The significance of the dissociation constant of an indicator In the chapter acid-base, I learned about indicators and how they are actually weak acids.

This is the equation given in our book:

$\ce{HIn + H_2O <=> H_3O^+ + In^-}$

$K_{in}= \left(\frac{[\ce{H_3O^+} ][\ce{In^-}]}{[\ce{HIn}]}\right)$

When $\ce{HIn}$ and $\ce{In^-}$ are of different colors

My question:

Further in my book, the concentration of $\ce{H_3O^+}$ at which $\ce{[HIn]=[In^{-}]}$ is calculated. Apparently, this is the concentration of $\ce{H_3O^+}$ at which the indicator changes color and thus this can be used to identify end point of a titration. However, how is this the point where the indicator "changes" color, I mean isn't the indicator changing color throughout the titration as the concentration of $\ce{H_3O^+}$ is changing continuously as the alkali is added to the acid for example
Why do different indicators have different $pK_{in}$ i.e. "change color" at different concentration of $\ce{H_3O^+}$

1. It all depends on the $\text{p}K_{\text{in}}$ of the indicator.

If we look at the Henderson-Hasselbalch equation:

$$\text{pH} = \text{p}K_{\text{a}} + \log \left( \frac{[\ce{A-}]}{[\ce{HA}]} \right) $$

You can see that when $\text{pH}=\text{p}K_{\text{a}}$, then logarithm part must equal 0. This occurs when we have equal parts of the protonated and unprotonated state, because when $[\ce{A-}]=[\ce{HA}]$, the ratio between them will be 1, and $\log 1=0$.

Every time you increase $\text{pH}$ by 1, to balance the equation, the logarithm must also be 1 higher, and this happens when the thing you take the logarithm of becomes 10 times larger. This means that for a small change in pH, we have a large change in the ratio between protonated and deprotonated forms.

Typically, with indicators, you say that the colour change happens when $\text{pH} = \text{p}K_{\text{a}} \pm 1$, because outside of that interval, you will quickly have almost purely $\ce{A-}$ or $\ce{HA}$.

  If you follow this far, let's try an example with an indicator with a high $\text{p}K_{\text{in}}$

You put an indicator with $\text{p}K_{\text{in}}=10$ into pure water, which has a $\text{p}K_{\text{a}}=7$. From these values, we can see that water is a stronger acid than the indicator. This means that $[\ce{H3O+}]$ just in pure water will be high enough, that the indicator is largely protonated:

$$\text{pH} = \text{p}K_{\text{in}} + \log \left( \frac{[\ce{In-}]}{[\ce{HIn}]} \right) $$

$$7=10+\log \left( \frac{[\ce{In-}]}{[\ce{HIn}]} \right)$$
$$\log \left( \frac{[\ce{In-}]}{[\ce{HIn}]} \right)=-3$$

Which means $[\ce{HIn}] = 1000*[\ce{In-}]$ - try plugging in 1/1000, 1/100, 1/1, 100/1 etc. into the logarithm on your calculator to see why this is so!

Anyway, we see that when the indicator has a $\text{p}K_{\text{in}}$ significantly higher than the $\text{p}K_{\text{a}}$ of the solution, it will mostly be protonated. Why? We will see this in part 2 of the answer.

  Another example. Let's see what happens if we have a low $\text{p}K_{\text{in}}$ indicator

You put an indicator with $\text{p}K_{\text{in}}=3$ into pure water ( $\text{p}K_{\text{a}}=7$). This time the indicator is a stronger acid. This means that $[\ce{H3O+}]$ is not high enough to keep the indicator largely protonated:

$$\text{pH} = \text{p}K_{\text{in}} + \log \left( \frac{[\ce{In-}]}{[\ce{HIn}]} \right) $$

$$7=3+\log \left( \frac{[\ce{In-}]}{[\ce{HIn}]} \right)$$
$$\log \left( \frac{[\ce{In-}]}{[\ce{HIn}]} \right)=4$$

Which means $10000*[\ce{HIn}] = [\ce{In-}]$ - Statistically, every time you find a protonated indicator molecule, you will have found 10000 deprotonated indicator molecules!

  Titration scenario: acetic acid titrated with NaOH, using an indicator with $\text{p}K_{\text{a}}=10$

Acetic acid has a $\text{p}K_{\text{a}}=4.7$. The acetic acid is a stronger acid than water, so it will have increased the concentration of $[\ce{H3O+}]$. To raise the $\text{pH}$, we need to lower this concentration. As we add $\ce{OH-}$ to the solution, the following reaction occurs:

$$\ce{H3O+ + OH- <=>> 2H2O}$$

We are removing $\ce{H3O+}$ from the solution. But the acetic acid was only partially deprotonated, and what happens if the remove $\ce{H3O+}$ from the following reaction?

$$\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$

Well, by the principle of Le Chatelier, acetic acid and water reacts further, to replenish the $\ce{H3O+}$. This goes on until the acetic acid is deprotonated to a large degree, and then the $\ce{OH-}$ very quickly raises the $\text{pH}$ of the water, since there is no acid to react with anymore.

When the $\text{pH}$ gets into the area of the $\text{p}K_{\text{in}}$ of the indicator, the colour change starts to happen. It happens swiftly, because we are not using very much of the indicator. This means that it doesn't take a lot of $\ce{OH-}$ to do to the indicator what we did to the acetic acid.

2. Indicators have different equilibrium constants because they have different acid/base energetics
Indicators have different values for $pK_{in}$ for exactly the same reasons as acids have different $\text{p}K_\text{a}$. It depends on the stability of the bond to the hydrogen that can be lost as a proton. If the bond is very unstable (as in HCl), the proton will lose its share of the electrons in the bond, and become free very easily, and the compound will have a low $\text{p}K$.

For a compound such as acetic acid, protons will become free, but protons from $\ce{H3O+}$ can also transfer back to the acetic acid. The more $\ce{H3O+}$ that exists in the environment, the larger the chance, that a deprotonated acetic acid molecule will collide with it and receive the proton.

For the indicator with a high $\text{p}K_{\text{in}}$, if there is a lot of $\ce{H3O+}$, then the indicator molecules will be re-protonated as fast as they can lose their protons. Thus, they overall stay protonated.

The equilibrium constants $K_{\text{in}}$, $K_\text{a}$, etc. basically tell you how likely the molecule is to shed its proton.


Link between centrosymmetry, dipole moment and chirality What is the relationship between the three concepts in the title?

I know all centrosymmetric molecules are achiral, but not all achiral molecules are centrosymmetric. Thus, there are no molecules that are centrosymmetric and chiral, but all other combinations exist.

What about having a dipole moment vs. centrosymmetry? And what about having a dipole moment vs. chirality? Is there a relationship between them, i.e. are some combinations possible and others not?

And if there is a link between these concepts, what is the proof? For example, for a centrosymmetric molecule, I could apply the inversion operation (which does not change the molecule as it has centrosymmetry) and then rotate it by 180$^\circ$, showing that there is a mirror symmetry. Thus, the molecule is achiral.

A molecule can't be chiral if it has an $S_n$ improper rotation (note that $S_1$ is equivalent to a mirror plane and $S_2$ is equivalent to inversion). Chirality is only possible in the $C_n$ and $D_n$, as well the less common $T$, $O$, and $I$, point groups, which only have rotation axes as symmetry elements.

A dipole moment can only occur along the primary $C_n$ rotation axis of a molecule (if it has one) and if there are any perpendicular mirror planes or rotation axes, it will have to be zero by symmetry. A $S_n$ improper rotation with $n>1$ also precludes a dipole. This limits the point groups that can have dipoles to $C_n$, $C_{nv}$, and $C_s$.

Centrosymmetry is equivalent to having the $S_2$ symmetry operation. As mentioned above, this eliminates the possibility of chirality or a dipole. However, centrosymmetry is just a special case and  molecule won't be chiral and/or won't have a dipole under other conditions as well. That is to say, centrosymmetry is sufficient, but not necessary for eliminating chirality and/or a dipole.

To summarize:
$$\begin{array}{|c|c|c|} \hline
\text{Condition} & \text{Exact Point Group} \\ \hline
\text{Chiral/Nonpolar} & I,O,T,D_{n} \\
\text{Achiral/Polar} & C_{nv}, C_s  \\
\text{Chiral/Polar} & C_n \\
\hline
\end{array}$$

Where I'm taking "exact point group" to mean the highest symmetry group the molecule is contained in. This delineation has some minor exceptions. Technically, molecules of the appropriate symmetry still might be nonpolar (or at least too small to measure) by an accidental cancellation. For chirality, while all molecules of these symmetries are formally chiral, they can appear achiral in practice if the enantiomers can readily interconvert, as occurs with nitrogen inversion.


What interesting things can be seen in spectra between 400-700nm? The visible spectrum (400-700nm) tends to give a very blurry spectral picture, as it is not a wide band relative to the typical peak width of features you can find there. Further, the vibrational modes tend to fall below it in IR, and the electron transitions that get physicists excited tend to fall above it in UV.

What are some examples of interesting features that a sightseeing student could find with a visible-spectrum spectrometer? They are quite common in low-budget labs but tend to only get used for measuring concentration with Beer's law.

Absorbance spectrum of colored solutions may be boring for a "sight seeing" student, as you call yourself. Emission spectrum is far more interesting for visual learners. Anything luminous, which you can see with your eyes, should have a spectrum in the visible range. This includes the sky, moon, a red hot stove, fluorescent lamps, computer screen, flames, bulb, LED etc.

  The visible spectrum (400-700nm) tends to give a very blurry spectral
  picture

I gather you are talking about absorbance spectrum of solutions but this is not universally true. If you ever had a chance to try holmium oxide "solution" placed in a spectrometer, its absorbance spectrum has very sharp features.

Emission spectrum on the other hand is far more interesting. One can see very sharp emission lines even with a CD spectroscope. All you have to look at sodium street lamps, white fluorescent bulbs or the moon (one can see very weak solar spectrum with dark lines).

If are looking for low budget "instruments" I would suggest that you play with a CD or DVD as a diffraction grating rather than using a lab based absorbance measuring instruments. That is just a black box.

https://hackaday.io/project/28529-portable-dvd-spectroscope


Is it safe to combine sodium hypochlorite (bleach) sodium triphosphate (TSP) for cleaning? I've heard this combination can be particularly effective at cleaning biofilms, perhaps because the TSP is a surfacant, allowing bleach to "attack" more layers of the biofilm.
However, I'm also aware one should be cautious when mixing anything other than water with bleach, as there are a wide variety of reactions that can occur. But I've seen nothing about bleach and TSP. Should it be safe?

Actually, the more complete answer is that Na3PO4 is a recommended stabilizer for bleach.
Also, per a ScienceDirect source, to quote:

Trisodium phosphate hydrolyzes to produce hydroxide ions. Phosphate acts as buffer...'coordinated phosphate pH control.'

where pH control prolongs the shelf life of chlorine bleach, as supported by the statement (same source):

One down-side of chlorinated products is a limited storage stability. This instability reduces the chlorine benefit which the detergent can deliver, but also reduces the benefit of the other active ingredients which are destroyed by bleach action.

Further, more advanced radical-based bleaching is suggested by the comment:

For dyeing with vinyl sulphone dyes, dyebath is prepared with trisodium phosphate (2%) and dye.

So, the scavenging of light (or transition metal contamination) induced radicals (which can lead to unwanted chlorate formation), is likely also an advantage of employing Na3PO4, in my opinion.


In the pH system of acids and bases, are there 10^-7 loose ions per liter (at pH of 7) or per mole of solvent? (Or per mole of total substance?) Most places tell me that the number on the pH scale tells you the number of dissociated ions, in moles, per liter of substance.    But, Wikipedia's article pH says this....:

... For a commonplace example based on the facts that the masses of a mole of water, a mole of hydrogen ions, and a mole of hydroxide ions are respectively 18 g, 1 g, and 17 g, a quantity of $10^7$ moles of pure (pH 7) water, or 180 tonnes ($18\times10^7$ g), contains close to 1 g of dissociated hydrogen ions (or rather 19 g of H3O+ hydronium ions) and 17 g of hydroxide ions.

But, if there are 10^-7 moles of hydrogen cations (protons) per LITER, and one mole of protons is approximately one gram (as Wiki's example says), then there is one mole (one gram) of protons per 10^7 liters of water, right?  One gram of hydrogen cations per 10,000,000 liters?
As opposed to what Wiki says, claiming one gram of loose hydrogen ions per 180 tonnes of water?
(One liter of water weighs about one kilogram, and there are a thousand kilograms in a tonne, correct?)
EDIT: Wiki's section on pH and all the incorrect numbers have been corrected.

In the pH system of acids and bases, are there 10^-7 loose ions per
liter (at pH of 7) or per mole of solvent? (Or per mole of total
substance?)

In a rigorous sense none of the options is correct, because pH being approximated by molarity is an approximation itself. The problem lies in the measurement itself and pH should be
$$pH= -log (a_{H^+})$$
The pH probe "senses" the activity of hydrogen ions not the molarity. This is why the pH of 5 M HCl is not equal to $$pH= -log (5)$$
There are no free or loose hydrogen ions in water, like a bare proton, rather this proton is attached to a network of water molecules.
Wiki is talking about an approximation. 1 ton is equal 1000,000 grams, so 10$^7$ mole/g x 18 g $\ce{H2O}$ is indeed 180 t. This part is correct so far. Note the author must be following metric ton. The whole world does not follow the US ton.
1 liter of water weighs ~ 1000 grams. Then 1000 g water contains 10$^{-7}$ moles of $\ce{H+}$, and 18x10$^{7}$ g water should have 1.8 x 10$^{-2}$ mol $\ce{H+}$ or 0.018 g $\ce{H+}$. I don't get 1 g.


How does Al-Hg react with carbonyl in presence of concentrated HCl? In Clemmensen reduction, $\ce{Zn-Hg}$/ $\ce{ conc. HCl}$ reduces the carbonyl $\ce{>C=O}$ to $\ce{>CH2}$, by the following mechanism(?):

What happens to the carbonyl group if we replace Zinc with any other metal such as Al or Na?

Will this reaction take place if X is $\ce{Al-Hg / conc. HCl}$ ?

In comparision to Clemmensen reduction, both $\ce{Na-Hg}$ and $\ce{Al-Hg}$ cannot be used as direct substitutes for $\ce{Zn-Hg}$.
Reaction with $\ce{Na-Hg}$ amalgam:
$\ce{Na-Hg}$ amalgam is also a good reducing agent and is capable of reducing ketones/aldehydes to alcohols1. However, Do not use conc. $\ce{HCl}$ with $\ce{Na-Hg}$ amalgam, as there would be a vigorous reaction between the acid and the sodium in the amalgam. Instead, use an organic solvent such as absolute alcohol.
The mechanism for this reduction is a bit different than standard Clemmensen reduction:

\begin{align}
\ce{2(C6H5)2CO +2Na &->2(C6H5)2C-ONa} \\
\ce{2(C6H5)2C-ONa&<=>(C6H5)2C(ONa)-C(C6H5)2-ONa\tag{1}} \\
\ce{2(C6H5)2C-ONa + 2C2H5OH &-> 2(C6H5)2C-OH + 2C2H5ONa} \tag{2a} \\
\ce{2(C6H5)2CO &-> (C6H5)2CO +(C6H5)2CHOH \tag{2b}}
\end{align}

Reaction with $\ce{Al-Hg}$ amalgam:
Aluminium amalgam, like Sodium amalgam, vigorously reacts with water, and you cannot use concentrated $\ce{HCl}$ with it. With ethanol as a solvent, it is a mild reducing agent, can reduce cycloalkanones and aldehydes to alcohols. Acyclic ketones remain almost inert2

In some cases, it behaves similarily to $\ce{Mg-Hg}$ amalgam and promotes dimerization of ketones to form pinacols. Thus, your reaction would not take place, and instead pinacol would be the major product.

This question and this one cover some other reduction uses of $\ce{Al-Hg}$, such as reduction of imines to amines.
References:

Bachmann, W. E. “The Mechanism of Reduction by Sodium Amalgam and Alcohol. I. The Reduction of Aromatic Ketones to Hydrols.” Journal of the American Chemical Society, vol. 55, no. 2, Feb. 1933, pp. 770–74. doi:10.1021/ja01329a051.
Troyansky, Emmanuil I., and Meghan Baker. “Aluminum Amalgam.” Encyclopedia of Reagents for Organic Synthesis, John Wiley & Sons, Ltd, 2016, pp. 1–6. doi:10.1002/047084289X.ra076.pub2.


Gaussian09 vs home made Hartree-Fock implementation I implemented a restricted Hartree-Fock (RHF) calculation in the STO-3G basis set, as described in Szabo and Ostlund's book [1]. I managed to reproduce the energies of all their calculations ($\ce{H_2}$, $\ce{HeH^+}$, $\ce{H_2O}$, $\ce{CO}$, $\ce{CH_4}$ and $\ce{FH}$) with precision. For example, for $\ce{N2}$ I found
$$
E_\text{tot}^\text{HM} = -107.49583784393724 \text{ a.u.}
$$
while in the book they report the following (p. 192)
$$
E_\text{tot}^\text{Szabo} = -107.496 \text{ a.u.}
$$
The situation is analogous for all other molecules I tested.

Now I am trying to implement the calculation of forces, in order to perform Born-Oppenheimer molecular dynamics. I will start to compute the gradient of the PES by finite differences (computationally costly but easy to implement). In order to check my implementation I wanted to use Gaussian09 (for the first time!) as a benchmark because we don't find many force calculations on standar books. In order to check that my input file is correct (i.e. the options can reproduce my "dumb" implementation) I performed a single point calculation for the $\ce{N2}$ dimer. Surprisingly I found
$$
E_\text{tot}^\text{G09} = -106.765838750 \text{ a.u.}
$$
which is quite different from my result. The input file I used is the following:

%NProcShared=2
%Mem=1GB
#P RHF/STO-3G SP Symmetry=None Units=Bohr Pop=None Guess=Core

RHF

0 1
N 2.074 0.000 0.000
N 0.000 0.000 0.000

I tested the same structure on the input file for $\ce{H2O}$:

%NProcShared=2
%Mem=1GB
#P RHF/STO-3G SP Symmetry=None Units=Bohr Pop=None Guess=Core

RHF

0 1
H +1.4305507125e+00 +0.0000000000e+00 +0.0000000000e+00
H -1.4305507125e+00 +0.0000000000e+00 +0.0000000000e+00
O +0.0000000000e+00 +1.1072513982e+00 +0.0000000000e+00

In this case I find
$$
E_\text{tot}^\text{G09} = -74.9629400524 \text{ a.u.}
$$
wigh Gaussian09, while with my program I have
$$
E_\text{tot}^\text{HM} = -74.962937313769473 \text{ a.u.} = E_\text{tot}^\text{Szabo}
$$
For $\ce{H2O}$ everything seems to work fine. In a similar way, I get matching results also for $\ce{CH_4}$, $\ce{FH}$ and $\ce{CO}$.

There is a problem with the $\ce{N2}$ molecule (or simply with the input file) I am not considering?

Which is the best set of options for Gaussian09 that can emulate a home-made Hartree-Fock program (written following closely Szabo's book [1])?

[1] A. Szabo and N. Ostlund, Modern Quantum Chemistry, Dover, 1996.

Diagonalization of the core Hamiltonian provides usually not the best guess for the SCF procedure to say the least, and thus, by default Gaussian uses a more sophisticated guess obtained by diagonalizing the Harris functional (Guess=Harris). With this default guess one get the same energy as OP reported:

 SCF Done:  E(RHF) =  -107.495842181     A.U. after    5 cycles

Now, Pop=None was a terrible idea, of course. It is always advisable to look at the final orbitals. Everything looks fine for the case of the default guess

 Alpha  occ. eigenvalues --  -15.51806 -15.51612  -1.44283  -0.72249  -0.57311
 Alpha  occ. eigenvalues --   -0.57311  -0.53949
 Alpha virt. eigenvalues --    0.28131   0.28131   1.12344

but results with the core Hamiltonian guess look suspicios

 Alpha  occ. eigenvalues --  -15.44896 -15.44853  -1.34440  -0.64250  -0.60670
 Alpha  occ. eigenvalues --   -0.52980  -0.16921
 Alpha virt. eigenvalues --   -0.07243   0.31047   1.20826

The energy of the first virtual orbital is negative which is odd. It is difficult to say what went wrong with the core Hamiltonian guess, but some clue can be provided if turning the symmetry back on (this is another very sensible default turning which off was a bad idea, I think).

The default guess yields the electronic state with a particular symmetry ($\sigma_{\mathrm{g}}$)

 Initial guess orbital symmetries:
       Occupied  (SGU) (SGG) (SGG) (SGU) (PIU) (PIU) (SGG)
       Virtual   (PIG) (PIG) (SGU)
 The electronic state of the initial guess is 1-SGG.

and converges within this symmetry

 Orbital symmetries:
       Occupied  (SGG) (SGU) (SGG) (SGU) (PIU) (PIU) (SGG)
       Virtual   (PIG) (PIG) (SGU)
 The electronic state is 1-SGG.

But the core Hamiltonian guess yields the electronic state with no symmetry

 Initial guess orbital symmetries:
       Occupied  (SGG) (SGU) (SGG) (SGU) (PIU) (PIU) (PIG)
       Virtual   (PIG) (SGG) (SGU)

and converges within this state with no symmetry

 Orbital symmetries:
       Occupied  (SGG) (SGU) (SGG) (SGU) (PIU) (PIU) (PIG)
       Virtual   (SGG) (PIG) (SGU)
 Unable to determine electronic state:  partially filled degenerate orbitals.

which is again suspicious.

So, there is something wrong with the core Hamiltonian guess for this system (I suspect the basis is too small for the core Hamiltonian guess to yield sensible results), thus, SCF spits out some "strange" numbers. Strictly, speaking there is nothing strange with the numbers. Remember that SCF usually converges to a local minimum, and which particular depends on which part of the wave function space the initial guess placed the system in. And besides, you're not even guaranteed to have a minimum, SCF might converge to a stationary point of any kind.


How to distinguish first excited state and second excited state? I am learning excited state calculation. In this paper, the author use "$D_0$","$D_1$" represent excited state of nitrate radical.

I know the symbol such as "$D_1$","$T_2$" means first excited doublet state, second excited triplet state. Can I get specific electron configuration of a molecule with these symbol,like ${}^2\!A_2$ symbol in this answer? Or first and second excited state are different in a range of energy? In other word, Can I write a electron configuration use symbols like "${}^1\!A_2$" for $D_1$ state of nitrate molecule?

Unfortunately, no. If you don't want to work it out by hand, the best you can hope for is that the authors explicitly specify orbital occupations with something that looks like an electron configuration. This paper doesn't, the previous paper sort of does.

Here is a paper that doesn't have symmetry labels, but does have configurations. For example, on page 2977, the SCF reference for $\ce{CuO}$ is given as

$$
\ce{ (O 2s)^2 (O 2p\sigma)^1 (O 2p\pi)^4 (Cu 3d\sigma)^2 (Cu 3d\pi)^4 (Cu 3d\delta)^4 },
$$

where the wavefunction symmetry is determined by the singly-occupied orbital (everything else is doubly occupied). $\ce{2p\sigma}$ probably lies along the bond, which would be taken as the $z$-axis when looking at a character table. Go to the $C_{\infty v}$ table, check for $z$, and discover that it's $\Sigma^+$, which is even mentioned in the paper. So there are symmetry labels given here, but the linear groups also have Greek characters for their irrep labels...

One interesting point is that non-Abelian groups are not used in practical quantum chemistry. $C_{\infty v}$ would be reduced to $C_{2v}$, and the $^{2}\Sigma^+$ state would become $^{2}\!A_{1}$.


Calculating new volume during expansion of an ideal gas Question
  $\mathrm{5~mol}$ of a gas is confined in a piston (volume $\mathrm{100~cm^3}$) at $\mathrm{400~K}$. $\mathrm{10~kJ}$ of heat energy is supplied, causing a reversible and isothermal expansion of the gas. Calculate the volume of the expanded, heated, gas. You may use the equation below (assume ideal gas behaviour).

  $$W = -nRT(\ln V_2-\ln V_1)$$

I substituted the values (after converting to SI units) into the equation and attempted to rearrange to find $V_2$, but work was unknown. I tried to use the enthalpy value (10 kJ) given but I was unsure of what to do with it.

I dont understand where you are getting stuck up since everything is already given to you and you just have to substitute stuff using:

$\Delta U= W+Q$, the basic conservation equation.

$Q$ is the heat supplied. Think about the change in internal energy ($U$) during an isothermal process.


Cyclic elimination reaction Which of the following will give cyclic elimination by heating?

  phenol
  propanoic acid
  trimethyl isopropyl ammonium hydroxide
  cyclohexyl propanoate

Cyclic elimination by heating is Syn Pyrolytic elimination reaction. This is given by ester, quaternary ammonium salt, and amine oxide. The reaction is according to Hoffman. This cancels out option (1) and (2).

Option (4) is ester and hence the answer to the question.

But why does the ammonium salt not give the reaction? What am I getting wrong?

Example of a similar tertiary ammonium reaction with mechanism

The thing is, it does react but there will be no cyclic product, in the end, I think that is the reason it is not the answer here.


Friend threw mercury in a fire, is this dangerous? My friend broke a mercury thermometer and threw the mercury in the fire. Is this dangerous? and what should she do about the fireplace?

Note that I generally do not use a lot of boldface in posts. Please recognise that it is required here.

Immediately get out of the room! Keep children and pregnant women away! Ventilate that room very well for a long amount of time! And never let your friend handle a mercury thermometer again!

Liquid mercury is not as dangerous as some people make it to be; a number of mercury compounds (e.g. dimethyl mercury) are much more toxic. However, that is because of the way most people would be exposed to mercury typically. If the thermometer breaks, there is a puddle of mercury on the floor and barring yourself doing anything stupid it will take a while for the mercury to fully evaporate. It’s inhaling mercury vapour that is most dangerous with respect to liquid mercury.

Mercury’s boiling point, according to Wikipedia, is some $350~\mathrm{^\circ C}$. A fire in a fireplace is typically hotter. Thus, by pouring mercury into the fire, your friend has chosen the most effective way to generate mercury vapour quickly from liquid mercury. Unless the fireplace is a closed oven, this vapour will now gladly move around the room with air currents, thus being easily breathed in and absorbed by the body.

Of all the ways to handle mercury, your friend chose the most dangerous and the most stupid!


Chemical composition of silver tarnish I had taken as a given that silver tarnish is a layer of silver sulfide on the surface of a utensil, and that cleaning it using a thiourea based agent like Tarn-X dissolves the sulfide layer.

However, while recently cleaning a variety of tarnished sterling and silver plated utensils, I noticed (1) the sulfureous odors emitted during the cleaning process varied considerably from one piece to the next; and (2) the tarnish on a few pieces could not be removed completely via the non-abrasive chemical process.

Knowing that many foods such as members if the cabbage and onion families have various sulfur compounds in them as flavor and aroma components, and that the sulfur containing amino acids cysteine and methionine are almost universally present in living organisms (including those we eat), I wondered: is silver tarnish a complex mixture of silver complexes that might reveal a history of the foods a person has eaten. Sherlock Holmes would know for sure.

The composition of Silver tarnish is Silver sulfide, to quote from Wikipedia:

Silver sulfide is an inorganic compound with the formula $\ce{Ag2S}$. A dense black solid, it is the only sulfide of silver. It is useful as a photosensitizer in photography. It constitutes the tarnish that forms over time on silverware and other silver objects. Silver sulfide is insoluble in most solvents, but is degraded by strong acids.

The last sentence is important as Silver cleaning is not accomplished with harsh chemicals, but via electrochemistry which leads to the hydrogen atom radical formation, which then reduces the $\ce{Ag2S}$ liberating $\ce{Ag}$ and $\ce{H2S}$.  Here is a short summary of likely mechanics:
$\ce{Al -> Al3+ + 3 e-}$
$\ce{H2O ⇌ H+ + OH-}$
$\ce{HCO3- ⇌ H+ + CO3(2-)}$
$\ce{H+ + e- ⇌  •H }$
Effectively, $\ce{•H}$ can operate as a $\ce{(e-, H+)}$ pair:
$\ce{Ag2S + 2 •H -> 2 Ag + H2S (g)}$
Here is a confirming source, which states to quote:

The nascent hydrogen may also act as a reducing agent as it evolves from the metal surface.

And also, per the same source, the formation reaction of Ag2S can be reversed:

In that case silver suffers attack by dissolved oxygen and present sulphide species according to Reaction (5):
$\ce{4Ag (s) + 2HS- (aq) + O2(g) ⇌ 2Ag2S (s) + 2 OH- }$ (5)

So working in air, it is advised to employ a base to drive the above equilibrium to the left. Also, in alkaline conditions expect the reaction:
$\ce{H2S + 2 OH- -> 2 H2O + S(s)}$
Which is also noted in the referenced source, to quote:

However, after the cleaning of heavily tarnished samples (immersed for 30 and 60 min) some small quantities of sulfur (~0.3wt.%) are detected on the surface by EDS analysis, and the colour is changed to light yellowish.

[EDIT] With particular reference to the thiourea system, it is described in this hydrometallurgy source.  A cited reaction equation is:
$\ce{ Ag_2S + (SC(NH)(NH)2)2 + 2 H^+ + 4 SC(NH2)_2 -> 2 Ag(SC(NH2){_2}){_3}^{+} + S_o }$
Silver metal can then be recovered with reductive electrolysis.


Do auto-catalytic aromatic compounds capable of forming self-replicating molecules exist? I want to know are there any aromatic compounds able to catalyze their own synthesis? Do aromatic compounds with molecules capable of self-replication exist?

P.S.I'm looking specifically for heterocyclic aromatic compounds (preferably containing nitrogen in the ring) capable of self-replication which could self-assemble into sheets of planar structures capable of self-replication. Is such thing possible and are such structures known in modern organic chemistry and/or nanotechnology?

I recall seeing a system that consisted of two planar components (and I am fairly certain those were aromatic) that could couple to form the product. The self-catalysis consisted of a product molecule acting as a template for two component molecules to absorb to, which was claimed to have increased the reaction rate.

Edit: I have found the system I was thinking about. The Asker may decide for themselves if this fits the bill. The example is given in K. Ariga, T. Kunitake: "Supramolecular Chemistry - Fundamentals and Applications", Springer, 2006, p. 31. doi: 10.1007/b137036. As of this writing, the page in question is available through Google Books and the reference list for the section (2.10) is on the publisher's page. I was not able to definitively select the correct reference and I do not have access to them, anyway. Note that Ariga & Kunitake seem to have up their mind on what to call it:

  In this system, [the product] acts as a template for self-replication. It may be surprising that a simple host-guest system like this mimics the fundamental activity associated with life, self-replication.


How do I extract cyanide from apple seeds? I'm working on a crime story about cyanide poisoning from apple seeds. I just would like to have an idea of what processes and extraction techniques might be involved in getting cyanide from the seeds. The character is supposed to have access to a high school chemistry lab; so I was thinking equipment and tools that can be found in a typical HS lab. I prefer a detailed answer, step by step, if possible. If not then just generalized terminologies would work.

Forget about the apple seeds, they contain about 1 to 4 mg amygdalin per gramm seeds (DOI).
Instead, collect apricot seeds during the right season, the amygdalin content varies though the year and can be as high as 5% of the dry weight of the seed (DOI).
It is probably advantagenous to break the husk with a nut cracker, a plier, etc. and cut the softer inner material to smaller pieces.
Extraction of amygdalin can be performed by

immersing the material in methanol and subsequent ultrasonification
Soxhlet extraction with methanol
reflux extraction in water in the presence of citric acid

A comparison of the extraction methods is given here.

Removal of the solvent in vacuum will yield a crude material with significant amounts of amygdalin. You might want to have a look at this article from the Western Journal of Medicine on the toxicity. Here, an $\mathrm{LD_{50}}$ of 522 mg amygdalin per kg body weight was estimated for the oral application to rats. The online resource of the U.S. National Library of Medicine gives a value of 405 mg/kg.

Further information on the health risk of apricot kernels are provided by of the German Bundesinstitut für Risikobewertung (Federal Institute for Risk Assessment) and the British Committee on Toxicity.

A note in the a German medical journal, Deutsche Ärzteblatt, (PDF) describes a case where boy of four years (110 cm, 18 kg body weight) was given apricot kernels during an alternative cancer treatment. Upon additional treatment with a single dose of 500 mg amygdalin, the kid showed agitation, spasms and the eyes started to roll.

I'll leave it up to your fantasy as a writer on how to apply the poison, but spicing some marzipan with it might help ;)


Effects of Groups on the Nucleophilcity of Nitrogen The reactivity of nitrogen mustards increases with increasing
  nucleophilicity of the central nitrogen atom. Select the most and least reactive from each of the following groups of nitrogen mustards.

My Attempt

I know that II will be the most reactive since the methyl group will donate electron density to the nitrogen atom due to inductive effects, making it more nucleophilic and hence more reactive.

I also thought that III will be the least reactive since the ester group pulls more electron density away from the nitrogen atom compared to the carbonyl group. I also thought in general that esters are stronger deactivators than carbonyl groups. However the answer is I.

After searching on the internet, it turns out that esters and carbonyl groups are roughly equally strong deactivators. So how am I suppose to determine which one will be less nucleophilic?

Your analysis of "pulls more electron density away" is a bit too simplistic. Both of them are approximately equally electron-withdrawing via the inductive effect, but in terms of resonance, which is the main factor here, they are quite different.

Compound I is an amide. Amides are non-nucleophilic on nitrogen because the N lone pair is delocalised into the C=O π* orbital. In fact, amides act as nucleophiles via the carbonyl oxygen and not the nitrogen.

In compound III, there is also the same delocalisation, but a way of looking at it is that the other oxygen (on the -OMe group) also wants to donate its lone pair into the C=O π*. In a sense this means that there is some degree of competition between the nitrogen and the oxygen, and the nitrogen lone pair is no longer delocalised to such a great extent.

Formally speaking, I think the best way of putting it is that the HOMO of compound III has a greater coefficient on nitrogen than the HOMO of compound I. You can also draw resonance forms to show that there is a smaller contribution from the resonance form with positively charged nitrogen.


Friedel-Crafts Alkylation of Pyrrole I thought that I already know how the Electrophilic Aromatic Substitution (EArs) for Aromatic rings work but then this reaction was shown to me and I was really sure that my professors didn't bothered teaching or even mentioning to us this reaction:

In my defense, at my University, our Org Chem professors always use benzene or benzene derivative rings for demonstration of EArS so I never knew how an acylation of a five-membered aromatic ring works.

Can someone please explain to me how this reaction works?

To my knowledge, Friedel-Crafts alkylation of pyrrole with aluminum chloride ($\ce{AlCl3}$) is unattainable. However, recent literature suggests F-C alkylation of pyrrole with other catalysts is possible. For example, an asymmetric Friedel–Crafts alkylation of pyrrole with nitroalkenes catalyzed by a copper complex of a bisphenol A-derived Schiff base is reported (Ref.1):

  Abstract: The asymmetric Friedel–Crafts (FC) alkylation of pyrrole with nitroalkenes was mediated by $\ce{CuBr2}$ and a novel bisphenol A-derived chiral catalyst at room temperature. The catalyst was found to be applicable for the asymmetric FC alkylation of pyrrole with a wide range of nitroalkenes, affording optically active alkylated pyrroles with enantioselectivities up to 94%. Furthermore, enantiomerically pure 3-nitro-2-arylpropanamides were prepared by the oxidative cleavage of the pyrrole rings in the FC products with $\ce{NaIO4/RuCl3}$ to demonstrate the synthetic application of the products.

For other example, see Ref.2 & 3. For a review, see Ref.4.

References:

Halide Sedef Özdemir, Engin Şahin, Murat Çakıcı, Hamdullah Kılıç, “Asymmetric Friedel–Crafts alkylation of pyrrole with nitroalkenes catalyzed by a copper complex of a bisphenol A-derived Schiff base,” Tetrahedron 2015, 71(19), 2882-2890 (https://doi.org/10.1016/j.tet.2015.03.059).
Yuan-Zhao Hua, Xing-Wang Han, Xiao-Chao Yang, Xixi Song, Min-Can Wang, Jun-Biao Chang, “Enantioselective Friedel–Crafts Alkylation of Pyrrole with Chalcones Catalyzed by a Dinuclear Zinc Catalyst,” J. Org. Chem. 2014, 79(23), 11690-11699 (https://doi.org/10.1021/jo5023712).
Hugh J. Anderson, C. W. Huang, “Pyrrole chemistry. X. Friedel–Crafts alkylation of some pyrrole and furan derivatives,” Canadian Journal of Chemistry 1970, 48(10), 1550-1553 (https://doi.org/10.1139/v70-252).
Hai-Hua Lu, Fen Tan, Wen-Jing Xiao, “Enantioselective Organocatalytic Friedel-Crafts Alkylations,” Current Organic Chemistry 2011, 15(24), 4022-4045 (https://doi.org/10.2174/138527211798109196).


Accelerated stability testing for chemicals: Shelf life / Expiry Date How does one use the results of Accelerated stability testing for chemicals to determine the typical Shelf life  and Expiry Date?

e.g. Suppose the typical storage is going to be at 25 C and the expected shelf life is 2 years (as printed on the labels or the best-before date).

Now in the interest of time suppose we test at a higher temperature how do we extrapolate those numbers to the typical shelf life at room temperature.

Any standard protocols for this? Also what's a reasonable high temperature to use for the accelerated tests. I am dealing with a fairly temperature stable & nonreactive material here. 60 C?

Stability testing seems quite common in the Pharma industry etc. so I'm sure there's norms out there.

Note that I'm not dealing with a drug / medicine / food etc. This is strictly for an intermediate which will not be consumed directly by the end user.

Yes, there are guidelines. For example from ICH and WHO for pharma, and from CIPAC for pesticides. Some can be downloaded freely.

Yes, there are guidelines. For example, from ICH and WHO for Pharma, and from CIPAC for pesticides. Some can be downloaded freely.
Accelerated stability tests are generally not accepted as replacements for full-scale stability tests, but can provide an indication of the formation of undesirable by-products etc.

As a rule of thumb, a very general rule, chemical reaction rates in-/decrease by a factor of 2 for every 10C the temperature changes. According to this rule, if the test temperature is raised from 20C to 30C, the test period could theoretically be halved. However, if you look through the temperature/time combinations in the guidelines, you will find this relationship hard to find :-).


Mass of produced carbon dioxide I am studying for a big test and I ran across something that I couldn't solve no matter how hard I tried. (reading through old tests)

Burning ethanol produces carbon dioxide ($\ce{CO2}$) and water ($\ce{H2O}$).

a) Write down a balanced chemical equation that illustrates the complete burning of ethanol.

b) What is the mass of produced carbon dioxide ($\ce{CO2}$), when completely burning 575 liters of ethanol?

NTP conditions apply. Ethanols density is 0.8 g/cm³.

MC = 12 g/mol, MH = 1 g/mol, MO = 16 g/mol

For the first part I came up with $\ce{C_2H_6O (l) + 3O2(g) --> 2CO_2 (g) + 3H2O (g)}$

Then I calculated that $M_\ce{CO_2}$ = 44 g/mol and $M_\ce{H_2O}$ = 18 g/mol

I also calculated that 575 liters of ethanol weighs 460 kg

I know that m = nM

However I am not sure what to do next.
I may also be using some weird terms here as this is my translation and I've never dealt with math/chemistry in English.

Ethanol's molecular mass $M(\ce{C2H6O})=46\ \mathrm{g/mol}$

$$n=\frac mM,$$
so
$$\frac{460\,000\ \mathrm g}{46\ \mathrm{g/mol}} = 10\,000\ \mathrm{mol}$$

So having 10 000 moles we can multiply that by the molecular mass of carbon dioxide and multiply again by two because there are two carbon dioxides.

$$10\,000\ \mathrm{mol}\times44\ \mathrm{g\ mol^{-1}}\times2=880\,000\ \mathrm g=880\ \mathrm{kg}$$


What is the "resonance hybrid" of benzene? In my book, to explain the structure of benzene, it referred to the idea of resonance:

  The idea of resonance is that the actual molecule is a definite structure that is a hybrid of two or more theoretical structures.

I partially understood this sentence, specifically how a molecule could have multiple structures, but am confused about what is meant by hybrid. Could someone please elaborate?

It means that what we usually consider to be the 'structure' of benzene is not the actual structure.

Although there are three π bonds in the structure of benzene, there is actually resonance, or electron delocalization. This means that the electrons are not localised in π bonds between two specific carbons, but distributed throughout the ring. This is one of the most important characteristics of an aromatic system.

The "localised" structures 1a and 1b above are resonance forms of benzene: neither 1a nor 1b are a true representation of benzene on their own; they are hypothetical, or "theoretical", structures.

The resonance hybrid is a weighted average of the resonance forms 1a and 1b. In this case, both forms contribute equally to the hybrid. So, each C–C bond has a bond order of (1+2)/2 = 1.5, which we can represent using dotted lines in 2 (each dotted line represents a fraction of a bond, so one solid and one dotted line means a bond order between 1 and 2).

More symbolically, we can also use the depiction with an inscribed circle 3. This means that the six π-electrons are shared between all six of the carbon atoms.


Is a temperature decrease required for hydrogen bonds to form in water? As I understand it, increasing the temperature of water causes hydrogen bonds to break.  Does this mean that a decrease in temperature is required for the formation of those hydrogen bonds?

You are right, in a sense.

A more accurate statement would be that as the temperature decreases, the average lifetime of a hydrogen bond increases. Thus, on average the liquid/solid/molecule will be more stable. This is because, as you point out, hydrogen bonds are weak enough to be broken by collisions due to thermal motions (translation and rotation). They are almost always immediately formed again (in liquid), but the more frequently they are broken, the less hydrogen bonding there will be on average.

One indicator of the degree to which hydrogen bonds stay formed is the average number of hydrogen bonds over time per water. In water, this number comes out to be around 3.4 hydrogen bonds, where it could have a maximum of 4.0. Two donors and two acceptor molecules.

Thus, you can rationalize the transition from liquid water to ice in this way by considering that in ice you come quite close to the number of 4.0 hydrogen bonds per water. As you say, then, when temperature decreases, more hydrogen bonds are formed in general. I would hesitate to call it a requirement though because hydrogen bonding can take place at any temperature, but is just rarer when the temperature increases.


How to name the compound with the formula Br3C-CHCl2? Name the compound:

How should I name $\ce{Br3C-CHCl2}$? Is it 1,1,1-tribromo-2,2-dichloroethane or 2,2,2-tribromo-1,1-dichloroethane?

The first name is correct (1,1,1-tribromo...).  Counting the substituents from either end, you would get 1 for the first substituent going from either direction.  As a tie-breaker, look at the next substituent, which would give you 1,1 going from the bromine end, and 1,1 going from the chlorine end.  Since the tie is still not broken, you go to the third substituent.  Going from the bromine end you would have 1,1,1 but 1,1,2 going from the chlorine side.  So, you number the carbon with the bromines number 1, giving you 1,1,1-tribromo-2,2-dichloroethane.

As a side note, if the carbon on the right had three chlorines instead of two, this method wouldn't work (you'd get 1,1,1,2,2,2 numbering from either direction).  In this circumstance you would use alphabetical order to break the tie, with bromo- having higher priority than chloro.  That would give you 1,1,1-tribromo-2,2,2-trichloroethane.  I know you didn't ask this, but it could prove useful to know!


Why does lemon juice reduce the "fish" odor of sea food — specifically fish? In studying about amines, I read that lemon juice is also used to wash fish because it reacts with the amines on and in the fish to convert the amines to its salt, just reducing the "fishy smell".
I was thinking that it must be the citric acid in lemon juice that somehow reacts with the amines to convert them to their salts.

What is the general chemical reaction of citric acid with amines in fish, and why does this reaction actually reduce the "fishy" smell?

There are 2 cases, both related to the acid-base reactions. Both are also partial reasons why so many fish recipes use lemon juice.
Fish, especially sea fish, naturally contain trimethylamine-N-oxide $\ce{(CH3)3N-O}$ that, after death, gets enzymatically reduced to trimethylamine $\ce{(CH3)3N}$, the source of ammonia-like fish odour.

Trimethylamine N-oxide is an osmolyte found in saltwater fishes, sharks, rays, molluscs, and crustaceans. It is considered as a protein stabilizer that may serve to counteract urea, the major osmolyte of sharks, skates and rays. It is also higher in deep-sea fishes and crustaceans, where it may counteract the protein-destabilizing effects of pressure.TMAO decomposes to trimethylamine (TMA), which is the main odorant that is characteristic of degrading seafood.
TMAO is biosynthesized from trimethylamine, which is derived from choline.

In acidic environment, TMA forms water soluble, non volatile, odourless trimethylammonium salts, what eliminates the volatile smelly trimethylamine.
$$\ce{(CH3)3N + H+ <=>> (CH3)3NH+ }$$
It is the direct analogy to the smelly volatile ammonia and odourless ammonium:
$$\ce{NH3 + H+ <=>> NH4+}$$

Another source of fishy smell/taste, especially of some sweet water species like carp or catfish, is geosmin, a polycyclic alcohol, being a derivate of decahydronaphthalene(decaline). It is released after death of some Streptomyces or cyanobacteria cells which gives a muddy or earthy taste, and the human smell sensors are extremely sensitive to it (5 ppt).
Geosmin is also responsible for smell of the soil after raining, of some beet roots and even of drinking water of surface origin. There is speculation the nose sensitivity for geosmin is a evolutionary response for the need of efficient searching for water.
Geosmin reportedly breaks down in acidic environments.


How is Negative Temperature Hotter than Infinite Temperature? Recently I have read articles stating that negative Kelvin has been achieved. I was a bit speculative at first as I though how can you get cold than 0 Kelvin, but after doing some research it makes sense on how this is possible.

However the one thing that I still not sure is that it stated that the temperature scale can't be thought as being linear but rather as a loop which goes in the following order, "+0 K, … , +300 K, … , +∞ K, −∞ K, … , −300 K, … , −0 K".

How can negative Kelvin be hotter than infinite temperature?

The other answer given is correct, but for completeness, I will give the thermodynamic, rather than statistical mechanics, explanation.

Thermodynamics gives us that $$\frac{\partial E}{\partial S}=T$$

This means that  when entropy--as a function of energy--increases, the temperature must also increase.

On a macroscopic level, this makes perfect sense because you're never going to think about a system reaching a maximum temperature because energy and entropy are always going to increase together.

Of course, it must be remembered that this definition of temperature was being given long before people were thinking about quantum mechanics. Even more, they were only concerned with temperatures because they realized that they could use a temperature difference to get work out of a system. A good example of this is the Carnot heat engine. The reason I reference the Carnot heat engine is that it demonstrates that when temperature was being defined mathematically, which was around this time, they were concerned with heat flow. We know that heat flows from a hot body to a cold body, so this means, almost exclusively, that heat flows from high energy to low energy.

For instance, in a system with only two energy states containing $N$ particles, the state of maximum entropy is when $\frac N2$ particles are in both states 1 and 2. The state of maximum energy, however, is when all $N$ particles are in energy level 2.

That means, if we push beyond the state of maximum entropy, energy will increase as entropy decreases, hence, $\frac{\partial E}{\partial S} < 0$.

So, negative kelvin can be "hotter" than infinite temperature because our definition of temperature has nothing to do with "hotness" as we feel on the macroscopic level, it is merely an exchange between energy and entropy.

Furthermore, negative kelvin is hotter than infinite temperature, not because it would feel hotter, but because heat flows from the negative temperature object to the positive temperature object because we can demonstrate mathematically that the negative temperature object is in a higher energy state. It makes sense to think about it that way because 90% of chemistry is just looking for the low-energy state.

In regards to the issue of our temperature scale going in a loop, there is a solution to this using the thermodynamic beta, $$\beta=\frac{1}{kT}$$Consider this graph which shows that temperature as a function of energy is asymptotic and thus discontinuous, while $\beta$ is continuous as a function of energy.

Also worth pointing out is that $\beta$ has units of energy, and is thus more physically relevant than temperature because energy is fundamental to all matter, while temperature is basically nothing more than a mathematical construct to describe "hot" and "cold".


What would you call the polymerised sheet-like polymer arising from Fittig reaction of 1,3,5-trichlorobenzene? If we dip a sample of chloroalkane in an ethereal solution, radicals form at the sites where chlorines are attached, which subsequently pair with other radicals.

Now, consider 1,3,5-trichlorobenzene:

If I attempted to perform a Wurtz-Fittig reaction with this, I suspect the product would be a two dimensional polymerised product. I drew a portion of it:

This polymer is quite interesting. Has this been observed before? Does it have a name?

While looking for the name of this molecule is my primary concern, it would be great if I could also get some information about its properties. This polymer looks like a "hollow" version of graphene, so I believe this would have some properties of graphene, although quite lighter. I think this could conduct electricity, due to conjugated $\pi$ systems of benzene rings. I also believe it would be quite porous, and probably allow small molecules like $\ce{H2O}$ to pass through.

Ignoring the complex challenges it would take to make such a structure, the name is rather unclever with the literature referring to it as "holey graphene". You could probably take this a step further with your hypothetical material and call it :

well-ordered holey graphene
or better yet, well-ordered monodisperse-C6-holey graphene.

with C6 alluding to the missing $\ce{C6}$ ring that would be present in perfect graphene. Materials science has very unrigid naming conventions. If you make it first, you can greatly influence the literature terms.

As far as conductivity goes, I agreed that it would likely be conductive or semi-conductive at a minimum similar to poly(9,10-anthracene diylidene).


Why does silicon not react with dioxygen? Why does silicon not react with dioxygen?

I'm doing an organic lab experiment. I'm heating up silicon, carbon, and sulfur individually then placing them in pure $\ce{O2}$. I don't really know about carbon chemistry but I thought the reaction might be related to the electronegativity of silicon being lower than that of sulfur/carbon causing the bond to be more polar or something?

Elemental sulfur commonly occurs in form of $\ce{S8}$, with relatively low melting point and, more importantly, relatively low bond energies. So, it easily breaks.

Carbon commonly occurs as graphite or structures composed of graphite-like fragments, but usually something we call carbon contains up to 10% of other elements, meaning it can produce some combustible gases on heating.

However, more importantly, both elements mentioned above produce gaseous oxides on oxidation, that leave the surface of the reacting compound, exposing it to further reaction.

Silicon, on the other hand, produces solid oxide on heating, that prevents further reaction. It is still possible to make a very fine dust of silicon to react with oxygen, but it is not an easy feat. For reaction to occur to bulk silicon, the products must be removed from the surface. Given that, bulk silicon will not give silicon oxide, but it can react with chlorine in reasonable conditions, giving volatile silicon tetrachloride.

Why silicon forms solid oxide with high melting point is another matter, with answer considerably more obscure. The simplest approach to rationalize it would be to consider atomic radius of carbon, sulfur and silicon and consider bonding in alternative structures for this compounds.

Carbon dioxide is an oxide with small central atom, that can form $\pi$-bonds effectively with oxide, so it does not form polymeric structure.

Sulfur dioxide is an oxide of larger atom, but it has a lone pair on central atom, so polymeric form would be considerably hindered. (on contrary, sulfur trioxide forms polymeric forms, but still has gaseous form  which is only marginally less stable).

Silicon dioxide in monomeric form has two $\pi$-bonds, and this is uncharacteristic for third row elements, and silicon is significantly larger than carbon, having enough space for four oxygens around it. So, it easily and effectively forms 3-d network of $\ce{SiO4}$ units with shared oxygens, that is quite hard to melt and vaporize, and, once formed on the surface of silicon, it prevents further reaction.

Still, formation of silicon dioxide films on surface of bulk monocrystalline silicon is essential for microelectronics.


Why doesn't beryllium, an elemental solid, obey Dulong and Petit's law? Beryllium has a molar heat capacity of $\pu{17.7 J mol-1 K-1},$ which is quite far from the $\pu{25 J mol-1 K-1}$ as predicted by Dulong and Petit's law. What is the reason for this discrepancy?

Not a  'Law' a such it is more like a 'rule of thumb' based on a limited set of data measure at high (i.e. room) temperatures. The heat capacity is not constant but varies with temperature and for metals reaches a limiting value of approximately $3R$ at high temperatures. (The limiting value is higher for molecules due to the many translational, vibrational and rotational energy levels).

The heat capacity is the rate of change of the energy with temperature, i.e. the slope  of a plot of energy vs temperature.  At low temperatures only the lowest few energy levels are populated, so the rate of change vs temperature is almost zero. As the temperature is increased, more energy is added, more levels become populated and $C_V$ rapidly increases because the rate of change of energy is large. At high temperature, very many levels are now populated and the rate of increase of energy with temperature is constant, and therefore $C_V$ levels off becoming almost constant.

In some metals, those with large energy gaps between levels, room temperature is insufficient for them to reach the point where the slope of internal energy increase with temperature becomes constant and so the heat capacity is smaller than $3R$. The value at higher temperatures (say 1000 C) should reach the limiting value.


Order of stability of substituted benzyl carbanion I think the order should be II > III > I. The +m effect should be more in case 1 than in 3, isn't it ? So, 1 should be more unstable than 3.

Someone please explain.

I'm not a chemist, but I doubt whether the mesomeric effect is "more in case I than in III" as you put it.

I'm under the impression that the mesomeric effects (alone) in both cases activate the benzene ring to the same extent (If you try drawing the resonance structures, you'd probably feel the same way).

However, I suppose in case I, the minor $-I$ effect of the $\ce{-OCH3}$ group makes I slightly more stable than III (because the negative charge is somewhat diminished).

So to me, $\mathrm{II > I > III}$ looks about right.


On the stabler tautomer in amide-imidic acid tautomerism In the section on acid-catalysed hydrolysis of nitriles, this webpage has the following line of reasoning;

...the more stable amide tautomer predominates the (following) equilibrium.

(credit)
How are the relative stabilities of these tautomers compared? I made up a reason of the resonance structure for the amide one having negative and positive charges on oxygen and nitrogen respectively, and vice versa for the imidic acid one, but I'd like clarification on that.
There's also another hand-wavy thing I thought about of the double bond being stable on the more electronegative atom, oxygen. Is that correct, too?

The relative stability of the two tautomers (two isomers differing how protons are bound) may be assessed from the equilibrium constant you may formulate for the two.  Deriving from above illustration, it would be

to indicate that this is dynamic, with a back and forward reaction at microscopic level.  Once formulated as such, you probably recall $\Delta{}_RG = -RT \ln{}K$ to relate between the sample temperature $T$, the universal gas constant $R$, the free reaction enthalpy $\Delta{}_RG$ and the equilibrium constant $K$.
Depending on the substrate, you may use spectroscopic evidence (e.g., $\ce{^1H}$-NMR spectroscopy) to assess the concentration of the two forms.  Equally, based on prior work by others, there are databases to either list these values, or provide a prediction (i.e., an educated guess) about $K$.  Knowing these equilibria is important to anticipate the likelihood of chemical reactions. It equally offers insight regarding the pharmaceuticals and their uptake, anabolism, interaction with the body (metabolism), and catabolism and eventual elimination, too.
One of the databases available to the public is Tautobase (2020JChemInfModel1085). It is implemented in the freely available DataWarrior program allowing you a search-by-(sub)structure.  With some luck, one of the 1680 pairs matches your substrate closely enough, too.  In the simplest case one may imagine, however, the amide is strongly favoured:

The database backs this particular entry with the publications 1998JAmChemSoc10359 and 2001JAmChemSoc2681.


Effect of halogen atom on Markovnikov's rule I know that the Markovnikov's rule is because of the stability of carbocations. If a halogen atom (electron-withdrawing group) is directly attached to the carbon atom carrying a positive charge, would it destabilize the carbocation (for example dichloroethene reacts with HBr)and 'break' the Markovnikov's rule?

Markovnikov's rule is just a simple rule for predicting the structures of  alkenes in some reactions.
It is not advisable to apply this rule to all cases,it is often helpful to check the relative stability of carbocation intermediates.

In your question the carbocation will be stabilised by +M
effect of halogen as halogens have lone pair of electrons.


What are the applications of quantum field theory to chemistry? I have a physics background and I learned how quantum field theory is applicable to condensed matter physics, and since some parts of condensed matter physics overlap with chemistry I was wondering if there are any direct applications of QFT to chemistry as well. Also what are the research areas in chemistry that use QFT, because I'm considering doing my graduate work in chemistry and it would be cool if I could use QFT in chemistry as well.

Have a look at this article: "The Physics behind Chemistry and the Periodic Table"
Pekka Pyykkö Chem. Rev. 2012 112 (1), 371-384
(http://pubs.acs.org/doi/abs/10.1021/cr200042e)
I think it answers your question and has most of the references to work done in that particular field. For most questions in chemistry the non-relativistic Schrödinger Equation is good enough. Even for systems with an appreciable amount of relativistic effects, the Dirac Equation is often not solved directly. So, the number of applications where Quantum Electrodynamics is used is really limited.


What causes polarization in noble gases? According to my knowledge, polarization occurs when in a compound there is a difference of polarity its constituents.

But in the case of noble gases, there is a single component. What exactly is polarization of noble gases and how does it occur?

Although this applies to all noble gases, consider helium. There are two electrons moving around the nucleus. The movement of those electrons is symmetrical and during a defined amount of time, the atom is not polarised. However, the momentary positions of those electrons can polarise the atom. This can happen when both electrons are relatively close, on the same one side of the atom. In this case a momentary dipole is formed, since there is separation of two charges, nucleus being positive and the pair of electrons being the negative pole.


Reaction of Grignard reagent with primary amines What does the reaction $\ce{CH3CH2NH2 + CH3MgBr ->}$ form?  Does it form a secondary amine or a tertiary amine or an alcohol? I can't find what a Grignard reagent does to a primary amine.

Here's the thing about Grignard reagents.  Though we typically want them to do nucleophilic addition to something, these powerful bases tend to go for protons -- a faster reaction -- if there is just about anything that might make the proton vulnerable.  Attaching the proton to an electronegative atom like nitrogen is enough, so primary and even secondary amines will react like that with Grignard reagents.  You just deprotonate the amine, the hydrocarbon part of the Grignard reagent ends up an an alkane (or an arene if there are aromatic rings), then on acid workup you get just the amine (or ammonium ion) back.  Yup, this is not a very productive synthesis.

Choose your substrate carefully with Grignard reagents to avoid this deprotonation and get the addition you want, or consider less strongly basic alternatives like an organocadmium compound (which itself can be made from a Grignard reagent plus a cadmium salt).


Heat capacity of a calorimeter Why do we mix water with different temperatures to calculated the heat capacity of the calorimeter?
Why can't we calculate it directly using just warm water?
Asked differently, would the following be wrong? (With $T_E$ the equilibrium temperature and $T_C$ and $T_W$ the initial temperatures of the caloriemeter and water)
$$
c_{C}=\frac{\Delta Q}{m_{C} \Delta T}=\frac{c_{W} m_{W}\left(T_{E}-T_{W}\right)}{m_{C}\left(T_{C}-T_{E}\right)}
$$

Karl's comment (see above) prompt me to suggest the following rough experiment which can be done in the classroom with minimum equipment:
Suppose you are required to find out the specific heat capacity of water. Weigh $m_1 \ \pu{g}$ of warm water at $T_1 \ \pu{^\circ C}$ to a Styrofoam cup with a lid. Also weigh $m_2 \ \pu{g}$ of ice at $T_2 \ \pu{^\circ C}$ to another Styrofoam cup. Then add ice to warm water (Note: Measure the temperature of warm water just before addition to make sure it is still at $T_1 \ \pu{^\circ C}$). Stir the mixture carefully while the coup is covered by the lid. Measure the temperature of the mixture until reading is steady. Suppose the steady temperature is $T_3 \ \pu{^\circ C}$. Ensure this time no ice is visible (meaning all dissolved). Now we are ready for the calculations:
Suppose the specific heat capacity of water is $c_{(\ell)}$ and the specific heat capacity of water is $c_{(s)}$. For novices, we can assume that $c_{(\ell)} \approx c_{(s)} = c$. Thus, heat lost by warm water at equilibrium:
$$Q_\text{lost} = m_1 \times c \times (T_1 - T_3) \tag1$$
For ice, if $T_2 \lt 0$, ice would gain heat first to become ice at $\pu{0 ^\circ C}$, then $\pu{0 ^\circ C}$ ice melts to $\pu{0 ^\circ C}$ water, and warm up to $T_3 \ \pu{^\circ C}$ water. If heat of fusion of ice is $L \ \pu{Jg-1}$, total heat gained by ice at equilibrium:
$$Q_\text{gained} = m_2 \times c \times (0 - T_2) + m_2 \times L + m_2 \times c \times (T_3 - 0) \tag2$$
At equilibrium, $Q_\text{gained} =  Q_\text{lost}$. Therefore, equations $(1)$ and $(2)$:
$$m_1 \times c \times (T_1 - T_3) = m_2 \times c \times (0 - T_2) + m_2 \times L + m_2 \times c \times (T_3 - 0) tag3$$
Only unknown in the equation $(3)$ is $c$, you can solve the equation $(3)$ for $c$.
Also note that you can avoid $m_2 \times c \times (0 - T_2)$ part by choosing $\pu{0 ^\circ C}$ ice, because it is easy to get $\pu{-20 ^\circ C}$ from freezer, it it kept in constant temperature.


Solve this mole fraction calculation with incomplete information I am currently trying to solve a problem that seems quite simple at the beginning but the complexity reveals itself soon thereafter. The problem is the following:

  You mix a known mass of water with a known mass of ethanol. Then you fill up the flask to the 50 mL mark with water, but cannot weigh the contents when you're done. What is the molar fraction of ethanol in this mixture?

My attempt at solving this problem is to basically calculate the initial molar fraction, and then stepwise add small virtual amounts of water. Then the new molar fraction is calculated, and somehow the excess volume (which is negative for the water-ethanol mixture) would also be included somewhere here. This is repeated until the final volume is reached.

An other attempt would be to simply neglect the excess volume. While this might be fine at the edges of the molar fractions, it would introduce a grave error towards the middle. Or is the error still negligible?

Are these reasonable attempts or am I completely missing something here?

I don't think you can solve that unless you know additional information like mass of water added or volume of water added, or the density of the final solution.

You could approximate by dividing mass of ethanol by density of ethanol to get volume of ethanol, and saying volume of water added is 50mL-(volume ethanol), but this is only an approximation because volumes are not additive when the two solutions are different.

The error could be estimated from this graph:

source: http://en.wikipedia.org/wiki/Partial_molar_property

or the data from the graph could be used to improve the accuracy of the answer

or use this density table: http://www.handymath.com/cgi-bin/ethanolwater3.cgi?submit=Entry


Reverse electro-plating byproducts Following this plan, I am wondering what the byproducts of this method are? At the site with the q-tip, the solution turned a bright yellow, with the q-tip turning a dark yellow, almost red. Holding the q-tip in place for about 10 seconds results in a small amount of smoke, and bubbles around the contact point. What are these colors, smoke, and bubbles? I don't know if the kettle is steel or aluminum, it's an old Budweiser keg, if that helps.

In case the link above dies, the method is etching metal using a vinegar and salt solution, combined with a 9 volt battery. Connecting the negative lead to a piece of metal, and using the positive lead with the solution results in a type of reverse electroplating, thus removing metal ions and etching the metal.

Not sure on the tags for this question, so feel free to adjust them, and the description of the etching method.

The yellow colour in solution is likely from iron ions from the metal being oxidized and dissolved:
$$\ce{Fe_{(s)} + 2e- -> Fe^{2+}_{(aq)}}$$ (Probably $\ce{Fe^3+}$, too)

The gas bubbles/smoke is the product of the other half of the reaction, the reduction of something in your solution. Because you have chloride in solution, chlorine gas is a possibility:
$$\ce{2Cl^{-}_{(aq)} -> Cl_{2(g)} + 2e-}$$

Hydrogen may also be produced depending on what you're using as an electrode on the q-tip.


Does hyperconjugation and induction always have the same effect? My professor does not believe in the hyperconjugative effect. Instead, we only consider inductive effects.

So do both effects have the same general effect? I know for methyl groups both have the same effect; both hyperconjugation and induction can be invoked to explain the electron donating tendencies of the methyl group.

Are there any cases in which the hyperconjugative effect and the inductive effect have opposite effects?

The inductive effect occurs through overlapping sigma bonds, whereas the hyperconjugative effect occurs through overlap of a sigma bond with a p orbital.

There are many different types of hyperconjugation and hyperconjugation has been used to explain many different chemical phenomenon.  A nice overview can be found here.

  Are there any cases in which the hyperconjugative effect and the
  inductive effect have opposite effects?

Since inductive effects are transmitted through the sigma system, they effect should be independent of rotation around the bond(s) involved.  On the other hand, the pi type of overlap used in hyperconjugation is very dependent upon rotation about the bond(s) involved.  In the diagram above, rotation about the $\ce{C-C}$ bond by 90° shouldn't affect any inductive effect of the methyl group upon the carbocation center, but such rotation would completely destroy the overlap required for hyperconjugation to occur.

As you might expect, hydrogen/deuterium isotope effects can be used to assess hyperconjugation $\ce{C-H}$

[Note: isotope effects are a subject unto themselves; a nice overview can be found here.  See p. 12 for their application to hyperconjugation.]

Here is a "thought experiment" where inductive effects and hyperconjugation should have different effects (I don't know if this or a similar experiment has been done).  Take the following two carbocations.

One would expect them to both have similar inductive effects upon the carbocation center.  The tris-ethyl carbocation, free to rotate about its various $\ce{C-C}$ bonds could also exist in conformations where the $\ce{C-H}$ bonds alpha to the carbocation center are correctly positioned for hyperconjugation to occur - it should show a secondary $\ce{K_{H}/K_{D} >1}$ since hydrogen is more electron donating than deuterium (see here for the reasoning behind this statement).  However, the $\ce{C-H}$ bonds alpha to the carbocation center in the rigid bicycloctyl carbocation cannot adopt a conformation that would allow effective hyperconjugative overlap of these $\ce{C-H}$ bonds with the cationic center - it should show a secondary $\ce{K_{H}/K_{D} =1}$.  Different secondary isotope effects would be expected in the solvolysis of these two systems.

Another case where inductive effects and hyperconjugation do not provide equivalent explanations relates to rotational barriers.  The origin of the rotational barrier in ethane has been a subject of investigation for a long time.  Inductive effects cannot be used to explain the barrier because, as discussed above, inductive effects are rotationally invariant.  An argument using hyperconjugation to explain the rotational barrier has been advanced and is the subject of much discussion (see the "Rotational barrier of ethane" section).  Whether the hyperconjugation argument is correct remains to be proven, but it is another case where hyperconjugation can be used to explain observations that the inductive effect cannot.


Are ethyl iodophenylundecylate and Ethyl-10-(Iodophenylundecanoate) the same compound? I have no understanding of chemistry, so these to chemical formulas mean nothing to me. What I am trying to ascertain is are these compounds the same? If not, what differences are there especially if injected into the CSF [which they were]?

Known as Ethyl Iodophenylundecylate:

CH3
I   (CH2)8 COOC2 H5
CH

Known as Ethyl-10-(Iodophenylundecanoate)

CH3
I -CH2 (CH2)6 CH2 COOC2 H5
CH

The "structures" you are trying to post and their respective names do not match, as I cannot spot a phenyl group in there. (Also it is hard to tell from this diagram.) But based on what you gave us, I believe these are the same compounds, where ethyl 10-(4-iodophenyl)undecanoate is the correct IUPAC name and Ethyl iodophenylundecylate is a shortened synonym. I also found trivial names for this compound, for example Pantopaque and Iofendylate.
You can access this molecules record on ChemSpider and you will find a couple more names attached to it. From the same page is the following structure taken:

(source: chemspider.com)


Does sodium cyanide react with water, and can it cause an explosion? According to some conspiracy theory, the explosion that happened last year in Tianjin (China) wasn't caused by sodium cyanide. Can sodium cyanide react with water, and can it cause an explosion if you have 700 tons of it?

Usually, I dislike conspiracy theories.1 However, this one (if it even is one) is true. Looking at the list presented on Wikipedia of the chemicals that were stored in that warehouse, why would you even suspect sodium cyanide as being a cause of the explosion? It is a stable ionic compound that can be put on the shelf, wet with water, dissolved in water, and won’t react with air with any measurable half-life. (Except that you possibly shouldn’t put it on a shelf because it is slightly volatile and highly poisonous.)

What was also in the warehouse and is an explosive was ammonium nitrate. In the 1921 explosion in Oppau (now part of Ludwigshafen), BASF workers tried to use dynamite to loosen a plaster of ammonium nitrate (it is hygroscopic and baked together). Mixtures containing at least $40~\%$ ammonium sulphate were considered stable and the mixture in question was a $50~\%$ mixture of the two according to the production data. After approximately 20,000 uses of dynamite where nothing happened, the entire 4,500 tonne lot exploded killing some 500 or 600 people and leaving thousands injured. The theory $40~\% = \text{safe}$ was later disproven in careful examinations. Apparantly, the explosion could be heard in Munich, some $300~\mathrm{km}$ away.

Another substance stored there and dangerous is calcium carbide $\ce{CaC2}$. When calcium carbide gets wet, it produces acetylene gas (and calcium hydroxide). The acetylene is flammable and can also explode in certain acetylene–oxygen mixtures.2 Luckily, no acetylene or carbide explosions are known enough to make it onto Wikipedia.

So far more likely than sodium cyanide having any primary effect would be one the following theory:

Fire (something must have ignited because the compounds I read there aren’t self-flammable).
Attempts to extinguish fire with water by the fire brigade.
Water reacts with calcium carbide liberating acetylene.
Acetylene burns.
The critical heat is reached, ammonium nitrate explodes.
Toxic sodium cyanide is liberated into the environment causing further severe health risks for man and animal alike.

1: Okay, that was a lie. I love listening to them and making fun of them.

2: A teacher at school talked about his time at uni and mentioned a professor that let a balloon of acetylene explode — he made a fuse long enough to ignite it from outside of the lecture hall (nobody was allowed inside), had the doors closed and my teacher said ‘it was the loudest bang I ever heard.’


What atom is always paramagnetic and why? Regardless of its electron configuration, it must always be paramagnetic when it's a single, neutrally charged atom:
  (A) Carbon (B)
  Nitrogen (C) Oxygen (D) Neon (E) Argon

At first, I eliminated D and E. Then, I tried to find the answer by checking the electron confiurations of the atoms; I saw the phrase "regardless of its electron configuration", but I didn't know what else I can do. Then, I found out that all three atoms (A, B, and C) have some unpaired electrons …

I checked charts with paramagnetic elements filled in; three first pictures on google said different things (check here), but after all the info I have found on the net I understood that C (oxygen) is the most likely answer. I don't understand why … I checked a few videos that said that if an atom has 1 or more unpaired electrons it is paramagnetic; then, A, B, and C are all paramagnetic. A few sites said we need to look not at the elements, but at the compounds (not oxygen, but $\ce{O2}$); the task said, anyway, we are looking at the single, neutrally charged atom.

Please tell if it is oxygen and if yes why it is paramagnetic but other elements are not.

Neon and argon are correctly eliminated, no questions asked. Henceforth, we do indeed need to pay attention to question semantics and the phrase ‘regardless of its electronic configuration’.

The key here is that following Hund’s rule will only give you one possible electronic configuration. For example for carbon, Hund’s rule will give $\mathrm{1s^2\, 2s^2\,2p}_x{}^1\, \mathrm{2p}_y{}^1 2\mathrm{p}_z{}^0$ or any other combination of two of the three equivalent p orbitals half-populated. However, in early organic chemistry courses you typically rapidly learn about the hybridisation approach which changes that electronic configuration to $\mathrm{1s^2\, 2s^1\,2p}_x{}^1\, \mathrm{2p}_y{}^1 2\mathrm{p}_z{}^1$ — four instead of two unpaired electrons. Likewise, it is possible to imagine an electronic configuration as follows: $\mathrm{1s^2\, 2s^2\,2p}_x{}^2\, \mathrm{2p}_y{}^0 2\mathrm{p}_z{}^0$. In this case, all electrons are paired and the atom is diamagnetic.

Performing this type of analysis with all the atoms at your disposal — carbon, nitrogen and oxygen — you will be able to succeed in pairing all electrons for carbon and oxygen but not for nitrogen since it has five valence electrons. Five, being an odd number, cannot be mapped pairwise. Thus, a nitrogen atom is always paramagnetic.


Forming ester from amide using protonated alcohol as catalyst In my level of understanding, if I want to make an amide from an ester, I should add:

$\ce{H3O+}$ for hydrolysis to make carboxylic acid
$\ce{SOCl2}$ to form acid chloride
$\ce{ROH}$, pyridine to form the ester as product

I think this is a good way since it is difficult to form a more reactive product compared to the reactant, but in one of the question, I really want to try to make the 3-step process 1-step:

My attempt is to use a protonated $\ce{ROH}$ as acid-catalst, this time, to form a methyl ester. I used methanol, and somehow I can get to the end.
[for the 4th intermediate, I searched online and found that $\mathrm{p}K_\mathrm{a}$ of $\ce{(CH3)2NH = 10.73}$, $\mathrm{p}K_\mathrm{a}$ of $\ce{CH3OH = 15.56}$, this means that $\ce{(CH3)2NH+}$ is a better leaving group (lower $\mathrm{p}K_\mathrm{a}$ --> lower basicity --> more stable in the reaction mixture)]
Can someone please kindly tell me what is wrong with my mechanism? I know it should be wrong, since the 3-step process is the most suitable way and yield the most.

There are several points that I'd throw out:

As a technicality, you probably can't buy protonated alcohols; you'd need to use the alcohol plus an acid like HCl in a non-aqueous solvent.

Now, once the amine drops off, it will mop up your catalytic acid, so you had probably better use at least stoichiometric acid, and possibly quite a bit more.

Amides are more stable than esters; you know that already. So, the reaction thermodynamics are not in your favour to begin with. Thus, in order to get a decent yield, you probably also need to use a big excess of alcohol (Le Chatelier's principle: more alcohol → more ester).

You'd probably also need to heat it quite a bit to get the reaction going. (The same is true for hydrolysis with aqueous acid.)

All in all, probably you could get it to work, but it might not necessarily be fun, and if your reactants had any other sensitive functional groups on it they would pretty much be destroyed by heating in strong acid.
In general, this is a tough reaction, and I suspect that if you could find a general way to do this cleanly then it would be quite a substantial achievement. Recently, Hie et al. have developed a Ni-catalysed procedure (Nature 2015, 524 (7563), 79–83), but it only works on aromatic  amides, so isn't applicable to your example substrate. (Essentially, both the group attached to the carbonyl carbon, as well as one of the groups on nitrogen, have to be aromatic. Despite that obvious limitation in scope, it's still in Nature, which maybe suggests something about the difficulty of the transformation.)


When surface area of electrodes is increased, why does the current in the galvanic cell increase? I did a lab in galvanic cells where I changed the surface area. As a result, the current increased, but I don't really understand why. I guessed it was because more are means that more ions from the anode are oxidized and more ions from the cathode reduced, more electrons are transferred and taking into account that current is the flow of electrons per unit time, the current increases.

Is that even correct?

Thank you all

Yeah, you're on the right track. In any battery reaction there is some rate-limiting step that determines the maximum power the battery can supply. Most of the possible rate-limiting steps involve the surface of the electrode.

As an example, imagine that the rate-limiting step is the diffusion of new ions from the electrolyte bulk to the cathode (or diffusion of ions away from the surface at the anode. The physics of diffusion in this case is governed by Nernst-Planck equation, where the quantity of interest is the ion flux (number of ions of unit area). However, the total current is determined by the total number of ions. To get from flux to total number of ions you have to integrate over the area of the electrode: if you double the area of the electrode you'll double the total number of ions (ie. current) if all else stays equal.

That's just one example that depends on the specifics of your cell, but many of the surface-related limited steps in a battery follow this type of behavior.


Ok to separate steel from aluminum with bronze? (Galvanic action) I have an aluminum hinge that I want to use with a stainless steel hinge pin. The problem is that I cannot use the steel pin directly because there would be galvanic action between the stainless steel and aluminum which will result in corrosion. Therefore I need to put some kind of bearing liner between them. Note that the hinge will experience strong forces, so using something soft like nylon will not work.

My preference would be for a bronze bushing, but I do not know if there is galvanic action between bronze and stainless steel, or bronze and aluminum.

What material can I use? Is there a simple explanation of how I can predict whether a metallic alloy will react galvanically with a different alloy? I have Mathematica, so if there is a formula I can write in Mathematica, that would be just as good.

-------------------------------- Sacrificial Anode?

One method I have heard of to protect from galvanic corrosion is a "sacrificial anode". If I attach a bar of zinc to the back of the aluminum hinge will that somehow protect it, and allow me to use a stainless steel hinge pin?

If you must use an aluminum hinge with a steel pin, anodize the aluminum. Anodizing builds up a coat of very hard, non-conductive $\ce{Al2O3}$, sapphire.

There are commercial metal fishers that will perform this, if you do not want to do so. There is a bit of an art in the process of creating a uniform coat of the required thickness: too thick and it cracks off; too thin and it wears off.


Why do noble metals like silver not form natural cubic crystals? Pyrite forms cubic crystals, both in theory and in nature:

However, the noble metals, like silver, that have cubic, face-centered lattices, do not seem to form cubic crystals in nature.

One possible explanation is that silver and other noble metals are invariably contaminated by trace amounts of other elements that prevent perfect crystal formation. For example, salt (NaCl) is also cubic, and when very pure deposits of salt are found, they may have large cubic crystals, but normal salt generally does not form such crystals presumably because of contamination.

Is this the explanation why gold and silver do not form cubes when they solidify? If so, imagine we somehow obtained perfectly pure silver as a gas or liquid, would it crystallize on cooling to a cube?

Native silver crystals can be found in cubic, octahedral, or dodecahedral forms. As you mention, silver often is found with gold, copper and other metals, and with gold, tends to hexagonal stacking, rather than cubic crystals.

Silver may also form pseudomorphs, in which the crystal shape is from its progenitor or its surroundings. For example, dyscrasite, $\ce{Ag3Sb}$, or acanthite, $\ce{Ag2S}$ may be replaced by silver in the form of the antimony or sulfur minerals.


NaCl water electrolysis without membrane, how measure the free chlorine concentration? I'm trying to made electrolysis using water and NaCl. I need at the end a concentration of 2 g/l. How can I measure the free/active chlorine? Do you know a way to "measure" the chlorine during the reaction without chlorine sensor? Using for example the measure of conductivity of the water, the electrode current and voltage and the temperature or the salt remaining?

The electrolysis is 'open' without membrane, a titan electrode is directly put in the water + NaCl.

The best way for determining the active chlorine amount ($\ce{ClO^-}$) in a bleach is to dilute it with water, then to add Potassium iodide $\ce{KI}$, and enough HCl solution to make it acidic. It produces $\ce{I_2}$ by the equation $$\ce{ClO^- + 2 H^+ + 2 I^- -> I_2 + H_2O}$$ Then the $\ce{I_2}$ so produced is titrated by a standard solution of sodium thiosulfate $\ce{Na_2S_2O_3}$, according to : $$\ce{I_2 + 2S_2O_3^{2-} -> 2 I^- + S_4O6^2-}$$

A quicker (but not so precise) way is to add 2 mL acetone to 50 mL bleach in a insulated flask, and to measure the change of temperature due to the reaction. The reaction produces chloroform $\ce{CHCl_3}$ and acetate. After one minute, the temperature increases to a stable final value. The change of temperature is proportional to the amount of chlorine. If $\Delta T $= 10°C, the concentration of active Cl is 5.6%. Generally speaking, the concentration of active Chlorine is : 0,56% x $\Delta T$


Refractive properties of materials? I've just read this article about meta-materials with 0-refractive index, which potentially may "permit" light with super-luminous rate /under rate is meant the propagation of light's phase, which doesn't transmit information => there is no valuation of the laws of physics/.

Generally speaking, the refractive index of materials depends on the (dominant) molecular interactions within them, on their morphous structure, etc. As you know, these molecular properties govern lots of other chemical features like the viscosity, the density, to some extend the aggregate state, and so on... On this account, it should not come as a surprise that there exists a clear correlation between the refractive index and the properties, mentioned above - e.g. fluids of higher density tend to have higher refractive properties; same goes for viscosity; and so on... Based on those remarks one is able (at least) to guess how they can improve / worsen the refractive index of a given fluid - discussed for instance in this research on aerosols.

My question, though, is how does the refractive index of alloys, polymers, crystals, and solids in general relate to their other chemical properties?

Furthermore, is there a way at least to guess what materials one should potentially try to use in order to create / synthesize a meta-material, like the one from the article, that breaks the limits?

I can only answer your question about the relation between refractive indices and other properties with two examples.

When looking at dispersion interactions between particles in solution the Lifschitz Theory connects the refractive indices with the Hamaker constants.
The information about refractive indices is even used to choose materials in a way to cancel out dispersion interactions between particles of different kind. (I will attach a paper URL in the next days, just heard it from our professor in the lecture)

Another application of refractive indices is X-Ray NEXAFS spectroscopy and  many more spectroscopic methods involving electron or high energy light radiation. By the Kramers-Kronig relation it is possible to connect absorption constants and refractive indices. Especially one can find for example absorption edges.

Then here I found some probably interesting papers regarding your question. At least their references will be a good starting point. Unfortanetly I have no access and could only read the abstract:

http://www.sciencedirect.com/science/article/pii/0020089165900205

http://www.sciencedirect.com/science/article/pii/0032395083902836


What is "silanol activity"? In a previous post about how pyridine failed to move much from the baseline of a TLC, I was told to look into "silanol activity". However, I was unable to find a textbook-style description. All the articles I've seen where silanol activity was discussed assumed it was a known prerequisite.
Could anyone explain silanol activity, or direct me to an online explanation please?
Furthermore, if it is referring to an acid–base reaction, would that imply that what I'm seeing on the TLC is no longer pyridine, but rather $\ce{C5H5NH+}$?

Silanol activity is perhaps an informal term which is quite frequently used in the chromatographic literature to explain the poor band shape of primary amines. Other amines show this too but to a lesser extent. That is why you will not find a formal textbook definition of silanol activity. One might also see the term silanophilic interactions used occasionally. If you look at the surface of silica gel, you will see that a silanol group, is like an alcohol analogue, except that silanol is slightly a stronger acid.
If we disperse silica particles in water, and let it stay there for sometime, the pH of water is < 7 (around 3–4, if I remember correctly). So the point to keep in mind is that silica is acidic due to its ionizable silanols.
There are different types of silanols as drawn in this paper: Sunseri, J.; Cooper, W. T.; Dorsey, J. G. Reducing residual silanol interactions in reversed-phase liquid chromatography: thermal treatment of silica before derivatization. J. Chromatogr. A 2003, 1011 (1-2), 23–29. DOI: 10.1016/S0021-9673(03)01070-7.

Now if you are doing chromatography of relatively strong organic bases, the silanol may be deprotonated by the amine (acid–base). The locally ionized silanol Si–O− groups can now act as a cation exchanger, which is a pretty strong interaction. The reason for seeing the streaking or tailing, as it is called in the formal literature, arises from those dual modes of retention. One mode could be regular partitioning or adsorption and the other
one is this additional cation-exchange behavior.
There are dozens of chromatographic tests for assessing the silanol activity of a given silica gel. Not all silica gels were created equal!
As your question

would that imply that what I'm seeing on the TLC is no longer
pyridine, but rather $\ce{C5H5NH+}$

Remember that ionization is an equilibrium process and hence dynamic. You are seeing the combined effect of both, pyridine and protonated pyridine.


Effect of increasing nitrogen concentration in the synthesis of ammonia For the reversible reaction
$\ce{N2 + 3H2 <--> 2NH3}$

When $\ce{N2}$ is increased, the the forward reaction is favored, increasing the yield of $\ce{NH3}$ and decreasing the yield of $\ce{H2}$. I read from multiple textbooks that the $\ce{H2}$ decreases in concentration to a greater extent than $\ce{NH3}$ increases in concentration, such that the final concentration of $\ce{NH3}$ is greater than that of $\ce{H2}$. Why is this so?

The final concentration of H$_2$ is lower than the concentration of NH$_3$ because for each NH$_3$ you produce, you need 1.5 H$_2$.  So the rate of consumption of hydrogen is higher than the rate of production of ammonia.

Here you can read about the multiple steps in the synthesis of ammonia, also called the Haber-Bosch Process.  As you mention, by increasing the concentration of nitrogen, you can increase the rate.  The rate limiting step is separation of the nitrogen atoms in N$_2$


How to to test the concentration of melatonin in a strong acid? If a small tablet of melatonin was coated with a substance to ensure a slow dissolution in water and was then placed in a strong acid, neglecting the limits imposed by maximum saturation, it would slowly but surely dissolve. I would imagine the simplest method for determining exactly how much has dissolved would be by testing its concentration, however the information available on the topic uses methods beyond my capacity to preform as a high school student. Is there any relatively simple method to determine the concentration?

Is there any relatively simple method to determine the concentration?

That depends on your equipment. Having the following at hand would be helpful:

a UV/VIS spectrometer
mortar and pestle
UV-grade ethanol (or methanol)
pure melatonin

How to proceed:

Prepare a stock solution of the pure melatonin in ethanol.
Prepare some dilutions for the expected concentration range.
Measure the UV absorption of all solutions at $\lambda = \pu{277 nm}$.
Draw your calibration curve (absorption vs concentration).
Powderize the tablet using a mortar an a pestle.
Dissolve the powder in a defined volume of ethanol.
Measure the UV absorption of the sample.
Determine the concentration in the sample using your calibration curve.

You have just applied the Lambert-Beer law. Others have done the same and reported the method in Pharmacophore 2014, 5, 252-257 (free PDF).


How to carry out cyclization of 1‐(2‐methylbutyl)‐2‐nitrocyclohexane? How the cyclization can take place as there is no other functional group than $\ce{NO2}?$

One way that I can think of is photochemical bromination which occurs selectively at tertiary center which then can be promoted to give cation in polar medium. Nitro group on being reduced to amino group will attack on carbocation, thus closing the ring.
The problem here is that there are three tertiary centers and radical close to functional group will be more stable than the one at the far tertiary center on side chain.
How would you handle this conversion?

The Hofmann-Loeffler-Freytag reaction as suggested by @user55119 in the comments above is the way to go.
First you need to reduce your nitro group. The reduction of alkyl nitro compounds is less simple than aromatic nitro groups and lower yielding in general. I would suggest the use of Al/Hg amalgam.
Acetylate the resulting amine.
N-chlorinate the amide - trichloroisocyanuric acid is good for this conditions
Irradiate the chlorinated amide to form the [5] ring similar to this example Baldwin gelsimicine synthesis
Finally hydrolyse off the acetyl group with cHCl.


Why are the oxide materials insulating? Why are the oxides usually insulating?  It would be nice to get an explanation with chemistry terms (e.g., electronegativity, electron affinity, etc...) and a different explanation involving the language of band structure (i.e. energy gaps).

To have conductuing valence zone requires to have infinite amount of ovarlapping valence orbitals with lesser amount of electrons. This is not usually possible for oxides, because they contain direct contacts O-O, but not M-M. O-atoms in oxides has completely filled electronic shell. Pure metals has electronic shells with much less electrons then available orbitals so they can allow electrons to move freely. This is the difference.


Chemical Reagent Test sensitive at low concentrations Would there be a spot test/tests that can detect either/both of the following molecules at low concentrations?

The likely concentrations are in the $10{-}\pu{50 mg/L}$ range in an aqueous solution. Ideally, I'd want something that does not need an exotic/expensive instrument.  e.g. a color change or precipitation or conductivity/pH change.

The background interference: The sample is Terrific Broth which has the following composition:

$\pu{12 g}$ Bacto tryptone
$\pu{24 g}$ Bacto yeast extract
$\pu{4 mL}$ Glycerol
$\pu{100 mL}\ \pu{0.17M}\ \ce{KH2PO4}$ and $\pu{0.72M}\ \ce{K2HPO4}$, sterile, to be prepared separately from the tryptone, yeast extract, and glycerol solution.

Do you have the pure compounds on hand?  If you do, the two target compounds can be crudely isolated by extracting the broth with diethyl ether, concentrating it down, and spotting on SiO2 TLC plate. Compare RF with that of the pure compounds.
Alternatively, the same general extraction methodology could prepare a sample for GC-MS or crude NMR, if such instrumentation is available