What is the reaction mechanism for the hydrolysis of Amifostine? I have to do research on a drug called Amifostine: 2-(3-aminopropylamino)ethylsulfanyl
phosphonic acid.

I'm supposed to say how the organic reaction works inside the body.  I understand that it's hydrolyzed by an enzyme into its active form; 2-((aminopropyl)amino)ethanethiol.

I don't get what happens to the phosphonic acid functional group. Does the water break apart the $\ce{S-P}$ bond because it is a weak bond? And what does the acid group become? $\ce{H3PO4}$?

You are expelling a thiol rather than an alcohol via alkaline phosphatase-catalyzed hydrolysis.  One can also view it as a free electron pair on "X" attacking the phosphorus to pop out its =O to an anion, then the charge poking back in to expel thiolate.


How much energy is stored when H+ is converted to H2? I'm doing a write-up for this for my students:

Source: Amirav Research Group (Facebook)

The energy of formation of isopropyl alcohol is $-318.2~\mathrm{kJ/mol}$ (kilojoules per mole), that of acetone is $-249.4~\mathrm{kJ/mol}$, so $68.8~\mathrm{kJ/mol}$ are gained on this end.

But I'm getting confused about the hydrogen end. I'm seeing that The energy of formation of $\ce{H+}$ is $0~\mathrm{J/mol}$ (it can be treated as a free proton), and that of $\ce{H2}$ is also $0~\mathrm{J/mol}$. Of course you would say that oxygen has $0~\mathrm{J/mol}$ as well, so since  water has enthalpy of formation of $241.83~\mathrm{kJ/mol}$, I'd expect it to yield that much energy in the combustion.

But hydrogen gas has an “energy density” as a fuel of $71~\mathrm{kJ/mol}$. So the treatment that $\ce{H2}$ and $\ce{O2}$ have $0~\mathrm{J/mol}$ enthalpy of formation and water has enthalpy of formation of $242~\mathrm{kJ/mol}$ seems to not square with that. Where's the rest of the energy?

Particularly, in the diagram above you see the charge separation as the electron pair moves to the hydronium, producing hydrogen. But a charge separation (J = coulomb volt) is energy, so moving the electrons to the hydrogen has to be adding energy to them, which seems to indicate that the $\ce{H+} = 0~\mathrm{J}$ and $\ce{H2}$ also = $0~\mathrm{J}$ things is wrong. But I don't know where to go from here, does anyone feel confident helping with this?

I'm seeing that The energy of formation of H+H+ is 0 J/mol0 J/mol (it can be treated as a free proton)

No. Nope. Noway.
Formation of proton from hydrogen molecule is extremely endotermic process. It requires to 1) dissociate H-H bond, which is, btw, 432 kJ/mol, and then ionization of hydrogen atom, which is around 1300 kJ/mol.

However. In solution hydrogen ion is bound to water molecules (or other solvent), which results in serious energy discount, around 1100 kJ/mol for the case of water.

However again :) this is of no significance here, since the dive into small details is not required here.

The gross process is $\ce{ Me2CHOH \space =\space  Me2CO + H2 }$, and since entalphy of formation of hydrogen is roughly zero, only change from isopropanol to acetone is relevant. Hence, the difference is 68.8 kJ/mol.


Why isn't cyclopropane considered unsaturated? Looking up "cycloalkanes" on wikipedia (link), I found this:

  In organic chemistry, the cycloalkanes (also called naphthenes, but distinct from naphthalene) are monocyclic saturated hydrocarbons.

Of course, that implies that cyclopropane is saturated. Later I looked up "saturated and unsaturated compounds" on wikipedia (link). Here's the definition they mentioned:

  In chemistry, a saturated compound is a chemical compound (or ion) that resists the addition reactions, such as hydrogenation, oxidative addition, and binding of a Lewis base.

Now the definition mentions that a saturated compound "resists addition reactions". I am aware that cyclopropane does undergo addition reactions, as mentioned here (link):

So finally, to me it seems these two wikipedia pages are contradicting each other. My question is - why is cyclopropane considered saturated if the above definitions are correct?

In short, there are two definitions with separate purpose.

IUPAC defines alkanes as

  "Acyclic branched or unbranched hydrocarbons having the general formula $\ce{C_nH_{2n+2}}$, and therefore consisting entirely of hydrogen atoms and saturated carbon atoms."

and notes notes about cycloalkanes, contrasting e.g., to unsaturated cycloalkenes:

  "Saturated monocyclic hydrocarbons (with or without side chains), e.g. cyclobutane. Unsaturated monocyclic hydrocarbons having one endocyclic double or one triple bond are called cycloalkenes and cycloalkynes, respectively. Those having more than one such multiple bond are cycloalkadienes, cycloalkatrienes, etc. The inclusive terms for any cyclic hydrocarbons having any number of such multiple bonds are cyclic olefins or cyclic acetylenes."

Thus, as in the earlier comment by @MaxW, cyclopropane equally is a saturated alkane because each carbon atom is connected with the next carbon atom only by a $\ce{C-C}$ single bond.  Or, as MSU condenses,

  "Alkanes and cycloalkanes are termed saturated, because they incorporate the maximum number of hydrogens possible without breaking any carbon-carbon bonds."

This set of definitions is not the one serving to determine organic structures from spectra, e.g. in Field's training book, where

  "[...] the molecular formular is reduced to $\ce{C_nH_m}$ and the degree of unsaturation is given by $$\mbox{Degree of Unsaturation} = n - \frac{m}{2} + 1$$ The degree of unsaturation indicates the number of $\pi$ bonds or rings that the compound contains."

(4th edition, p. 4; emphasis added by mine.)


Finding pH after dilution? Assume that $\ce{pKa}$ is 4.76

  Mix $\pu{50 mL}$ of $\pu{0.1M}$ $\ce{HA}$ and $\pu{50 mL}$ $\pu{0.1M}$ $\ce{NaA}$. This is solution (i). Calculate concentrations of $\ce{HA}$ and $\ce{A-}$, measure the $\ce{pH}$.

  Then, mix $\ce{10 mL}$ of solution (i) and $\ce{90 mL}$ of water. Calculate $\ce{[HA]}$ and $\ce{[A-]}$, and measure the $\ce{pH}$.

How do I approach these calculations? I have 13 of these to do, these are the first two.

Equation is:
$$\ce{pH} = \ce{pKa}+\log{\left(\frac{\ce{[A-]}}
 {\ce{[HA]}}\right)}$$

For the first one, I got,
$$4.76 + \log{\left(\frac{0.1}{0.1}\right) = 4.76}$$
But I don't know if that's correct. How would I approach the subsequent problem when water is added?

@Tyberius: I think you're neglecting Berry Holmes' comment above. Dilution is not always inconsequential to the pH, especially when concentrations get quite low. As my analytical chemistry teacher once put it, if you squeeze a lemon into a lake...

@notiarogers: Tyberius already showed you that you miscalculated the formal concentrations of $\ce{HA}$ and $\ce{A^-}$. Don't skip any steps: $50 \space mL$ of $0.1 \space M$ solution contain $50 \cdot 0.1 = 5 \space mmol$ of substance; the final volume is $100 \space mL$, therefore the final concentration is $5 \space mmol / 100 \space mL = 0.05 M$.

The formula used above (for the calculation of the pH of an aqueous buffer solution made from a weak acid + its salt with a strong base) is only an approximation. Its main assumption is that the concentrations of acid $\ce{HA}$ and anion $\ce{A^-}$ are equal to the nominal ones (0.05 M, as explained above). In reality, when you dissolve $\ce{HA}$ in water, some of it already dissociates (forming some $\ce{A^-}$), and when you dissolve $\ce{NaA}$ in water, it dissociates to $\ce{Na^+}$ and $\ce{A^-}$, and some $\ce{A^-}$ already reverts to $\ce{HA}$ (which, BTW, is why the former solution is slightly acidic and the latter is slightly basic). So things are quite a bit more complicated than it seems: even your initial solutions don't contain 'only' $\ce{HA}$ or 'only' $\ce{A^-}$.

Now, it may be that whoever is asking you to do this calculation wants you to stick to such approximation, and in fairness the final concentration after dilution is probably still large enough, unless you want a very precise result.

In case they want you to do a more complicated calculation, you can write an equation for each equilibrium, one for each mass balance and one for the charge balance. Your equation is the dissociation equilibrium of $\ce{HA}$: $$\ce{K_A=\frac {[H^+] \cdot [A^-]}{[HA]}}$$
You can add the dissociation equilibrium of water: $$\ce{K_w=[H^+] \cdot [OH^-]}$$The mass balance of all acid-related species is:$$\ce{[HA] + [A^-] = C_{HA} + C_{NaA}}$$ where $\ce{C_HA}$ is the nominal concentration of $\ce{HA}$, i.e. 0.05 M as calculated above, and $\ce{C_{NaA}}$ is the nominal concentration of $\ce{NaA}$, i.e. also 0.05 M in this case.

The charge balance is: $$\ce{[H^+] + [Na^+] = [OH^-] + [A^-]}$$
And as $\ce{[NaA]}$ is a strong electrolyte, we know that: $$\ce{[Na^+] = C_{NaA}}$$

If you put all these equations together and eliminate all unknowns except $\ce{[H^+]}$, you get a 3rd degree equation in $\ce{[H^+]}$ and known constants alone, which you can solve (probably best numerically) to find the pH.

The exact solution would look very different from the approximate one you were using, which was essentially:$$\ce{[H^+] = K_A \cdot \frac{C_{HA}}{C_{NaA}}}$$

I can tell you that the pH after your 10-fold dilution is only slightly higher than 4.76 (it's 4.763). You would have to dilute 100-fold to start seeing some difference (4.79); and frankly that looks to me like a minor difference for most practical applications.

Note: we are neglecting the difference between activity and concentration; but I guess we have seen enough maths for one day...


Which of the following reactions doesn't represent an oxidation-reduction Which of the following reactions doesn't represent an oxidation-reduction?
$$
  \begin{align*}
    \ce{2 P + 5 HClO + 3 H2O &-> 2 H3PO4 + 5 HCl} \\
    \ce{Zn + 2 HCl &-> ZnCl2 + H2} \\
    \ce{Mg + CuSO4 &-> MgSO4 + Cu} \\
    \ce{NaOH + HNO3 &-> NaNO3 + H2O}
  \end{align*}
$$

I want to find which of those reactions is not an oxidation-reduction reaction, but I'm not sure how to do this and how to find the oxidation numbers, should I memorize them?
Is there some rules like when bases and acids react this can't be an oxidation-reduction reaction — the last reaction implies this — so I can know if the reaction is an oxidation-reduction reaction or not when I look at it without searching for every oxidation number?

If there is an element in the reactants and it is not an element in the products then yes, oxidation or reduction has occurred.
Acid-base reactions are pretty much non-redox reactions. But it is always handy to learn the oxidations numbers.
Extra sources
Do non-redox reactions exist?


Internal Standard Methodology A tablet of weight $\pu{200 mg}$
was analysed for its morphine content using a HPLC system in conjunction with a single wavelength UV detector operating at $\pu{260 nm}$.
The tablet was prepared for analysis by crushing and extraction using a solvent system similar to the HPLC mobile phase i.e. $70\%$ acetonitrile, $30\%$ water. $\pu{50 mg}$ of codeine was added as the internal standard and the mixture made up to $\pu{50 mL}$. The resulting solution was analysed by HPLC using the instrument conditions given below. A standard solution was obtained by taking $\pu{50 mg}$ of each component, dissolving in mobile phase and making up to $\pu{50 mL}$.
Using the results obtained below, calculate the morphine content of the unknown tablet (both as %w/w and mg/tablet).
Standard Solution Analysis
\begin{array}{c|cc}\hline
&\textbf{Retention Time (min)} & \textbf{Peak Area} \\ \hline
\textbf{Morphine} & 2.00 & 153000 \\
\textbf{Codeine} & 3.25 & 212000\\ \hline
\end{array}
Analysis of tablet solution
\begin{array}{c|cc}\hline
&\textbf{Retention Time (min)} & \textbf{Peak Area} \\ \hline
\textbf{Morphine} & 2.02 & 17600 \\
\textbf{Codeine} & 3.29 & 212500\\ \hline
\end{array}

I understand the use of the internal standard methodology, however I am unsure which values I should use in the equation as although I have been given the internal standard solution I also have been given the second standard solution of both codeine and morphine.
The question is asking to find the content of morphine in the sample tablet (200 mg) via the use of an internal standard solution of codeine (50 mg) which was made up to 50 mL ([codeine] = $\pu{1gL^-1}$). This was given as well as another standard solution of codeine (50 mg) and morphine (50 mg) made up to 50 mL ([codeine] = $\pu{1gL^-1}$).
The equation I have been given is
$$D_\textrm{rfx} = \frac{A_x}{ A_\textrm{is}} \times \frac{C_\textrm{is}}{ C_x}$$
where subscripts and symbols have the following meaning: $x$ = unknown, $\text{is}$ = internal standard, $A$ = peak area, $C$ = concentration.
First we need to calculate the detector response factor ($D_\text{rf}$) of the standard solution - but which standard solution?
Assuming that the standard solution conc. of both morphine and codeine are $\pu{1gL^-1}$, this gives a $D_\text{rf}$ for the standard solution of 0.722.
Using the calculated $D_\text{rf}$  value, the $D_\text{rf}$ equation can be rearranged to find the concentration of codeine in the sample tablet.
However my question is which values need to be used for the $A_x$ and $A_\textrm{is}$ - do they need to be both morphine or just the two values from the sample peak areas?

However my question is which values need to be used for the $A_\mathrm x$ and $A_\mathrm{IS}$ - do they need to be both morphine or just the two values from the sample peak areas?

To answer just this part of the question,
\begin{align}
A_\mathrm{x} &= A_\mathrm{unknown} \mathrm{(Morphine)} \\
A_\mathrm{IS} &= A_\mathrm{standard} \mathrm{(Codeine)} \\
C_\mathrm{x} &= C_\mathrm{unknown} \mathrm{(Morphine)} \\
C_\mathrm{IS} &= C_\mathrm{standard} \mathrm{(Codeine)} \\
\end{align}
Looking at your picture, it looks like you successfully calculated your $\mathrm {D_{rf}({morphine})}$ (though it's next to a misleading picture, as the peak area of codeine is larger than that of morphine when their concentrations are equal).
Moving down, you've successfully solved your $\mathrm{D_{rf}}$ equation for unknown concentration, but the substitution is a little off. When you're dealing with the dissolved tablet solution, $A_\mathrm{unknown}$ is indeed $17.6$ but $A_\mathrm{standard}$ would be the codeine area of $212.5$, not the earlier morphine value.
If I had to guess, the confusion is between the terms "internal standard," a known spike of (usually) a single standard (here codeine) and "standard solution," which is an reference standard of known morphine and internal standard concentrations. Two common uses of reference standards are as initial calibrants and a good way to check for interference between analytes.
Getting back to the solution, once you have your unknown concentration, in units of $\pu{g/L} = \pu{mg/mL}$, you will simply multiply by the $\pu{50 mL} = \pu{0.05 L}$ volume to have the morphine concentration of the solution and thus of the whole $\pu{200mg}$ tablet. %w/w is also easy at that point.


IUPAC name of aldehyde with two functional groups I am confused as to what suffix to use for the following compound:

Should it be carbaldehyde or just -al, like should it be propane-1,2-dicarbaldehyde or should it be propan-1,2-dial?
Or maybe it's 1-methylethane-1,2-dial? I am pretty sure its either of these three or maybe something else.

You have correctly identified the principal characteristic group as aldehyde.
According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), simple mono- and dialdehydes are named substitutively using the suffix ‘al’.

P-66.6.1.1.1 Mono- and dialdehydes derived from alkanes are named substitutively using the suffix ‘al’ added to the name of the parent hydride with elision of the final letter ‘e’ of the parent hydride before ‘a’.

For example: pentanal and pentanedial, respectively. Thus, the compound that is given in the question is a butanedial.
The terminal locants of aldehydes, i.e. ‘1,4’ in case of butanedial, are not cited.

P-14.3.4.1 Terminal locants are not cited in names for mono- and dicarboxylic acids derived from acyclic hydrocarbons and their corresponding acyl halides, amides, hydrazides, nitriles, aldehydes, amidines, amidrazones, hydrazidines, and amidoximes, when unsubstituted or substituted on carbon atoms.

Also the locant ‘2’ of 2-methylbutanedial is not cited.

P-14.3.4.3 The locant is omitted in monosubstituted symmetrical parent hydrides or parent compounds where there is only one kind of substitutable hydrogen.

Therefore, the preferred IUPAC name is methylbutanedial.

By way of comparison, the Blue Book contained the similar example ‘2-chlorobutanedioic acid’, which was later corrected to ‘chlorobutanedioic acid’.


Mechanism for Amino-Imidazole Ring Fission A reaction I've tried several times in the past is the hydrogenation of nitroimidazole derivative shown below. The amino product is extremely unstable and rapidly decomposes.

This instability is noted in may pieces of literature and the explanation tends to only go far as saying 'ring fission' occurs.

Question: What is the mechanism of this ring fission? To be clear, I am not interested in getting this reaction to work.... It is no longer a target of mine.

Perhaps the entire mechanism is not determinable, but I presume the amine of one molecule attacks another molecule's imidazole ring? Where would this be? How is fission occuring?

EDIT: Possibly something to do with the apparently favoured tautomerism between these two forms:

http://imgur.com/tZMLWkG

I'm afraid the answer to how this ring fission occurs is... it doesn't, you just made some mistake.

Here and here the authors used the same reaction as you and got the product in good yields. I think you might not realized that the product is a very hydrophilic amine (or salt, depending on workup). If you did an extraction of your reaction mixture then the product surely ended up in the aqueous phase.


How does Le Chatelier's principle work in a gaseous system? My textbook said Le Chatelier's principle normally applies to gases but there is no explanation.

I did some Googling and it said if you increase the pressure, it will shift to the side with less moles of gas.

What does this really mean? Could someone clarify Le Chatelier's principle for me?

Le Chatalier's principle states that if a stress is applied to a system at equilibrium, then  the system, will shift its equilibrium position in order to bring the system back to its equilibrium.

So the stress applied to the system could be temperature, pressure and the concentration of one of the components in the system.

I will choose to explain about pressure because apparently you have the problem understanding that.

Pressure.
If you add pressure to a system then the stress will be to decrease the pressure by shifting the equilibrium position towards the side with less number of particles.
If you  remove pressure from a system, then the stress will be to increase the pressure of the system by shifting the equilibrium position towards the side with the higher number of particles.

So it is true that when you increase the increase the pressure in a system, then the system is stressed to reduce the stress. In order to do that, the system's equilibrium will shift to the side with less number of particles.

For example:

2Fe$_2$O$_3$ + 3C -----> 4Fe + 3CO$_2$

So if you add pressure to this above system, then the stress is to reduce the pressure. On your reactants side the sum of the coefficients is 5 and on your products side, the sum is 7. Since you need to bring it down then the shift will be to the side with the least sum which is to the left-hand side. Therefore, the shift will be to the left-hand side.

Hope this helped you with the problem.


What type of acid is found in tea? We were trying the litmus paper test and we found out that tea is acidic. So the question came in my mind and my science teacher could not give me any satisfactory answer: what acids are found in tea?

Some of the major components of tea are polyphenols. They are fairly acidic and may dominate the pH of tea.

For some examples of the types of polyphenol that are common see this site on tea chemistry:

  The major flavanols [a specific type of polyphenolic compounds] in tea are: catechin (C), epicatechin (EC), epicatechin gallate (ECG), gallocatechin (GC), epigallocatechin (EGC), and epigallocatechin gallate (EGCG).

EGCC has this structure:

from which some degree of acidity is likely to result from all those phenolic OH groups.


How do I control the negative temperature in an experiment? If one were to adjust the positive temperature in an experiment, they would use a heating plate, or a stovetop.

How would one adjust negative temperatures in an experiment? (Not keeping the temperature constant, but decreasing and increasing the temperature at my will in specific amounts).

What I tried so far: "Spraying" liquid nitrogen at my work surface until it cools to the desired temperature.  It is really inaccurate though.

For temperatures below room temperature cooling baths are used.
These baths consist of a solvent + something that has a low temperature already (ice, dry ice, or liquid nitrogen). A comprehensive list is given below:

Although all of these temperatures are technically achievable, in practice we rarely prepare most of them due to either cost (acetonitrile is quite expensive to be using for a cooling bath), toxicity (a benzene/CO2 bath is not ideal) or difficulty (a methanol/LN2 bath forms a very thick slurry which is difficult to work with).
Most reactions below room temperature are therefore conducted at 0 (ice/water) or at -78 (dry-ice/acetone), with in-between temperatures achieved by adding dry ice slowly to acetone until the desired temperature is reached.
In addition to using this kind of cooling baths, there are machines that will allow you to dial up a given temperature and they will cool the solvent to within a few degrees. These are incredibly useful but their size and cost restricts them from everyday use (most labs don't have one, and those that do won't have enough for everyone to use one everytime they need cryogenic conditions)


Do core electrons have molecular orbitals? Valence electrons are associated with molecular orbitals and hybridizations. Do core electrons have molecular/hybridized orbitals, or the original atomic orbitals?

When valence atomic orbitals start to interact with orbitals on another atom or molecule (e.g. bond formation), the valence electrons respond by forming molecular orbitals with the electrons on the other atom or molecule.  The core electrons also respond to this bonding interaction.  Their energy levels and spatial distributions are perturbed.  The more deeply buried the core electrons, the smaller the perturbation - but the core electrons will also be affected by the bonding interaction.  If one were able to write out the exact wave function to describe the new molecule that was formed, the wave function would include the core electrons.  Their importance would be much less than the valence electrons, but they would be involved in describing the total bonding situation.


Atmospheric methane to carbon I read an article by popsci that talked about huge fans forcing methane over a catalyst.

Now that got me wondering, can't we produce hydroxyl radicals with powerfull uv lasers blasting the sky, creating more hydroxyl radicals to break down the methane.

Would this be effective enough to help create enough hydroxyl radicals to help break down methane or would it be a waste of energy against too small an offset?

It would be a total waste of energy.

Remember that the Sun is sending orders of magnitude more energy than humanity can produce. A significant fraction of that is UV, so the Sun is already breaking up more Methane than humanity could do by adding laser diodes or whatnot.

We might be able to do more at the ground, where the Sun's UV is mostly filtered.
It would still be an enormous feat to merely reach the Sun's UV level.

However, there is a big reason NOT to do it at the ground: UV does not only break up methane, but also oxygen. Part of that will recombine to ozone, which is an irritant (in the medical sense, i.e. extended-time exposure will cause damage no matter what).
(I do not know whether the oxygen bonds are stronger or weaker than the methane bonds. If methane has the weaker bonds, one could use an UV wavelength below the oxygen energy. You'd still break up bonds of any atmospheric molecule that has weaker bonds than methane, such as pheromones and whatnot; it is entirely unclear what consequences that would have.)


Birch Reduction Of Anthracene Which ring of anthracene is reduced during Birch Reduction? What will the final product look like? Why?

My teacher said that the more electron deficient ring should undergo the reduction but I couldn't figure it out still. Any ideas?

There are two ways of rationalising this. One way is to look up or qualitatively determine the frontier orbitals (HOMO and LUMO) of anthracene, the other is to think about the stability of various radical anions.

But first, a quick note on the Birch reduction and its mechanism. As you may know, the $\ce{Na/NH3}$ system effectively works as a single electron transfer reagent. The electron is added to the π system’s LUMO, and the resulting radical anion is trapped by a proton (donated from a proton source; e.g. added ethanol) to give a radical. This radical is still electron-deficient and thus susceptible to another single electron transfer reduction resulting in a second anion which again is captured by a proton resulting in the reaction product.

Frontier orbital analysis

A great resource for frontier orbitals is the WOrbit page of Hans-Ulrich Wagner from the LMU Munich. Among the systems whose orbitals you can look at is, luckily, anthracene. For posteriority, I have included the HOMO and LUMO pictures in the image below.

Figure 1: LUMO and HOMO of anthracene as calculated on the WOrbit page linked above.

The LUMO is of importance; after all, we are adding an electron. We see that the MO is most localised on carbons 9 and 10 of the central ring — thus, this is where the first electron will be added in a localised picture, and this is where the first proton will capture.

Scheme 1: First reduction sequence.

As I drew in scheme 1, this gives us the radical 3, of which I have only shown one resonance structure. Applying the same analysis for the second step will tell us that the second electron will be added exactly at that position where I have drawn the radical (data not shown). Thus, we find that the central ring is reduced as depicted in scheme 2.

Scheme 2: Second single electron transfer and subsequent reduction of radical 3 to 9,10-dihydroanthracene 5.

Stability of radical anions

If we consider the $+I$ effect on substituted carbons, we see that we only need to consider the three structures shown in scheme 3 below; reversing charge and radical in structure 2 would result in the negative charge being on the tertiary carbon which is much less favourable when compared to the secondary one.

Scheme 3: Possible radical anions of anthracene.

At first, it looks like 6 and 7 might be better stabilised structures: 6 effectively represents a methylnaphthalene radical and 7 a naphthylallyl radical. However, if we take a closer look at 2 we see that it can be rearranged in a manner shown in scheme 1 above. If you consider the above drawn structure 3 as a radical anion (i.e. before protonation), notice that the radical is now a diphenylmethyl radical. This diphenylmethyl radical wins in favourability against both naphthyl derivatives.

As a general rule, if you can preserve the aromaticity of a benzene ring the structure is ‘better’ than one where you can’t. And two benzene rings as are present in structures 3 and 4 (and 5) are better than the naphthyl rings of structures 6 and 7. Thus, it is most favourable to reduce the central ring.

Note, by the way, that depending on the conditions employed, 5 is not inert towards further Birch reduction. Mehda, Sen and Ramesh report a triple Birch reduction of anthracene leading to the very interesting tricyclic triene 8 in $80~\%$ yield (scheme 4).[1]

Scheme 4: Synthesis of 1,4,5,8,9,10-hexahydroanthracene.[1]

Reference:

[1]: G. Mehta, S. Sen, S. S. Ramesh, Eur. J. Org. Chem. 2007, 423. DOI: 10.1002/ejoc.200600610.


name of a pharmaceutical drug used in dentistry that has Ionic bonding? and how this bounding is made? Hello I am looking for this information and I did not find anything, the closest thing I have found it is Fluorocarbon, but I think it is wrong, does anybody know?.
Help really appreciated I am still looking for this

The fluorocarbons that you mentioned do have polarized $\ce{C-F}$ bonds, but no ionic bonds. If you think ionic, think salts.

I'm not a pharmacologist, but what about the local anestethics used at the dentist? Lidocaine (1), mepivacaine (2), and articaine (3) are frequently used.

All these compounds have tertiary and secondary amino groups. (I've marked the relevant nitrogen atoms blue). When these compounds are treated with hydrogen chloride, the amines get protonated and the neutral compounds are converted to ionic hydrochlorides, $\ce{R3N ->[\textrm{HCl}] R3NH+ Cl-}$.


Explain the selectivity in bromination of this compound with two substituted benzene rings As I know, when $\ce{Br2/FeBr3}$ is added to acetophenone ($\ce{C6H5-CO-CH3}$), Br will substitute a H from the meta position which leads to 3-bromoacetophenone. So why doesn't a Br substitute a H from the right side benzene (connected to the carbonyl group)?

Because that is not the most electron rich aromatic. The ring at other end of the molecule is, because it is an aniline. The first, and probably the second, bromination will go there. If you keep throwing in Bromine you will eventually get bromination of the Acetophenone ring.


Why does ammonia act as an acid in only the first of these two reactions? $$\ce{2Na(l) +2NH3 (g) ->2NaNH2(s) + H2(g)}$$

$$\ce{CH3Cl (alc) +NH3 (alc) -> CH3NH2 (alc) +HCl(alc)}$$

Doesn't ammonia donate a proton in both cases?

Ammonia has a $\mathrm{p}K_{\mathrm{a}}$ value (a description what that means can be found in this answer of mine) of 33 and is thus not acidic at all - you'd need a $\mathrm{pH}$ value around 30 to deprotonate $\ce{NH3}$ to any appreciable amount! This means that $\ce{NH3}$ very rarely acts as a proton donor itself.

So, what happens in your first reaction is the following: It is widely known that when you dissolve $\ce{Na}$ in liquid ammonia you get solvated electrons (in the form of electride complexes). But electrons have of course very strong reducing power, so after a while they attack $\ce{NH3}$ abstracting one of its $\ce{H}$ atoms under the liberation of hydrogen gas. The important point here is that there is no dissociation of $\ce{NH3}$ here, i.e. no reaction $\ce{NH3 <=> NH2^{-} + H^{+}}$, so $\ce{NH3}$ does not act as a proton donor in the same sense a Bronsted acid would.

Your second reaction is an $\mathrm{S}_{\mathrm{N}}2$ reaction where $\ce{NH3}$ acts as a nucleophile. So, it doesn't act as a proton donor in this reaction neither. What it does is that it replaces (substitutes) the $\ce{Cl}$ in $\ce{CH3Cl}$, thus forming a methylammonium salt, $\ce{CH3NH3^{+}Cl^{-}}$.

Now, the $\mathrm{p}K_{\mathrm{a}}$ of an alkylammonium cation lies somewhere around 10 (the $\mathrm{p}K_{\mathrm{a}}$ of $\ce{NH4^{+}}$ is 9.24). So, it is much more acidic than $\ce{NH3}$ and also more acidic than the surrounding alcohol (I assume that "(alc)" means that the reaction is done in alcohol), which has a $\mathrm{p}K_{\mathrm{a}}$ between 15 and 17, and thus it (the ammonium salt) acts as a proton donor.


What is the hybridisation of trifluoromethyl free radical? Having three σ bonds in a similar manner to $\ce{CH3^·}$ free radical, $\ce{CF3^·}$ should also have $\mathrm{sp^2}$ hybridisation. However, if we look at its shape, it is pyramidal and not planar like $\ce{CH3^·}$ free radical (which is $\mathrm{sp^2}$-hybridised), which signifies that $\ce{CF3^·}$ should have $\mathrm{sp^3}$ hybridisation.

But how is this possible because the three σ-bonds will bond with three hybrid orbitals? Where does the third p-orbital come?

If its hybridisation is $\mathrm{sp^3},$ then why is it?

Hyperconjugation between the orbital that contains the lone pair and the $\ce{C-F}$ antibonding orbital contributes to the pyramidalization of $\ce{CF3^·}$ radical:

In the planar geometry the orbitals are perpendicular and no overlap occurs.

References

Carey, F. A.; Sundberg, R. J. Advanced Organic Chemistry, Part A: Structure and Mechanisms, 5th edition.; Springer: New York, 2008. ISBN 978-0-387-68346-1.


Reactivity of Grignard reagent I had the following question at an exam:

What is the fastest reacting compound towards nucleophilic addition with Grignard reagent?
(a) Acyl chloride
(b) Aldehyde
(c) Ester

It is obvious that ester would be the slowest, but among acyl chloride and aldehyde, acyl chloride would have a highly electrophilic, but a bit sterically hindered carbon atom and the resonance will also decrease the electrophilicity, whereas the aldehyde will have a less sterically hindered and moderately electrophilic carbon atom.
Which one will prevail and how to judge?
What would happen if instead acyl chloride there were acyl bromide?

The answer will be (a) Acyl chloride.
The acyl carbon in an acyl chloride is connected to the chlorine atom. While you have acknowledged this fact, you have missed the point that the +M (resonance) effect of Chlorine is overcome by the -I (inductive) effect. Thus, the electrophilicity of the acyl carbon is actually greater than the electrophilicity of the carbonyl carbon in an aldehyde. This jump in electrophilicity is so significant that the difference in steric hindrance is neglected.
In fact, this the major reason why acyl chlorides are considered to be one of the best electrophiles in all of organic chemistry.
If Acyl bromides were used, then they would have an even greater electrophilicity and it would be the answer to this question.
Hope this answers your question!


Understanding Moseley’s law from the Rydberg-type equation? To explain the characteristic X-ray emission peaks for various elemental targets, a formula was developed which was similar in construct to the Rydberg equation for H-atom as derived by the Bohr-model.

$$\tilde\nu=R\left(\frac{1}{n_\mathrm{f}^2}-\frac{1}{n_1^2}\right)\left(Z-\sigma\right)^2$$
where $\tilde\nu$ is the wavenumber, $\sigma=1$ for the $K_\alpha$ line and $7.4$ for the $L_\alpha$ line. My question is:
Although Bohr model is wrong in light of the modern quantum mechanical concept of atomic structure, why does the energy for transition (and X-ray emission) of complicated atoms with complicated screening effect be represented by a simple adjustment of the $Z$ factor in the Rydberg equation with the value of $R$ used in case of H atom? Why isn’t there a need of complicated corrections (instead of simple subtraction of an experimentally obtained $\sigma$) to predict the X-ray wavelength?

The concept to understand here is that most theories in physics and chemistry are obtained by observing then theorizing, this results in the theories and models being made in such a way that they can explain most (if not all) observable phenomenon.

So, when Bohr made his model of the atom, it had to explain the the phenomenon of "X-ray emission" otherwise it wouldn't be accepted, because, X-ray emission is an observed phenomenon of atoms. Following on, it was discovered that the Bohr-model couldn't explain all observable phenomenon as equipment got better and more things could be seen (notably the splitting of spectral lines in magnetic fields). So any theory that followed on to explain this splitting of spectral lines had to explain X-ray emission as well, without disturbing the results of Bohr's theory.

A suitable analogy would be the discovery of special relativity by Einstein, whatever discovered had to explain why Newton's Law had stuck around for hundred's of years without people realizing it was wrong. (And it did, the correction at velocities we know are less than one part in a million and that's why we didn't think Newton was wrong)

So the deeper question is, even a one part in a million error is still an error and so we should know that the theory is wrong even early on... this is not the case as most instrumental errors on their own are much larger in this regard (at least in Newton's time), and the same applies here in that older instruments weren't usually sensitive enough to notice the change brought on by quantum mechanics on X-ray emission to that of Moseley's rough equation. (Though now it is usually possible to detect differences to a good degree.)

So to the final part of your question, there is a horrendous equation waiting if you want to plug in exact values to find the exact wavenumber, but most like the instrument barely notices a difference, so in a sense, it doesn't matter which theory we use. (It's very well possible a hundred years down the road, another genius comes up with an even smaller correction that can be observed only with futuristic devices)


Phosphate buffer from phosphoric acid and K2HPO4? Is it possible to make a solution of ph 7 phosphate buffer solution using phosphoric acid and $\ce{K2HPO4}$ ? How can I calculate the amount of $\ce{K2HPO4}$ needed for 1L of phosphoric acid ?

Edit:
I have 50 mL of 0.2M $\ce{H3PO4}$ solution. How can I convert this solution into 50 mL of pH 7 buffer solution by adding (only) $\ce{K2HPO4}$ ? How can I calculate the weight of $\ce{K2HPO4}$ considering all the equilibria present in the $\ce{H3PO4}$ solution and by the application of Henderson-Hasselbalch equation ?

I suggest you first consider the following reaction:
$$\ce{H3PO4 + 3K2HPO4 -> 2HPO4^{2-} + 2H2PO4- + 6K+}$$

It appears, that transforming all $\ce{H3PO4}$ to equal amounts of $\ce{HPO2-}$ and $\ce{H2PO4-}$
requires 3 mole equivalents of $\ce{K2HPO4}$. If you add 3 mole equivalents of $\ce{K2HPO4}$ you will end up in a situation where the concentration of $\ce{[HPO2^{-}] = [H2PO4^{-}]}$, i.e. at the $\ce{pH} = pK_{a2} = 7.21$. Then, I suppose you use the $\ce{HH}$-equation to figure out the rest.

At pH = pka2 = 7.21 the concentration of [H2PO4(-)] = [HPO4(2-)] = 0.40 M. This is because we have added 3 mole equivalents of K2HPO4 to 50*0.2 = 10 mmole of phosphoric acid, i.e. we have reached a total concentration of phosphoric acid protolytes of (3*50*0.2 + 50*0.2)/50 = 0.80 M.

Now, since we wanted to reach pH = 7.0, we have theoretically added too much of K2HPO4. We suppose the excess amount is equal to x.

At pH = 7.0: [HPO4(2-)] < [H2PO4(-)].   At  this pH, only HPO4(2-) and H2PO4(-) are present in significant amounts in the solution. We can then calculate the following:
7.00  = 7.21 + log ([HPO4(2-)] - x/[H2PO4(-)]) = 7.21 + log (0.4 - x)/0.4) => x = 0,1533.

Now, initially we had 50*0.2 mmole of phosphoric acid. To reach pH = 7.0 we should then add 3*50*0.2 - 0.1533*50 mmole =  30 - 7,66(5) = 22,34 mmole of K2HPO4 or 3.8(9) gram.

Please check the calculations.


amino acid pka of carboxylic acid Why is the pka of the carboxylic acid part of alpha amino acids (pka around 2.3 for most amino acids) quite a bit less than the pka of a normal carboxylic acid, say acetic acid (pka =4.76)?

The amino group in $\alpha$-amino acids is electron-withdrawing via the inductive effect. Mesomeric (resonance) effects cannot occur here because there is no conjugation of the nitrogen lone pair with the carboxyl group. There are two equivalent ways of looking at how the EWG affects the acidity:

It stabilises the conjugate base via withdrawal of electron density from the negatively charged oxygens;
It polarises the $\ce{O-H}$ bond further than in the unsubstituted carboxylic acid, making the proton more prone to dissociation.

Therefore, the $\mathrm{p}K_\mathrm{a}$ of glycine is $2.34$, significantly lower than the unsubstituted acid (acetic acid, $\mathrm{p}K_\mathrm{a} = 4.76$). Similar effects are seen in 2-chloroacetic acid ($\mathrm{p}K_\mathrm{a} = 2.86$).

As for the comparison between glycine and 2-chloroacetic acid, one thing to bear in mind is that glycine exists as a zwitterion at neutral pH. Therefore, the relevant acid-base equilibria are:

For $\mathrm{p}K_\mathrm{a1}$, which you are interested in, note that the substituent is no longer merely a $\ce{-NH2}$ group but is now a $\ce{-NH3+}$ group. Despite nitrogen and chlorine having extremely similar Pauling electronegativities, the positively charged nitrogen is now more electron-withdrawing than a simple uncharged chloro substituent.

One could draw some parallels with aromatic chemistry: aniline is rather unreactive (compared to chlorobenzene) in electrophilic substitution under acidic conditions, where it is protonated. Of course, the reactivity of aromatic rings is a kinetic effect whereas $\mathrm{p}K_\mathrm{a}$ is a thermodynamic effect, but the underlying rationale is the same: $\ce{-NH3+}$ is more electron-withdrawing than $\ce{-Cl}$.


Why does potassium form peroxides but sodium does not? As we go down the periodic table, Group 1 alkali metals hold their single outer valence electron more loosely, and so reactivity increases. Below potassium (Na, Li), we can store the metal in oil with no difficulty. Above potassium (Rb, Cs), we need to exclude all oxygen to prevent reaction in air. Potassium (K) is the unique element where we can technically store it in oil, but it is highly likely to react with any available oxygen to form explosive peroxides. I'd like an explanation of why this is; i.e. why is the reactivity favorable for the formation of peroxides. Thanks!

If you look at the standard Gibbs free energies for the alkali metal peroxides, actually, lithium should form it most easily. The standard Gibbs free energies of formation are -564.8, -430.1, -418.4, -351.5, and -359.8 kJ/mol for $\rm Li_2O_2, Na_2O_2, K_2O_2, Rb_2O_2,$ and $\rm Cs_2O_2$, respectively.

If we're looking at reaction of an alkali metal with oxygen to form explosive compounds, perhaps we should look at superoxides. The superoxides are powerful oxidizing agents, and they can explode when mixed with water, acids, organics, or powdered graphite.

At room temperature, oxygen readily reacts with K, Rb, and Cs to form superoxides ($\rm MO_2$), while Li and Na don't. There is a direct correlation between superoxide stability and electropositivity of the metal concerned (source: Cotton & Wilkinson's Advanced Inorganic Chemistry). $\rm NaO_2$ can be obtained by high pressure and temperature reaction of $\rm Na_2O_2$ with $\rm O_2$, and $\rm LiO_2$ won't form at room temperature.


Why does the derivation Gibbs-Helmholtz equation hold when the enthalpy is temperature-dependant The derivation shown here assumes H and U are independant of temperature, right? But what if $H(T) = H_{0}+C_V \Delta T $?

Then $dG = C_V\, dT - SdT$, right? So
$ \displaystyle \left( \frac{\partial G}{\partial T}\right)_p=C_V -S$ ?

$$\begin{align}
\left(\frac{\partial (G/T)}{\partial T}\right)_p &= \frac{T(\partial G/\partial T)_p - G(\partial T/\partial T)_p}
{T^2} \\[8pt]
&= \frac{T(C_V-S) - G(1)}{T^2} \\[8pt]
&= \frac{TC_V-TS-G}{T^2} \\[8pt]
&= -\frac{H_0}{T^2}
\end{align}$$

as desired (since $G = H_0+C_V\Delta T - TS$).

But this way we see that the H in this equation would only be containing the constant term? Of course we insert H(T) to get thee exact G, but where in my thinking process did I go wrong? How would this derivation be made correctly, assuming that H changes with Temperature?

The derivation of the van't Hoff equation does not require that you specify the T-dependence of H. It arises from computing the derivative (limiting slope) of $G/T$ so you need to consider a polynomial expansion of $G/T$ to linear terms in $T$ around the temperature of interest (that is, an infinitely  small interval dT). At the risk of redundancy:

$$\begin{align}
 (d(G/T))_p &= ((1/T)dG-G(dT/T^2))_p \\
&= ((1/T)(VdP-SdT)-(H-TS)(dT/T^2))_p \\
&= (-(S/T)dT-(H/T^2-S/T)dT)_p \\
&= (-(H/T^2)dT)_p
\end{align}
$$

The answer is therefore found in differential Calculus.

Another comment explains that, if

$G(T)=H(T) -TS(T)=H_0  +C_p \Delta T -TS(T)$

where $H(T)$ is expressed as a linear polynomial in T ($C_p$ and $p$ assumed constant) then

$dG=C_p dT -SdT-TdS$

Since for a process at constant pressure,  $dq = C_p dT=TdS$ (since $C_p = (dH/dT)_p$ and $dq_p= dH$; also for a reversible process $dq= TdS$), it follows that

$\left(\frac{ \partial G}{\partial T} \right)_p=-S$

and the van't Hoff expression provided in the linked answer also follows.


Avogadro - How can bonds be hidden and only show atoms? I am using Avogadro for visualizing some cluster systems. Is there a way to hide the bonds of the system, ie. only showing atoms?

Thanks!

Not in the current version, as far as I know.

If you need it, you could either make a feature request on the Sourceforge page, or you could do something with extensions, such as making a new version of the ball and stick renderer removing the code that renders bonds, or making an extension that adds a button to delete all the bonds.


What are these crystals on my champagne cork? I opened up a bottle of champagne that was left in a wine cooler for approx. 8 years. To my astonishment there were small crystals at the underside of the cork.

The crystals are colorless, odorless and have no specific taste at all. What is going on here?

Image:

Close up:

They are potassium bitartrate crystals (source).

The crystals form because the potassium bitartrate is not very soluble.

Since solubility is a function of temperature, when wine is chilled the solution can become saturated, causing the precipitate to form in the bottle. Stabilizers are often added to prevent this.

In your case, the precipitate is on the cork, and so what I think has happened is that some wine soaked into the cork, perhaps while the bottle was moved. Then later, the bottle was stood upright or on its side where the wine could not reach the cork. As the wine evaporated back into the bottle, the local concentration of potassium bitartrate increased until it reached the saturation point, at which point the crystals grew. Since the crystals take a very long time to dissolve once formed, they will persist even if wine reaches the cork again.

The good news is, they are harmless (especially good news given that you tasted them - please don't make tasting strange crystals a habit!). They are made of the same substance that cream of tartar, which is a common baking ingredient, is made of.


Toothpaste and water I noticed when brushing my teeth that toothpaste causes water to spread away from it. If you put a bit of toothpaste water mixture on your finger and touch it to a wet sink you will see the water on the sink travel away from your finger.

My question is what chemical in toothpaste is causing this to happen and in particular what property of the chemical is causing it.

Drops or patches of water in your sink are basically water that is not evenly spread out. The surface tension of the outermost water molecules acts a bit like a "balloon" containing the inner molecules. To a small degree it is thus possible for the water molecules to be stacked on top of each other because the water's surface tension is stronger than the gravitational pull.

Toothpaste contains detergents like sodium dodecyl sulfate which act as a surfactant reducing the surface tension of water.
When you coat your finger in a mixture of water and a weak detergent like toothpaste or a stronger one like soap and touch the water drops in the sinks, the water mixes with the detergent. This lowers the surface tension and gravity spreads the drop out. The water is thus not really moving away from your finger, but the drops are being spread out under and around your finger.


Fractional distillation of crude oil and resulting fuel ratios I understand from the diagram below that different fuels condense at different temperatures. But I don't know anything about crude oil (and very little about chemistry in general).
Could I simply refine 100% of crude oil to gasoline by heating it up to 150 °C (value taken from diagram).

Put another way, are the ratios of end products from refining crude oil fixed?

Crude oil is a complicated mixture consisting of a variety of different hydrocarbons and the particular composition is very variable depending on where it is found.

Distillation is merely a means of separating those components so it can do nothing to create more of one of them. And gasoline is, essentially, just a mixture of those components with similar properties and boiling points. There is little you can do by distillation to create more gasoline.

However refiners do have access to chemical means to alter the composition. Catalytic cracking, for example, converts constituents into different chemicals and this can allow a refinery to create more gasoline or diesel by breaking down the higher-boiling components into smaller parts. But this is much more complicated than distillation.


Why benzene takes in more delocalized electrons from its substituents even though it has six of them Benzene has six delocalized electrons in its structure then where does it places the new electrons which it gets from other substituents like oh, etc. Why does it accepts those new ones?  Please give an elaborated answer.

Using atomic orbitals to construct molecular orbitals involves some shifting and redrawing because of overlaps. Benzene is symmetrical, so whether you draw six carbon atomic p-orbitals or squash and spread them out, you get the idea. Now, when you look at phenol, the C-O bond is unsymmetrical, so if you want to be more precise, you say that oxygen draws some electron density from the carbon thru the sigma system (because oxygen is more electronegative).

The oxygen becomes richly negative at carbon's expense, but this isn't fair(!), so it gives back a little electron density thru the pi system of molecular orbitals that it doesn't care about so much (because the pi lobes are farther from the oxygen nucleus). If you draw molecular orbitals, draw the sigma bond with more electron density toward oxygen, and draw the pi system with electron density a wee bit toward carbon.

The energy diagram of the molecular orbitals of benzene will be shifted by putting a hydroxyl on the ring. One one hand you can consider the six orbitals of the aromatic system merely being shifted in the space above the carbon ring, or, on the other hand, you could consider the three filled pi orbitals to remain essentially the same while oxygen donates some electron density to the unfilled (anti-bonding) orbitals (ignoring the sigma system, which just gets shoved around a little).

So the bottom line is that you are not putting whole electrons into someplace new, but swelling up some electron balloons while squashing some others.


How to calculate the number of water molecules of a hydrate based on mass loss upon heating? A $\pu{4.250 g}$ sample of $\ce{Na2SO4.nH2O}$ the sample loses $\pu{2.388 g}$ upon heating. What is $n$ for this hydrate?

Do I first determine the mass of the water than the amount of substance of the sample and than the whole number ratio?

Determine what the $\pu{2.388 g}$ is.

   It's the amount of water lost from the crystal structure, assuming the sample was heated enough for this purpose.

What does 'mole' mean? Isn't it just an indicator of the number of atoms/ions/molecules? So, what would be the relation between the quantity sodium sulfate and water molecules?

   Yes, 'mole' is an indicator of quantity. Take a look at the formula: $\ce{Na2SO4.nH2O}$. What does that mean? It means in a unit, there are $n$ water molecules trapped inside the crystal framework, which consists of sodium and sulfate ions.

How do you proceed into finding the amount of substance (in mole)?

   You need to somehow 'remove' any unknown variables to calculate the amount of substance, so you need to remove $n$. Try subtracting the amount of water from the weight of the whole sample:
 $$m(\ce{Na2SO4.nH2O}) - m(\ce{H2O}) = m(\ce{Na2SO4})\\ \pu{4.250 g} - \pu{2.388 g} = \pu{1.862 g}$$

How many grams of anhydrous $\ce{Na2SO4}$ are there? What is the formula to convert that into moles? How many moles of $\ce{Na2SO4}$ are there?

   The formula is $\displaystyle \mathfrak{n}=\frac{m}{M}$ where $m$ is mass, $M$ is molar mass and $\mathfrak{n}$ is the amount of substance. Thus $\displaystyle \mathfrak{n}=\frac{\text{mass of anhydrous sample}}{\text{molar mass of sodium sulfate}} =\frac{1.862}{142} = \pu{0.0131 mol}$.

You found how many moles of sodium sulfate are there. Isn't that going to be the same for water? Plug into the formula; and find $n$.

   We know that for in any unit of Gluber salt, there are 2 sodium ions, one sulfate ion, and $n$ molecules of water. Thus, it's safe to assume that $\mathfrak{n}$ for water is $n$ times that of what we found for sodium sulfate.
 $$\begin{align}0.0131\times n &=\frac{2.388}{18} \\ &= 0.132\bar{6} \\ \Rightarrow n&=10\end{align}$$


Is the phenyl cation or ethynylium more stable? I recently gave an exam and I received an interesting question to which I believe the answer is wrong.

Given two compounds, a benzene molecule with carbocation at one of its constituting carbon atoms and an ethyne carbocation. Now according to me even though ethyne carbocation is pretty unstable due to the sp hybridized carbon, but if a charge is placed on a carbon in the benzene ring, too much strain is caused on the ring which destabilizes it more. I believe that the stability of the ring is a more dominating factor when carbocations are involved,  but the answer was in favour of the ethyne carbocation.

Could someone please clear this doubt if whether any other factors were taken into consideration or my basic premise is wrong?

Preamble
As trivial as this question might seem, it is not. It should certainly never be asked in at an exam level lower than basic quantum chemistry.
Any of the tempting explanation schemes will certainly fail. A bonding picture in the Lewis formalism is almost impossible; hybridisation descriptions in these compounds is extremely complex.
As a general tip: The smaller the molecule, the more complicated is the bonding.
What is stability?
This question seems so simple, yet is incredibly complex. As a quantum chemist, I see any molecule as stable if it is an energy well (stationary point) on the potential energy surface (and this might already be a simplification). This concept does not help in any way when you want to compare stabilities, because it is just one property of a particular arrangement of nuclei in a particular state.
There is no such thing as absolute stability. We cannot measure it. It is an abstract concept, that we like to think that we understand it. An example from the real world: ice is stable below 0 °C and vapour is stable above 100 °C. Yet ice is not more stable than vapour.
Comparing stability, we always need a framework of multiple variables or constants and a reference point.
It is therefore often assumed, that the boundary conditions are the same (exactly). Even on a purely theoretical level this it is extremely difficult to find these exact boundary conditions. Sooner or later you have to introduce approximations, which eventually crash your system, especially regarding the reference point.
It is difficult, but not impossible.
In order to compare the relative stability of those two cations we can assume a hypothetical hydride transfer between them, hence building a closed reference frame. In computational chemistry we use the concept of isodesmic reactions for such purposes. It is obviously not free of any bias and a complete hypothetical, but it can give us some more insight.
In this case we can propose the following hypothetical isodesmic reaction:

Note that the number of bonds of the same type are the same on each side are equal. Close enough at least. One could argue, that we break a hydrogen sp²-carbon bond and form a hydrogen sp-carbon bond, and vice versa, therefore not having the same types of bonds on the left and right side. But, this is probably as close as we can get.
Having set up a concise set of parameters, we can now go ahead and perform a few quantum chemical calculations. I decided to use my personal standard: DF-BP86(D3BJ)/def2-SVP level of theory.
And we run into more problems
The singlet, closed shell molecules benzene (despite aromaticity) and acetylene are no problem. But when we come to the cations, things get interesting.
The phenyl-cation does not propose a big threat either, it is big enough to compensate the empty orbital in the σ- and π-system.
The high symmetry of the acetylene molecule make it impossible to treat on this level of theory. The linear form is - by my definition - not stable. At least not at this level of theory, at least not in the singlet state. What we do need is a multi-reference method like CASSCF at least. And then we still don't know which state is preferred and if we can still use our proposed hypothetical reaction. Because if the spin state is different on either side, the whole thing falls apart.
Results
Nevertheless while we're already at it, let us look at some of the results and clear up some misconceptions.
Benzene and the Phenyl-cation
Upon ionisation the system needs to compensate the change with a change in geometry. This is a purely electronic effect. In benzene all carbons are trigonal coordinated, quite nicely having three sp² hybrid orbitals; it is one of the very few examples, where symmetry dictates that. Removing a hydrid ion also means removing electrons from such an orbital, freeing up more stable s-character. As a result the bond angle widens to accommodate more s-character in the carbon-carbon bonds. Now this leads to a visibly distorted geometry - a delicate equilibrium of multiple factors.

The π-system remains largely undisturbed. The σ-bonds compensate almost completely for the change. The cationic carbon is now better described as being between linear and trigonal, hence having hybridorbitals between sp and sp².
Acetylene and the Ethynyl carbocation
Acetylene is a wonderful molecule by itself. It has extremely high symmetry, which is quite rare for four atoms. With that comes a large number of degenerate orbitals. And a highly complicated bonding situation.
You might want to regard the carbons as having sp-hybrid orbitals, and that is fine as an approximation, but the deeper you dig, the creepier it gets. (Also see Bent's rule.)
Disturbing this molecule in any way, results in breaking the symmetry. The linear version of HCC+ still has high symmetry and now we have a problem. Upon removal of the hydrid ion, the remaining electron density will try to use as much s-character as possible. As a result part of the s-character is tried to be redistributed into the π-system. These orbitals now do not match the spacial symmetry anymore and the molecule is not stable in its linear form in the singlet state. This is not the whole truth though, but it gets much more complicated and this will certainly be too much for this answer.
A stable structure on this level of theory in the singlet state is an almost three-membered ring. It is probably best to think of it as a protonated dicarbon. All bonds are elongated compared with acetylene.

Since we have a significant different bonding situation now, the previous assumptions of the same type of bonds left and right break down and we can no longer compare these two molecules in the given reference framework.
(Without respect for actual science, we could still do it. Then we would know that the left side of the equation is about 600 kJ/mol more stable than the right.)
Conclusion
The phenyl-cation might be stable, it is a local minimum at DF-BP86/def2-SVP, while the acetylene-cation does not exist. The HCC-cation is better thought of as a protonated dicarbon molecule.
If I understood you correctly, then your assumption, that the "strain" on the ring is bigger than the instability of the orbitals in the acetylene cation, is indeed wrong.


Why is the hydrolysis of the conjugate base of a weak acid neglected in buffer solutions? Consider a solution which contains a weak acid and the salt of its conjugate base with a strong base.
e.g. $$\ce{CH3COOH + H2O <=>[$K_\rm{a}$] CH3COO- + H3O+}\tag1$$
$$\ce{CH3COO- + H2O <=>[$K_\rm{h}$] CH3COOH + OH-}\tag2$$

Now, $\displaystyle{K_\rm{a} = \frac{[\ce{H3O+}] [\ce{CH3COO-}]}{[\ce{CH3COOH}]}}$ and $\displaystyle{K_\mathrm{h} = \frac{K_\rm{w}}{K_\mathrm{a}}}$.

Now, my notes say that for the equilibrium concentration of undissociated acid, the acid coming due to hydrolysis of the salt is neglected and the equilibrium concentration is approximately equal to the initial concentration.

It is also mentioned in a book that

  Assume that the extent of protonation of acetate ions and
  the deprotonation of acetic acid molecules is so small that the concentrations of both
  species are nearly the same as their initial values.

Why is this so? If $K_\rm{a}$ is low, then $K_\rm{h}$ should be high. I understand that the deprotonation of acid will be low as it is a weak acid so that we can neglect the $\ce{CH3COO-}$ coming through the acid dissociation but how can we neglect the hydrolysis of the $\ce{CH3COO-}$ coming from the salt?

Edit $-$

My calculation for equilibrium concentrations.

$\ce{[CH3COOH]_\mathrm{eq}} =$ initially added acid $-$ deprotonated acid $+$ hydrolysed salt

$\ce{[CH3COO-]_\mathrm{eq}} =$ initially added salt $-$ hydrolysed salt $+$ deprotonated acid

$\ce{[H3O+]}_{\mathrm{eq}} =$ deprotonated acid $-$ hydrolysed salt

Let the concentration of

initially added acid $= c_1$

initially added salt $= c_2$

deprotonated acid at eq. $= x$

hydrolysed salt at eq. $= y$

Thus, $$K_{\mathrm{a}}  = \frac{[\ce{H3O+}] [\ce{CH3COO-}]}{[\ce{CH3COOH}]} = \frac{(x - y)(c_2 - y + x)}{(c_1 - x + y)}$$

Now, assuming $x << c_1, \, c_2$,

$$K_{\mathrm{a}}  = \frac{(- y)(c_2 - y)}{(c_1 + y)}$$

$\mathrm{p}K_\mathrm{a} = 4.75$, therefore $K_\mathrm{a} = 10^{-4.75}$. Thus,

$$K_\mathrm{a} = \frac{[\ce{AcO^-}][\ce{H+}]}{[\ce{AcOH}]}$$

So, let's assume $[\ce{AcOH}] = \pu{1.0 M}$. Just to get a feel, how far the acid is dissociated:

$$K_\mathrm{a} = \frac{x \cdot x}{\pu{1.0 M}} \to x = \sqrt{10^{-4.75}}~\pu{M} = 0.00422 \approx 0.4\%$$

where we have assumed that the contribution from the dissociation of water is negligible, and therefore $[\ce{H+}] = [\ce{AcO-}] = x$.

Now, what happens if I add $\pu{1.0 M}$ $\ce{AcO-}$? Then, effectively, we have:

$$K_\mathrm{a} = \frac{[\ce{AcO-}][\ce{H+}]}{[\ce{AcOH}]} = \frac{\pu{1.0 M}
 \cdot [\ce{H+}]}{\pu{1.0 M}}$$

Therefore, we have:

$$[\ce{H+}] = K_\mathrm{a} \frac{\pu{1.0 M}}{\pu{1.0 M}} = \pu{10^{-4.75} M}$$

If we compare the two cases, before addition of acetate, we had a $\mathrm{pH}$ of

$$[\ce{H+}] = \sqrt{10^{-4.75}} \to \mathrm{pH} = 2.375$$

and after addition of acetate, we have

$$[\ce{H+}] = 10^{-4.75} \to \mathrm{pH} = 4.75$$

Meaning that the solution got more basic after the addition of acetate, as excpected.


What color is solid methane? Is it blueish, perfect white or some other color? I am interested because some celestial bodies are said to be covered with methane ice.

Your question is not very specific. So the answer is it depends. Think of water as an analog.

Methane "snow" would be white. (Due to scattering of light).
Solid methane under sufficient pressure would form a clear colorless solid.
Methane "snow" (which would be on surface" of some celestial body) could be any color if contaminated by even small amounts of other hydrocarbons. A reddish color seems to be typical. See ref: http://www.caltech.edu/news/astronomers-find-ice-and-possibly-methane-snow-white-distant-dwarf-planet-1714


Must a good leaving group be a weak base, or a weak nucleophile? In LG Wade's textbook, he has written that a good leaving group must be a weak base. Since substitution reactions (not too sure about elimination reactions) tend to be kinetically controlled, i.e irreversible, why has the concept of a weak base (which is defined for thermodynamically controlled reactions) been used here? Why does the leaving group have to be a weak base rather than a weak nucleophile?

It is a common misconception that good leaving groups should be weak nucleophiles. However, there is a simple counterexample: iodide is known to be both a good nucleophile and a good leaving group in $\mathrm{S_N2}$ reactions.

However, it does make sense for the leaving group to be defined in thermodynamic properties. Consider the reaction equation for a nucleophilic substitution following an $\mathrm{S_N2}$ mechanism:

$$\ce{Nu- + R3C-Lg -> [Nu\bond{...}CR3\bond{...}Lg]^\ddagger -> Nu-CR3 + Lg-}\tag{1}$$

Nucleophilicity, i.e. the ability of $\ce{Nu-}$ to attack the reaction centre is, as you noticed, a kinetic property: the nucleophile has to approach the electrophilic carbon atom from a certain direction in a certain angle with appropriately constructed orbitals. The leaving group itself does play a slight role here in determining the energy of the $\unicode{x3c3}^*$ orbital and thus whether it is energetically favourable for the nucleophile to attack.

However, the leaving group’s principal characteristic is to diffuse away after the reaction. If it is diffusing away, its kinetic properties no longer influence how the reaction proceeds as it is not attacking anything anymore. Rather, a favourable overall reaction enthalpy is achieved if $\ce{Lg-}$ is able to stabilise the negative charge well (or lone pair, if e.g. $\ce{Lg} = \ce{NR3+}$). This is indeed a thermodynamic property, not a kinetic one. And stabilising a negative charge directly correlates with being a weak base.


HCl/H2O with alkene: Which one reacts? I think that the initial step forms a tertiary amine, which then in the presence of the ketone and trace acid undergoes nucleophilic addition. But then, for the $\ce{HCl/H2O}$ in the last step, which one reacts with the double bond in 1G?

I think the last step  is for the Hydrolysis.  After reaction with $\ce{LiAlH_4}$, the amide is reduced to amine and thus when it is reacted with a ketone it forms an imine($1G$) like this below,

which quickly tautomerises to form the most stable Enamine as this

After  that when it is reacted with $\ce{CH_3Br}$, the nitrogen's lone pair conjugates and the nucleophilic attack occurs from the adjacent carbon(like Michael addition) with attached methyl group. Thus you have the intermidiate substituted proiduct as,

But this imine  is hydrolysed in the presence of $\ce{H^+/H_2O}$ to give you back the original ketone but with an extra methyl group attached to it. So I think your final product $1H$ should be this,

and also you get the amine after hydrolysis which is actually the  same as $1F$.


Can molecules diffuse upstream when emptying a cup of tea? I was drinking some tea, and I randomly decided to pour some into the toilet in an unbroken stream; can particles travel "upstream" quickly enough to reach the mug of tea? How fast can diffusion occur in this system?

What if I pour it into highly concentrated toxic waste?  Given that even trace amounts could make it toxic, should I not drink my tea?

Particles/solutes are always diffusing in all directions. The question to ask is, "are they diffusing fast enough to overcome being swept along by the flow?"

Short Answer: No. Don't try it, though.

Long Answer:

One governing equation that can be used to answer this question is the following:

$$
\overline v\!\left(\vec r\right)\cdot \overline\nabla C_i\!\left(\vec r\right) = D_i\nabla^2C_i\!\left(\vec r\right)\tag{1}
$$

This is the (simplifed) steady-state mass transport balance for species $i$ at a given point $\vec r$ in space. The left-hand side represents transport due to convection; the right-hand side encapsulates transport due to diffusion. $\overline v\!\left(\vec r\right)$ is the fluid flow velocity, $C_i\!\left(\vec r\right)$ is the concentration, and $D_i$ is the diffusivity. For concision, the explicit dependence on $\vec r$ will be omitted below.

Given that we're most interested in the transport in the direction of the flow of the tea, if we approximate the stream as a cylinder of uniformly flowing liquid with velocity $v_z$ and assume that concentration variations perpendicular to the flow are unimportant, Eq. $\left(1\right)$ becomes:

$$
v_z{dC_i\over dz} = D_i{d^2C_i\over dz^2} \tag{2}
$$

if we choose the $z$-axis as lying along the stream of tea.

The most useful way to use Eq. $\left(2\right)$ is to non-dimensionalize it. That is, to re-write it so that all of the variables and coefficients have no units. For this problem, $z$ and $C_i$ are the variables to be non-dimensionalized, whereas $v_z$ and $D_i$ are parameters to be grouped into non-dimensional groups.  So, let's define:

$$
\eta = {z\over L}; \Theta = {C_i\over C_i^*} \tag{3}
$$

$L$ is some length scale representative of the system of interest. Here, the best choice is probably the height you're holding the cup over the surface of the toxic waste. $C_i^*$ is a representative concentration of a chemical species of interest, such as the nastiest player in that concentrated toxic waste. To substitute these definitions into Eq. $\left(2\right)$, I'll rewrite them as $z=\eta L$ and $C_i = \Theta C_i^*$:

$$
{v_zC_i^*\over L}{d\Theta\over d\eta} =
  {D_iC_i^*\over L^2}{d^2\Theta\over d\eta^2} \tag{4}
$$

The terms with derivatives in Eq. $\left(4\right)$ are now dimensionless. To make the coefficients non-dimensionless, divide both sides by $D_iC_i^*/L^2$:

$$
{v_zL\over D_i}{d\Theta\over d\eta} = {d^2\Theta\over d\eta^2} \tag{5}
$$

The new grouping of parameters in Eq. $\left(5\right)$ has a special name: the Péclet number, abbreviated $\mathrm{Pe}$.

$$
\mathrm{Pe}{d\Theta\over d\eta} = {d^2\Theta\over d\eta^2} \tag{6}
$$

The magnitude of $\mathrm{Pe}$ indicates the relative importance of convection relative to diffusion in a problem like this one. If the Péclet number is large $\left(\mathrm{Pe}\gg 1\right)$, it means that convection dominates the mass transfer behavior; if it's small $\left(\mathrm{Pe}\ll 1\right)$, then diffusion is the key phenomenon. If it's in the vicinity of unity $\left(\mathrm{Pe} \sim 1\right.$, or approximately $\left.0.3\leq \mathrm{Pe}\leq 3\right)$, then both diffusion and convection must be considered together.

So, can pouring tea into toxic waste poison your tea?

Well, setting aside the questionable wisdom of bringing a drink into a toxic waste area in the first place, what are some representative numbers for $D_i$, $v_z$, and $L$?  The upper limit for the diffusivity of a solute in water is about $2\times 10^{-6}\,\mathrm{cm^2\over s}$ (see, e.g., here). Since we want to know if diffusion can outrace convection, we should give it the best chance possible and selecting this maximum value will do that. (That is to say, maximizing $D_i$, which is in the denominator of the definition of $\mathrm{Pe}$, will tend to minimize $\mathrm{Pe}$.)

Similarly, selecting the minimum practical values of $L$ and $v_z$ will also tend to minimize $\mathrm{Pe}$, giving us a worst-case scenario for whether or not you face a gruesome death from toxic waste poisoning.  Let's say, optimistically(???), that one inch is the minimum height you can hope to pour your tea from without it splashing back into the cup. If we (rather loosely) approximate the  velocity of the tea stream at the toxic waste surface as that of a freely falling particle, the velocity will be $v_z=\sqrt{2gL}$.

So, our estimated Péclet number is (Google Calculator result):

$$
\mathrm{Pe} = {v_zL\over D_i} = {\sqrt{2gL^3}\over D_i} = 8.96\times 10^7
$$

This is a huge Péclet number. There's no way any of that toxic waste will diffuse up into your tea. (Please don't test this, though. Splashing, breathing toxic fumes, falling in, etc. are very real possibilities.)

As a further note, observe that in combining terms to form Eq. $\left(5\right)$, the reference concentration $C_i^*$ cancelled out completely. So, in this case, the answer is the same regardless of how concentrated that toxic waste is.


Black & Green Nickel Oxide-Supplemental A previous question (April 2017) asked about black $\ce{NiO}$ did not supply any answer. Sadly, such a referral only deters colleagues from commenting on the current question.

There is a great deal of literature on cation defects in non-stoichiometric nickel (II) oxide. There is much on charge-transfer in defective crystal structures. But, I can't find any attempt to marry the two together - which may explain how the deep colour (black) is observed.
Of course, it may simply be that I have not  discovered it yet. I did find a question on how/why copper (II) oxide is black. This one has semiconductor properties; so, $\ce{CuO}$ and non-stoichiometric $\ce{NiO}$ may be black for similar reasons?
Can anyone comment?

The non-stoichiometric form of nickel (II) oxide and copper (II) oxide are black, with the same square-planar (at metal) solid-state structure, (with crystal defects), so the metal centres are bridged by oxides. The assignment of oxidation state is +II; therefore, could consider a fragment of the structure as: -O-Cu(II)-O-Cu(II)-; but, with redox-active metals, this also:
-O-Cu(III)-O-Cu(I)- thus. Electrons can drift through the structure giving rise to intervalence electronic transitions, with strong involvement of oxygen atoms. The influence of pie-donating oxygen prevents the formation on Cu(0), the native metal.

Intervalence transitions can give rise to intense colours (e.g. black); normally, forbidden to colours arising from d-d electron transitions. (This, also seen with Prussian Blue, where Fe(II) & Fe(III) are bridged by cyanides.) A discrete Cu(II) centre would have a weak, less-intense, colour, arising from said d-d transitions.

Stabilisation of highly-polarising Cu(III) & Ni(III) is achieved by the oxide, which is a very electronegative and pi-donating ion, stimulating low activation-energy electron-flow.

Transitions between p and d orbitals (as in permanganate: metal and oxide) give rise to very intense colours. These extended oxide lattices are also susceptible to being defective, so missing ions; or, ions in the wrong location in the lattice. This also gives rise to colour. An example of this is seen in zinc (II) oxide, when heated. In this case it is not mixed valence; but, loss of oxygen from the lattice, that gives rise to defects that allow lower energy (than in "complete" ZnO) electronic transitions to occur and absorption at the UV-end of the visible region. On cooling the lattice picks up oxygen, from the air, becoming complete again, reverting to its original white from the, high-temperature, yellow colour.


Reduction of Organic compounds I've encountered in chemistry quite a lot of reducing agents. However one question strikes me quite often and  any research done on the same hasn't been fruitful. This is in context to organic chemistry. Now for example, in reductive ozonolysis, we often use ozone in combination with zinc and water. Zinc's presence ensures reductive results and not oxidative ones. Now, some sources(not books) which I have seen, say that some sulphur compounds would do the same like $\ce{Me2S}$. Also, in case of 1,3- dinitrobenzene,  $\ce{Sn + HCl}$ converts both the $\ce{-NO2}$ groups to $\ce{-NH2}$ groups. Now if we use $\ce{NH4HS}$ or some other nitrogen and sulphur containing ammonia like compounds, we get reduction again, but only for one group out of the two. So my question is, how do we :

Decide that there is going to be reduction
Know the extent of reduction (partial/full)

Can some insight into the inorganic properties of the elements provide an answer like d-orbital availability, lone pair, electronegativity, size/charge ratio? I've tried reasoning but haven't gotten anywhere. Or is it that memorization of the reagents is the only way to know it's function.
Note: This question is only in context to some less commonly used reagents, so as to recognize it's function if we ever see one in problems present in question books.

I will try to explain the difficulty of this problem by using a simpler reaction, the reduction of nitrobenzene, which is already difficult enough. The reduction of dinitrobenzene should produce more different possibilities.
Nitrobenzene can be reduced into at least five different compounds. It depends on the choice of the reducing agent. I am afraid nobody can justify the particularly mechanisms used by a given reducing agent to get to each final result.
With ($\ce{Sn + HCl}$) or with ($\ce{Fe + HCl}$), it gives aniline, $\ce{C6H5NH2}$.
With sodium methoxide ($\ce{NaOCH3}$), it gives azoxybenzene, $\ce{C6H5N(O)NC6H5}$
With zinc, it can produce three different products:

With zinc + $\ce{NH4Cl}$ at elevated temperature, it gives
phenylhydroxylamine, $\ce{C6H5NHOH}$.

With zinc and an alcoholic solution of $\ce{NaOH}$, it gives azobenzene
$\ce{C6H5N=NC6H5}$.

With zinc and an aqueous solution of $\ce{NaOH}$ at $\pu{60 ^\circ C}$, it give hydrazobenzene $\ce{C6H5NH-NHC6H5}$.


Why are we not supposed to heat Erlenmyer flasks? Chemistry says that we are are we supposed to heat boiling flasks, and not Erlenmeyer flasks.  However, I can't find any information about this on Google.  Is there a particular reason for this?

Is it just because we want to separate the purposes for our glassware, or is there an actual valid reason?

This really depends on what you mean by heating.

One of the main uses for Erlenmeyer flasks is for recrystallization, which of course involves heating the container and contents. Typically this is only moderate heating, such as over a steam bath or with a hair dryer. Some manufacturers specify that their Erlenmeyer flasks are ideal for heating applications.

Traditionally, boiling flasks have more uniform and thicker walls, which allow for much greater heating limits, and a more uniform heating distribution. They are designed to handle more intense localised heating, and manage thermal and physical shock distribution. Erlenmeyer flasks usually have thinner walls and greater wall thickness variation, and are less able to handle sudden thermal changes, such as heating with a high energy source like a Bunsen, or heat gun. Using them in this way may well lead to cracking, fracturing or worse, and is unsafe.

So, although Erlenmeyer flasks can be heated, they are not specifically designed for high temperature, constant heating applications. For this, you should use a boiling flask.


Why sodium sulphate has water absorbing capability? At first I thought maybe sodium sulphate in contact with water produces sulphuric acid which absorbs water but I do not think it is actually a valid reason.

Sodium sulphate reacts readily with water at room temperature to form hydrates up to sodium sulphate decahydrate, $\ce{NaSO4\cdot10H2O}$. $$\ce{NaSO4 + 10H2O -> NaSO4\cdot10H2O}$$ This means that $\ce{NaSO4}$ can absorb up to 10 mol of water for every 1 mol of salt that is used, making it one of the most effective drying agents in terms of sheer capacity.

Using data from this source we can calculate the Gibbs energy change for the above reaction at different temperatures. $$\mathrm{\Delta G_{20^\circ C} = -1.33~kJ~mol^{-1}}$$ $$\mathrm{\Delta G_{30^\circ C} = 1.28~kJ~mol^{-1}}$$

Since the entropy change for the reaction is negative, we can see that the cooler the solution you are trying to dry is, the more effective $\ce{NaSO4}$ will be as a drying agent.


How to properly store acids at home? I bought different kinds of acid for experiments and home usage that I stored in secured containers into an IKEA cabinet.

I just realized that despite the fact that all the containers are properly closed, the metal parts of the cabinets are completely rotten. I am quite surprised because there shouldn't be much vapor in there.

What is the correct way to store acids at home? Should I purchase some specific container? How can I prevent vapors from escaping? Perhaps the problem is somewhere else and my containers are not good enough for my acids.

The acids I have: $\ce{H2SO4}$ (30%), $\ce{HCl}$ (32%), $\ce{H3PO4}$ (85%), $\ce{HNO3}$ (10%), $\ce{CH3COOH}$ (98%).

Note: the chemicals are actually stored in my electric/electronic workshop which is a separate room at home and always locked when I have visitors.

First I'd locate the bottle which causes the problem. Usually HCl is #1 suspect, but to be sure you can put a vial with smelling salts (aqueous solution of $\ce{(NH4)2CO3}$) or ammonia in the box with acids; white coating of $\ce{NH4Cl}$ on the bottle signifies the leak.

It's also a good practice to store acids in glass bottles with a proper joint (teflon ring) and a screw cap (e.g. Merck's SafetyCap). I would strongly recommend to get the proper bottles as soon as possible. Plastic bottles are only used to reduce the production and transportation costs, they are a poor choice for a long-term storing of chemicals. Even thick plastic is prone to diffusion, whereas glass is not; also plastic is, well, plastic, and is prone to mechanical deformations, so it's also tricky to maintain an impermeability of a bottleneck-cap joint over time.

In the meantime I'd wrap the necks of the bottles with parafilm, and/or pour some baking soda (sodium bicarbonate) on the bottom of the container where you store the bottles. This should help to neutralize the vapors before they reach the furniture and your nose, but it's a temporary solution.


Galvanic corrosion: Copper, iron and aluminum in an automotive cooling system I am building a small copper heat exchanger I am going to plumb into my coolant system. I'm afraid that the copper will corrode my aluminum radiator or iron engine through galvanic action via the coolant. How was this prevented on older vehicles with iron engines and copper radiators? Or is this less of a problem then I expect it will be?

If the copper components are not in direct contact with the aluminum or iron parts, then there should be no electrolytic corrosion. This would require that non-conductive rubber hoses are used for coolant lines (avoid those with steel spring cores to prevent collapsing on the vacuum side, lest the rubber crack through and allow metal-to-metal contact). Also mount the copper parts on rubber shock mounts, rather than directly to the chassis.

Actually, corrosion was not well prevented in the past. For example, some Renault engines had an aluminum block with steel cylinder liners (shrunk with liquid nitrogen and dropped in place to expand for a very tight metal-to-metal seal, if I remember correctly). This short-circuited cell corroded rapidly. Modern antifreeze contains corrosion inhibitors such as silicate/phosphate mixtures that form an insulating film on metal and should help prevent any issue with the copper parts. Use a good quality ethylene glycol antifreeze with inhibitors.


Using Hess's law to find molar enthalpy of formation without an initial reaction I'm having issues relating Hess's law to this question:

  The given chemical equation represent the combustion of ammonia and the combustion of hydrogen

  $$
\begin{align}
\ce{4 NH3 + 3 O2 &->6 H2O + 2N2} &\qquad &ΔH_1 = \pu{-1516 kJ} \tag{1} \\
\ce{2 H2 + O2 &-> 2 H2O} &\qquad &ΔH_2 = \pu{-572 kJ} \tag{2} \\
\end{align}
$$

  What is the molar enthalpy of formation for ammonia?

  A. $\pu{-100 kJ}$; B. $\pu{-50 kJ}$; C. $\pu{50 kJ}$; D. $\pu{100 kJ}$.

I'm not sure how I would apply the  product–reactant equation here, or rearrange the equation to get my answer since I don't have an initial reaction to compare it to. How would I go about answering this question?

Since you are asked to find the enthalpy of formation for ammonia, it's convenient to write the equation of ammonia synthesis normalized for $\pu{1 mol}$ of $\ce{NH3}$:

$$\ce{0.5 N2 + 1.5 H2 -> NH3}$$

To obtain this equation and apply Hess's law, the following linear combination of the provided equations should be used:

$$
\begin{align}
\ce{4 NH3 + 3 O2 &->6 H2O + 2N2} & ΔH_1 &= \pu{-1516 kJ} &&|\cdot(-0.25) \tag{1} \\
\ce{2 H2 + O2 &-> 2 H2O} & ΔH_2 &= \pu{-572 kJ} &&|\cdot 0.75 \tag{2} \\
\hline
\ce{0.5 N2 + 1.5 H2 &-> NH3} & ΔH_3 &= -0.25ΔH_1 + 0.75ΔH_2 \tag{3}
\end{align}
$$

So the final answer is

$$ΔH_3 = -0.25\cdot(\pu{-1516 kJ}) + 0.75\cdot(\pu{-572 kJ}) = \pu{-50 kJ}$$

The catch here is that you are asked to find molar enthalpy of formation, and calculating for the "convenient" equation with integer coefficients

$$\ce{N2 + 3 H2 -> 2 NH3}$$

you eventually would obtain the doubled value of $\pu{-100 kJ mol-1}$ referred to $\pu{2 mol}$ of ammonia (I almost fell for it, too).


How to test for fluoride levels in drinking water? I would like to test for fluoride levels in the water.

Are there any methods or test kits available to test the levels of fluoride in home water supply?

See also: Is there any at home test for fluoride levels in your water, and what is the best water distillation unit?

A good reliable set-up is pricey.  Here's a link to a reliable, quality probe for fluoride ion detection in water, it costs \$740  (US).  The meter that it plugs into sells for \$790 (US) if you get the less expensive hand-held device.  There also manuals and instructions on how to actually use the gear, at the website as well.


Which of the resonance structures is more stable? I am quoting a rule on resonance from Organic Chemistry by T.W. Graham Solomons, Craig B. Fryhle, 12th edition, page 25.

  Structures in which all the atoms have a complete valence shell of electrons
  (i.e., the noble gas structure) are more stable.

I tried to apply to the following.

According to this rule structure 4 must be most stable. But structure 4 is not aromatic, while structure 1 is aromatic. Since aromatic species are more stable then nonaromatic species, structure 1 must be more stable.

My question is which of the structures is more stable?

There are few things you should keep in mind while deciding comparative stability of resonance structures:

Non-polar structures are more stable than dipolar.

Resonance structures with a greater number of covalent bonds are more stable than those with lesser number. Thus, nonpolar structure of buta-1,3-diene is more stable than any other resonating structures.

Resonance structures with similar charges on adjacent atoms are insignificant due to electrostatic repulsion and thus are unstable.

Structures with positive charge on electropositive element and negative charge  on electronegative element are more stable.

Structures in which charge is delocalised are more stable than those in which there is charge separation.

Structures with complete octet are more stable.

All the atoms in a molecule taking part in resonance should be co-planar. because this enables effective overlap of p-orbitals and delocalization of electrons.

But when you have to decide among these properties itself, you can use the following observational facts:

Although during comaprison, there may be a positive charge on the more electronegative atom(O in this case) with complete octet, but it will be preferred over a structure with an electropositive atom ( C in this case ) having positive charge and incomplete octet.

But when number of covalent bonds are same, positive charge on C atom with incomplete octet is more preferred than positive charge on more electronegative O atom.

A stable neutral structure is a major contributor of a resonance hybrid.

An important note: when you are comparing the stability of the resonance structures , you are actually trying to figure out the major contributor to the resonance hybrid (that which best describes it). Technically, all the resonance structures are contributors to the same true structure of the ion. Since all of your mentioned structures represent the same species, they cannot have have different energies, and therefore they cannot have different stabilities.
Your question is actually asking "which contributor is more important in describing the structure and behavior of the hybrid?"
So fact number 1 and 2 makes it very clear that structure 4 in this case will be the most stable.


Meta-directing nature of anilinium cations The comments on this answer seem to detail a trend in which $\ce{NR3+, NHR2+, NH2R+}$ all show meta-directing nature towards electrophilic substitution, with the slightly stronger para-direction of $\ce{NH3+}$ being the odd one out, as cited from March's Advanced Organic Chemistry.
The 2020 edition has this to say, on pages 617 and 618 under "Aromatic Substitution, Electrophilic" (emphasis mine);

Groups that lack an unshared pair on the atom connected to the ring are electron
-withdrawing with a (+) charge or (δ+) dipole. In this category are, in approximate order of decreasing deactivating ability, $\ce{NR3+}$, .... and $\ce{NH3+}$. Also in this category are all other groups with a positive charge on the atom directly connected to the ring... and many groups with positive charges on atoms farther away, since these are often still powerful −I groups. The field-effect explanation predicts that these should
all be meta directing and deactivating, and (except for $\ce{NH3}$
+) this is the case. The $\ce{NH3+}$ group is an anomaly, since this group directs para about as much as or a little more than it directs meta, often because the ammonium ion is in equilibrium with the amine. The $\ce{NH2Me+, NHMe2
+}$, and $\ce{NMe3+}$ groups all give more meta than para substitution, with the percentage of para product decreasing with the increasing number of methyl groups.

@ron's explanation coupled with the equilibrium argument makes a pretty good reason for why aniline behaves this way.
But why do the other groups meta-direct? The field effect mentioned above seems to refer to inductive effects, which should make them para-directing.

This does not answer your question exactly, but I hope this can shed some light on the anomalous behaviour of anilinium cations.
First of all, it is not true that the para-product is always favoured with the $\ce{NH3+}$ ion, it depends on the reaction conditions. There are two effects to consider here—1) kinetic/thermodynamic control, and 2) rate determining step of the reaction. If the reaction is under thermodynamic constrol, the most stable isomer will form regardless of the directing property of the substituent. If it is under kinetic control, then you have to look at the rate determining step of the reaction. The electrophilic aromatic substitution is a two step process in general, and if the first step is rate determining, then the nucleophilicity of each postion of the ring will be the main factor. If the second step is rate determining then the energy of the intermediate will also matter. Obviously the ortho and para intermediates will be unstable as the positive charge is delocalized next to the electron withdrawing group. So, it's not always true that inductive effect  by itself cannot be meta-directing.
Anyway, back to your original question. Let's consider the nitration of aniline in presence of conc. $\ce{H2SO4}$.

Brickman and Ridd[1] has noted the relative ratios of products for different sulfuric acid concentrations.
$$\begin{array} {|r|r|}\hline \text{conc. of }\ce{H2SO4}(\%) & 85 & 92.4 & 96.4 & 100 \\ \hline ortho(\%) & 6 & - & - & - \\ \hline meta(\%) & 34 & 53 & 58 & 64 \\ \hline para(\%) & 59 & 47 & 42 & 36 \\ \hline  \end{array} $$
As you can see, the meta product becomes more favoured as the concentration of $\ce{H2SO4}$ increases.
What about the N-methylated anilines? In another paper, Brickman, Utley and Ridd[2] have mentioned the percentages of meta and para products (ortho does not form at high acid concentrations, probably due to the steric effect).
$$\begin{array} {|r|r|}\hline \text{conc. of }\ce{H2SO4}(\%) & 90.9 & 96.2 & 99.8 \\ \hline ortho(\%) & - & - & -  \\ \hline meta(\%) & 61 & 67.5 & 70.3 \\ \hline para(\%) & 39 & 70.3 & 29.7 \\ \hline  \end{array}$$
$$\text{For N-methylanilinium ion }\ce{Ph-NH2Me+}$$
Here, we again see the same trend that the increasing sulfuric acid concentration increases the meta-product percentage. However, the meta-product is now the major product.
$\ce{Ph-NHMe2+}$ also shows similar results. The N,N,N-triemthylanilinium ion $\ce{Ph-NMe3+}$ shows $89\%$ m-product and $11\%$ p-product at $98\%$ conc. $\ce{H2SO4}$.
In the two papers, the authors mention that the analysis of acid concentration vs. rate plot seems to indicate that the substitution predominantly happens on the anilinium cation i.e. the protonated state, so there is no possibility of the para-product arising from the reaction of the deprotonated compound. Sulfuric acid is a pretty strong acid, so it's unlikely that there is much deprotonated aniline left at such high concentration. The p/m ratio of aniline nitration changes by a factor of less than $2$, when the acid concentration is changed from $90$ to $98\%$, which reduces the proportion of deprotonated form by a factor of $31$. So, it's not possible that the para-product is coming from aniline, because then the rate profile would have been much more sensitive to the concentration of acid.
They have also ruled out significant hyperconjugative effects, by comparing the rates of the benzylammonium variants of all of these anilines. So, the effect must be mostly from inductive effects.
After a lengthy discussion, it is noted that the $\ce{-NH3+}$ group is much more solvated than the $\ce{-NHMe2+}$, $\ce{-NMe3+}$ etc. groups. So, it is possible that the solvent forms hydrogen bonds to the $\ce{-NH3+}$ group and allows the positive charge on it to spread away. This means that the inductive effect from $\ce{-NH3+}$ is actually weaker than the methylated groups. So, the methylated anilinium ions are simply more strongly averse to reaction on ortho/para positions than anilinium.
However, the authors note after that[2]:

The above explanation  accounts for the major differences in the substituent effects recorded in Table 3 but it is not completely satisfactory, for the extent of  para-substitution in the anilinium ion is  more than would  be  expected if the effect of replacing $\ce{-N+Me}$, by $\ce{-N+H}$, were simply to "scale down" the effect of the positive pole;  such an effect would  never  cause the para-position to be more reactive than one meta-position. This anomaly is also illustrated by the comparison of  the anilinium ion with the benzyltrimethyl-ammonium ion: the latter ion is more reactive by a factor of ca. 25 but gives less para- substitution. It appears that the protonated poles give significantly higher p/m ratios than expected  from their reactivity relative to the methylated poles.
The reason for this is not entirely clear ...

So, it's not completely known why aniline gives more para-products than its N-methylated derivatives.
This is only considering the nitration reaction. If you consider other reactions, there are varied directing nature. For example the sulfonation reaction (done with $\ce{H2SO4}$ and $\ce{SO3}$) produces almost entirely the para-product[3] (sulfanilic acid), and this has been attributed to the fact that the sulfonation is reversible (thermodynamic control). I haven't been able to find any data on halogenation or other substitutions on the anilinium ion.

M. Brickman, J.H. Ridd, J. Chem. Soc., 1965, 6845-6851
M. Brickman, J.H.P. Utley, J.H. Ridd, J. Chem. Soc., 1965, 6851-6857
J.J. Jacobs Jr., D.F. Othmer, A. Hokanson, Ind. Eng. Chem., 1943, 35, 3, 321–323


How to determine the valence electrons of D-block elements in their mono atomic form? For s and p block we can determine the valence electrons easily,but what about d-block elements. How to determine their valence electrons?

Based on the Aufbau principle, they would all be filling the (x-1)d sublevel, so they would ALL have 2 valence electons in the xs sublevel.

Since the Aufbau priniciple fails to account for the fact that some elements will be more stable with incomplete valence xs sublevels and additional electrons in the (x-1)d sublevel, you just need to memorize the transition metals that are more stable with 1 valence s electron:

Chromium, Possibly Nickel, Copper, Niobium, Molybdenum, Ruthenium, Rhodium, Silver, Platinum and Gold,

And the two extra specials:

Palladium which prefers to fill the x-1d sublevel and not put any in the xs sublevel.  It has 18 in its highest energy level, level 4, but only the 10 4d electrons are considered to be valance electons.

And Lawrencium which has 7s2 (5f14) xp1, or 3 valence electrons, putting 1 electron in the xp sublevel rather than in the (x-1)d sublevel.  It turns out to be a d-block element that is not filling the d-block.

Notes
This will give you the number of electrons in the highest energy level for transition metals.  Transition metals in the highest d sublevel (but not in the highest primary energy level) can provide valence.  Many sources will refer to all of the electrons above the configuration of the prior noble gas to be valence electrons.

On one hand, most of the oxidation states for transition metals have been observed, up to the point where they would have lost all of those d electrons.

On the other hand, the d electrons in transition metals are not de-localized in metallic bonding of transition metals (to the best of my knowledge).


Electroless Plating of Zinc onto Copper in NaOH Solution I’m researching how the classic “alchemy” demonstration works in which zinc is plated onto a copper penny in a solution of sodium hydroxide. Most online resources refer to the electroplating of copper with zinc, which is not what I’m interested in. The best article I’ve been able to find on the electroless plating of zinc onto copper is from the Royal Chemistry Society. However, I have several questions. The three reactions this article describes are:

$\ce{Zn(s) + 2NaOH(aq) + 2H2O(l) -> Na2Zn(OH)4(aq) + H2(g)}$, which is the “formation of sodium zincate”

$\ce{Zn(s) -> Zn^2+(aq) + 2e- }$, which is the “electrode reaction” at the “zinc electrode” and is “followed by the complexing of the zinc ions as $\ce{Zn(OH)4^2- (aq)}$”

$\ce{Zn(OH)4^2- (aq) + 2e- -> Zn(s) + 4OH- (aq)}$, which is the “electrode reaction” at the “copper electrode”

I really don’t understand how these three reactions come together to describe the plating of the zinc onto the copper. Specifically, I don’t understand:

1) What is the role of zincate, $\ce{Zn(OH)4^2- (aq)}$, in the reaction?

2) On the molecular level, how are the zinc and copper actually attached together?

3) How is this an electrochemical cell (i.e. how is electricity involved)? Where can I learn more about electrochemical cells?

An explanation is provided in this reference. The idea is that the electrode reactions are

$\ce{Zn(s) + 4OH^- -> [Zn(OH)_4]^{2-} + 2e-}$
and

$\ce{[Zn(OH)_4]^{2-} + 2e- -> Zn(alloy) + 4OH^-}$

So the zinc powder remaining will go into solution and the zincate is reduced on the copper surface to form a form of zinc rich brass. Brass is an alloy and the atoms are held together via metallic bonding.