When synthesising the porphyrin itself, if I used sodium carbonate to neutralise propionic acid used in the reaction, I would get a (green, assumed) sodium complex which I could separate from the free porphyrin again by column chromatography.

Remember though, that many complexes are very polar and thus difficult to separate on standard phase columns.


Platinum Orbitals in Cisplatin Forming Bonds with LCAOs of Chlorine My inorganic chemistry professor placed two questions on the exam review that he did not cover in class; I haven't been able to figure out how to answer them and could really use some help in learning how to think about this problem.

The two questions are:

  Which orbitals of $\ce{Pt}$ can form a $\sigma$ bond with the LCAO $\left[\Psi((\ce{Cl}')_{\mathrm p_z})+\Psi((\ce{Cl}'')_{\mathrm p_z})\right]$ of the two $\ce{Cl}$ atoms in cisplatin?

  Which $\mathrm d$ orbital of $\ce{Pt}$ can form a $\pi$ bond with the LCAO $\left[\Psi((\mathrm{Cl}')_{\mathrm p_z})-\Psi((\mathrm{Cl}'')_{\mathrm p_z})\right]$ of the $\mathrm p_{z}$ orbitals of the two $\ce{Cl}$ atoms in cisplatin?

How do I arrive at the corresponding oritals?

I figured it out!

We have to start by looking at d orbital shapes:

To form bonds, the d orbitals of platinum must have the same symmetry (i.e. shape) as the molecular orbital that's approximated by the LCAO.

We start with the bonding orbital, which looks like this: (sorry about the pencil-and-paper phone picture).

We can see we need a d orbital with exactly one lobe in the zx plane (x is pointing to the right in the above). The only d orbital that fits the requirement is the $\textrm{dz}^{2}$.

Moving on  to the anti-bonding orbital, which I can't post because I don't have at least 10 reputation, we see that it has four lobes with two nodal planes. It looks like a $\textrm{d}_{xz}$ orbital, in fact.

We can see we need a d orbital in the zx plane that has two lobes. Clearly, this is the $\textrm{d}_{xz}$ orbital.

Easier than I thought once I got the concepts down!


What made my solution cloudy? I may know, but I would like an expert opinion and a possible solution?

I had an ethanol product from molasses, brown sugar and yeast which was clear. I then added (and feel all information is important):
1 vanilla bean
1 cinnamon stick
whole allspice berries
whole black peppercorns
1/2 star anise
1/8 tsp grated nutmeg
3, 2-inch pieces ginger
3-inch strips orange zest

I let that sit for 2 weeks and remained clear. I then poured that solution through a coffee filter into beaker and it remained clear.

Then I took tap water, boiled it and poured it into the bottle with the used spices and let that sit for a day. I then used that water to dilute the ethanol solution to a lower percentage. Again, running it through a coffee filter.

Final product, cloudy.

The only thing I can think of is my water. We get our water from a well so it could be high in mineral content.

Would water be the problem?
And solution to make it clear again?

You likely formed an emulsion. Many of the flavor compounds in the products you mentioned are negligibly soluble in water but soluble in ethanol. Once you add the ethanol extract to water the much of the ethanol from the extract will mix with into the water but the flavor compounds in the extract will not dissolve in the water, which leaves micro-sized spheres of flavor floating around the water.  these micro-sized spheres have a different refractive index of light and thus disperse the light making it cloudy.  The same thing will happen if you mix water and oil with soap.  If you look at Gatorade or PowerAde sport drinks all of them are cloudy because of the emulsion produced from micro-size pockets of flavor in the drink.


Why are hydrogenation reactions exothermic? I learned that all reactions that yield hydrogen are endothermic (such has reforming) and reactions that use up hydrogen are exothermic (FCC cracking, hydrogenation, etc.). But why is this so? I know what exothermicity/endothermicity means, but I haven't found a specific explanation for the correlation with hydrogen generation or consumption yet.

Let's take a look at an example of a hydrogenation reaction, namely the hydrogenation of ethene to give ethane

$$\ce{CH2\bond{=}CH2\ (g) + H2 (g)-> CH3\bond{-}CH3 (g)}$$

And now let's determine the enthalpy change of this reaction. By Hess's Law, the enthalpy change of the reaction is essentially the same as the enthalpy change of breaking all the bonds of the reactants and forming all the bonds in the molecules of the product. We can see that

1 $\ce{H\bond{-}H}$ bond has been broken
1 $\ce{C\bond{=}C}$ bond has been broken
1 $\ce{C\bond{-}C}$ bond has been formed
2 $\ce{C\bond{-}H}$ bonds have been formed

The other bonds remain unchanged by the end of the reaction

The following is the list of bond enthalpies of the aforementioned bonds, obtained from the Internet

$\ce{H\bond{-}H}$ bond: +432 kJ/mol
$\ce{C\bond{=}C}$ bond: +614 kJ/mol
$\ce{C\bond{-}C}$ bond: +347 kJ/mol
$\ce{C\bond{-}H}$ bond: +413 kJ/mol

Breaking bonds gives a positive enthalpy change since it is an endothermic process, and vice versa for bond formation. We can now estimate the enthalpy change of the reaction

$\Delta$$H$ = 2(-413) + (-347) + 614 + 432 = -127 kJ/mol

As indicated by the negative value of the enthalpy change, the reaction is exothermic.

In fact, the enthalpy changes of hydrogenation reactions of other alkenes are also similar (though more substituted alkenes may have a lower magnitude of heat of hydrogenation, but still the reaction is exothermic). You may notice that for other alkenes, the way you calculate the enthalpy change is the same since they involve the same bond breaking and bond forming processes aforementioned (any substituents bonded to the carbons with the $\ce{C\bond{=}C}$ bond remain unaffected throughout the reaction and will still be present in the product being bonded to the carbon).

To simply put, the energy released from bond forming is larger than the energy required (absorbed) for bond breaking, and that is why you get net energy released from the reaction.


Calculating Standard Gibbs Energy of Dissolution The solubility of mercury (I) iodide is $\mathrm{5.5\ fmol/L}$ in water at $25\ \mathrm{^\circ C}$. What is the standard Gibbs energy of dissolution of the salt?
  The reaction is
  $$\ce{Hg2I2(s) -> Hg2^2+(aq) + 2I- (aq)}$$

Here is what I think I need in order to solve the question:

$$\Delta G=-RT\ln(K)$$ where $K$ will be the solubility constant. When the question says “solubility of mercury (I) iodide is $\mathrm{5.5\ fmol/L}$” I don’t exactly understand what this means.
The units make it seem like a concentration to me.

You are correct that you need $\Delta G^\circ =-RT\ln{K}$ and that the equilibrium constant in this case is $K_{sp}$ the solubility product constant, which is defined for a process like this one to be:

$$\ce{A_{m} B_{n} (s) <=> mA^{n+}(aq) + nB^{m-}(aq)}$$
$$K_{sp} = \ce{[A^{n+}]^{m} [B^{m-}]^{n}}$$

In your case $K_{sp}=[\ce{Hg2^{2+}}][\ce{I-}]^2$

You are given the solubility (maximum obtainable concentration) of $\ce{Hg2I2}$ in water at $25 \ ^\circ \text{C}$. How can you use that information to find $[\ce{Hg2^{2+}}]$ and $[\ce{I-}]$ when the solution is saturated?


Formal charges can mislead when writing arrow-pushing mechanisms We know opposing charges attract and like charges repel. By this logic then, in the two examples on the left, the curved arrows would be show the formation of an oxygen-oxygen bond. However, these mechanisms are incorrect, and instead the mechanisms on the right take place.

Can anyone explain this with some more examples?

Arrow pushing is a tool that helps you think about the possible reactions available for a given species or set of reactants: where electrons are likely to move in order to form new bonds and break old ones. But you still have to look at the species that you get after doing the arrow pushing and consider whether they will be favorable or not. Sometimes you can get a reaction between a (formally) negatively charged atom and a positively charged atom - for example between methoxide and tert-butyl cation. Do the arrow pushing and see how that works out.

The reactions you have labeled as incorrect can be seen (via arrow pushing) to produce species that have too many bonds or electrons around a single atom (that was what was meant by the term "hypercoordination" - you could also say hypervalent - for example, having 10 valence electrons around oxygen), whereas the ones labeled as correct lead to stable or at least normally bonded species.


Help me understand electrochemistry and batteries I have a brief understanding of how batteries work, but there are still some things that I can't put together. Therefore I would like to ask for help in understanding how batteries work. Every time I try to learn about batteries, I end up seeing the Zinc and Copper cell, but I can't connect with any of these materials. Therefore I would like to ask what would happen if instead of Zinc and Copper, one would make a battery made out of... Steel and Aluminum, for example. Or any materials that are randomly or wrongly chosen.

I have chosen a random example because I think I can understand better if a system doesn't work rather than having a system that works. So now the questions:

Would such a system work if the electrolyte would be a solution that would allow the transfer of ions?
If not, why would this not work?
What can one do/change to make such a system work?
What would happen at an atomic level with both electrodes?

By work, I mean store energy. Numbers such as electrode potential and cell potentials would also be useful.

You can't "connect" with Zn/Cu? You'll need to be more clear, I don't understand what you mean. (You seem to be implying that you understand Al and steel in a more comprehensive way than zinc or copper, but that doesn't seem plausible to me.)
The zinc half cell reaction is Zn°(s) + H2O + H+ → Zn++(aq) + 2e- + H2O + H+ (pH << 7 ) or eliminating the "by-standers" Zn°(s)→ Zn++(aq) + 2e-. The chemical reaction at the copper electrode (and it would be the same reaction for many electrically conductive and inert electrodes, such as Pt (platinum), C (graphite), Au (gold), Aq (silver), and many more) is 2H+(aq) + 2e- → H2↑ (pH << 7) where I've 'canceled' the by-standers again. There are two chemical reactions going on, one at each electrode. The simple analogy is two cups with a tube between them. If the cups and the tube are all filled with water, then water will flow from one to the other if one is higher (and of course one lower) than the other. The water flows because the gravitational potential energy is greater for the water in the higher cup. Well, energy flows if the half-cell reaction (the reaction on one electrode) is more energetic than the other. Please refer to http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Tables/EStandardTable.htm or http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/electpot.html or https://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page) for tables of just a few of the possible redox reactions (note that most of the half-reactions listed are inorganic, with a couple of exceptions. There is an enormous number of organic redox reactions which aren't generally tabulated.) The hyperphysics page links to a further discussion. Aluminum is always (at least, on this planet's surface) covered by a insulating layer of Al2O3. I don't know what would happen if you connected a piece of Al to a piece of "steel". First because I don't know what your electrolyte is, second because commercially available aluminum is, just like steel, an alloy of several elements. In other words, neither is a pure (or nearly pure) element, unlike the copper or zinc you can get. But, there's no reason why you couldn't substitute a piece of stainless steel for the Cu electrode, or even carefully open up a zinc-carbon battery's carbon electrode (if you can find one).=============== The important facts are that each half-cell reaction has a "potential" and the larger the difference is between the two half-reactions, the greater the current flow and voltage will be between electrodes. Since many of the half-cell reactions won't work in the same electrolyte, salt bridges (see Wikipedia) can be used to connect the fluids. (BTW, there's another reason why the use of steel can be a problem, there can be various reactions on the surface of the steel (corrosion) in which one part of the surface acts as an anode, and another part acts like a cathode. If an electrode is "too reactive" in a particular electrolyte, these side reactions can decrease the amount of electricity flowing to the other electrode. Cu/Zn is popular because all you need is a copper wire, a lemon (or even some vinegar (and salt)), and something that's either made of zinc, or has been galvanized (like a galvanized nail, which is coated with a layer of (mostly) zinc). While it doesn't seem like the hydrogen half-reaction involves the copper, actually the voltage (potential) of that reaction depends on the nature of the surface of the metal, sanding the surface until it is 'bright' will increase the amount of H2 produced. So, to sum it up, first you need two different oxidation/reduction half-cell reaction components, then you need to connect them in an electrolytic system so that the circuit is complete. An important fact, but sometimes overlooked in beginning discussions of half-reactions, is that the reaction should be considered reversible with the arrow → reversed ← as well as changing the sign of the potential (multiplying it by -1, so if it is negative it becomes positive, and vice versa). Also note that many tables assume an aqueous environment, and often an acidic or basic pH. The tendency of any "randomly chosen" electrodes to work, depends on the system they create (including your choice of electrolyte and pH). In a sense, half-reactions are a clever way of separating chemical reactions, like putting a paddle wheel into a stream of flowing water, to use some of the energy being release by the chemical reaction. If you didn't keep the two reactions separate, all that would happen is the chemical reactions would generate heat instead of electricity (and heat). In other words, putting zinc metal into acid will generate hydrogen gas (as long as a surface, maybe even the zinc) exists for the gas to form on), and won't generate any electricity (because the electrons aren't being forced through the circuit). This answer is a little disorganized, sorry.


Why H2S2 is hydrogen disulfide and not hydrogen persulfide? Context: I was checking this question and in the answer, the compound $\ce{FeS2}$ was named "iron disulphide" (it was previously named "iron persulfide" but later changed after responding to the comments).
The structure of "hydrogen disulfide" is quite analogous to hydrogen peroxide i.e. containing the sulfur-sulfur linkage bond ($\ce{-S-S -}$) similar to peroxide bond. The bond angles of both compounds are also similar. So, why name it "disulfide" instead of "persulfide"?
In a group, compound are named in a similar manner in order to follow a certain fashion/trend. For e.g. methane, silane, germane (group 14). So, why $\ce{H2O2}$ is hydrogen peroxide and $\ce{H2S2}$ is hydrogen disulfide? Why isn't it named hydrogen persulfide as per analogy?
Related:
Why is H2O2 named hydrogen peroxide?

TL;DR: "Hydrogen persulfide" is technically a very crude mixture of various hydrogen polysulfides ($\ce{H2S_n ; n\geq2}$). It has long sulfur chains and is formulated as $\ce{H-(S)_{n}-H}$. To avoid confusion, each members of the mixture is named according to number of sulfur atoms. So, $\ce{H2S2}$ is hydrogen disulfide.
Long answer:
It is prepared by heating sodium sulfide ($\ce{Na2S.9H2O}$) for three hours at 100° C. The crystalline sulfide melts and the sulfur is gradually dissolved to a deep-red solution of which the solute has various kinds of sodium polysulfides. This solution is mixed with rather less than its own bulk of water and is then introduced as a thin stream into a mixture of dilute hydrochloric acid and ice which dissolves the alkali and gives a crude oily mixture which is referred to as "hydrogen persulfide".
Individual constituent of the crude mixture can also be separated. Treatment of hydrochloric acid to the distillate amounting to approximately one-sixth of the original bulk yields a solution which is called hydrogen trisulfide, $\ce{H2S3}$. A more volatile liquid, hydrogen disulfide, $\ce{H2S2}$, can be condensed in a second receiver cooled by a mixture of ether and solid carbon dioxide.
The crude persulfide can also be formed if a large proportion of sulfur or sulphur dioxide either as gas or in solution is reduced by hypophosphorous acid.
The so-called hydrogen persulfide, are yellow liquids at r.t. They are unreactive towards acids but are very sensitive towards alkalis causing vigorous decomposition. It is stable only for few hours if stored in vessel. Even at r.t., they readily decompose into hydrogen sulfide and sulfur. The compounds burn with a blue flame, giving water and sulfur dioxide. The addition of alcohols induces rapid decomposition producing sulfur. Ketones, nitrobenzene, aniline and pyridine also catalyze the decomposition. It is miscible in benzene, toluene, chloroform, bromoform, carbon disulfide, ether and heptane, giving relatively stable solutions. As such, it finds its use in vulcanization.
Source: http://sulphur.atomistry.com/hydrogen_polysulphides.html


reaction between copper metal and aqueous sodium bicarbonate I wanted to know what reaction would occur between copper metal and sodium bicarbonate?

I believe that once in sodium bicarbonate solution, the copper metal will go into solution as copper ions and replace the sodium ions. Is this wrong?

Also, if it is indeed true, what would the reaction between copper ions and bicarbonate ions be?

First, reduction potentials.

Theory page
https://en.wikipedia.org/wiki/Reduction_potential?wprov=sfla1

Table of standard electrode potentials
https://en.wikipedia.org/wiki/Standard_electrode_potential_%28data_page%29?wprov=sfla1

Standard electrode potentials of sodium is -2.70V while copper (I) is 5.20V, single electron reduction of copper (II) has 1.59V, full reduction potential is 3.37V

Higher the reduction potential, more the tendency for the species itself to reduce. Reduction potential of Copper is much higher than Sodium.

So conclusion is no.

Lets say if copper can replace sodium. By how much? Under room conditions, RT/F*lne is approximately 0.06V, so the amount of copper ions that will replace sodium will be orders of magnitude 71.5 to 115 times dilute. In other words, non existant.


Can carboxylic acids be oxidised further? I know that compounds like alcohols can be further oxidized to aldehydes (or ketones) and to carboxylic acids. But, is carboxylic acid the best oxidized form of a given compound?
What happens when carboxylic acids further reacts with oxidizing agents like $\ce{KMnO4}$?

In carboxylic acid, the carbon atom of the COOH group is already in high oxidation state, namely +3. In order to reach a higher oxidation state (+4), it requires breaking the C—C bond, typically to form molecular CO2. Thus, oxidation of carboxylic acid in strongly oxidizing conditions is known as a decarboxylation reaction.


Does OH- attack carbonyl carbon in aldehyde during aldol reaction? Why does $\ce {OH^-}$ not attack the carbonyl carbon in an aldehyde during the aldol reaction and instead, deprotonate the $\alpha$ hydrogen to form an enolate intermediate?

Also, why does the enolate intermediate attack the carbonyl carbon of another aldehyde once it is formed?

As noted by Jan here, both pathways do proceed, i.e. the hydroxide attacks the carbonyl carbon and also deprotonates the $\ce {\alpha-H}$. To give you a rough gauge of the extent of each reaction, I will now provide some data from Carey & Sundberg (2007).

In general, the hydration of carbonyl compounds (aldehydes, ketones) is unfavourable, i.e. the equilibrium constant has a value that is less than 1. Only in cases where there are inductive (e.g. presence of halogen substituents) or mesomeric effects (e.g. presence of phenyl substituents) operating will the value be much greater than 1. Also, it is noted that the hydration proceeds very quickly, i.e. the rate of reaction is pretty fast. Note that this is reversible.

Having a greater number of steps, amongst other factors, results in the aldol reaction being slower in general. The equilibriums of each individual step in the aldol reaction mechanism are likely to not establish themselves as quickly as that for the hydration of the carbonyl.

From Carey & Sundberg (2007), p. 684:

  In general, the reactions in the addition phase of both the base- and acid-catalyzed mechanisms are reversible. The equilibrium constant for addition is usually unfavorable for ketones. The equilibrium constant for the dehydration phase is usually favorable because of the conjugated $\alpha$, $\beta$-unsaturated carbonyl system that is formed. When the reaction conditions are sufficiently vigorous to cause dehydration, the overall reaction can go to completion, even if the equilibrium constant for the addition step is unfavorable.

So what happens? In the aldol condensation reaction, the dehydration step is what essentially moves the whole equilibrium forward. As the carbonyls react in this aldol pathway and are depleted, the hydrates would convert back to this carbonyl form to react in this aldol pathway. Eventually, if the reaction were to go to completion, the amount of hydrates present would go to zero and all the carbonyls would have been converted to the $\alpha$, $\beta$-unsaturated carbonyl.

Reference

Carey, F. A.; Sundberg, R. J. Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer, 2007.


Metal reactivity with hot water VS HCL Does being able to react with hot water take more metal reactivity than being able to react with HCl? To me, it seems like a metal would need to be more reactive to be able to react with hot water since HCl is a very strong acid and so should theoretically be able to react with more metals, but I am not so sure.

Edit: Context: (Question) Al reacts with hot water (boiling pt), Mg does not react with hot water but reacts with HCl, which is more reactive? According to the reactivity trends, Mg is more reactive than Al, but according to the information given in the question, it seems logical that Al is more reactive since it reacts with hot water but Mg does not. So is Mg or Al more reactive if we were to assume that everything in the question were true?

you might as well check the reactivity series.There is no unique and fully consistent way to define the reactivity series.The reactivity series is sometimes quoted in the strict reverse order of standard electrode potentials, where it is also known as the "electrochemical series":according to the series which has been defined there is no reason behind it.It is experimentally defined and according to the reactivity series magnesium is placed above aluminium.You might as well check this link for more clarity https://en.wikipedia.org/wiki/Reactivity_series#Reaction_with_water_and_acids.In your question it is given that magnesium doesn't react with hot water but it does actually vigorously.
 Magnesium  reacts very slowly with cold water, but rapidly
in boiling water, and very vigorously with acids.Aluminium reacts with acids and steam but less vigorously than magnesium.Magnesium DOES react with hot water even more vigorously than aluminium . Please check the link given.
Yes another factor is the formation of oxide layer on the surface of aluminium which reduces its reactivity


What makes trans fats more harmful than saturated ones? Why are by many trans fats considered to be worse for you than fully saturated fats?

I am asking specifically which of their structural features accounts for this condemnation. They won't form clumps any larger nor will their melting temperature be higher than that of saturated fats, unless I'm mistaken, so it must be due to the double bonds' reactivity in trans configuration. But is it different from that in cis one?

It's sometimes suggested that trans fats increase low-density lipoprotein (LDL) and decrease high-density lipoprotein (HDL) - could that solely result from the properties of the trans configuration of the molecules?

First of all, it's important to distinguish between the types of fatty acids that may form the cell's membrane:

Note that both cis- and trans-fatty acids are unsaturated, the difference being the conformation around the double bond. A trans-fatty acid is more-or-less a straight line, whereas the cis-fatty acid has a noticeable "kink" at the point of the double bond.

Now, the suggested molecular mechanism behind the health risks associated with trans-fatty acids has to do with a property of the cell's membrane called membrane fluidity. This property affects the cell's ability to send and receive signals, and it greatly affects the function of many proteins whose structure and role are dependent on the membrane's state. The membrane has to be fluid enough in order to serve its purpose properly.

(If you don't have time to read a lengthy answer, you can skip to the last paragraph now. Otherwise, keep going...)

One of the main functions of the cell's membrane is to separate the inside of the cell from the outside world, in order to control the cell's internal environment, help it avoid damages from the surrounding environment and deal with harmful substances or unfavorable living conditions. However, the cell needs to communicate with the outside world through different signaling mechanisms which involve sending and receiving chemical signals, as well as adapt itself to different living conditions.
In order to do that, the cell has to have a membrane with a dynamic structure, i.e. one that will be able to adjust itself to the surrounding environment. Evolution decreed that this may be best achieved by a lipid bilayer.
This image shows a portion of a cell membrane, which is made of many fatty acids in a formation that is considered to be the most stable energetically, due to the hydrophobic/hydrophilic interactions of the fatty acids among themselves and with the surrounding water. This image also shows how cholesterol is embedded here and there within the bilayer.
The measure of a membrane's fluidity results from the type of fatty acids that form it and the membrane's cholesterol content.

To put it simply, the more straight fatty acids in the membrane, the more rigid (i.e. less fluid) it will be. The more "interruptions" to the hydrophobic interactions between the fatty acids' tails (by "kinks" in the tails or cholesterol), the more fluid the membrane will be.

The above picture shows two distinct states of the membrane. On the right-hand side the membrane contains mostly straight (i.e. saturated and/or trans) fatty acids, which results in a more rigid and viscous membrane. This state makes it more difficult for the cell to communicate with its surroundings (chemical signals cannot enter the cell easily, proteins in the membrane are less mobile, etc.).
On the left-hand side, the membrane contains more unsaturated fatty acids of the cis conformation. The "kink" that characterizes cis-fatty acids is the main reason this membrane is more fluid - it makes it harder for the fatty acids' tails to stick together due to hydrophobic attraction.
Cholesterol serves the same purpose as those "kinks". by disrupting the attraction between fatty acids' hydrophobic tails. This is a very important role of cholesterol in virtually all of the cells in the body, which makes cholesterol a very important molecule in the body.

To summarize, trans-fatty acids are considered unhealthy because when consumed, they will mostly remain intact (i.e. will not be metabolized; as this form of fatty acids is not found in nature, the body's enzymes have a limited ability to catabolize them), and then they can penetrate cellular membranes, form rigid hydrophobic bonds with the other fatty acids and decrease the membrane's overall fluidity.

See Diet and HD - article from Stanford university.

Other references may be found in most introductory biochemistry texts or online (be sure to read only scientific texts from reliable websites such as the one given above).


How does vinegar get rust off these coins As experiment I found some really nasty rusty coins on the street. They were green. After looking on a website I found vinegar could eliminate the rust. It worked pretty well.

I know this is a redox reaction. Can you tell which ones? I know these are pesos of various denominations from Dominican Republic from the past 10 or 20 years. So that might help identify the metal.

The vinegar is household apple cidwe vinegar from the supermarket. Low concentration.

As previous comments said, this isn't a redox reaction.

The patina or corrosion is made of metal salts -- for copper, it's copper oxide, sulfide, carbonate, and perhaps other things. Vinegar readily converts all these to soluble copper acetate.

Most coins are made of metal alloys, and those alloys sometimes contain metals that are attacked by acid. Nickel and zinc are common components. Soaking in vinegar will dissolve these metals (they'll displace hydrogen, again forming soluble metal acetates). This damages the coin's surface, making it porous, and eventually reducing the coin's weight. (It also breaks collectors' hearts.)

Adding salt to the mix forms an equilibrium with hydrochloric acid, which attacks metals even more aggressively. Please don't do this unless you don't mind ruining your coins.