How does existence of alpha and beta form of glucose prove that it exists as a cyclic structure My book says that

  Glucose is found to exist in two different crystalline forms which are
  named as $\alpha$ and $\beta$.

Next it says that

  This behaviour could not be explained by the open chain structure for
  glucose.

I am not able figure out how does existence of two crystalline forms prove that glucose exists as a cyclic structure? Are there no noncyclic compounds which exist in more than once crystalline forms?

And why can't it be explained by the open chain structure?

I think I am missing something, but what?

Thanks!

Edit:

If we take the example of glycine, it exists in two different crystalline forms, but doesn't have a cyclic structure like that of glucose. So why does existence of two crystalline forms for glucose seen as a supporting evidence of its ring form?

According to Wikipedia:

  Glucose is usually present in solid form as a monohydrate with a closed pyran ring (dextrose hydrate). In aqueous solution, on the other hand, it is an open-chain to a small extent and is present predominantly as α- or β-pyranose, which partially mutually merge by mutarotation.

Glucose predominantly occurs in nature in the form of the D‐enantiomer, which is generally believed to exist in three crystalline forms: $\alpha$‐D‐glucose monohydrate (Figure 1A)(Ref.2), anhydrous $\alpha$‐D‐glucose (Figure 1B)(Ref.2), and anhydrous $\beta$‐D‐glucose (Figure 1C)(Ref.1,3). Both anhydrous $\alpha$‐D‐ and $\beta$‐D‐glucose crystals are orthorhombic while $\alpha$‐D‐glucose monohydrate crystals are monoclinic (see Fig. 1A-C). However, a fourth form, which is metastable in solution phase at $\pu{38\!-\! 50 ^{\circ}C}$ and thought to be a hydrated form of $\beta$‐D‐glucose, has been reported as well (Ref.1,4).

The crystal structure of $\beta$-D-glucose published in 1960 (Ref.5) clearly showed the exsistence of pyranose ring system. As in the inserted box in Figure 1 state that, in aqueous solutions, 99% of D‐glucose exists as a mixture of the $\alpha$- and $\beta$-forms (approximately 62%  $\beta$ and 38% $\alpha$ when equilibrated at $\pu{31 ^{\circ}C}$ (Ref.1). Recent NMR study using fully $\ce{^13C}$ labelled glucose (Ref.6) clearly showed $\alpha/\beta$ ratio of $37/63$, which is almost identical to this literature value (Figure 2):

References:

F. W. Schenck, "Glucose and Glucose-Containing Syrups," In Ullmann's Encyclopedia of Industrial Chemistry: Ullmann's Food and Feed, Vol. 2; B; Elvers, Ed.; Wiley‐VCH Verlag GmbH & Co. KGaA: Weinheim, Germany, 2017, pp. 781-802 (https://doi.org/10.1002/14356007.a12_457.pub2).
W. B. Newkirk, “Manufacture and Uses of Refined Dextrose,” Industrial & Engineering Chemistry 1924, 16(11), 1173-1175 (DOI: 10.1021/ie50179a028).
G. R. Dean, “Optical-Crystallographic Properties of $\beta$-D-Glucose,” Anal. Chem. 1973, 45(14), 2440–2441 (DOI: 10.1021/ac60336a005).
G. R. Dean, “An unstable crystalline phase in the D-glucose-water system,” Carbohydrate Research 1974, 34(2), 315–322 (https://doi.org/10.1016/S0008-6215(00)82906-7).
W. G. Ferrier, “The crystal structure of $\beta$-D-glucose,” Acta Cryst. 1960, 13, 678-679 (doi: 10.1107/S0365110X60001588).
T. Richter, S. Berger, “A NMR method to determine the anomeric specificity of glucose phosphorylation,” Bioorganic & Medicinal Chemistry 2013, 21(10), 2710–2714 (https://doi.org/10.1016/j.bmc.2013.03.008).


When calculating the distance between molecules in a solution, is it reasonable to approximate a sphere around each molecule? I am attempting to calculate the average distance between molecules in a particular concentration of a solution. My first instinct was to treat the solution as a system of spheres surrounding the solute molecules, the size of which would be dictated by the concentration. Obviously such a calculation of the distance between molecules is primarily dimensional analysis, but is it a realistic treatment of the system, or is there a more appropriate way to arrive at the distance? A spherical treatment, in my mind, has the drawback of not accurately reflecting the distribution of solvent in the solution: a matrix of identically-sized spheres has vacant space, which a solution clearly would not have.

The distribution of molecules in solution is not uniform but has a distribution. The calculation of nearest neighbour distances from one molecule to another has been worked out a long time ago see,  S. Chandrasekhar, Rev. Mod. Phys. 1943, v15, p1.
Let $w(r)$ be the probability that the nearest neighbour occurs between distance  $r$ and $r+dr$. This must be the probability than no molecules exist closer to the victim molecule that one occurs at a distance $r$ and that this molecule exists in the shell $r$ to $r+dr$. Thus,
$$ w(r)=\left(  1-\int_0^r w(r)dr \right)4\pi r^2 n$$
where $n$ is the average number of molecules / unit volume. (1 mol/m3 $\equiv 10^3N_A/10^{27} = 0.6023 $ molecules / nm$^3$).
This looks hard to solve because $w(r)$ is also inside the integration, so it is necessary to isolate the integral first (by dividing both sides by $4\pi r^2 n$) and then differentiating both sides to find
$$\displaystyle \frac{d}{dr}\left(\frac{w(r)}{4\pi r^2 n}\right)= \frac{d}{dr}\left(1-\int w(r)\right) =-4\pi r^2 n \frac{w(r)}{4\pi r^2 n}$$
and the last step we effectively multiply by 1. The result is
$$ w(r)=4\pi r^2n\exp(-4\pi r^3n/3)$$
The average distance between molecules is
$$\langle r \rangle =\int_0^\infty rw(r)dr =\Gamma (4/3)/(4\pi n/3)^{1/3} =0.55396n^{-1/3}$$
where $n$ is in number of molecules/volume then $n^{-1/3}$ is in units of length.


Chiral centres with stereoisomeric substituents The structure of one particular isomer of 4-chlorohepta-2,5-diene is shown here. I am confused about how two of the propenyl groups attached to the central carbon have the same atoms with different configurations.

More specifically:

Is the carbon shown in the image is a chiral carbon or pseudo-chiral carbon?

Will this compound show optical isomerism, i.e. does it have an enantiomer?

Are the two alkenyl groups considered to be different from each other, or the same?

If the attached side centers were chiral centers (e.g $\ce{-CHClCH3}$) with different configurations, instead of geometrical isomers, then would this compound show optical isomerism?

Let's address your concerns one at a time:

The structure 1 you have drawn is chiral with $\ce{C4}$ having four different groups attached to it. $\ce{C4}$ has the (S)-configuration with the CIP priorities Cl>(Z)-propenyl>(E)-propenyl>H.
The structure has an enantiomer of the (R)-configuration. Just switch any two groups: Cl, (Z)-propenyl, (E)-propenyl or H. Each enantiomer will have the same specific rotation but opposite in sign.
The two propenyl groups are different stereoisomers, just as different as the (Z)- and (E)-2-butenes.
If the two propenyl groups were changed to -CHClCH3, say, (2R,4R)-2,3,4-trichloropentane 2a ($\ce{CH3CHClCHClCHClCH3}$), then the central carbon, $\ce{C3}$, is chirotopic but non-stereogenic. That is to say, if the chlorine and hydrogen at $\ce{C3}$ were switched as in 2b, the compound would still be chiral but not a different stereoisomer. Rotate 2a by 180o about a vertical axis and you will obtain 2b. They are identical.


How Does Benzyl Alcohol Produce a Peak at m/z = 79? I have understood the mechanism for converting ion 1 (benzyl alcohol) to ion 2 (m/z = 107) but not the mechanism for the conversion of ion 2 to ion 3 (m/z = 79).
Please elaborate the mechanism for the conversion of ion 2 -> ion 3.

Interesting question.

The cycloheptatriene - norcaradiene tautomerization is a common equilibium.  Usually the norcaradiene is present in small amounts, but as long as some is present the reaction can rapidly proceed through this intermediate.

In the diagram below, the norcaradiene can undergo hydrogen shift with concomitant decarbonylation to produce the m/z = 79 peak in the mass spectrum.


Calculating the chloride concentration What are the concentrations of $\ce{Cl-}$ in the following solution? $1.9\ \mathrm{g}\ \ce{MgCl2}$ is dissolved in water to make $1\ \mathrm{L}$ total solution

$$48.31\ \mathrm{g/mol}\ \ce{MgCl2}$$

$$(1.9~\mathrm{g}\ \ce{MgCl2})(1\ \mathrm{mol}/48.31\ \mathrm{g}) = 3.932\times 10^{-2}\ \mathrm{mol}\ \ce{MgCl2}$$

$$\ce{MgCl2 -> Mg+ + Cl2-}$$

$$(3.932\times 10^{-2}\ \mathrm{mol}\ \ce{MgCl2})((1\ \mathrm{mol}\ \ce{Cl2})/(1\ \mathrm{mol}\ \ce{MgCl2})) = 3.932\times 10^{-2}\ \mathrm{mol}\ \ce{Cl2-}$$

then

$$(3.932\times 10^{-2}\ \mathrm{mol}\ \ce{Cl2-})/(1~\mathrm{L}) = 3.932\times 10^{-2}\ \mathrm{M}\ \ce{Cl2-}$$

Is that correct? Or does the $2$ go infront of the $\ce{Cl}$, so the mole ratio is $((2\ \mathrm{mol}\ \ce{Cl-})/(1\ \mathrm{mol}\ \ce{MgCl2}))$?

The first thing we need to do is address your chemical equations.  $\ce{MgCl2}$ dissociates in water as follows:

$$\ce{MgCl2 -> Mg^{2+} + 2Cl-}$$

Now you can see that one mole of $\ce{MgCl2}$ dissociates to give two moles of $\ce{Cl-}$. So all we have to do then is calculate the number of moles of $\ce{MgCl2}$ from the given mass and its molecular weight, then multiply that by 2 to get the number of moles of $\ce{Cl-}$:

$$\mathrm{\frac{1.9\ g\ \ce{MgCl2}}{ 95.2\ \frac{g\ \ce{MgCl2}}{mol\ \ce{MgCl2}} } = 0.0200\ mol\ \ce{MgCl2}}$$

$$\mathrm{0.0200\ mol\ \ce{MgCl2}\ *\frac{2\ mol\ \ce{Cl-}}{mol\ \ce{MgCl2}} =0.0400\ mol\ \ce{Cl-}}$$

Then since the volume is 1L, the $\ce{Cl-}$ concentration is just $\mathrm{0.040 \frac{mol}{L}}$.


What will be the major enol form of methyl acetoacetate? I presume the best enol form would have a double bond between the two $\ce{C=O}$ so they're conjugated, but I'm not sure how the resonance would play out.

The enol forms of beta-diketones and beta -ketoesters are stabilized
by intramolecular hydrogen bonds in its enolic form. Enol structure with greater conjugation is more stable and amajor enol.

update :

In organic chemistry terms, conjugation is used to describe the situation that occurs when π systems (e.g. double bonds) are "linked together".
A conjugated system requires that there is a continuous array of "p" orbitals that can align to produce a π bonding overlap along the whole system.If a position in the chain does not provide a "p" orbital or if geometry prevents the correct alignment, then the conjugation is interrupted (broken) and therefore lost at that point. Due to extra π bonding interactions between the adjacent π systems there is an overall stabilization of the system .This is referred to as the delocalisation energy or the resonance energy or conjugation energy

source:

http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch10/ch10-1-1.html

According to MO theory,1,3-butadiene ,comprised of four overlapping p orbitals are mathematically combined to produce four molecular orbitals.As seen in figure , the four π electrons of a conjugated diene occupy the two lowest energy MOs (the bonding MOs).Specifically, the two electrons that occupy ψ1 are delocalized over four carbon atoms. This delocalization accounts for the observed stabilization energy .

source :
Organic chemistry ,David R. Klein.page no.773

ref :Advanced Organic ,Chemistry FIFTH EDITION ,Part A: Structure and Mechanisms FRANCIS A. CAREY
and RICHARD J. SUNDBERG


How to determine which combination of substances will give a buffered solution given target pH and Kb of each substances What combination of substances will give a buffered solution that has a pH of 5.05? (Assume each pair of substances is dissolved in 5.0 L of water.) (Kb for NH3 = $1.8 × 10^{–5}$; Kb for C5H5N = $1.7 × 10^{–9}$)

  a)1.0 mole NH3 and 1.5 mole NH4Cl

  b)1.5 mole NH3 and 1.0 mole NH4Cl

  c)1.0 mole C5H5N and 1.5 mole C5H5NHCl

  d)1.5 mole C5H5N and 1.0 mole C5H5NHCl

According to my text book it is choice C but I do not understand why and here was my thought process:

For starters I change the $K_b$ into $K_a$ to get : $5.6*10^{-10}$ for Ammonia and for the other I got: $5.9*10^{-6}$

It seems to me that this has to do with the Henderson Hasselbach Equation so I set up the following:

$5.05=5.9*10^{-6}+log{\frac{Base}{acid}}$

$5.05=5.9*10^{-10}+log{\frac{Base}{acid}}$

because I already have two substances.

But from here I feel like that wasn't the best route I should take because I got stuck right here and would like some assistance.

Using the Henderson-Hesselbach equation is a good idea. As you want the buffer's pH to be acidic, then using the $\ce{NH3/NH4+}$ buffer is not a good idea, as that would be basic, so we should use the $\ce{C5H5N/C5H5NHCl}$. Then, by Henderson-Hesselbach, we have that $\ce{5.05 = pH = pKa + \log \left(\frac{[C5H5N]}{[C5H5NH+]}\right)}\implies \ce{5.05 - pKa = 5.05 - 5.23 = \log \left(\frac{[C5H5N]}{[C5H5NH+]}\right)} \implies \ce{\frac{[C5H5N]}{[C5H5NH+]} = 10^{-0.18}\implies \frac{[C5H5N]}{[C5H5NH+]} \approx 0.66\approx \frac 23}.$ Thus, we want a buffer solution where the ratio of the base to the acid is approximately 2/3, which answer choice C satisfies.


Is it right to say that fluorescent lamps are actually fluorescent? Fluorescence occurs rapidly ($\approx 10^{-6}$ to $10^{-9}$ s) but phosphorescence is slow, and is therefore observable ($> 10^{-4}$ s). So, is it right to say that fluorescent lamps actually fluoresce when they actually do is to phosphoresce? Or am I wrong in saying that?

Fluorescent lamps do primarily work by fluorescence. According to this Wikipedia article:

  "The inner surface of the lamp is coated with a fluorescent (and often slightly phosphorescent) coating..."

First off, an electric current excites the outer electrons of gas atoms (typically mercury) within the lamp to a higher energy state. When the atoms relax back to their stable state, UV light is emitted. This UV light is then absorbed by a fluorescent material coating the inside of the lamp, which in turn fluoresces in the visible region of the EM spectrum. Some confusion could arise here as this coating is called a phosphor, even though the type of luminescence is fluorescence, not phosphorescence.

The rate at which the luminescence occurs is not important because there is a continuous supply of electric current. If it were necessary to provide light for any significant period of time after the power source was removed, then some type of phosphorescence would be necessary due to it's much longer lifetime as compared to fluorescence, as you pointed out. However, this is clearly not a requirement for a standard fluorescent lamp.


Number of bonds in a Lewis Structure I'm asked to find the Lewis Structure  of sulfuric acid, $SO_4H^-$. After struggling for a while, I sumply looked it up.

How can this be? Sulfur has 6 electrons in its valence shell, so it should form bonds so as to collect 2 more electrons and satisfy the octet rule. So, surely it should form 2 bonds?

But it's involved in no less than six bonds here. Huh?

The octet rule only applies firmly to elements below main energy level 3 (in fact, there are rare instances in which even that isn't entirely accurate, but molecular orbtal theory needs to be invoked to explain those situations). At main energy level 3, which Sulfur occupies, unfilled d orbitals become available, and those d orbitals serve to accommodate the additional electrons in Sulfur's bonds. In reality, Sulfur is a much larger atom than Oxygen, and the π orbital overlap between the two elements is fairly poor, giving the bonds significant single-bond character. You might therefore see resonance structures in which single bonds are drawn instead, with formal positive charges assigned to Sulfur and formal negatives assigned to Oxygen.


How to rationalise the planar structre of I2Cl6? $\ce{I2Cl6}$ is a dimer of $\ce{ICl3}$. $\ce{ICl3}$ is T-shaped, with $\ce{I}$ being $\mathrm{sp^3d}$ and having two lone pairs.

I read in Inorganic Reaction Mechanisms by R. K. Sharma that $\ce{ICl6}$ is formed by the extra two $\ce{Cl-I}$ bridge bonds. What I don't understand is, why does the bridge bond form? And is it a coordinate bond or something like a banana bond (as in $\ce{B2H6}$)?

Source: Sharma, R. K. Inorganic Reaction Mechanisms; Discovery Publishing House: 2007.

The bonding situation in the compound $\ce{(ICl3)2}$ is by far more complex than what is depicted in this book. The molecule itself has very high symmetry, i.e. $D_\mathrm{2h}$, that needs to be satisfied in the molecular orbital/ valence bond description.[1,2] That is why one cannot distinguish between the bridging chlorine bonds.

Analysing this bonding situation in the simple concept of Lewis-structures is not trustworthy and will lead to wrong conclusions. I have performed a short gas-phase optimisation at the DF-BP86/def2-SVP level of theory and analysed the resulting wave function with the Natural Bond Orbital theory. Covalent treatment of any of the bonds seems to be wrong. In this case I assumed all single bonds. Due to symmetry reasons, we only need to be concerned with two bonds. The outer $\ce{I-Cl}$ bond is polarised 93% towards the chlorine, but only occupied by 1.25 electrons. The remaining 0.75 electrons can be found in anti-bonding lone pair orbitals. This is not even close to expected Lewis structure behaviour. The bridging $\ce{I-\mu{}Cl}$ bond is still 81% polarised towards the chlorine, and has an occupation of 1.92 electrons. This only tells us, that this bond has a higher covalent character.

The analysis with the Quantum Theory of Atoms In Molecules (QTAIM) gives us more tools to analyse the electron density. In here the picture is quite obvious. All of the bonds in the molecule are of highly ionic character. The electron density at the bond critical point of the outer $\ce{I-Cl}$ is with $0.08~\mathrm{e\,a.u.^{-3}}$ very low and the Laplacian is positive at this point. The electron density at the bridging bond is even lower with $0.04~\mathrm{e\,a.u.^{-3}}$, again with a positive Laplacian.

(Laplacian distribution, solid blue lines indicate charge depletion $∇^2ρ<0$, dashed blue lines indicate charge accumulation $∇^2ρ>0$, red spheres are bond critical points, pink spheres are ring critical points, black lines are bond paths, dark red lines are zero flux surfaces.)

This finding is fairly consistent with the observation, that the melt of $\ce{(ICl3)2}$ conducts electricity, which according to Wikipedia is a hint for partial ionic dissociation.
$$\ce{(ICl3)2 <=> ICl2+ + ICl4- }$$

However, the planarity of this compound in solid state can be attributed to the possibility of having a somewhat quite extensive pi-bonding system as part of the small covalent bonding. For final conclusions about this, the results are not sufficient.

What can be said, judging from the molecular orbitals of $\ce{(ICl3)2}$, is that apart from the occupied core d-orbitals of iodine, no other d-orbitals are involved in any bonding - as it is expected. Shown here are the occupied non-core orbitals of $\ce{(ICl3)2}$. It is noteworthy, that the HOMO is fully antibonding.[3]

Finally a short warning:
While the assumption made by VSEPR may lead to the approximately correct conclusion for the interhalogenide compounds $\ce{XF3}$, where $\ce{X$=$Cl, Br, I}$, that the molecules are T-shaped, it is doubtfully the correct conclusion for the much less electronegative chlorine. This can be mainly explained by Bent's rule. Even in the monomeric compound the bonding situation is somewhat more complicated than you would expect.

Notes and References

The point group $D_\mathrm{2h}$ specifies that there is a main symmetry axis $C_2$ which passes through the centre of the ring. Additionally there are also two $C_2$ axis perpendicular to this and a horizontal mirror plane $\sigma_\mathrm{h}$. This means the molecule is planar. There are also non-planar molecules that can be described by this point group.)
As Max pointed out (Thank you), the structure is taken from the solid state. K. H. Boswijk and E. H. Wiebenga,  Acta Cryst. 1954, 7, 417-423. It mentions that it was purified by sublimation. This is no absolute proof of a gaseous dimer, but at least very suggestive. For our purposes we assume that the gas phase calculations are sufficient.
Orbitals obtained at the DF-BP86/def2-SVP level of theory with Gaussian09 rev. D. The energy increases row-wise from left to right and from top to bottom. The last shown structure is the molecule itself.


Synthesis of 2-benzylcyclohexan-1-ol from cyclohexanone There are no conditions given and a mechanism is required.

I first thought it could be along the lines of Claisen rearrangement but that needs an aromatic ring. Since the Claisen rearrangement can't be used I considered a ($\mathrm{S_N2}$) reaction of benzyl bromide by the alpha carbon followed by reduction of the carbonyl.

Making a carbon-carbon bond at the alpha position of a carbonyl suggests using some kind of enolate chemistry. In contrast to an SN2 type reaction with an alkyl halide (which can give over-alkylation and competitive O-alkylation), an aldol condensation is a reliable way to construct the requisite C-C bond. Here cyclohexanone reacted with benzaldehyde under basic conditions gives an enone. Two subsequent reduction steps, hydrogenation followed by hydride addition to the ketone, give the target compound.


How can sodium chloride melt ice or keep it frozen? In European countries, they use $\ce{NaCl}$ or $\ce{KCl}$ to melt ice during the winter season.
In Asian Countries, they use $\ce{NaCl}$ to keep the ice without melting, for example in ice cream and beer boxes.

I asked my chemistry teacher about this and his answer was,

  when you add ions to the ice, its melting point goes (i.e. freezing point of water) to -9 to -19 degree Celsius, depending on the amount of ions. The reason behind that is, when water becomes ice its water molecules re-arrange to specific shapes (see the link for the video at bottom). After we put ions to it, ions disturb that shape as they come in-between the water molecules. If water wants to be ice again, it should be like -9 degree Celsius because of these ions disturbing them to rearrange as a solid.

I also asked my Physics teacher about this and his answer was,

  When we put salt into water, its temperature goes down to somewhere around $\pu{-6 ^\circ C}$, so that it takes more time to come to $\pu{0 ^\circ C}$ and begins melting

Link to the videos are as follows,

Salt melts ice (YouTube)
Salt reduces the temperature of ice (YouTube)

Could you give me a correct explanation for this question?

The apparent contradiction is resolved as follows: in both cases, addition of salt makes ice melt faster, which is important in the street and negligible in a beer box (or an old style cold bath used in the lab). But in doing so, it also causes the temperature of the ice and its surroundings to go well below $0^\circ\rm C$, which is important in the lab and negligible in the street.


Breaking/reducing water adhesion with an electric field Does anyone know if the presence of an electric field can help to reduce water adhesion to a surface?

On the question relating to breaking/reducing water adhesion with an electric field, I suspect, the answer is that magnetic fields can have an impact.
Support, here is a 2008 paper published in Journal of Applied Physics, The surface tension of water under high magnetic fields, to quote in part:

The surface tension of water-air interface was measured by means of a surface-wave resonance method under magnetic field up to 10 T. It was found that the surface tension increased linearly with an increase in the square of the magnetic field.

Of note, background on the concept of surface tension, per Wikipedia, to quote:

At liquid–air interfaces, surface tension results from the greater attraction of liquid molecules to each other (due to cohesion) than to the molecules in the air (due to adhesion).
There are two primary mechanisms in play. One is an inward force on the surface molecules causing the liquid to contract.[1][2] Second is a tangential force parallel to the surface of the liquid.[2] The net effect is the liquid behaves as if its surface were covered with a stretched elastic membrane.

So, per the cited paper, surface tension increased linearly with the square of the magnetic field. Per Wikipedia, surface tension relates directly to cohesion, as opposed to adhesion.
As such, the question likely correct cites increased magnetic fields as reducing water's adhesion (as it apparently relates to promoting surface tension/cohesion).


Form of the MOs of butadiene We know that in butadiene there are 4 MOs. This is the MO with one node:

My question is: Why, with one node, can't we have something like this ?

A good way to think of this is by looking at forming butadiene from the MOs of ethene, which are an in-phase and an out of phase combination of p orbitals. If we looked at all the combinations of these two MOs of ethene, we obtain the MOs for butadiene


Hydrolysing large molecules to reveal amino acids Consider Oxytocin:

I am asked to hydrolyze Oxytocin and reveal 5 amino acids which are the result of this hydrolysis.

I must admit I have no idea where to begin. I know that generally amino acids are on the form of $\ce{NH_2-CH(R)-COOH}$, so perhaps looking at these double bonds to oxygen may be a starting point, as I assume they will become the acid group? But I really don't know how to move forward from there!

You are on the right track.  $\ce{NH_2-C(H)R-COOH}$ is the formula for an alpha amino acid, one where the amino group of the molecule is attached to a carbon atom one carbon atom away from the carboxylate carbon.  This is the most common type of amino acid found in biochemistry.  (A beta amino acid would be $\ce{NH_2-C(H)R^2-C(H)R^1-COOH}$).

When amino acids condense into a polymer or oligomer they do so by forming peptide bonds, also called amide bonds.

$\ce{R^1COOH + NH2R^2 -> R^1CONH2R^2 + H2O}$

So to find the places in oxytocin that would be hyrolyzed into amino acids, look for individual carbon atoms that are attached to both (i) an oxygen atom via a double bond and (ii) a nitrogen atom while (iii) not being attached to any other heteroatoms.  (Item (iii) is generally necessary to exclude other functional groups such as carbamides or carbamates, but these groups are not found in oxytocin.)  For oxytocin, your impulse to look just at carbon double-bonded to oxygen (item (i) only) is OK, but be warned that this procedure is not precise enough in general, because molecules with e.g. ketone moieties will have C=O bonds but will not be hydrolyzed.

The red-circled amino group is a free amino group:  the carbon attached to it is not bound to a carbonyl oxygen, so it is not a place where hydrolysis will happen.  (However, look one carbon atom to the left of C atom attached to the circled N....)

@DavePhD is right; there are more than five amide bonds in oxytocin.  I think there are a total of ten.  Eight are the result of amide bond formation between alpha amino acids and two are the result of amide bond formation between an acid side chain and unsubstituted ammonia.


Help with two containers problem, Xenon In two containers we have Xenon. The volume of the first container is 2780 L, temperature is 300 degree Celsius, and n = 200 mol. Volume of the second container is 4000 L, temperature is 1300 degrees Celsius and n = 100 mol. Temperature in both of the containers is constant. How much Xe do we have to transfer from one container to another so that the pressure in both containers is the same?
I have calculated pressure in container one and that is p1 = 342817.92 Pa, and in container two p2 = 326979.2275 Pa. Then I have substracted this two values which gives me 15838.6925 Pa. After that I have divided this value with two, to figure out how much do I have to decrease pressure in container 1, and that gives me 7919.34625 Pa.
After substracting this value from p1 I get the new pressure in container 1 which is 334898.5738 Pa which is the same as the pressure in container 2. From the new p1 I have calculated new n1 = 195.3798 mol, so I have to transfer 4.620 moles, or 606.593 grams to container 2.
My question is: When I take into the consideration the new n2 = 100 mol + 4.620 mol = 104.620 moles, and then I try to check if the p2 new is the same as p1 new I don'5 get the same result. p1 new = 334898.5738 Pa, and p2 new = 342086.14 Pa. Is my calculation wrong?

I have calculated pressure in container one and that is p1 = 342817.92 Pa, and in container two p2 = 326979.2275 Pa. Then I have substracted this two values which gives me 15838.6925 Pa. After that I have divided this value with two, to figure out how much do I have to decrease pressure in container 1, and that gives me 7919.34625 Pa.

You cannot generally just average both pressures. As $pV=nRT$, then ( for $\mathrm{T}$ and $\mathrm{V}$ constant ) :
$V \cdot \mathrm{d}p=\mathrm{d}n \cdot RT$
$\frac{\mathrm{d}p}{\mathrm{d}n}=\frac{RT}V$.
So pressure averaging would work only in the special case if $\frac{T_1}{V_1}=\frac{T_2}{V_2}$. Therefore, a weighted average must be used.

I suppose the scenario is vaguely defined. It can be the polytropic  expansion, or it may be the "locally isothermal" expansion. But for the former, it has like infinite number of possible solutions, depending on the rate of the transfer and of heat exchange.
I consider below the latter, i.e. isothermal conditions in each of containers.

We can consider the ideal behaviour, where
$$p=\frac{nRT}V$$
For equality of pressures, if $\Delta n$ is the moved molar amount of xenon from the container 2 to the container 1, the following equality must be true:
$$p=\frac{(n_1 + \Delta n)RT_1}{V_1}=\frac{(n_2 - \Delta n)RT_2}{V_2}$$
From that, we can easily calculate $\Delta n$:
$$(n_1 + \Delta n)RT_1V_2=(n_2 - \Delta n)RT_2V_1$$
$$\Delta n \cdot R(T_1V_2 + T_2V_1)=n_2RT_2V_1 - n_1RT_1V_2$$
$$\Delta n=\frac{n_2T_2V_1 - n_1T_1V_2}{T_1V_2 + T_2V_1}$$
Now you plug in the numbers here in the final formula, instead of working with literal numbers all the way:

In two containers we have Xenon. The volume of the first container is 2780 L, temperature is 300 degree Celsius, and n = 200 mol. Volume of the second container is 4000 L, temperature is 1300 degrees Celsius and n = 100 mol. Temperature in both of the containers is constant. How much Xe do we have to transfer from one container to another so that the pressure in both containers is the same?

$$\Delta n=\frac{(\pu{100 mol})(\pu{1573.15 K})(\pu{2.780 m3}) - (\pu{200 mol})(\pu{573.15 K})(\pu{4.000 m3})}{(\pu{573.15 K})(\pu{4.000 m3}) + (\pu{1573.15 K})(\pu{2.780 m3})}$$
In the sense below, if to be precise, we should round the temperature values for presented formula, because temperature was not known in 2 decimals accuracy ( otherwise there would be $\pu{300.00 ^{\circ}C}$, resp. $\pu{1300.00 ^{\circ}C}$), internally still computing with non rounded values.
$$\Delta n=\frac{(\pu{100 mol})(\pu{1573 K})(\pu{2.780 m3}) - (\pu{200 mol})(\pu{573 K})(\pu{4.000 m3})}{(\pu{573 K})(\pu{4.000 m3}) + (\pu{1573 K})(\pu{2.780 m3})} \approx \pu{-3.18 mol}$$
The negative value means the $\pu{3.18 mol}$ of xenon is to be moved from the first container to the second one, as we have formally assumed the opposite direction.

Internal full calculator precision of computation still makes good sense, to avoid artefacts of rounding errors.  But for presentation of final or intermediate values of quantities from the real world, the values have to reflect error propagation rules. As input data and constants have limited accuracy.


Is the ethyl cation really more stable than benzylic and allylic carbocations? Source: Concepts of Organic chemistry by O.P. Tandon, page no. 235

My book appears to state that the ethyl cation (a primary carbocation) is more stable than both allyl and benzyl carbocations.

I knew that cation stability depends on the following factors: resonance > hyperconjugation > +inductive effect.

Can someone please clarify whether the ranking shown above is true?

I believe this to be a simple typo in the book. In the original image you'll see that in the surrounding schemes $<$ and $\to\text{Stability}$ are given. Only in this line the relations are reversed, while the stability indicator remains the same.

However, since we're at it, we can at least try to put some values to the task. I'll adopt the same scheme used in the comparison of the t-butyl cation and the benzyl cation, or the radicals. I calculated the isodesmic reactions of the form in $\eqref{isodesmic}$ at the DF-B97D3/def2-TZVPP level of theory.
$$\ce{ R+ + CH4 -> RH + H3C+ }\tag{1}\label{isodesmic}$$

According to this we'll find the following order in decreasing stability: t-butyl, benzyl, allyl, (ethyl). Note that the ethyl cation is a non-classical cation, basically a proton coordinating to the π-bond of ethene; see also: Which carbocation is more stable, the ethyl- or 1-propyl-carbocation?.

\begin{array}{llr}
  \ce{R+}  &  \ce{RH}  &  \Delta G / \pu{kJ mol-1}\\\hline
  \ce{H3C+}           &  \ce{CH4}         &      0.0  \\
  \ce{[H2C=CH2]H+}    &  \ce{H3C-CH3}     &   -197.9  \\
  \ce{H2C=CH-CH2+}    &  \ce{H3C-CH=CH2}  &   -259.3  \\
  \ce{H5C6-CH2+}      &  \ce{H2C-C6H5}    &   -343.0  \\
  \ce{(H3C)3C+}       &  \ce{HC(CH3)3}    &   -370.9 \\\hline
\end{array}

(Side note: I was unable to remove a small imaginary mode from the isopropyl carbocation. That has almost no influence though.)

(I won't attach geometries or absolute energies this time, because that would exceed the character limit.)


Does cathodic protection work for air oxidation? I'm not expert in cathodic protection, but does it work with metals under air oxidation? Is it possible to protect a metal from air oxidation by applying an electrode to it? Or it has to be inside an electrolytic solution like sea water?

Not according to Wikipedia (current version of that page as of writing this answer) https://en.m.wikipedia.org/wiki/Cathodic_protection

  Galvanizing, while using the electrochemical principle of cathodic protection, is not actually cathodic protection. Cathodic protection requires the anode to be separate from the metal surface to be protected, with an ionic connection through the electrolyte and an electron connection through a connecting cable, bolt or similar.

So, the portions of your material exposed to air won't actually be receiving any "cathodic protection" formally per se, though galvanization may still help out quite a bit.


Points needed to come up with a calibration curve Usually when you are doing your masters, what are the minimum points that you can have when doing a calibration curve?

There is not really a simple answer to this—it depends on what you're doing and what level of error your application can tolerate. If you've characterized the response of your instrument and you know that it's linear within a certain range but the response has a variable offset, even a single-point calibration may be sufficient. On the other hand, if you just built an instrument and have no idea of its response characteristics but need very accurate results, you're best to take measurements at many different points (and over a length of time to see how stable the response is) to accommodate the possibility of non-linear response in the range of interest.

Occasionally, if you need to follow a measurement standard (e.g. the ITS-90 standard for temperature), you may be constrained to doing it a certain way so your data can be directly compared with that of other labs but in general, do enough to get the results you need, but not so much that you're just wasting your time. This might mean taking a few well-separated calibration points to get a ball-park value for an unknown sample, then bracketing the value with more calibration points near it to get a better measurement.


Why is the OO-H dissociation is unlikely for the Caro's Acid? Let us consider the Caro's Acid: $\mathrm{H_2SO_5 \equiv SO_3H-OOH}$.
It is widely known that the hydrogen tied with the $\mathrm{-OO-}$ group is pretty much unlikely to dissociate:
$$ \text {Terminal dissociation: }\mathrm{SO_3H - OOH \rightleftharpoons H^+ + [SO_3-OOH]^-} \\
   \text{An unlikely step: }\mathrm{[SO_3-OOH]^- \rightleftharpoons H^+ +[SO_3-OO]^{2-}} $$
The first time I encountered that fact I was presented a pretty inconclusive piece of argumentation: it was stated that $ \mathrm{-OO-} $ group is way more electronegative than $ \mathrm{-O-} $, which somehow brings about an obstacle for the dissociation.

Unfortunately, it was a long time ago and I cannot remember the exact details. I stumbled upon this problem not so long ago. Having a better background in chemistry, I attempted to apply the Sanders' electronegativity balance theory, but the result was pretty inane in terms of our chemical reality: $\chi_{eq}(\mathrm{H_2SO_4}) \approx 2.89894; \chi_{eq}(\mathrm{H_2SO_5}) \approx 2.96803 $. In other words: $ \chi_{eq}(\mathrm{H_2SO_5}) > \chi_{eq}(\mathrm{H_2SO_4}) $.

What am I doing wrong? Is it possible to tweak my attempt to use Sanders' theory? Is there any other approach to this problem?

One possibility is the formation of an intra-"molecular" hydrogen bond between the peroxy-bonded hydrogen and a third oxygen atom.  The structure shown in Wikipedia for the acid suggests this interaction.  The involved hydrogen is curled up into the five-membered ring formed by such a hydrogen bond, while the other hydrogen atom is freely exposed and ready to launch.  Potassium peroxymonosulfate corroborates shows only the non-peroxo hydrogen dissociated.

Hydrogen peroxide does not form this sort of intramolecular hydrogen bond, and as we expect from induction considerations the proton dissociates from this molecule more readily than from water.


How was pH measured back in the day if you had nothing to calibrate to? We learn about indicators but these seem awfully useless if you have no clue what pH the color change occurs at. How did early chemists figure this out? I guess more broadly, how was the hydronium content even measured in the first place?

"Hydronium" content can be easily determined by simple acid base titration. Keep in mind that there is nothing fundamental about pH. It is just a convenient scale for expressing hydrogen ion concentration. For example, a kilogram is a recent invention for expressing mass measurement, it does not mean that ancient people did not weigh anything before kilogram was known.
In the same way, acid-base titration is more than 300 years old, with litmus among the earliest indicators. pH is a relatively new concept. Of course, the titrations were not analytically precise and accurate 300 years ago.
Historically, the simplest way to do acid base titration was to use potash (naturally occurring mineral $\ce{K2CO3}$). One could continue titration until the evolution of bubbles stopped.* No indicator needed. This is the way vinegars were assessed. Note that in those time moles, molarity, did not exist. There was the concept of equivalent weights, i.e., this weight of X will react this that weight of Y completely. The concept of equivalents is still taught in some countries, although it has vanished from Western curricula.
Kolthoff (father of modern analysis) has written books on acid base indicators. Check internet archive. Your question is worth an entire essay.
*From Bishop, Indicators: International Series of Monographs in Analytical Chemistry


Conditions for spontaneity of redox reaction among redox couples with equal standard reduction potentials A question from the university course:

Could a redox reaction spontaneously occur among the redox couples $\ce{A_{ox}}/\ce{A_{red}}$ and $\ce{B_{ox}}/\ce{B_{red}}$ within non-standard circumstances if they have the exact same standard reduction potentials? So then, under which conditions?

The lecturer used the Nernst equation as follows for this question:
$$\Delta E = \Delta E^{\circ\prime} + \frac{RT}{zF} \ln{\frac{[\ce{A_{ox}}]}{[\ce{A_{red}}]}} - \frac{RT}{zF} \ln{\frac{[\ce{B_{ox}}]}{\ce{[B_{red}}]}}$$
From my understanding he is calculating the energy that is needed to run the reaction spontaneously, but I don't understand how he came to this equation.
If the reduction potentials of two half-reactions are the same, is it still possible for them to run spontaneously under certain conditions?

Yes, it is possible to construct two half cells with the same standard reduction potentials. For example, we can half cells made of copper dipping in copper sulfate solution. The caveat is that if both the cells are under standard conditions, the resulting electrode potential would be zero.
The trick is to use different concentration of copper ions in each half cell. With the help of Nernst equation you can determine which side will be the anode and which side will be the cathode. Read more about "concentration cells" as they are called.


What is going on when pen stops working on a section of paper? Ever had a ballpoint pen that would all of a sudden hit a 'dead spot' on a sheet of paper and refuse to write? It's clearly not the pen, because when moved off to the corner it would draw perfect circles instantly. Move it back to the spot, and it stops working again. Usually I can get something blotchy if I write out the word a dozen or so times while pressing hard.  What's going on with the paper and ink here?

Some pages of my notebook are like this and it's very annoying. My best guess is that it got splashed and something about the paper changed, but my guess probably isn't a very good one since it doesn't have the signature crinkly waviness of wetted paper.

If in doubt, try it out.

I just spilled some olive oil on a sheet of paper and wiped it off again.

Result: The ballpoint pen does not write on the oily spot.

Did you fat-finger the paper?

EDIT

The experiment above confirmed the finding in the question but didn't give an explanation.

Actually, two hours later it is still not possible to write on the oil-soaked and more transparent parts of the paper with a ballpoint pen, while a felt-tip pen (here: an Edding, ther German equivalent to a Sharpie) leaves perfect lines on every part of the paper.

Paper is fibrous material with a porous structure - without that and the capillary effects coming with it, fountain pens wouldn't be able to write on it.

For ballpoint pens, this probably isn't so important - the ink is very viscous.

But paper also has a rough surface - and this might be crucial, as already suggested by Lord Stryker in his comment. The rough paper surface provides friction, allows the tungsten carbide ball to roll, take up ink and spread it on the paper.

On a smooth or slippery surface - and this seems to be the case for the oily regions on the paper - the tip of a ballpoint pen only slides but does not roll. As a result, no ink is transfered to the paper.


For an electron arrangement diagram, is it necessary to show the inner electron shells or is it customary to only show the valency electrons? Which electron arrangement diagram is correct? The one that shows all electron shells or the one that only shown the valence electron shell?

Both are correct: for $\ce{Cl-}$, the inner set of electrons would be implicit, as $\ce{Cl}$ has 17 electrons. The point of drawing these orbital diagrams, especially in a molecular context, is to show the nature of bonds and to show how the octet/duplet rules are being followed.

If you have difficulties in using the second diagram (finding it harder to read or are confused by the similarities in $\ce{H}$ and $\ce{Cl}$), you should use the first diagram.


What possibilities are there for refining aluminium on the moon? Engines burning aluminium and liquid oxygen have been investigated for use on the moon because there is plenty of both elements in the soil - about 13% and 40% of highland soil, respectively. Propellant thus has the potential to be sourced from the moon directly, but it isn't practical unless a refining process can be found that uses the moon's environment to extract pure aluminum efficiently using local resources. A viable way to extract the oxygen is known already, using a pyrolysis process.

Is there a way of taking advantage of the moon's environment to do this effectively? You have at your disposal:

A hard vacuum
A month-long day that is strong sunlight for two weeks then a two week night - it is estimated that a furnace using concentrated sunlight could reach 3000 K.
1/6 Earth gravity
No moisture - chemicals are reduced
Powdery soil that is loose to a depth of 10 or 15 cm
Some soil samples from Apollo 16 were a quarter alumina by weight

A source of fuel in space is a prerequisite for a larger ongoing human presence there. A rocket launched from Earth can't manage to be more than about 5% payload. Shipping fuel from Earth to any off-world base by rocket would be prohibitively expensive (even if the rocket is reusable). This is why there has been so much interest in sources of water ice in space - it can be electrolysed into liquid hydrogen and liquid oxygen, which makes excellent propellant. But you have to go to the asteroid belt to get it (there are indications there might be significant reserves in craters at the moon's poles in permanent shade, but this is inconclusive). Once there is a way to refuel in space, all sorts of things become possible that are beyond our reach right now. An option that can compete with the price of mining asteroids has business potential. All such undertakings are going to require vast initial outlays that won't pay off for a decade or more. The process being considered here only has to meet that standard.

This question is a partner to a question at SE's Space Exploration site - How feasible is it to use aluminum and liquid oxygen as future propellant sourced from the moon?. I realized it was largely a chemistry question and so decided to ask about it here. There are two references in the other question: one from Wickman Spacecraft & Propellant, and one from the Artemis Data Book. I'm afraid neither discusses the chemistry and I haven't found anything else. The AIAA has several papers on it in their catalogue.

My theory was to use electromagnetic separation, aka the Calutrons used in the Manhattan Project.  Fine powder feedstock (check), into an arc discharge to vaporize and ionize (lots of good designs for high volume ion sources out there), electrostatic acceleration, then electromagnets to mass separate.  Power with a solar panel array.  No moving parts (except for the powder feed).  No vacuum systems needed.  And you separate all of the elements at once. Now, it only works during sunlight, but during the lunar night you could do maintenance and collect the separated elements.


Creating energy profile diagrams for publication Is there automated software to create energy profile diagrams like the ones below?

One could use something like Illustrator or Inkscape, but they aren't automated, are expensive (Illustrator), are difficult to use, and wouldn't be practical for anything with more than a dozen energy levels. These diagrams don't look like they were hand-drawn, but I could be mistaken.

Ideally, it would be written in Python, but something in LaTeX or TikZ would be workable as well.

More specifically, I think this could be improved upon quite a bit.

PyEnergyDiagrams

I use my own script. I have posted it on GitHub now.
This is the final result:

It is quite easy to use, on the GitHub readme there is a small tutorial. You have to create an instance of ED class and then add the energy levels to it.

from energydiagram import ED
diagram = ED()
diagram.add_level(0,'Separated Reactants')
diagram.add_level(-5.4,'mlC1')
diagram.add_level(-15.6,'mlC2','last',)
diagram.add_level(28.5,'mTS1',color='g')
diagram.add_level(-9.7,'mCARB1')
diagram.add_level(-19.8,'mCARB2','last')
diagram.add_level(20,'mCARBX','last')

To add the links between the levels you can have a look to the IDs of the level using this the argument of the function plot show_IDs=True:

diagram.plot(show_IDs=True)

And then you can use:

 diagram.add_link(left_ID_levetobelinked,right_ID_levetobelinked)

Export to pgf,pdf and svg

You can export it just saving the plot from the GUI button or programmatically in different formats for publications using matplotlib backend.
 I prefer to export the plot in .pgf   and then I import it in the LaTeX document this allow a full control over fonts and appearance of the figure.

For reaching better quality you have to change a little bit the autogenerated .pgf file.

The first thing you have to do is to find and replace − with -.

Later you can follow the procedure in this answer so that the font of the figure matches with the font size and type of the document.


Diluting a sodium chloride solution with specific glassware I have a $\pu{0.3 M}$ $\ce{NaCl}$ stock solution and need to create a $\pu{0.01 M}$ $\ce{NaCl}$ solution that is $\pu{50 mL}$. The only tools I am given are volumetric pipettes that are $10.0, 5.0, 1.0 \text{ and }\pu{0.5 mL}$ and $\pu{50.0 mL}$ volumetric flasks. I understand the physical process where I would measure out a specific amount and place it in the flask using the pipette(s) and then add the required amount of water.

I need to use the dilution equation $M_1\times V_1 = M_2\times V_2$ which I did and I got $\pu{1.666667 mL}$ of solution that needs to be taken from the $\pu{0.3 M}$ stock solution to create the $\pu{0.01 M}$ solution. The only problem is that I can't do that since the materials I have do not allow me to measure out that amount.

I tried to "convert" the $\pu{0.3 M}$ to various other concentrations that would hopefully get me down to $\pu{0.01 M}$ but have failed so far. I tried to find concentrations that would result in a whole number or a value with $.5$ at the end after dividing by the product from $M_1\times V_1$ but failed every time. In addition, I combined the two concentrations and found the midpoints between them and tried to use those values, but they did not work either. Where is a good place for me to start?

Assuming you can use a beaker (or just one of those 50 ml volumetric flask for mixing), you could first dilute some of the 0.3 M solution to 0.1 M by adding 5 ml of stock to 10 ml of water. Now you have a 0.1 M solution of NaCl which you can calculate how to dilute using the M1*V1=M2*V2 equation. Doing this, I calculate that to make 50 ml of 0.01 M solution, you need 5 ml of the 0.1 M solution you made, which you can do with a pipette.


Example of an achiral molecule without a plane of symmetry or inversion center? Some websites, such as this textbook, mention how generally a plane of symmetry or inversion center in a molecule is enough to consider it achiral. Nonetheless, this source clarifies

but if these symmetry elements are absent the molecule should be checked carefully for an  S$_n$  axis before it is assumed to be chiral.

Furthermore, Wikipedia notes the definition of chirality as "a molecule which has no S$_n$ axis for any value of n is a chiral molecule."
Thus, are there any molecules lacking plane of symmetry/center of inversion but that are achiral due to presence of an axis of improper rotation? Are there are also examples of vice-versa, where a molecule has a plane of symmetry/center of inversion, but lacks an axis of improper rotation and is thus chiral?
edit: This post does cover some definitions which is helpful, but the point of this question was to see an actual example.

Part 1

Are there any molecules lacking plane of symmetry/center of inversion but that are achiral due to presence of an axis of improper rotation?

Yes, although such examples are very rare. In general, the point groups $S_{2n}$ $(n \geq 2)$ are where you should look. These molecules have a $S_{2n}$ rotation axis, but no plane of symmetry, and no inversion centre.
Molecules in the $S_6$ and $S_8$ point group are are exceedingly rare, and generally need to be constructed specifically to satisfy these criteria. The best examples, therefore, come from the $S_4$ point group. You can find several examples on these websites:

http://gernot-katzers-spice-pages.com/character_tables/S4.html
http://csi.chemie.tu-darmstadt.de/ak/immel/tutorials/symmetry/index7.html#Sn

In particular, one of the nicest examples is 2,3,7,8-tetramethyl-spiro[4.4]nonane from the second link. In the graphic below, I've taken the liberty of using four different coloured atoms in place of the methyl groups. The reader should therefore bear in mind that the third image is the same thing as the first, even though the colours are different.

The XYZ coordinates that I used for the pictures above are as follows. You can download these and view them in software such as Avogadro or ChemCraft, although I don't think you can perform a "reflection" per se (I manually swapped the colours between the second and third pictures to achieve this effect).
C         -2.32545        1.12231        0.91714
C         -2.19292        2.47894        0.24035
C         -0.83350        2.34003       -0.44571
C         -1.75438        0.10430       -0.11277
H         -3.31782        0.88874        1.35769
H         -1.62382        1.17900        1.77968
C         -0.98281        0.99078       -1.13371
H         -1.65567        1.19844       -1.99603
H         -0.05915        0.56089       -1.57585
C         -2.87426       -0.73718       -0.79187
C         -0.83536       -0.95866        0.55714
C         -2.33811       -2.16138       -0.80738
H         -3.27791       -0.35456       -1.75318
H         -3.74616       -0.78244       -0.10108
C         -1.65652       -2.23999        0.55910
H         -0.36215       -0.67792        1.52187
H          0.00882       -1.17785       -0.13479
B         -1.40058       -2.29939       -1.82101
H         -3.13035       -2.92594       -0.95624
O         -2.60331       -2.18962        1.57227
H         -1.04117       -3.15429        0.69961
H         -2.24954        3.33056        0.95160
F         -3.19594        2.63169       -0.70627
N          0.17857        2.27493        0.50135
H         -0.60168        3.16745       -1.14998

Part 2

Are there are also examples of vice-versa, where a molecule has a plane of symmetry/center of inversion, but lacks an axis of improper rotation and is thus chiral?

An improper rotation $S_n$ is defined by a rotation about $360/n$ degrees, followed by reflection in a plane that is perpendicular to that rotation axis.
A plane of symmetry ($S_1$) and an inversion centre ($S_2$) are special cases of an improper rotation ($S_n$). It is easier to convince yourself of the $S_1$ case: according to the definition above, $S_1$ means rotation through $360^\circ$ followed by reflection in a plane. Since rotation through $360^\circ$ obviously does nothing, this is the same as a reflection in a plane.
So, the answer to this question is no. if a compound has either a plane of symmetry or an inversion centre, that automatically means it has an improper rotation axis.


Do Moore's tables really exist? I recently have a lesson about quantum chemistry, and in one lesson we spoke about hyperfine stucture of elements.

The teacher give us something which looks to be an extract of Moore's tables here for the carbon. He said there are thousands of this tables which give the energy levels of many atoms.

But I search on the net and I only found things about electronic such as transistors et caetera... can you tell me the real name of this tables or how can I find them ?

Thank you in advance

Yes, Charlotte Emma Moore and her tables really exist(ed).

See Atomic Energy Levels, 2nd edition.


Which has the higher chemical shift E/Z alkene? If you have a simple alkene for example 1-methyl-But-2-ene

then you will have two methyl groups on one side of the double bond, which one of these will have the highest. Will the one on the same side as the rest of the chain of butene have the higher ppm or will it be the one in trans with the butene.

By the way, on my spectra, these two methyl groups are not equivalent. I need to know if the E or Z will have a higher chemical shift and why, thanks.

Are you talking about 1H NMR? In the general case, it is not so easy to say - it depends on all substituents, with geminal substitution usually contributing most to change in chemical shift. trans and cis substitution will affect chemical shifts differently, but will depend on the electronegativities of all groups concerned.

For your example, there will be very little difference between chemical shift for these two methyl groups, but the methyl trans to the alkyl chain is probably slightly downfield shifted. Coupling constants for the two methyl groups would be slightly different. In acyclic systems, 4J couplings from a methyl to a cis or trans proton can range from about -3 to +2.5 (but usually 0.8 to 2), and usually |Jtrans|>|Jcis|. Of course, nOe information would also be useful.

If you are talking about 13C NMR, a methyl peak trans to a carbon chain will be downfield shifted from a methyl group trans to a proton.


Ions and Polyatomic Ions Before a sodium atom and a chlorine atom combine together to form the ionic compound $NaCl$, they are just neutral atoms. However, when an electron is transferred from the outer energy level of the sodium atom to the chlorine atom, both of these gain a charge on them; the sodium atom or ion has a charge of $1+$ and the chlorine ion has a charge of $1-$. Both of these ions attract each other and an ionic bond is formed between them.

  But what happens when an atom of an element say $Ca$ combines with a polyatomic ion say $OH^-$. Is this calcium atom an ion, is it already a charged atom when it interacts with the hydroxide ion or does it gain the charge by any transfer of electrons to the $OH^-$ ion and becomes $Ca^{2+}$ to bond with the $OH^-$ ion?

I think the polyatomic ions say $OH^-$ is already an ion and not an uncharged compound of elements that gains charge as in the case of $NaCl$, and the bond between an ion and a polyatomic ion is due to the attractive forces between them if already they have opposite charges.

I risk not directly answering the question as I have not perfectly understood it.

In your first example, you consider the reaction between two neutral atoms, $\ce{Na}$ and $\ce{Cl}$. An electron is transferred generating $\ce{NaCl}$ where the species are now charged ($\ce{Na^+ \  Cl^-}$). Then you follow up with a hypothetical reaction between a neutral atom ($\ce{Ca}$) and an already charged species ($\ce{OH^-}$). The two reactions are fundamentally different, but not because only the latter has polyatomic ions; the difference is in the very nature of the reactions you constructed (two neutral species reacting in the former, compared to one neutral species and an ion reacting in the latter).

The most direct way to obtain $\ce{NaCl}$ is the reaction between sodium metal (composed of neutral atoms, $\ce{Na^0}$) and chlorine gas (composed of neutral diatomic molecules, $\ce{Cl2}$). One then has:

$$\ce{2Na^0_{(s)} + Cl2_{(g)} → 2NaCl_{(s)}}$$

To obtain $\ce{Ca(OH)2}$, a similarly simple reaction would be that between calcium metal (composed of neutral atoms, $\ce{Ca^0}$) and water (composed of neutral molecules, $\ce{H_2O}$), where the water serves as a source of hydroxyl ions according to the equation:

$$\ce{Ca^0_{(s)} + 2H2O_{(l)} → Ca(OH)2_{(s)} + H2_{(g)}}$$

Now imagine you want to force a reaction between a neutral atom and an ion. There are many cases of this, but we can stick to something similar to what you wanted. In order to expose a substance directly to ions, a conceptually easy way is to add it to a molten salt. For now, let us consider pumping sodium vapour into molten caesium hydroxide ($\ce{CsOH}$). You seem to expect some interaction between the neutral $\ce{Na^0}$ and charged $\ce{OH^-}$, but in fact it is $\ce{Cs^+}$ that reacts with the sodium, according to the following equation:

$$\ce{Na^0_{(g)} + CsOH_{(l)} → NaOH_{(l)} + Cs^0_{(g)}}$$

Here we have simple exchange reaction. The hydroxide ion is actually only a spectator, and can be omitted without loss of understanding in an ionic equation:

$$\ce{Na^0_{(g)} + Cs^+_{(l)} → Na^+_{(l)} + Cs^0_{(g)}}$$

If we consider the reaction between sodium and molten caesium chloride instead, where chloride ions would be present, we have again a simple exchange reaction:

$$\ce{Na^0_{(g)} + CsCl_{(l)} → NaCl_{(l)} + Cs^0_{(g)}}$$

Indeed, it can be described by the same ionic equation above.

(Disclaimer: I actually am not sure the reactions between sodium and caesium compounds works as above. I don't have any understanding of the relative electrochemical potentials in molten salts, though I believe I've read that the reaction with molten chloride is a legitimate way to obtain metallic caesium. But that is mostly beside the point, either way. I avoided exemplifying with calcium and calcium hydroxide because I think the reaction wouldn't work, and the hydroxide is thermally unstable)


When to include or exclude lone pair counting electrons according to Hückel's rules? When counting the number of pi electrons, in furan only one lone pair was included, but the other one was not.

In the cyclopentadienyl anion (2nd aromatic compound in the picture) the lone pair was counted.

When to include and exclude lone pairs in Hückel's rules?

As written on Chemistry Libre Texts,

  Perhaps the toughest part of Hückel's Rule is figuring out which electrons in the compound are actually π electrons. Once this is figured out, the rule is quite straightforward.  π  electrons lie in p orbitals and  sp2 hybridized atoms have 1 p orbital each. So if every molecule in the cyclic compound is sp2 hybridized, this means the molecule is fully conjugated (has 1 p orbital at each atom), and the electrons in these p orbitals are the π electrons.

In furan, there are two lone pairs, one in the sp2-orbital, and the other in the p-orbital. The latter is hence the set of π electrons.

In the cyclopentadienyl anion, the lone pair is present in the p-orbital, as all the sp2-orbitals are involved in bonding.

Thus, π electrons and not just lone pairs are to be counted in Huckel's Rule.


On the use of hexane in the organic reduction by DIBAL-H In some variations of the organic reduction by DIBAL-H, I've seen hexane used, as in this equation from this website;

I've looked it up, but I haven't seen any explanation for why hexane's used there. It's also seen in the action of DIBAL-H on alkyl nitriles to give aldehydes with ammonia.
Can anyone help me with why it's used in both cases?

Unlike other reducing metal hydrides (e.g., $\ce{NaBH4}$ and $\ce{LiAlH4}$), diisobutylaluminum hydride (DIBAL-H) is a liquid at room temperature and dissolve in many hydrocarbons such as toluene and hexanes, which also have very low freezing points. For example, hydrocarbons toluene and hexanes both have freezing points around $\pu{-95 ^\circ C}$. Thus, they are good solvents for reaction involving reduction by DIBAL-H at low temperatures such as the example given in the question ($\pu{-70 ^\circ C}$). As shown in the equation, the low reaction temperatures help stop the reduction at aldehyde level. At room temperature with excess DIBAL-H, the reduction may continue further to give corresponding alcohols or even corresponding alkanes.
DIBAL-H reduces alkyl or aryl nitriles to their corresponding imines at low temperatures and resultant imines converted to corresponding aldehydes upon acid work-up. The same principle mentioned above apply here as well. Here, at high temperatures, imines reduced further to give corresponding amines.
Note: To see the mechanism of the reduction, see here.


Do ions also form special electron configurations? (just as Copper and Chromium do) Do ions also form special electron configurations? (just as Copper and Chromium do)

I know that because copper has 1 electron in its s-orbital (highest) because it wants to have a full d-orbital (d^10)

But do ions also act this way?

Check out the NIST database for electronic configurations for atoms and ions in the gas phase. If you play around a bit, you will find that yes, some ionizations will not follow the expected trends from the aufbau principle. For example, search for rhenium and compare the configurations between $\ce{Re^{11+}}$ and $\ce{Re^{17+}}$, and you'll see some of them are unexpected (and rather hard to explain). Osmium shows an anomaly going from $\ce{Os^{+}}$ to $\ce{Os^{2+}}$. There are other elements with ions containing aufbau anomalies, but I can't remember many from the top of my head.


Nomenclature for stereoisomer example possible wrong? I understand the principal of the Cis-Trans naming system (at least I think so), until I saw this example in my learning material. My teacher doesn't know either so I thought I would post it here.

This is the molecule:

There are two ethyl groups both on the bottom side of the double bond, so it could be cis right?

Apparently not, the book says this is the answer:

Could someone please explain the reasoning behind this? This particular example is unusual because there are three ethyl groups attached to the double bond so I cannot determine which one should be the parent chain. The book doesn't explain.

Unless I'm very much mistaken, this molecule should not be labelled with any stereochemical descriptors, whether they be relative (cis/trans) or absolute (E/Z). The double bond has three identical ethyl substituents, therefore there is no stereoisomerism possible. By way of explanation, imagine if the ethyl group and the hydrogen on carbon #4 were interchanged: the resulting molecule would be completely identical to the one pictured. This is always the case when an alkene bears two identical substituents on the same carbon atom.

To correctly use cis/trans or E/Z nomenclature, it is necessary that there not be two identical groups on the same carbon of the alkene; if there are, the labels are meaningless, since interchanging the two groups results in an identical molecule.


DCC coupling of N-hydroxysuccinimide and 2,2-dimethoxyacetic acid I am forming N-hydroxysuccinimide ester (1) by DCC coupling with N-hydroxysuccinimide and 2,2-dimethoxyacetic acid using distilled THF as solvent.

What I see on $\ce{^1H}$-NMR is three major singlets:
$\delta~\pu{4.840 ppm}~(1\ce{H}),$ $\delta~\pu{3.432 ppm}~(7.26\ce{H}),$ $\delta ~\pu{2.731 ppm}~(6.36\ce{H})$ and some impurities.
While the chemical shifts are well-aligned with theoretical predictions (in my opinion), integration ratios do not quite make sense. I wish I could purify it to have a clearer NMR spectrum and see whether I actually have the desired product, but I don't think my ester will stay stable in silica in column chromatography. Any suggestions?

If peaks for "some impurities" are appear around $\delta \: \pu{1.8 ppm}$, I assume you have residual THF in your crude product. The other multiplet of THF is probably burried under $\delta \: \pu{3.43 ppm}$ resonance. That would confirmed by your impurity integration, which should be equal to 1H unit equivalence. If I'm right, your 6.36H unit equivalence is actually overlap one of $\ce{CH3}$-resonance and one of $\ce{CH2}$-resonance.

On the other hand, the two resonances at $\delta \: \pu{3.43 ppm}$ and $\delta \: \pu{2.73 ppm}$ would very well be representing two $\ce{CH3}$-groups and (two $\ce{CH2}$-groups + part of THF impurity), respectively, as well.


Subtle implications of quantum numbers Question:

  Given the subshells $\ce{1s, 2s, 2p, 3s, 3p}$ and $\ce{3d}$, identify those
  that meet the following descriptions:
  a) has $l=2$
  b) Can contain two electrons with spin $m_{s}=\pm\frac{1}{2}$.

For a): from my understanding $l=n-1$. Therefore $\ce{3s, 3p}$ and $\ce{3d}$ should be correct but according to my solutions manual, the answer is only $\ce{3d}$. Why is this so?

For b) The answer is $\ce{2p, 3p}$ and $\ce{3d}$. Why are these the answers?

a) You are wrong. $l=2$ simply means $\ce{d}$-orbitals, thus, only $\ce{3d}$ is the right answer.

b) Assuming that the meaning of the question is to specify subshells that can contain two electrons in the same spin state, i.e. both with $m_s = +1/2$ or both with $m_s = -1/2$, $\ce{s}$-orbitals should indeed be excluded.


Is it the Na+ or the Cl- that gives table salt its taste? What gives it that salty taste? I'm wondering if it's the $\ce{Na+}$ or the $\ce{Cl-}$. I know that $\ce{KCl}$ also tastes salty, and other salts have different flavours,  but where does the saltiness in $\ce{NaCl}$ come from? I mean, it makes sense that it's the $\ce{Cl-}$ ion, but it could also be that $\ce{K+}$ and $\ce{Na+}$ have similar flavours.

The salt receptors are sodium ionic channels, i.e. peptides which are selectively permeable for Na ions [1].
Ionic current causes receptor activation and neural responce.

Anion plays minor role here, imposing upon salty taste. The result may appear acidic, bitter and even sweet.
For example, I believe that sulphates are overally bitter.
Carbonates or alkali probably just burn the receptors and may have random taste.


Why is Nitric Acid more reactive and caustic than Sulfuric or Hydrochloric? Nitric acid reacts much more vigorously with living tissue and is a obviously more immediately hazardous on contact than sulfuric or hydrochloric.

What is the reason for this?  Is it due to the oxidizing potential of the nitrate anion?

Does the greater reactivity of nitric acid also hold for reactions of acids with metals? Perhaps the factors related to the acid anions that determine dissociation are not the same ones that determine their reactivity?

Nitric Acid is indeed what most often is referred to as a "strong acid" (at ambient temperatures), and its $\mathrm{pK_a} \approx -1$, i.e nitric acid in diluted solution is almost always fully dissociated except in extremely acidic solutions.

First of all, how does it compare to other "strong acids", for example sulfuric acid

$\ce{HNO3 + 2 H2SO4 ⇌ NO2+ + H3O+ + 2HSO4−}$

Clearly, in this reaction sulfuric acid acts as the "acid", while nitric acid acts as the "base", hence sulfuric acid is indeed a stronger acid.

Reactions of dilute nitric acid with some metals are typical of an acid, namely:

$\ce{Mg + 2 HNO3 → Mg(NO3)2 + H2\uparrow}$

Other metals that react in this fashion are manganese and zinc.

You do correctly note that a special feature of nitric acid's reactivity is its strength as an oxidising agent, and this is also evident in its reactions with metals, for example. However, do note, that the exact details depend on the concentration of the acid and the nature of the metal

For example, nitric acid can oxidise non-active metals such as copper and silver, however, the products depend on temperature and the acid concentration.

For instance, copper reacts with dilute nitric acid at ambient temperatures with a 3:8 stoichiometry:

$\ce{3 Cu + 8 HNO3 → 3 Cu^2+ + 2 NO + 4 H2O + 6 NO3−}$
The nitric oxide produced may react with atmospheric oxygen to give nitrogen dioxide.

However, with more concentrated nitric acid, nitrogen dioxide is produced directly in a reaction with 1:4 stoichiometry:

$\ce{Cu + 4 H+ + 2 NO3− → Cu^2+ + 2 NO2 + 2 H2O}$

Metals like chromium, iron and aluminium readily dissolve in dilute nitric acid, but the concentrated acid forms a metal-oxide layer that protects the bulk of the metal from further oxidation; thus, concentrated nitric acid serves to passivate these metals towards further reactions. Common metals that are passivated by concentrated nitric acid are iron, cobalt, chromium, nickel, and aluminium.


Relation between molar mass and van der Waals force In the book that I'm reading it is mentioned higher molar mass results in stronger van der Waals force, but didn't continued why.

I wonder why is that so? molar mass is actually dependable on the number of protons and neutrons, how can it be effective on the van der Waals force?

From chemwiki.Ucdavis:

  The more electrons a molecule contains, the higher its ability to become polar. Polarizability increases in the periodic table from the top of a group to the bottom and from right to left within periods. This is because the higher the molecular mass, the more electrons an atom has. With more electrons, the outer electrons are easily displaced because the inner electrons shield the nucleus' positive charge from the outer electrons which would normally keep them close to the nucleus.

This seems totally legitimate. However, (as chemwiki subtly points out) the main factor in work here is the volume occupied by electrons. Polarizability increase is mainly due to a volume increase 1. So, all you need to take into consideration is whether a molecule is bigger than the other in terms of comparing the stronger London dispersion force.

Simply put, the ability of a molecule to get polarized is called polarizability. See Wikipedia for more info.


Question regarding the mechanism of oxymercuration-demercuration of alkenes Organic Chemistry by David R. Klein

The text in between says that the more substituted carbon bears a partial positive charge. Why not the other one? I am asking this considering inductive effect, although I am myself not too convinced by my inductive effect explanation.

This reaction, in my opinion is an extension of Markovnikov’s rule

To understand why Markovnikov’s rule works, consider the
structure and stability of carbocations and about the general nature of reactions
and transition states.

The first point to explore involves structure:

  A great deal of experimental evidence has shown that carbocations are
  planar. The trivalent carbon is sp2-hybridized, and the three
  substituents are oriented toward the corners of an equilateral
  triangle.

Because there are only six valence electrons on carbon and all six are used in the three s bonds, the p orbital extending above and below the plane is
unoccupied.

The second point to explore involves carbocation stability. Thermodynamic measurements show that,
indeed, the stability of carbocations increases with increasing substitution so that
the stability order is tertiary > secondary > primary >methyl.

Now here comes the question…

Why are more highly substituted carbocations more stable than less highly substituted
ones?

There are at least two reasons.

Part of the answer has to do with inductive effects (as you noted in question) , and part has to do with hyperconjugation.

  Inductive effects, result from the shifting of electrons in a s bond
  in response to the electronegativity of nearby atoms. (electrons from
  a relatively larger and more polarizable alkyl group can shift toward
  a neighboring positive charge more easily than the electron from a
  hydrogen

. Thus, the more alkyl groups there are attached to the positively charged carbon, the more electron density shifts toward the charge and the more inductive stabilization of the cation occurs)

Hyperconjugation, is the stabilizing interaction between a p orbital and properly oriented C- H sigma bonds on neighboring carbons that are roughly parallel to the p orbital. The more alkyl groups there are on the carbocation, the more possibilities there are for hyperconjugation and the more stable the carbocation


LC-MS: Can a "bad" peak still be interpreted I have done an untargeted metabolomics study using LC-MS. There is one peak that matches to a biologically relevant compound but analyte eluted at the beginning of the LC run and the peak shape is far from ideal.

Do I assume the peak is an artifact, or can I say that there is a compound with that m/z in my sample? Ideally I would run the experiment again with standards, but I cannot do that due to time constraints. Someone else will follow up on the interesting compounds that I've identified and I wouldn't want to send them in the wrong direction.

Below is the peak in question, and following it is an example of a "good" peak from the same dataset.

This is low-quality data.  It's very difficult to say whether this signal is meaningful for several reasons.

It appears as soon as the LC-MS starts recording data at very low retention times.  That means it could be an artifact of column regeneration.  There is no sign that the data contains a "void volume" peak, meaning it is unclear whether the low-RT signal is from sample injection, from the previous sample, or some other artifact.

You don't say whether you randomized sample injection order.  Therefore the differences between the red and black sets of samples could arise because of changing conditions in the LC-MS instrument over time (e.g. column fouling, sample aging in the autosampler) or from legitimate differences in the samples that are biologically meaningful.

You don't show the full mass spectrum for your proposed analyte, so it is possible that the single m/z you showed here has an isotopic pattern consistent with your analyte, or inconsistent with it.

Thus, to answer your questions:

It basically wrong on the basis of info presented to claim that you detected a "biologically interesting" metabolite on the basis of the information presented.

It is also bordering on misleading to say you detected "a molecule" with the given m/z at all.  Maybe you could say you detected an unknown signal at m/z of xyz and are unsure of its origin.  That's about it.


Synthesis of Pyrene from Benzene I know of this ring expansion reaction that can fuse a benzene ring with another benzene ring:

Performing this reaction twice gives us phenanthrene

How would one go about synthesizing pyrene from phenanthrene or benzene? I couldn't find any literature online, as most pyrene is obtained via fractionating and cracking petroleum. Synthesizing it from phenanthrene also turns up a dead end, as the 9 and 10 positions on it are most reactive, and I would not obtain pyrene in that way. Diels-alder reaction of phenanthrene is also not possible, as highlighted in this answer.

Petroleum cracking and incomplete combusion of organic compounds may give pyrene in a complicated mixture as discussed in else where. However, if you looking at a method to prepare pure pyrene or its derivative you may need to go through [2.2]metacyclophane procedure (For review: Ref.1):

As seen in the above image, [2,2]metacyclophane can undergo transannular ring closure to yield products with the 4,5,9,10-tetrahydropyrene (THPy) skeleton (e.g., Ref.2), which may then be dehydrogenated to pyrene (Ref.3).
The synthesis of [2,2]metacyclophane(s) can be achieved from m-xylene. This can be in two steps: First, the bromination of two methyl group using NBS to achieve 1,2-bis(bromomethyl)benzene and then subjected the brominated m-xylene to Wurtz reaction (e.g., Ref.4). Recent approach of some substituted [2,2]metacyclophanes is described in Reference 5 starting with substituted m-xylenes (The electronic supplementary material is attached for readers benefit).
The synthesis of highly substituted pyrenes via cyclisation of bis-alkynylbiphenyls has been also achieved (e.g., Refs.6-8). It is worth noting that these starting biphenyls were made using simple benzene derivatives (Ref.1):

References:

Juan M. Casas-Solvas, Joshua D. Howgego, Anthony P. Davis, “Synthesis of substituted pyrenes by indirect methods,” Org. Biomol. Chem. 2014, 12, 212-232 (DOI: 10.1039/C3OB41993B)(PDF).
Fritz Vögtle, Peter Neumann, “Stereochemistry of [2.2]Metacyclophanes,” Angew. Chem., Int. Ed. Engl. 1972, 11(2), 73–83 (https://doi.org/10.1002/anie.197200731).
Peter P. Fu, Ronald G. Harvey, “Dehydrogenation of polycyclic hydroaromatic compounds,” Chem. Rev. 1978, 78(4), 317–361 (https://doi.org/10.1021/cr60314a001).
Virgil Boekelheide, Rodger W. Griffin Jr., “Synthesis of [2.2]metacyclophanes,” J. Org. Chem. 1969, 34(6), 1960–1961 (https://doi.org/10.1021/jo01258a095).
Marco Blangetti, Helge Müller-Bunza, Donal F. O'Shea, “First asymmetric synthesis of planar chiral [2.2]metacyclophanes,” Chem. Commun. 2013, 49(55), 6125-6127 (https://doi.org/10.1039/C3CC42275E)(Electronic Supplementary Material).
D. Barney Walker, Joshua Howgego, Anthony P. Davis, “Synthesis of Regioselectively Functionalized Pyrenes via Transition-Metal-Catalyzed Electrocyclization,” Synthesis 2010, (21), 3686-3692 (DOI: 10.1055/s-0030-1258238).
Matsuda Takanori, Moriya Taisaku, Goya Tsuyoshi, Murakami Masahiro, “Synthesis of Pyrenes by Twofold Hydroarylation of 2,6-Dialkynylbiphenyls,” Chemistry Letters 2011, 40(1), 40-41 (https://doi.org/10.1246/cl.2011.40).
Mareike M. Machuy, Christian Würtele, Peter R. Schreiner, “2,6-Bis(phenylethynyl)biphenyls and Their Cyclization to Pyrenes,” Synthesis 2012, 44(9), 1405-1409 (DOI: 10.1055/s-0031-1290754).


Why is the E2 protein important for protein degradation? In the ubiquitination process, the E2 protein transfers ubiquitin from E1 to E3. Why can't E1 work with E3 directly to tag some protein with ubiquitin? Going further, why can't one of the E(1-3) proteins just do it all by itself?

E1: ubiquitin-activating enzyme

E2: ubiquitin-conjugating enzyme

E3: ubiquitin-protein ligase

Pretty hard question, "why", as we've not been watching for the millions of years it took to build that system... I think the better way to look at it is to name the advantages this system has and hypothesize on it's selection. BTW in biology a good answer to why is always because.

I would guess though that given the function of ligases, it has been selected against due most probably to lack of selectivity. E1s are very small and already need to do both activation of Ub consumming atp and doing a somewhat major conformation shift to accept ligation of the activated Ub. The amount of Protein Protein surface usually required to maintain the exquisite selectivity that is required for such a system can be assumed to be too great for the size of the E1s.

Then if you ask why it did not evolve in some kind of huge multidomain protein; I would probably argue to the point that you need to produce a very large amount of it all the time, and that it must be somewhat difficult to target that protein for proteosomal degradation / make sure it's folded properly. Having a smaller footprint might help in keeping the domains ordered and reducing entropy hence making sure the protein stays functionnal longer.

Then why not the E2 doing it directly, and why do you need the E3 in between... Well again, I would assume this is also a question of selectivity. If you need to regulate thousands of cytosolic proteins and control their regulators at the same time, you need some modularity, you need the ablity to create multi protein complexes that bring different layers of regulation and cannot assume that one protein can do that. The term E3 is also some horrible bin where people put anything. E3 are very often large and sometime huge multiprotein complexes, think of the APC complex, or in the most well documented cases, the Cullin complexes.

The way E3s evolved is pretty cool actually, to recruit E2-Ub a protein just needs to get a domain through some mutation (RING, U-Box, HECT...) then it can target whatever it used to bind to a degradation complex. That is some pretty neat and quick trick to create diversity rapidly if a system actually does something that brings an evolutionary advantage.

I see you left out the E4 which may play a role in Ub chain propagation but as they pretty much function as E3 anyways, there is debate as to why we would need to use that terminology.

This is by no mean some definite answer since I could be wrong on most of these things. I hope I helped you to further your reflection on the subject.


Why dishwashers are corrosive to aluminum? When I put aluminium parts into my dishwasher, they get tainted and they are sometime covered with white dots similar to salt.

I don't think it is aluminium oxide.

What are the chemical reactions that have an effect on aluminium but not other metals?

I think it is just alkalinity. Unlike many other metals, aluminum is amophoteric, so it dissolves in both strong base and in strong acid. Most dishwashing detergents are fairly alkaline, probably at pH 11 or 12 or so. That and the heat may be enough to corrode Al.  Most other common metals are attacked by strong acids but not by alkali.


White granular substance from restaurant with interesting properties - what is it? It isn't salt or sugar.

Was in a shaker at a restaurant - caused burning/bleeding sensation after consumption. Made up of small white granules that seem moist - all sticking together. The restaurant did not know what it was.

Tested properties:

Dissolves in water and vinegar.

When dissolved in water there was little effect. In vinegar, it produced substantial heat.

When mixed with lemon juice, the powder turned orange, heated up enough to crack the plastic container it was in, and gave off smoke.

No other information - based on these observations can any suggestions be made as to what it is?

You kept some of the unknwon as evidence, didn't you?  And a witness, too? Because if you get harmed, document it, and seek both medical as well legal counsel.

Turning to the chemistry: Likely, while dissolving X in water, you felt some heat, too?  Was the colorless solution obtained, when diluted further with additional water and then applied on skin a bit "soap like", felt slippery?

Vinegar is acidic, because of the acetic acid present.  Hence, mixing / dissolving the unknown proceeds with evolution of heat is indicative for the unknown acting as base that is neutralized. Such reactions generate heat. Similar token with lemon juice, equally reacting acidic, because of the presence of citric acid (even a more acidic than acetic acid): heat evolution possibly because of the neutralization of X (as a base) with it.

An additional indication that X is something basic were the action on diluted neutral juice of red cabbage.  If basic, the addition of X would yield a blue/greenish stain, in contrast to acids turning it red.

Speculation: it could be remaining NaOH.

Because sodium hydroxide actually (NaOH) is a base, that dissolves with ease in water, it reacts with the emanation of heat while dissolved in acids.  (Even if I am surprised the lemon juice experiment heated up enough to shatter the container.)  Even while preparing a dilute solution of NaOH in water, say 1 M as commonly used in the chem lab, the heat generated is sensible across the glass wall of the container used.
Because sodium hydroxide actually is used as a cleaning agent, to remove oily residues, to remove fat and grease (in other instances, to prepare lye soap) from pans (provided they are not made of aluminum).
Because NaOH is hygroscopic, in other words, the little white grains of it quickly look wet and stick to each other when exposed to humid air.
Because in pure form, sodium hydroxide is corrosive; and aqueous solutions of it are hazardous, too (pointing to your bleeding sensation).

Thankfully, such accidents in restaurants occure not often; yet there are examples ref1, ref2.  Bars ref3 use caustic soda to rinse their pipes of beer & other beverages, too, connecting the counter with the stock, often out of sight, in the basement.


Steam rises when the heat supply stops When water boils and you suddenly shut the heat off, a lot of steam arises from it just after you shut the heat off. Why this happens, I thought that maybe water condenzes very fast in small droplets so they get flown up?

Even water below boiling point has high partial vapour tension.

The gas containing mostly vapour, being hot and having low average molecular mass, has significantly lower density than air. Therefore, it raises up.

At contact with air, it quickly cools down and the vapour condensates.

The droplets are small enough to be taken upwards by the strong thermal vertical convection.


Crossing at critical temperature of a gas This is written in the book 'Treatise on Thermodynamics' by Max Plank-

the crossing of the critical temperature at any pressure other than the critical pressure differs in no way from the crossing of any other temperature.

But from the phase diagram of the states, it seems that if we change the temperature at a pressure greater than the critical pressure, then on crossing the critical temperature the gas will liquify where as crossing at other temperatures the gaseous state won't change.
So is the statement in the book wrong?

At the critical temperature/pressure, all properties of liquid and gaseous phase  mutually converge to the same values. There is just one phase at these temperatures, neither liquid, neither gas. Or, both at the same time.
Changing  pressure at temperature above the critical temperature causes continuous changes of the supercritical phase properties without the edge of the gas/liquid phase transition.
Changing  temperature at pressure above the critical pressure causes continuous changes of the supercritical phase properties without the edge of the gas/liquid phase transition.