Keto-enol tautomerism in a bicyclo compound Question

Which one is more stable- the keto or enol form. The compound given is this:

Solution

The enol form would violate Bredt's Rule and would therefore be unstable. Hence the keto form is more stable

My doubt
But I am not entirely convinced. Can't we have alpha hydrogens migrating from either of the carbon atoms I've indicated by arrows? Probably they're expecting us to assume that the alpha hydrogen would migrate from the bridge-head carbon, but I just wanted to know if there's any loophole in my concepts. Any help would be appreciated.

Most of the times, the keto-tautomer is lower in energy than the enol-tautomer of a given compound. In all cases this is also very dependent of the actual conditions.
There are few cases, where the enol-tautomer can be dominant. One of the most common examples is malondialdehyde, and, of course, its derivatives.
I am fairly certain that the original question is trying to trick you into thinking this way; and testing your 3D awareness. Or something like this; I'm not sure.
With these structures it's always a good idea to visualise then. I recommend a molecule kit for hands on approaches, but clay and matches are great, too. Even chestnuts and toothpicks are great tools. If you are technically inclined, there are great programs, such as Avogadro (avogadro.cc), which can do basic modeling.
For the following, I have used gfn2-xtb as an optimizer; that is a semi-empirical level of theory, see xtb-docs.readthedocs.io.

You can clearly see that if you migrate the bridgehead-hydrogen, the resulting lone pair would not be able to form resonance with either of the carbonyl groups. That is known as the empirical observation named Bredt's rule.
On the other hand you are quite correct that you could form a tautomer from the other α-carbon. This carbon can easily distort to have planar coordination and the lone pair can be in resonance with the carbonyl groups. (Provided of course that the framework is localised bond orbitals; and that the model is a valid approach.)

On this level of theory, the enol form is about 64.5 kJ/mol higher in energy than the keto-form. So while there certainly is a theoretical possibility, the likelihood is very small.
(Just for giggles: the bridge tautomer is 246.6 kJ/mol higher in energy.)

Appendix

Keto-Tautomer
17
symmetry c1
C       -0.232194517     -0.455059284     -0.108971891
C        0.435695625      0.521858137      0.853516654
H        0.456616659     -1.075279085     -0.674274662
C       -0.652035075      0.924669592      1.832141805
O        1.572314637      0.893979605      0.826125495
C       -1.888025507      0.221717603      1.265101729
H       -0.393829692      0.553384752      2.825493466
H       -0.740240162      2.008868613      1.896308913
C       -2.240914528      0.890781562     -0.065758217
H       -2.719883618      0.149774125      1.962761267
C       -1.080374859      0.487832943     -0.956781929
H       -3.167688540      0.496107389     -0.485905106
H       -2.336524789      1.975006913     -0.012902862
O       -0.862082222      0.837168772     -2.079995835
C       -1.289181703     -1.103984849      0.787745832
H       -1.992066775     -1.719717994      0.228564710
H       -0.851574934     -1.695108793      1.590840632

Enol-Tautomer
17
symmetry c1
C       -1.252989400     -1.048025267      0.775079043
C       -0.235131476     -0.354975263     -0.135182749
C        0.316010839      0.694441373      0.817231157
C       -0.666608713      1.048338273      1.645393318
C       -1.895763330      0.268945971      1.228776635
C       -2.331880748      0.853259922     -0.128837161
C       -1.153613858      0.494907935     -1.014278443
O        1.568858497      1.186824174      0.741278311
O       -0.952399200      0.822318914     -2.150244908
H        0.493388572     -0.958265049     -0.672260873
H        2.056130434      0.748941273      0.034835696
H       -0.640184178      1.790739752      2.416980529
H       -2.686913718      0.203266237      1.970705846
H       -3.231277420      0.373221210     -0.518677687
H       -2.502755602      1.928070454     -0.108020068
H       -1.934290934     -1.708837720      0.239732785
H       -0.777205764     -1.591075191      1.590309569

Bridgehead-enol-tautomer
17
symmetry c1
C       -1.375174317     -1.103606241      0.843045628
C       -0.248454950     -0.489532946     -0.004463589
C        0.359288830      0.442230533      0.837984097
C       -0.605216725      0.791584850      1.950720322
C       -1.905351826      0.244045543      1.301925006
C       -2.166063721      0.867188742     -0.087057960
C       -0.935024925      0.450040319     -0.933532366
O        1.456979798      1.143062588      0.565484195
O       -0.594744149      0.929934770     -1.975783635
H        1.590334610      1.860425911      1.199389402
H       -0.422954467      0.294599984      2.906280842
H       -0.604773151      1.872841199      2.115837534
H       -2.773451234      0.269475098      1.961354663
H       -3.086355462      0.469282331     -0.513923137
H       -2.229917288      1.956721084     -0.092669863
H       -2.077588583     -1.677749691      0.243281460
H       -0.985680441     -1.728640075      1.644336401


Is it possible to react two distinct carboxylic acids to form a dicarbonyl ester? By a dicarbonyl ester I mean something like this:

where R1 and R2 are alkyl substituents and the reacting carboxylic acids are:

The standard term for such molecules is acid anhydride, as they can be viewed as the product of a condensation reaction between two carboxylic acids, with concomitant loss of $\ce{H2O}$. Asymmetric anhydrides (i.e., ones that decompose into two distinct carboxylic acids if hydrolyzed) can certainly be prepared. One possible route is to synthesize an acid chloride from carboxylic acid #1 (thionyl chloride, $\ce{SOCl2}$, being one common reagent used to make the acid chloride) and then react it with the carboxylate salt of carboxylic acid #2. In that reaction, the carboxylate functions as a nucleophile and attacks the carbonyl carbon of the acid chloride, with chloride anion expelled as the leaving group.

I don't think it's an efficient laboratory method to prepare asymmetric anhydrides by simple dehydration of stoichiometric mixtures of distinct carboxylic acids. I suspect that would probably yield a (roughly) statistical distribution of products, at least in the ideal case. In reality, various factors might skew the distribution one way or another.


Pipette pumps durability According to your experience, in terms of durability and longevity of the tools, which one of these types of pipette pumps is better and why?

Tri-valve pipette bulb
Syringe pipette pump
Electronic pipette pump

These pipette pumps are mentioned in the YouTube video Bio-Rad Laboratories — Using Serological Pipets.

Syringe pipette pump.

Tri-valve is not as easy to use as syringe one. And the little metal sphere in it can be thrown while using. The syringe is more comfortable, it is just required to have 3 types for different volumes.

Additionally, when the pipette is broken at the top or has similar problems, you can somehow adapt the syringe, like inserting pipette deeper or tightening the syringe. It is also not expensive and easy to find from others when you need more.


Smallest unit of Sodium Chloride Since ionic solids are not molecular, can a single sodium chloride unit exist and if so, what is such a unit called? I am aware of the concept of "formula units", but as far as I can tell, this is a notional term and does not describe any physical structure. Is "unit cell" the correct term and can a single unit cell exist?
Just to be sure people understand the question: I am talking about solid sodium chloride and not an aqueous solution of it.

A “formula unit” would be closer to the answer than a “unit cell”. A unit cell is used in crystallography to define the smallest repeating unit of the crystal that has all the symmetries of the overall crystal structure. For NaCl, there are a variety of different ways to visualize the unit cell (2 interpenetrating FCC lattices, FCC with a 2-atom basis, etc. ), but regardless, there are four formula units of NaCl in the NaCl unit cell (4 Na and 4 Cl atoms).


What are examples of group 14 elements forming dπ - pπ bonding? While studying the Group 14 elements, I came across the anomalous behaviour of carbon. In that there was a point:

  Carbon forms only p$\pi$-p$\pi$ type bonding while other elements in the group form p$\pi$-p$\pi$, as well as d$\pi$-p$\pi$ type bonding.

I am searching for molecules in which  group 14 elements form d$\pi$-p$\pi$ bonds.

For $d\pi-p\pi$ bonding we need to consider other elements of group 14 except $\ce{C}$. (i.e, $\ce{Si, Ge ,Sn}$ etc.) which have got a vacant $d$ orbital of proper symmetry and energy to come into play bonding with attached group  having lone pair of electron in p orbital of comparable energy.
    For example $\ce{SiF4}$. $\ce{GeCl4}$ etc.
In tetrahedral geometry the orientation of vacant  d orbital of the central atom (dxy. dyz,dxz) may overlap with filled p orbital of the ligating atom  and hence $d\pi -p\pi$ bonding is possible.


Are the coefficients and exponents of contracted gaussian same for cartesian and pure form? The cartesian form of contracted gaussian is something like
$$\phi^{CGF}=\sum_{n} s_nx^iy^jz^k \exp(-\alpha_nr^2)$$
and these is also a pure or spherical harmonic version (which I don't know the details about).
Are the coefficients and exponents same between the two, different for $\ge d$ orbitals, or different for all orbitals.

They are the same.

Your equation can be rewritten as

$$ \phi_\mathbf l^\text{CGF}(r,\Omega)=\frac{x^{l_x}y^{l_y}z^{l_z}}{r^l}\sum_ns_nr^l\exp(-\alpha_nr^2)\equiv\Phi_\mathbf l(\Omega)R_l(r) $$

where $l=l_x+l_y+l_z$ is the total angular momentum, $\Phi_\mathbf l(\Omega)$ is the angular part, and $R_l(r)$ is the radial part, and they are separable not only in mathematical form, but also for practical purposes. (This discussion omits the norm, which is not separable to radial and angular contributions.) When specifying a Gaussian basis set in quantum chemistry, one talks only about the radial part, independent of the angular part. The angular part is really considered an implementation detail, because its form usually doesn't have any significant effect on the results.

For completeness, two most common angular forms are Cartesian:

$$ \Phi_\mathbf l(\Omega)=\frac{x^{l_x}y^{l_y}z^{l_z}}{r^l} $$

and spherical harmonic:

$$ \Phi_\mathbf l(\Omega)=Y_l^m(\theta,\varphi) $$


How can I solvent exchange from water to alcohol? Given an aqueous colloidal solution, is there a way to transfer the colloidal particles into an alcohol (assuming the colloidal particles are also stable in the alcohol)? I see in the literature that lyophilizing the aqueous solution and re-dispensing the colloidal particles is an option, but is there a method to perform this all in the wet phase?

Since it's a colloid, rather than a true solution, put the water-based mixture in a dialysis membrane of appropriate pore size and immerse it in concentrated alcohol. The water will dissolve into the alcohol, while alcohol slowly seeps in to displace the water, but the colloidal particles cannot get through the pores. Depending on how little water is desired in the final product, this step may be repeated multiple times.

For some materials, inexpensive cellophane sausage casing may be suitable.


Comparison of acids to dissolve phosphates I am currently reading this German paper on the dissolution of phosphates by acids and chelants.
The authors do a "fast-test" of the dissolving power of different acids which goes as follows:

For each test, increasing quantities (10-50ml) of 0.1 n acids are
added to 100 mg of tricalcium phosphate and agitated for 1 hour. The
residue is filtrated and weighed.
For 0.1 n hydrochloric acid 20ml suffice for complete dissolution of
100mg tricalcium phosphate. Citric acid [...] shows >90% dissolution
with 40ml.

Not being a chemist I am having trouble interpreting this. What quantity is effectively being controlled by specifying 0.1 n acid? The paper is from the 60s. I thus suspect that n stands for normality but I am not sure.
What exactly is being measured?

The speed at which different acids dissolve the material? (Why then vary the volume of acid?)
The quantity of acid needed to dissolve a given of material? (Measured in what exactly? What does the volume effectively stand for? Moles?)
Something else?

Edit 1
J M's answer makes this much clearer. I still have some problems though:

Regarding normality: Does this ensure that the same concentration of protons is in each solution? I assume this is not the same as "constant PH" as some acids may be strong acids while others may be weak?

If even more citric acid was added, there would still remain 10% residue? Why did they not try with 50ml citric acid? With this test there are two variables that change: volume and residue. Does this not make it difficult to compare the results of different acids? I would understand a test like this: how much acid is needed to dissolve the material 100%. Then one could compare easily.

Regarding weak acids: Once one of the available protons from the citric acid has reacted, should not another previously non-dissociated acid molecule dissociate (to keep the dissociation proportion in equilibrium) and keep everything going to dissolve the material without any residue? Given infinite time, why do not 30ml citric acid also dissolve 10 mg of 100 mg of tricalcium phosphate?

I am still not quite sure however what the point of this test is. Should it not be reasonably easy to calculate this using the reaction equations and $\mathrm{p}K_\mathrm{a}$ of the acids? (See reaction equation below.)

Thought experiment: If not the acid quantity was the limiting factor but the amount of solution what would limit the reaction: i.e. infinite supply of acid molecules (high acid concentration) and infinite supply of phosphates but finite volume for the solution to take up the reaction product: Would the reaction be limited by the solubility of the reaction products? I assume, in this case a strong acid would be advantageous, right?

Reaction Equation
If I understand the papers I read correctly, one can dissolve 15 moles of $\ce{Ca_3(PO_4)_2}$ with 60 moles of $\ce{HCl}$ or 20 moles of citric acid (disregarding the fact that it is a weak acid?!):
$$\ce{ 15 Ca_3(PO_4)_2 + 60HCl -> 15Ca(H_2PO_4)_2 + 30 CaCl_2 }$$
$$\ce{ 15 Ca_3(PO_4)_2 + 20 CitH_3 -> 10 Ca(Cit)_2 + 25 Ca(H_2PO_4)_2}$$

N, or Normality is a way of standardizing the # of acid equivalencies a particular acid has across multiple acids.  Acids that have only 1 acidic hydrogen for example $\text{HCl}$, the Normality will equal Molarity.

Normality is defined by:

$$
Normality = \frac{c_i}{f_{eq}}
$$

Where $c_i$ is the molar concentration of the acid, and $f_{eq}$ is the equivalency factor.  In the cases of something like $\text{H}_2\text{SO}_4$, the equivalency factor is 0.5 because it is a diprotic (two proton) acid.

This test also is showing the effect of weak vs strong acids.  $\text{HCl}$ is a strong acid, so it dissociates essentially 100%, while citric is a weak acid.  A weak acid exists in equilibrium between the dissociated acid species, and the neutral species.  This equilibrium is defined by the acid-dissociation constant $K_a$, or it's logarithmic version $pK_a$.

For citric acid though, it is triprotic, meaning there are 3 acidic hydrogens, so it's $f_{eq}$ value is $1/3=0.3\overline{3}$

Your numbered questions:

1)  The speed is being measured to a certain point.  However, the chance is that all of these reactions are complete at this point.  The hour timepoint is to normalize everything to the same time frame.  But as indicated, the reactions are likely complete at this point.  What is really being measured is your second point.

2)  Absolutely, the quantity of acid needed to dissolve the material is being tested.  As you use more and more volume of acid, you are providing larger amounts of acid to the reaction.  This helps it dissolve more of the powder (as there is more acid now present), and also increase the rate of the reaction.

They standardize the reaction to normality because if they used molarity, 1 mole of citric acid has 3 acid equivalents, while $\text{HCl}$ has only 1 acid equivalent.


Effect of the counterion on the colour emitted by a metalic salt Could someone please enlighten me on what would be the theoretical effect of a counter ion on the colour of the flame emitted by a metallic salt in a flame test?
Eg. NaCl - what would be the effect of the non-metal on the general colour of the flame of the compound?

I am looking into specific groups of metallic salts based on 3 metals: Sodium, Potassium and Copper II.
I have searched throughout the literature but have not found an answer yet.

Recall the flame emission experiment is basically an atomic emission experiment. We don't say it is a compound emission experiment. The job of the flame/plsama or any high temperature medium is to decompose the compound into the constituent atoms or even ions for an experiment. Once a dilute metal salt solution is aspirated into the flame, the compound decomposes into constituents and then to the atoms. For example, you start with NaBr, NaF or NaI or NaNO3, in the flame you will get one and only emission at 589 nm from the sodium atoms. All of these compounds will make the flame bright yellow. So what happened to the non-metallic anion? Provided the flame temperature is high, one should "see" emission from non-metallic atoms as well. Our eyes cannot see the emission bc it in the deep UV region.

Sometimes the anion interferes with the atomization process such as phosphate for calcium. It forms very thermally stable compounds in the flame. You will see the signal of calcium go down quickly if phosphate ion is present.

Forget about copper as atomic emission, I guess you are thinking about the green flame of copper compounds. This only happens in low temperature flames. That color originates from molecular compounds of copper in the flame, and the anion may play a role. You need a very high temperature flame and copper emits in the UV.


Side products during the electrolysis of Copper Sulfate? Using good quality reagents, during the electrolysis of $\ce{CuSO4}$ (at $\mathrm{15V ~ \& ~ 2A}$), I have noticed a significant amount of Hydrogen generated at the negative electrode in addition to copper deposition.

This generation occurred at all voltages, which surprised me. I expected a only the reduction of copper in the reaction. The solution was blue, so I know that copper sulfate was still in solution. At first, I thought that the copper might be reacting to the sulfuric acid being generated, but the copper that has fallen off the electrode is not bubbling.

I am using a platinum and copper electrodes.

Is $\ce{H2}$ generation happening as a result of the weak sulfuric acid in solution?

Can you calculate the products of simultaneous electrochemistry reactions?

There are a few key factors that I expect are contributing here:

$\ce{CuSO4}$ concentration – If your $\left[\ce{Cu^2+}\!\right]$ in solution is low, then that will disfavor deposition, as permeakra noted in a comment. You want at least $25$ to $100~\mathrm{g\over L}$ of copper in the electrolyte.
Cell voltage – $15~\mathrm V$ is an exorbitantly high voltage for copper electrodeposition. In order to avoid gassing at the cathode, you'll want to operate down in the $1.5$ to $2.0~\mathrm V$ range, tops.
Electrolyte conductivity – you don't mention anywhere that you've added other constituents to try to increase the conductivity.

If this is a home experiment, I would try dissolving the $\ce{CuSO4}$ in table or pickling vinegar instead of plain water. That will give you more charge carriers in solution, and also decrease the $\mathrm{pH}$, both of which should help.
If this is a school or work experiment with access to proper lab equipment and chemicals disposal services, I would use $1\%$ to $10\%~\ce{H2SO4}$ as the base electrolyte. That will really get your conductivity up, and your cell voltage down.

Finally, on the methods side, I would recommend choosing your voltage in order to initially target a current density in the vicinity of $\sim\!10~\mathrm{mA\over cm^2}$, as a value where I would expect to see little to no $\ce{H2}$ evolution. You can always nudge the voltage up from there as needed.

  Is $\ce{H2}$ generation happening as a result of the weak sulfuric acid in solution?

Almost certainly not.

  Can you calculate the products of simultaneous electrochemistry reactions?

Not with much certainty. Most of the time the best one can do is make qualitative arguments about what reactions might be happening.  Detailed modeling of all reactions occurring at an electrode is still prohibitively difficult, as far as I know.


Why does Calcium (specifically Calcium Carbonate) inhibit the Citrate buffer system? Is it because citrate chelates calcium and then is unable to react with other chemicals being added? If so can someone explain this reaction to me, or if not provide what the reason is? If anyone has any studies they've found on it that would be incredibly helpful! Thank you!

"Inhibited" is not probably the right term, as there is no catalyst inhibition, affecting the reaction kinetics.

If we consider about neutral $\mathrm{pH}$ range, the citrate buffer keeps $\mathrm{pH}$ by equilibrium reactions

$$\ce{HA^2- <=> H+ + A^3-}$$
$$\ce{HA^2- + OH- <=> H2O + A^3-}$$

with $\mathrm{p}K_\mathrm{a3}=6.4 $

Carbonate as a base spends the buffer capacity

$$\ce{CO3^2- + 2 HA^2- <=> CO2(g) + H2O + 2 A^3-}$$

If it was sodium carbonate, sodium ions would be just bystander ions.

But calcium ions form chelate with citrate 3:2, decreasing concentration of free hydrogen citrate + citrate ions.

So the overall calcium carbonate effect is being base and chelating citrate.


How many structural isomers does 1,2-dimethyl-cyclohexane have? Can someone tell me how many structural isomers exist for 1,2-dimethyl cyclohexane? My teacher said there is one structural isomer but I believe there aren't any because there is no other way to define the structure of the above compound such that they have the same molecular formula but a different structural formula. Also, we need two or more compounds to define isomerism, so how can a single molecule be ever defined as an isomer if there isn't any other compound to compare it with?
Also can structural isomers be defined if the position of substituents is already defined as in the case of above compound where the positions of the two methyl groups are defined beforehand (i.e. 1, 2)?

My teacher said there is one structural isomer

I think (hope) you are talking about stereo isomers (i.e. same Lewis structure, different substance).
There are two ways of counting. Either you can ask "How many substances are stereo isomers of cis-1,2-dimethylcyclohexane?" Or you can ask "How many stereo isomers exist of 1,2-dimethylcyclohexane?". The correct answers are off by one.
This is like asking "how many siblings do you have" vs. "how many siblings are there in your family (including you)".

How many structural isomers does 1,2-dimethyl-cyclohexane have?

The molecular formula for this substance is $\ce{C8H16}$. There are lots of substances with that formula but different Lewis structure.


Attack of ethoxide on epichlorohydrin I eliminated the possibility of the first two options since they included an alcohol and the solvent has not been specified so we do not know whether it is polar protic or polar aprotic.

As for the next two options I drew both of their mechanisms are they correct?

The answer given is D. Is it because the carbon next to oxygen is more electrophilic?

In general, you can expect an epichlorohydrin to be attacked nucleophilically only on the outer epoxide carbon. First, let me present the synthetic argument: In their total synthesis of RK-397, Burova and McDonald used epichlorohydrin as the starting point for the synthesis of their C10–C16 fragment. The fragment synthesis is shown in the scheme below which is taken directly from their publication.

They start with an enantiopure epichlorohydrin which they open using vinylmagnesiumbromide. The resulting alcohol is recyclised under elimination of chloride by distilling from $\ce{KOH}$. Starting from (R)-epichlorohydrin, their sequence exclusively generates (R)-33. Epoxidation gives 35 as a mixture of diastereomers which they separate using hydrolytic kinetic resolution with the Jacobsen cobalt catalyst.[1] This procedure acts only on the epoxide and selectively cleaves the (R)-epoxide to a diol. If there had been substantial amounts of (S)-33, 36 and 37 would each have been isolated as a mixture of diastereomers; however, both were isolated as single diastereomers.[2]

If there had been any direct attack of the vinyl Grignard on the $\ce{C-Cl}$ bond with direct displacement of chloride, the opposide enantiomer (S)-33 would have been formed (see scheme 1 below; 33 would have been inert to $\ce{KOH}$ and thus simply been co-distilled). This would have led to the aforementioned mixture of diastereomers further down the synthesis. Since no mixtures were observed, we can safely conclude that the attack only happens under cleavage of the $\ce{C-O}$ bond.

Scheme 1:  Hypothetical attack on $\ce{C-Cl}$ gives the opposite enantiomer; would lead to 36 and 37 being isolated as a mixture of diastereomers.

The synthetic argument already establishes that epoxide-cleavage must be orders of magnitude faster than chloride displacement. This does allow the question: why? One general argument is that the epoxide is highly strained, so a ring-opening attack will release the ring strain. Furthermore, $\ce{C-Cl}$ bonds are rather strong compared to other halogens (especially $\ce{C-I}$) and thus not as easily cleaved. Furthermore, we are likely looking at some hyperconjugation effects: the orbitals of the $\ce{C-Cl}$ bond and the $\ce{CHR2-O}$ bond can overlap as shown in scheme 2; this interaction leads to the chloride displacement attack being further disfavoured.

Scheme 2: Possible overlap of the $\ce{C-O}$ and $\ce{C-Cl}$ orbitals shown here as $\sigma_{\ce{C-O}}$ and $\sigma^*_{\ce{C-Cl}}$

Coming back to your question, since the outer epoxide carbon is marked and since that is where ethanolate will attack, only (D) is correct and your proposed mechanism checks out.

References:

[1]: S. E. Schaus, B. D. Brandes, J. F. Larrow, M. Tokunaga, K. B. Hansen, A. E. Gould, M. E. Furrow, E. N. Jacobsen, J. Am. Chem. Soc. 2002, 124, 1307–1315. DOI: 10.1021/ja016737l.

[2]: S. A. Burova, F. E. McDonald, J. Am. Chem. Soc. 2004, 126, 2495–2500. DOI: 10.1021/ja039618+.


Nomenclature of unusual nitrogen functional group I am trying to create a proflavine derivative containing a 1,2,4-oxadiazole-5-one functional group, the first step produces an interesting functional group that I have no idea how to name:
$$\ce{[Pro]-NH-CH=N-OH}$$
I am trying to find literature on the above functional group so I can analyze proton NMR to see if I have actual created the above molecule.

According to Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred IUPAC name of the compound $\ce{H2N-CH=NH}$ is methanimidamide. The name formimidamide may be used in general nomenclature. The name formamidine is no longer recommended.
The letter locants ‘N’ and ‘N'’ are used to differentiate among the different nitrogen atoms. Thus, the compound $\ce{R-NH-CH=N-OH}$ would be named as N'-hydroxy-N-…methanimidamide.


In the simplest terms, what is exchange current density? what is exchange current density, and ideally how does it relate to overpotential

I know it's related to the amount of dynamics in a reaction, so it can tell us how simple or complex a reaction mechanism is, but I don't have a clear idea of what exactly it means

Roughly speaking, it is a measure of how "dynamic" the "dynamic equilibrium" is when ancelectrode is at its equilibrium potential.

Let's say you are dealing with a copper electrode in a slightly acidic copper sulfate solution, a common situation in copper electroplating.  If you apply a high positive potential to this electrode, then of course the copper dissolves.  Upon measuring the rate of this dissolution, as the current density you are producing, you find that this current density increases exponentially as you crank up the potential, as long as you don't get any interference from mass transfer limitations or other electrode reactions.

Now switch to a highly negative potential.  Now copper deposits, and again (in the absence of mass transfer limits or other reactions) the current density varies exponentially with how much you are driving the electrode potential downwards.

What happens at intermediate potentials? The copper dissolution goes along with its exponential rate law and the deposition follows its rate law, but now these reactions are competing against each other so you macroscopically see only the difference between the plating and fissolution current densities.  You might have something like $10 \mu A/cm^2$ for dissolution and $1 \mu A/cm^2$ (these are just numbers I grabbed for this argument, not necessarily true kinetics), in which case you see a difference of $9 \mu A/cm^2$ dissolution.

At the equilibrium potential, the two current density numbers generated by the opposing rate laws are the same so you would not see any net reaction, even though in fact both are occurring together at the atomic level.  That common amount of current density being "exchanged" between the opposing reactions when you macroscopically see equilibrium, is the "exchange current density".


Particle size of oxides of nitrogen and sulfur pollutants and HEPA filters I am thinking about buying air purifier as pollution is high in my city.
Going through the specifications of HEPA filters, I found that they block particulate matter of 2.5. PM 2.5 contains dust and other fine particles.
Googling tells me that their size is around 2.5 micrometers.

My question is: Do pollutants such as sulfur oxides and nitrogen oxides (and related compounds) fall into this category, specifically in regards to size?

If the high efficiency particle air (HEPA) filter says it blocks PM 2.5 that's good as pm 2.5 particles are the ones your lungs cannot easily dispel and cause haze.  Unfortunately sulfur oxides and nitrogen oxides are gasses, so no particle filter will remove them.


Excitation of electron in Niels Bohr's atomic model Professor taught us that an electron gains or looses only those energies which are equal to difference in two energy levels. That is $E_1 + \Delta E = E_2$ or $E_1 + \Delta E = E_3.$
What if we give, say, $\pu{11.3 eV}$ and the remainder is $\pu{-2.2 eV},$ so the electron jumps to $n = 2$ where the energy is $\pu{-3.4 eV}$ and the remainder $\pu{-1.2 eV}$ is released?
I'm really confused with the working here.

Quantum mechanics received its name from the concept of the "quantum" which is Latin for "how much" or "amount" and is used to mean "small discrete piece or fragment". Classical physics prior to the development of QM described the energy of bodies as a continuum of possibilities. The novelty introduced by QM was the idea that within the continuum only a subset of the possible energies are in fact allowed, corresponding to allowed states. This means by extension that the energies of a body in different possible states differ by discrete rather than continuous amounts. This discreteness becomes increasingly evident for small and confined systems, where the wave-like properties of matter become more evident, as demonstrated for instance by the self-interference of the electron wavefunction in the double-slit scattering experiment.
Electrons in atoms fulfill all of the conditions for observation of QM phenomena. The particles involved are subatomic (very small) and are confined (by a Coulombic potential). The result is that electrons in atoms exist in a discrete number of allowed states, each state defined by a fixed energy and wavefunction that can be interpreted as the probability distribution of the electron.
When an atom is excited by a photon (a discrete light particle with a fixed amount of energy) it absorbs the entire photon. All of the photon's energy is acquired by the atom. However, such an excitation implies a transition between possible states of the atom. It is necessary that the energy of the photon match the difference in energy of the atomic states. In the case where the two states are bound (if an electron is not excited into the continuum of ionized free states) the result is that only a subset of photon energies can be absorbed by the atom.

As noted by Shoubhik R Maiti, photons can also be scattered by atoms and in the process transfer only part of their energy to the atoms. However the principle and meaning of quantization of electronic states remains the same. Electronic states are quantized (discrete) and so only particular amounts of energy can be absorbed (quanta corresponding to differences in energies of discrete levels).


Bond angles in NH3 and NCl3 The bond angle in a molecule is inversely proportional to the electronegativity of the surrounding atom if the central atom is same. This also happens with $\ce{NH3}$ and $\ce{NF3}$, as bond angle in $\ce{NH3 > NF3}$.

Why then is the bond angle in $\ce{NCl3 > NH3}$?

As Tan Yong Boon stated, it is Bent’s Rule with which we can explain the lower bond angle of $\ce{NF3}$ when compared to $\ce{NH3}$. The rule as stated by Henry Bent:

  Atomic s character concentrates in orbitals directed towards electropositive substituents.

Fluorine is more electronegative that hydrogen, and the $\ce{N−F}$ bond would have greater p character than the $\ce{N−H}$ bond. And more s character leads to large bond angles. Thus, the bond angle is greater in $\ce{NH3}$ than in $\ce{NF3}$.
Now, consider $\ce{NCl3}$. Clearly, $\ce{Cl}$ atom islarger in size than the central atom, nitrogen. Hence the higher bond angle here is due to the steric crowding caused by $\ce{Cl}$ atoms.(More pronounced than the electrongativity of $\ce{Cl}$ atom). They repel eachother and hence bond angle increases. This is not the case with small $\ce{H}$ atoms around the nitrogen in $\ce{NH3}$. Therefore, the $\ce{Cl-N-Cl}$ bond angle of $\ce{NCl3}$ is greater than $\ce{H-N-H}$ bond angle of $\ce{NH3}$.


Which of the compounds give the same SN1 and SN2 products? Which of the following compounds give same SN1 and SN2 product?

I think (b) should be the answer, but the given answer is (c). I think so as it is the only compound where carbon is not chiral. While in others racemic mixture in SN1 and enantiomer in SN2 are formed, in (b) there is no question about this. Please tell me if I got it right.

For starters, none of your products show chirality in any way, so it does not matter whether stereoinformation is lost or whether you convert a racemic mixture into a racemic mixture where every single molecule if you had labelled it pre-reaction now had the opposite stereochemistry — in both cases, both reactants and products are racemic mixtures. So you cannot take the presence or absence of chiraliy to distinguish here.

What we can say, is that every compound can take part in an $\mathrm{S_N2}$ reaction (some faster, some slower) but not every one will easily take part in an $\mathrm{S_N1}$ reaction. Specificly, (a) and (d) would give a secondary carbocation intermediate while (b) would even give a primary carbocation. These cations are very unstable and would immediately undergo Wagner-Meerwein type rearrangements towards more stable — read tertiary or resonance stabilised — carbocations. Or, to quote one of my professors from uni:

  Exactly one single unstabilised secondary carbocation exists, and that is the prop-2-ylium cation. (Because it cannot rearrange to a more stable one.)

Thus, only (c) will form a stable carbocation without any rearrangement happening: It would be an allyl cation which is well resonance stabilised. But now we have to dig a step deeper, because while all other reactants will certainly not give the same product in both cases, it could be that the correct answer is ‘none of these’, i.e. that somehow we can distinguish between products — because once we have the allyl cation, we could let the nucleophile attack on carbon 1 or carbon 3. Thankfully, though, both attacks give the same product here, because there are no other asymmetries present in the molecule, so we can transform one into the other by rotation.

And, never forget, the $\mathrm{S_N2}$ reaction could also follow the $\mathrm{S_N2'}$ mechanism — this, too, would switch the identities of the two carbons, so we can truly say that both mechanisms will lead to the same product.


Is there an actual distinction between through-space charge transfer and excimer emission? Variety of books and literature reviews consider excimer (or exciplex) emission something distinct from through-space charge transfer. Is this actually the case?
This question arises, because the more literature I read on this subject the less sure it seems to have any distinctive characteristics one from another. Mainly because the same ''evidence'' (e.g. the bathochromic shift of emission, lifetime increase, concentration studies) is used to argue for excimer or exiplex emission in one study, while other authors propose that such evidence supports CT emission.

The excimer / exciplex ground state is wholly repulsive, molecules $\ce{M  and M^*}$ diffuse together to form excimer $\ce{MM^*}$ the excimer or exciplex if formed from different species. The complex does not exist in the ground state. Emission is observed from both $M^*$ and $MM^*$. ($MM^*$ can also diffuse apart before fluorescence) The reason the excimer/exciplex emission is broad and structureless is because the ground state is repulsive. The exciplex is formed from different species so there has to be some charge transfer character, i.e. potential energy is sensitive to solvent polarity. Intramolecular excimers and exciplexes were first synthesised and studied in the late 1970's.
CT complexes are intimate in the ground state (via $\pi$ stacking etc.) and the CT absorption is generated upon absorption $(DA) + hv\to (D^+A^-)$. Any emission is then from this CT state plus normal fluorescence from either D* or A* should they be isolated in solution (as DA is in equilibrium with D and A) and then excited.
For details of kinetic see J.B. Birks, 'Photophysics of Organic Molecules' 1970, publ Wiley, and Organic Molecular Photophysics' 1975, publ Wiley. M. Gordon & W. Ware 'The Exciplex' 1975, Academic Press.


Derivation of relationship between Gibbs free energy and electrochemical cell potential Why is $\Delta G = -nFE$?

I don't understand what the motivation is behind this definition. Was it derived or just given? The textbook provides no justification for this equation. In fact, much of the book associated with the Gibbs free energy provides no justification and just says, 'This is how it is. Now go and solve some problems.'

There are two ways to understand this equation.

One is to realise that (reversible ideal case) $\Delta G = W_\text{non-exp}$ (non-expansion). Therefore, in an ideal chemical cell, if the potential difference between the electrodes is $E$, to move one mole electrons across the external circuit will be $FE$, which must be equal to the decrease in gibbs free energy of the system. Hence for $n$ mole electrons transferred at the same potential, $W_\text{non-exp} = \Delta G = -nFE$.

The fact that $\Delta G = W_\text{non-exp}$ can be derived as under:
\begin{align}
\mathrm{d}S
  &= \frac{\delta q}{T}  && \text{(reversible case)}\\
\mathrm{d}U
  &= \delta q + W_\text{non-exp} + W_\text{exp}\\
W_\text{non-exp}
  &= \mathrm{d}U - W_\text{exp} - \delta q \\
  &= \mathrm{d}U + p\,\mathrm{d}V - T\,\mathrm{d}S
  && \text{(const. $p$ and $T$)} \\
  &= \mathrm{d}H - T\,\mathrm{d}S \\
  &= \mathrm{d}G
\end{align}

Another approach is to use
$$\mu_\mathrm{electrochemical} = \mu_\mathrm{chemical} + zF\phi,$$
where $\phi$ is the electric potential at the point, $z$ is the charge on the species. Using this and
$$\mathrm{d}G_{T,P} = \sum_{i=1}^n \mu_i\,\mathrm{d}n_i,$$
we get the required equation for $z = 1$ for an electron, and considering the electrochemical potential of both the electrodes. I am unsure of a rigorous derivation for the electrochemical potential but Wikipedia has some basics on it.


Structure of carbon fiber What is the structure of carbon fiber? Wikipedia doesn't give much insight.

  The atomic structure of carbon fiber is similar to that of graphite, consisting of sheets of carbon atoms (graphene sheets) arranged in a regular hexagonal pattern, the difference being in the way these sheets interlock. Graphite is a crystalline material in which the sheets are stacked parallel to one another in regular fashion. The intermolecular forces between the sheets are relatively weak Van der Waals forces, giving graphite its soft and brittle characteristics.

The Wikipedia description begs the question of how exactly are the sheets of graphene interlocked. Are there pi-stacking interactions?

The key difference between graphite and carbon fibers is the degree (or range) of ordering.  In graphite there are large, extensive planes of carbon stacked one atop another resulting in long-range order. In the fibers, there are small ribbons of graphite-like (it is not all carbon, due to the way the fibers are prepared there is still some nitrogen present in the molecular array) material oriented along the fiber axis and there is only short range order.  As a consequence graphite with its extensive 3 dimensional planar array of ordered carbon atoms is crystalline.  Whereas the fibers, being ordered over a much smaller range,  are amorphous.  The amorphous nature of the fiber allows the smaller ribbons to bend and fold, and mechanically interlock with one another.  This mechanical interlocking is what serves to increase the fiber's strength.  See here for a very nice and concise description of how the fibers are prepared.  The pictures in the link make clear how the short range ordering in the fibers results in a highly folded and interlocked structure.  Further heating of the fiber will convert it to graphite.


Historically, how was the connection between NMR peaks and hydrogen atoms made? A large part of the basis of this question may be due to the student who didn’t pay attention in the history section of the NMR course (and not during the technical details bit either) but I find the question intriguing nonetheless.

My basic knowledge of the history of NMR spectroscopy is that in the very old, almost prehistoric days, Bloch and Purcell performed some experiments that applied radiofrequency to a sample and by chance got some radiofrequency signal returned. Then I have a big black box and arrive at modern spectrometers that tune themselves into hydrogen spectra, carbon spectra and see beautiful Fourier-transformed spectra on my PC screen (unless it’s carbon, in which case the baseline is, of course, fuzzy).

I imagine that there had to be a lot of research invested until it was realised that each signal in the Fourier-transformed spectrum corresponds to a hydrogen atom, that they are shifted around depending on how electron-rich or -poor they are, how couplings work and so on. I can imagine that most of this stemmed from trial and error: a certain combination of magnetic field and radiofrequency gives a signal, Fourier transformation seems to happen anywhere, and once you have rationalised that ‘signals’ correspond to ‘hydrogens’ in a certain type of spectrum (likely the one that gave most resonance anyway) you’re half done and the rest is a walk in the park.

But there is one step that seem very unintuitive:
Realising that ‘signals’ correspond to hydrogens.

There is another important step, namely figuring out that solutes can be analysed if hydrogen-free ($\ce{CCl4}$) or at least protium free (deuterated) solvents are used. Now either it was figured out first that hydrogen is the cause for the signals, in which case choosing the correct solvents is a non-issue, but capturing signals from the solute may be. Or it was figured out that using certain solvents gave ‘cleaner’ spectra, in which case it might also be interesting to know how the solvents were chosen.

I realise that there may be a good deal of speculation, retrofitting, inflationary use of the term serendipitously, anecdotical evidence, storytelling and bad memories involved, but is it known how the discoverers of NMR spectroscopy realised that a ‘signal’ is a hydrogen atom in the molecule observed?

TL;DR: Scientists knew that electrons had a spin, and thought, "wouldn't it be nice if atomic nuclei did too?" When they passed molecular hydrogen through a homogenous magnetic field (in a modified version of the Stern–Gerlach experiment) and subjected it radio waves, they found energy was absorbed at precise frequencies. This set the stage for the usefulness of $^1\ce{H}$ NMR in the distant future.
Taken from A Brief History of Nuclear Magnetic Resonance:

In the mid-1920s it became apparent that many features in atomic spectra could be accounted for only if certain atomic nuclei likewise possessed spin and a magnetic moment. Refinements of the Stern-Gerlach experiment verified this concept by 1933.
In 1939 Rabi et al. made a major improvement in beam techniques by sending a stream of hydrogen molecules through not only the inhomogeneous magnetic field required for deflection, but also through a homogeneous magnetic field, where they were subjected to radio frequency electromagnetic energy. Energy was absorbed by the molecules at a sharply defined frequency, and the absorption caused a small but measurable deflection of the beam. This was the first observation of NMR, and Rabi received the Nobel Prize in 1944. However, such studies were limited to nuclei in small molecules under very high vacuum in a molecular beam, the deflection of which served to detect the resonance.


When does Hartree-Fock fundamentally fail? The Hartree-Fock approximation (HFA) works by assuming each electron sees the effective electric field from all the other electrons as some self-consistent field.

The HFA is known to give pretty decent results, but often gets the last quantitative bit of accuracy wrong. This last bit is attributed to mysterious many-body or "strongly-correlated" effects.

There are many examples of HFA quantitatively giving wrong answers (energies are wrong, incorrect numerical results,  etc.). But, are there general phenomena that HFA gets totally wrong? I'm thinking of something where the chemistry/physics is fundamentally different from what HFA would tell you. Said differently, when do "strong correlations" actually matter and fundamentally change things in a chemical context?

I've tried reading up on this topic in the physics literature, and it is incomprehensible despite being an ex-physicist myself. Most of the discussion is about high temperature superconductors and complicated transition metal oxides. I can't understand the basic phenomena without all the specific details.

Restricted HF (RHF) fails anytime you have partial occupancy due to (near) degeneracy of the HOMO. For example, $\ce{H2}$ dissociates to a mix of $\ce{2H}$ and $\ce{H- + H+}$ and the dissociation energy is way off. Similarly, singlet $\ce{O2}$ is not well described by RHF if you can get it to converge at all.

The solution is to use more than one MO wavefunction (determinant) and this is sometimes referred to as "non-dynamical" (rather than "strong") correlation.

Unrestricted HF (UHF) can usually treat these systems qualitatively correct.


How do I perform a partial optimisation in GAMESS? Sometimes it is necessary to perform a partial optimisation of a molecule, i.e. keeping certain variables constant. One example might be to pre-optimise a transition state, taking advantage of not having to compute the force constants in the initial set-up.

For example, let's consider the transition state of the Diels-Alder Reaction of ethene and butadiene, where we want to freeze the distance between the two molecules.

How can I achieve this in the software suite GAMESS?

Unfortunately GAMESS is not as user friendly as other quantum chemistry packages, so when it comes to running non-standard calculations a little bit more effort is necessary.

First of all I recommend optimising the reactants by themselves, so that you have a somewhat valid idea what the bond lengths on your level of theory are going to be. After that you should assemble your best guess for the structure. In this example I have chosen a separation between the molecules of about 220 pm.

The coordinates of this structure are:

! DIENE
C       -0.486294808     -0.431915730     -1.395523790
C       -1.166676574      0.550508040     -0.705577944
C       -1.166676574      0.550508040      0.705577944
C       -0.486294808     -0.431915730      1.395523790
H       -0.352988862     -0.347708199      2.465648783
H       -0.084717264     -1.289523730      0.864198908
H       -1.712490163      1.338624758      1.223514576
H       -1.712490163      1.338624758     -1.223514576
H       -0.352988862     -0.347708199     -2.465648783
H       -0.084717264     -1.289523730     -0.864198908
! DIENOPHILE
C        1.559305232     -0.085874351     -0.690651129
C        1.559305232     -0.085874351      0.690651129
H        1.514911897     -1.022640981      1.232100569
H        1.603704743      0.851022589      1.232175888
H        1.603704743      0.851022589     -1.232175888
H        1.514911897     -1.022640981     -1.232100569

There are approaches, where you would simply fix the Cartesian coordinates. Here we only want to fix two bond lengths, 1(C)-11(C) and 4(C)-12(C). If we were to fix the coordinates of these atoms, we would also fix 1(C)-4(C) and 11(C)-12(C). Therefore a better approach is to use internal coordinates, in GAMESS the keyword group $ZMAT is used.
Since there are a lot, it is tricky to set them up. We will therefore use the implemented automated procedure. Starting with a dry run, i.e. EXETYP=CHECK, so that we can check what the program actually does. We need to define $3\times N-6=42$ internal coordinates, thus NZVAR=42. Also note, that we do not use symmetry in this case. Because of the nature of the structure we need to add some bonds already manually, NONVDW(1)=1,11, 4,12.
This is the resulting input:

 $CONTRL
   SCFTYP=RHF
   RUNTYP=ENERGY
   EXETYP=CHECK
   COORD=UNIQUE
   ICHARG=0
   MULT=1
   NZVAR=42
   $END
 $BASIS
   GBASIS=STO
   NGAUSS=3
   $END
 $ZMAT
   AUTO=.TRUE.
   DLC=.TRUE.
   NONVDW(1)=1,11, 4,12
   $END
 $GUESS
   GUESS=HUCKEL
   $END
 $DATA
   C6H10
   C1
   CARBON      6.0     -0.486294808        -0.431915730        -1.395523790
   CARBON      6.0     -1.166676574         0.550508040        -0.705577944
   CARBON      6.0     -1.166676574         0.550508040         0.705577944
   CARBON      6.0     -0.486294808        -0.431915730         1.395523790
   HYDROGEN    1.0     -0.352988862        -0.347708199         2.465648783
   HYDROGEN    1.0     -0.084717264        -1.289523730         0.864198908
   HYDROGEN    1.0     -1.712490163         1.338624758         1.223514576
   HYDROGEN    1.0     -1.712490163         1.338624758        -1.223514576
   HYDROGEN    1.0     -0.352988862        -0.347708199        -2.465648783
   HYDROGEN    1.0     -0.084717264        -1.289523730        -0.864198908
   CARBON      6.0      1.559305232        -0.085874351        -0.690651129
   CARBON      6.0      1.559305232        -0.085874351         0.690651129
   HYDROGEN    1.0      1.514911897        -1.022640981         1.232100569
   HYDROGEN    1.0      1.603704743         0.851022589         1.232175888
   HYDROGEN    1.0      1.603704743         0.851022589        -1.232175888
   HYDROGEN    1.0      1.514911897        -1.022640981        -1.232100569
   $END

The resulting auxiliary outputfile .dat has now the group that we will use from here on in. And the .log file will give us the exact value of the internal coordinate which we want to freeze.

                     --------------------
                     INTERNAL COORDINATES
                     --------------------

                 - - ATOMS - -         COORDINATE      COORDINATE
 NO.   TYPE    I  J  K  L  M  N        (BOHR,RAD)       (ANG,DEG)
 ----------------------------------------------------------------
   1 STRETCH   1 11                     4.1406435       2.1911343
...
  10 STRETCH   4 12                     4.1406435       2.1911343
...

We can now modify our $ZMAT group accordingly. Unfortunately the input gets very long, if you would like to retain some readability. Note that we create redundant internal coordinates, so we need to increase the number of NZVAR, too, to read all of them in.
Technically we do not have to set the generation of coordinates to false, but it helps remembering what you have done if you come back in a month or so, therefore AUTO=.FALSE.. Now we add the coordinates we would like to freeze with IFZMAT(1)=<list> and the corresponding values FVALUE(1)=<list>.

 $CONTRL
   SCFTYP=RHF
   RUNTYP=OPTIMIZE
   COORD=UNIQUE
   ICHARG=0
   MULT=1
   NZVAR=89
   $END
 $BASIS
   GBASIS=STO
   NGAUSS=3
   $END
 $ZMAT
   AUTO=.FALSE.
   DLC=.TRUE.
   IZMAT(1)=
   1,   1,  11,
   1,   1,  10,
   1,   1,   9,
   1,   1,   2,
   1,   2,   8,
   1,   2,   3,
   1,   3,   7,
   1,   3,   4,
   1,   4,   6,
   1,   4,  12,
   1,   4,   5,
   1,  11,  16,
   1,  11,  15,
   1,  11,  12,
   1,  12,  14,
   1,  12,  13,
   2,   1,  11,  15,
   2,   1,  11,  16,
   2,   1,   2,   3,
   2,   1,   2,   8,
   2,   1,  11,  12,
   2,   2,   3,   7,
   2,   2,   1,  10,
   2,   2,   1,  11,
   2,   2,   1,   9,
   2,   2,   3,   4,
   2,   3,   2,   8,
   2,   3,   4,   5,
   2,   3,   4,  12,
   2,   3,   4,   6,
   2,   4,   3,   7,
   2,   4,  12,  13,
   2,   4,  12,  14,
   2,   4,  12,  11,
   2,   5,   4,  12,
   2,   5,   4,   6,
   2,   6,   4,  12,
   2,   9,   1,  10,
   2,   9,   1,  11,
   2,  10,   1,  11,
   2,  11,  12,  13,
   2,  11,  12,  14,
   2,  12,  11,  15,
   2,  12,  11,  16,
   2,  13,  12,  14,
   2,  15,  11,  16,
   3,   1,   2,   3,   4,
   3,   1,   2,   3,   7,
   3,   1,  11,  12,  14,
   3,   1,  11,  12,   4,
   3,   1,  11,  12,  13,
   3,   2,   1,  11,  15,
   3,   2,   1,  11,  16,
   3,   2,   3,   4,   5,
   3,   2,   3,   4,  12,
   3,   2,   3,   4,   6,
   3,   2,   1,  11,  12,
   3,   3,   2,   1,  11,
   3,   3,   4,  12,  13,
   3,   3,   4,  12,  14,
   3,   3,   4,  12,  11,
   3,   3,   2,   1,   9,
   3,   3,   2,   1,  10,
   3,   4,  12,  11,  15,
   3,   4,  12,  11,  16,
   3,   4,   3,   2,   8,
   3,   5,   4,  12,  11,
   3,   5,   4,  12,  14,
   3,   5,   4,  12,  13,
   3,   5,   4,   3,   7,
   3,   6,   4,  12,  11,
   3,   6,   4,  12,  14,
   3,   6,   4,  12,  13,
   3,   6,   4,   3,   7,
   3,   7,   3,   2,   8,
   3,   7,   3,   4,  12,
   3,   8,   2,   1,   9,
   3,   8,   2,   1,  10,
   3,   8,   2,   1,  11,
   3,   9,   1,  11,  16,
   3,   9,   1,  11,  15,
   3,   9,   1,  11,  12,
   3,  10,   1,  11,  12,
   3,  10,   1,  11,  15,
   3,  10,   1,  11,  16,
   3,  13,  12,  11,  15,
   3,  13,  12,  11,  16,
   3,  14,  12,  11,  15,
   3,  14,  12,  11,  16,
   IFZMAT(1)=
   1,   1,  11,
   1,   4,  12,
   FVALUE(1)=
   2.1911343,
   2.1911343,
   $END
 $GUESS
   GUESS=HUCKEL
   $END
 $DATA
   C6H10
   C1
   CARBON      6.0     -0.486294808        -0.431915730        -1.395523790
   CARBON      6.0     -1.166676574         0.550508040        -0.705577944
   CARBON      6.0     -1.166676574         0.550508040         0.705577944
   CARBON      6.0     -0.486294808        -0.431915730         1.395523790
   HYDROGEN    1.0     -0.352988862        -0.347708199         2.465648783
   HYDROGEN    1.0     -0.084717264        -1.289523730         0.864198908
   HYDROGEN    1.0     -1.712490163         1.338624758         1.223514576
   HYDROGEN    1.0     -1.712490163         1.338624758        -1.223514576
   HYDROGEN    1.0     -0.352988862        -0.347708199        -2.465648783
   HYDROGEN    1.0     -0.084717264        -1.289523730        -0.864198908
   CARBON      6.0      1.559305232        -0.085874351        -0.690651129
   CARBON      6.0      1.559305232        -0.085874351         0.690651129
   HYDROGEN    1.0      1.514911897        -1.022640981         1.232100569
   HYDROGEN    1.0      1.603704743         0.851022589         1.232175888
   HYDROGEN    1.0      1.603704743         0.851022589        -1.232175888
   HYDROGEN    1.0      1.514911897        -1.022640981        -1.232100569
   $END

The optimisation should look something like the following animation; and took 19 steps in my example.

Now you could use the results of this calculation to run a full fledged transition state search. Remember, that you have all necessary information to restart a run in the auxiliary output, i.e. the .dat file. You should always keep this one until you are completely done with your calculations.

If you want to calculate along and wonder what the optimised geometry is, here are the results of the final step:

----- RESULTS FROM SUCCESSFUL RHF      GEOMETRY SEARCH -----
----- COORDS, ORBS, GRADIENT, AND APPROX. HESSIAN -----
 COORDINATES OF SYMMETRY UNIQUE ATOMS (ANGS)
   ATOM   CHARGE       X              Y              Z
 ------------------------------------------------------------
 CARBON      6.0  -0.4847330011  -0.4122786646  -1.4086312155
 CARBON      6.0  -1.1902020846   0.5288839653  -0.7019679480
 CARBON      6.0  -1.1902020983   0.5288839622   0.7019679173
 CARBON      6.0  -0.4847329728  -0.4122786820   1.4086312270
 HYDROGEN    1.0  -0.4076405265  -0.3274824211   2.4860530248
 HYDROGEN    1.0  -0.4019303863  -1.4244097276   1.0396872383
 HYDROGEN    1.0  -1.5467259917   1.4139143889   1.2140730563
 HYDROGEN    1.0  -1.5467259527   1.4139143906  -1.2140730901
 HYDROGEN    1.0  -0.4076405381  -0.3274824126  -2.4860529903
 HYDROGEN    1.0  -0.4019304494  -1.4244096950  -1.0396872193
 CARBON      6.0   1.5587691569  -0.0858507824  -0.6884831136
 CARBON      6.0   1.5587691834  -0.0858507803   0.6884831288
 HYDROGEN    1.0   1.9167366148  -0.9561957837   1.2206602417
 HYDROGEN    1.0   1.6758814772   0.8473264236   1.2212666939
 HYDROGEN    1.0   1.6758814263   0.8473264124  -1.2212666800
 HYDROGEN    1.0   1.9167365785  -0.9561957779  -1.2206602275
--- OPTIMIZED RHF      MO-S --- GENERATED AT Thu Dec 14 23:42:52 2017
E=     -230.0376739583, E(NUC)=  227.2020852580
 $VEC
...

Disclaimer: There is another way of freezing coordinates. I might add that at another point as another answer. It is a weeny bit more complicated, however offers also some more flexibility.


The energy gap between a pi-conjugated system with (2 bonding and 1 anti-bonding orbital) and (1 bonding and 2 anti-bonding) orbitals I asked a question previously about "why" it is the case the expanding the size of pi-conjugated systems decreases the required energy to excite an electron from a HOMO to a LUMO band: Why does the energy gap for π - π* transitions shrink with the size of the pi-conjugated system?.  The user "Philipp" (and others) very patiently and kindly explained the reason for this by pointing out the expansion in the number of possible bonding and anti-bonding configurations for a polymethine-like pi-conjugated system.

However, there's something I'm missing that has to do with what is probably my lack of knowledge about chemistry.  If we look at this image: http://i.stack.imgur.com/Ti3cx.png (provided by "Philipp") why is that that we have an energy gap between $\Psi_2$ and $\Psi_3$ smaller than the energy gap for a simpler, smaller pi-conjugated system where the HOMO corresponds to a single pair of p-orbitals in a bonding conformation and the LUMO corresponds to the same single pair of p-orbitals in an anti-bonding conformation (i.e. this system: http://i.stack.imgur.com/9cscv.png)?

My guess that this would be because the bonding interactions that exist in the larger pi-conjugated system mitigate the penalty for having the one anti-bonding interaction?

Update --- Could we justify the above guess using something like an overlap integral (if we know the shape of the two bonded p-orbitals forming a pi-bond)?  If so, this would be quite intuitive.

If you want to really understand why energy levels of conjugated $\pi$ systems are the way they are I suggest you take a good look at the Hückel theory as @ssavec suggested - this method is only an approximation but because of its simplicity it is really good for helping you to understand the underlying concepts.
Nevertheless, doing as you suggested and looking at the overall overlap between the $\text{p}$ orbitals in a given molecular orbital does indeed seem to work although I don't have a strict justification for that.

From Hückel theory it is known that the coefficients of the $\text{p}$ orbitals in a given molecular orbital can be calculated as

\begin{equation}
  c_{j,r} = \sqrt{\frac{2}{N + 1}} \sin\Bigl(\frac{r j \pi}{N + 1}\Bigr)
\end{equation}

giving the coefficient $c_{jr}$ for atom $j$ in molecular orbital $r$ of a conjugated system of $N$ atoms (so that $j$ and $r = 1, 2, 3, \dots , n$).

Let's apply this to butadiene.

The $\pi$ molecular orbitals of butadiene are $\Psi_{1}$, $\Psi_{2}$, $\Psi_{3}$, and $\Psi_{4}$.
They are built from a linear combination of 4 atomic $\text{p}$ orbitals ($\phi_{1}$, $\phi_{2}$, $\phi_{3}$, and $\phi_{4}$), so

\begin{equation}
  \Psi_{r} = c_{1,r} \phi_{1} + c_{2,r} \phi_{2} + c_{3,r} \phi_{3} + c_{4,r} \phi_{4}
\end{equation}

When you calculate the atomic orbital coefficients for each molecular orbital $r$ then you get to the following picture:

where the numbers above the different $\text{p}$ orbitals are the coefficients.

Doing the same for ethene you get the following:

Now, you need to compare the overall overlap in the HOMOs of ethene and butadiene.
Since it is already assumed in the Hückel theory only the overlap between neighboring $\text{p}$ orbitals is considered.
Furthermore, it is useful to set the overlap integral of two "normal" $\text{p}$ orbitals (both with coefficient $c = 1$) to 1 (its exact value doesn't matter because only the relative amounts of overlap in different MOs are important for this comparison), so that you can calculate the overall overlap in a molecular orbital simply by multiplying the coefficients of neighboring $\text{p}$ orbitals and summing over these products.
So, for the overlap in the ethene and butadiene HOMOs you get:

\begin{align}
  S_{\text{HOMO, ethene}} &= 0.707 \cdot 0.707 \\
  &= 0.500 \\
  \\
  S_{\text{HOMO, butadiene}} &= 0.600 \cdot 0.371 + 0.371 \cdot (-0.371) + (-0.371) \cdot (-0.600) \\
  &= 0.307
\end{align}

There is less overlap and thus less bonding in the HOMO of butadiene than in the HOMO of ethene which would imply that the HOMO of butadiene is higher in energy than that of ethene.

Now, what about the LUMOs:

\begin{align}
  S_{\text{LUMO, ethene}} &= 0.707 \cdot (-0.707) \\
  &= -0.500 \\
  \\
  S_{\text{LUMO, butadiene}} &= 0.600 \cdot (-0.371) + (-0.371) \cdot (-0.371) + (-0.371) \cdot 0.600 \\
  &= -0.307
\end{align}

There is more overlap and thus more bonding (or better to say less antibonding) in the LUMO of butadiene than in the LUMO of ethene which would imply that the LUMO of butadiene is lower in energy than that of ethene.

So, the HOMO of butadiene is higher in energy than that of ethene and the LUMO of butadiene is lower in energy than that of ethene.
Thus, the HOMO-LUMO gap in butadiene is smaller than that in ethene.


Does there exist a dye that will primarily bind with water in a water/oil emulsion? I'm wondering if there exists a type of dye that will bind primarily with water when introduced to a water/oil emulsion. That way if the water is centrifuged to the bottom of the vial, it will be dyed showing better visibility for the meniscus line.

The main application of this would be for centrifugation of clear oil that has suspended water. If the oil is clear and spun out, the water may separate but seeing the distinct meniscus can be difficult. If a dye in introduced and binds to the small water content within this emulsion, once it's spun out, ideally you'd see the dyed water at the bottom. The content of water is probably 0.5 - 1.0%

I'm also somewhat aware of some water by distillation methods using dyes to detect small amounts of distilled water, but I'm unfamiliar with these and I can't find any helpful resources online about it.

I suppose it would need to be a hydrophilic dye that also resists binding to hydrocarbons?

You are asking for water-soluble dyes. Typical applications are dyeing methods in microscopy and food coloring.

Water-soluble dyes for microscopy: e.g. Eosin Y, Methylene Blue, Ethyl Green.

See also this table: Wikipedia in German: Lebensmittelfarbstoff - Deutschland.


(E)-2-bromo-1,4-difluoro-3-iodobut-2-ene or (E)-3-bromo-1,4-difluoro-2-iodobut-2-ene I am struggling with naming this structure either (E)-2-bromo-1,4-difluoro-3-iodobut-2-ene or (E)-3-bromo-1,4-difluoro-2-iodobut-2-ene.

I am aiming at the question whether there is also a prioritization for IUPAC nomenclature given by the order of atomic numbers as we have it for the CIP nomenclature. Or do we just have a prioritization given by the alphabetic order in the case of halides?

As per the official IUPAC guidelines, namely Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book):

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(e) saturation/unsaturation:
     (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
     (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

The directly relevant part here is rule (g), which states that the substituent cited  first gets the lowest locant. Since substituents are cited in alphabetical order, this indirectly means that the lowest locant is assigned to the substituent which comes first alphabetically, i.e. 'bromo'.
Note that this is only rule (g), and that in general rules (a)–(f) should be applied first (their full text can likely be found elsewhere on Stack Exchange). In this case, rules (a)–(d) are irrelevant and rules (e)–(f) do not distinguish between the two proposed numbering systems, so (g) is used as the tiebreaker.
In any case, the preferred IUPAC name of the above molecule is (2E)-2-bromo-1,4-difluoro-3-iodobut-2-ene. [Technically, the locant of the double bond must be specified with the stereodescriptor, hence (2E)- instead of just (E)-, although in practice this is often ignored when there is only one double bond.]


What is the relationship between the molecular weight of a polymer and that of polystyrene? So, I read in a similar chemistry thread that you can get Mn and Mw of the polymer of the interest by comparing the result of the sample obtained from GPC/SEC with the standard known molecule weight polymer like polystyrene.
I don't understand how this works. You compare what of the polymer with what of the polystyrene to determine the molecular weight?? There has to be a constant factor when determining relative values between two things, and what is it in the case of polystyrene and the polymer of the interest?
What do you mean by polystyrene equivalent value? What is the point of determining the relative molecular weight of a polymer based on the polystyrene equivalent value??

Thank you!

You have to have something to calibrate the column with. Polystyrene is typically used since it can be made cheaply (well, relatively cheaply) with a pretty small molecular weight distribution.

Yes, there will be differences between the true molecular weight of a polymer and what the comparison to polystyrene gives you. But does that really matter? Why do you need to know the absolute molecular weight, when a relative comparison works just fine?

It's similar to how we determine the mass of something. At present, there is that infamous block of Pt/Ir in Paris that is the standard kilogram. All mass measurements are made through a series of comparison to that mass.


Electronic configuration in [Ni(CN)4]2- $\ce{[Ni(CN)4]^2-}$ is a low spin , inner orbital complex having $\mathrm{dsp^2}$ hybridisation. According to Crystal Field Theory, $\ce{CN^-}$ is a strong field ligand that causes pairing of electrons i.e electrons occupy the orbitals of the lower $t_{2g}$ level completely before they occupy the higher $e_{g}$ level. If I am not wrong, Hund's rule of maximum multiplicity is followed while filling the orbitals of $t_{2g}$ or $e_{g}$.

$\ce{Ni^2+}$ has a configuration of $[Ar]3d^8$. If we write down the electronic configuration of $\ce{Ni^2+}$ in this complex, 6 of the d-electrons would occupy the $t_{2g}$ level and the remaining 2 would occupy then $e_{g}$ level.

If we search for diagrams of electronic configuration of $\ce{[Ni(CN)4]^2-}$, many of them show the last 2 occupying the same orbital in $e_{g}$ which seems to contradict Hund's rule in my view. But I am pretty sure that the the diagrams on the Internet are correct according to experimental data of magnetic moments and other features.

My question is whether strong field ligands can pair up electrons within orbitals of $t_{2g}$ and $e_{g}$ like in this case where the 2 $e_{g}$ electrons are paired up within one orbital instead of letting them occupy two separate ones. Or is there some other factor at play?

Your thinking is correct. Within a set of degenerate orbitals (i.e. equal in energy), the orbitals will fill one electron at a time, and only after all have one electron will they start to pair up.

The situation with $\ce{[Ni(CN)4]^{2-}}$ is that it has square planar geometry, so the two orbitals that are $e_g$ in an octahedral complex are separated in energy. This diagram from Chemistry LibreTexts shows it nicely:

Because of the separation in energy, the $\mathrm{d}_{xy}$ fills completely (two electrons) before any electrons fill the $\mathrm{}_{x^2-y^2}$.


Is it possible to create a 100% identical copy of a chemical? Is it possible to create a 100% identical copy of a chemical? For example, a 100% identical copy of vitamin A would have the same shape, the same formula, and the same structure.

I agree with most of what the commenters and @jerepierre have written, with one or two exceptions.

  Is it possible to create a 100% identical copy of a chemical?

If your question were asked in an introductory chemistry class, I believe most instructors would answer with a simple "yes", for the reasons outlined above.  On the other hand, if your question were asked in an advanced chemistry class, or perhaps in a court of law, then I believe the answer would likely be "no".  A frustrating dichotomy, but such is life.

Let me explain the "no" answer.  To begin with, realize that simple molecules carry deuterium and $\ce{^13C}$ at levels found in natural abundance. Next, note that the way an organism would synthesize a vitamin, hormone, protein, etc., is likely different than the way a chemist might synthesize it in the laboratory. The differences would lie in the raw materials used in the process as well as the actual synthetic pathway.

Each step in a synthesis that requires breaking (or making) a carbon-hydrogen bond or a carbon-carbon bond will result in isotopic fractionation.  Simply put, $\ce{C-H}$ bonds and $\ce{^12C-^12C}$ bonds are easier to break than the corresponding $\ce{C-D}$ bonds and $\ce{^13C-^12C}$ bonds.  Therefore, if we repeatedly break some number of (they can be different or the same) C-H bonds and/or C-C bonds, then our final product will be enriched in molecules with lower deuterium and $\ce{^13C}$ content, the degree of enrichment being dependent upon the difference in the number of carbon-hydrogen and carbon-carbon bond breaking and bond making steps between the synthetic process as carried out within the organism and as carried out on the lab bench.  There may be further fractionation if the starting material used by the body has required C-H and/or C-C bond making and breaking and has, therefore, also suffered isotopic fractionation.  Such fractionation can easily be monitored through comparison of the mass spectra of natural and synthetic compounds.

In addition to this isotopic differentiation between synthetic and natural products, there will also exist a difference in reaction by-products and contaminants created in these two different synthetic processes.

Both methods (isotopic fractionation and by-product/impurity differences) could be used to determine if, for example, the testosterone in an athlete's body was produced naturally within the body or synthetically outside the body.  Also, we are probably all aware of various reports were a synthetic version of a drug (often illicit, but not always) naturally produced by a plant or metabolite has produced death upon administration due to toxic by-products produced in the lab synthetic route - clearly a different side-effect profile.

So in answer to your question, no, it is not possible to synthesize a natural product in the lab that will be 100% identical to the same material produced in a living organism.


Fried potatoes absorbing oil Suppose we bring a pan and put in it some oil and turn on the heat, before the oil could get very hot and nearly boiling, we put on the potatoes that start frying.

Someone told me that if the potatoes were put in oil before it gets really hot, then they will absorb more oil than if they were put in oil after it gets really hot.

Is this true?? If yes then WHY?

Note:(given that in both cases they stay the same period of time immersed in oil).

It is an example of the Leidenfrost effect. Ideally the water being expelled from the food would actually keep the oil from contacting the food itself by forming a barrier of steam (vaporizing water) around the food.


Why are radical intermediates are more stable on tertiary carbons? Is this because there's more electron density? More carbons around means the tertiary carbon "has more electrons around it"?

Before thinking about why tertiary radicals are 'stable', you need to have some kind of understanding about the structure of radicals.

Carbon centered radicals adopt similar geometries to carbocations: three coordinate and planar. Whereas carbocations have an empty p-orbital, radicals have a singularly occupied p-orbital. Overall the geometry is usually described as sp2 if we're thinking about hybridisation. (*)

With the structure of the radical now defined, we can think about what gives the tertiary radical such stability, and to do that we need to discuss what hyperconjugation is.

A good place to start (other than a textbook) might be the IUPAC definition from the Gold book:

  ...The concept of hyperconjugation is [...] applied to carbenium ions and radicals, where the interaction is now between σ-bonds and an unfilled or partially filled π- or p-orbitals...

Essentially, in the sp2 geometry, adjacent C-H bonds are able to interact with the vacant p-orbital, donating a small amount of electron density in order to stabilise the electron deficient p-orbital.

Individual C-H bonds to not exert a particularly strong stabilising effect, however, in a t-butyl cation/radical, there are 9 adjacent C-H bonds, each donating a very small amount of electron density, the sum of which amounts to a net overall stabilisation relative to primary radicals.

Whilst this explains the stability of the tert-butyl radical, there are other mechanisms of stabilisation for radicals which carbocations are not able to participate in. Due to the single electron in the p-orbital, radicals can also be stabilised by resonance, in a similar (but not identical) way to lone pairs or carbanions. A classic example is a radical generated next to a C=O group or C=C group.

Both of these effects discussed (hyper conjugation and resonance) are thermodynamic effects. The stabilisation is lowering the energy, making the radical less reactive.

(*) In the case of radicals, this is a bit misleading, theres a lot of spectroscopic and computational evidence to suggest that radicals deviate from planarity towards a pyramidal structure (the idea of hybridisation and sp2 doesn't hold up well)- especially when electron withdrawing groups such as fluorine are bonded to the radical centre. From the point of view of undergraduate chemistry this is largely irrelevant, however, given that the planar model explains the features of radical reactions that we're usually interest in, as organic chemists rather than theoreticians (EPR studies are one place in which the planar model falls down, the measured values don't correlate well to a completely flat carbon centre).


Is it possible to create a ketone using ozonolysis, when you have an alkene with no alkyl groups? Can Butan-2-one ($\ce{CH3-CH2-CO-CH3}$) be created from Butene ($\ce{C4H8}$) using ozonolysis?

I don't think that's possible.

The general reactions scheme for ozonolysis is as follows:

Consider the case of butene:

Herein $R_1$ and  $R_3$ can be identified as methyl groups, and similarly $R_2$ and $R_4$ are plain old hydrogens. Following the reaction scheme, one would naturally expect Ethanal as a product (if you use $\ce{Zn}$, ethanoic acid may be obtained if you use something else).

I'll drop by later and write out a mechanism perhaps.


Does SiO₂ have polar covalent bonds or coordinate covalent bonds? First off, I know $\ce{SiO2}$ is a network solid. I am looking at its intramolecular forces between the $\ce{Si}$ and the $\ce{O}$ molecule. On the one hand, it seems to me that this should be a polar covalent bond, but on the other, both electrons are given to the $\ce O $ by $\ce{Si}$ which makes it coordinate covalent.

The bonds are polar covalent.

'Given' is a strange term in this context and generally only of relevance for ionic compounds.

Coordinate covalent bonds imply the donation of two electrons to a bond with another atom.  In the case of $\ce{ SiO2 }$, The structure is actually tetrahedral, with an oxygen between and two silicon atoms, and where each oxygen forms two bonds and each silicon forms four bonds.  In these bonds, each of the four valence electrons for silicon is shared singly with a single electron from one of the oxygens.  Thus, in each bond, one of the electrons is from silicon and one from oxygen, which is why we don't normally consider this a coordinate covalent or dative bond.

This is one of those examples where the empirical formula, $\ce{ SiO2 }$, is highly misleading.


What makes protons give the property of acidity? Brønsted Lowry definition of an acid says that an acid gives away protons.

What I don't understand is what makes dissolved protons grant the property of acidity? Like, why would dissolved protons eat up materials?

My question applies to bases too, I suppose. And I guess you could include Lewis-acids and bases in your argumentation...

It's a question that has been bothering me for long. By the way I am not chemist(nor do I study chemistry).

While it is true that oxidizing strong acids like nitric and sulfuric are more corrosive than a non-oxidizing strong acid like hydrochloric, the latter still is able to dissolve materials (e.g, metals, minerals, proteins) and cause burns to the skin. A strong acid like HCl can (in the case of human skin or other biological materials) catalyze the hydrolysis of proteins, fats, and carbohydrates down to their water-soluble components. This is what happens in your stomach, where there is an abundance of hydrochloric acid, as well as enzymes to speed up the process.

The most widely accepted measure of how strong an acid is, its its acid dissociation constant, not the pH of a solution of it.


Why is there no salt bridge for hydrogen–silver bromide cell? What is the reason there is no salt bridge for this cell?
$$\ce{Pt(s) | H2(g) | HBr(aq) | AgBr(s) | Ag(s)}$$

The function of a salt bridge or a diaphragm is to avoid mixing of 2 different electrolytes or electrolysis products from the respective half-cells, while allowing ion electro-migration between them.
This cell does not need a salt bridge as it uses the only electrolyte $\ce{HBr(aq)}$ and the is no need to separate the $\mathrm{AgBr}$ electrode from gaseous hydrogen.


Does the distance between an electron and the nucleus change for excitation and relaxation? According to the Bohr model, when an electron moves from a ground state to the excited state, it will absorb energy, but what about the distance between electrons and nucleus, will it be changed? Why ? How?

Also, if there is the distance change when from a ground state to the excited state, will there be also any change in distance when electrons move from excited state to ground state ?

Yes.

The Bohr model shows the atom as a small, positively charged nucleus surrounded by orbiting electrons.

Electrons undergo exothermic or endothermic reactions during ground state or excited state. When the electrons move from ground state to excited state, it means that they gain more energy, resulting in higher energy levels (higher energy shells), and this causes the electrons to orbit further from the nucleus. The same explanation is applicable when electrons move from excited state to ground stage- the electrons lose energy and this results in lower energy levels (lower energy shells), causing the electrons to orbit nearer to the nucleus.


Why do diamond and graphite have different structures, when they are both composed of carbon? Both diamond and graphite are covalent networks and are both made entirely from carbon, but why does diamond have a three dimensional network of strong covalent bonds which makes it hard, whereas graphite has flat layers of carbon atoms which makes it a weak object and breakable.

This seems strange to me, because they both are made of the same atoms and both are covalent networks.

Chemical structures are a tradeoff of several factors, including the conditions on how they were formed.

The stability of any given chemical structure depends on the ease with which any specific reaction can turn it into something else. Both graphite and diamond are very stable structures which basically means they are hard to easily convert into something else. Another way to think about this is that to convert either into a different allotrope (different versions of the same element) of carbon requires a lot of bonds to be broken and rearranged and this requires a big input of energy. That's why both are stable.

But why is there more than one stable allotrope of carbon? Note that there are more allotropes than just graphite and diamond: buckminsterfullere is also a stable allotrope of carbon with a completely different structure that involves molecules of 60 carbons not a covalent network solid. The basic answer is that there are a lot of ways to arrange the atoms to give a stable structure (partially because there are a lot of options for stable C-C bonds given the chemistry of carbon). This is related to the same reason that organic chemistry is a big subject: C-C bonds are very "stable" and can be organised to give a huge variety of structures.

Not every arrangement of C-C bonds is equally stable: many will fall apart into other structures because there are mechanisms that make the rearrangement easy. But there are no easy mechanisms for rearranging the bonds in diamond or graphite into something more stable. Buckminsterfullerene is only formed from a plasma of carbon atoms when it is allowed to cool: you can't just make it by simple routes from other allotropes of carbon.

Both graphite and diamond have very similar energies of formation at normal conditions, though graphite is marginally more stable (the particular arrangement of bonds has a slightly lower energy than the arrangement in diamond under normal conditions). But this is not true at all conditions. Under extreme pressures, diamond is more stable. And this is how diamond is formed: it crystallised from solution in hot melted volcanic rocks deep in the earth containing lots of carbon and under high pressure. In these conditions, diamond is the more stable way to arrange carbon atoms and the conditions are such that the less stable form will covert to the more stable form because there is enough energy around to break the C-C bonds until the most stable structure is created.

Normally (by which I mean at atmospheric pressure) carbon rich compounds can be pyrolised to give pure carbon in the form of graphite. Here the heat drives off the non-carbon atoms and the carbon atoms rearrange to the form stable at lower pressures.

Once formed neither major allotrope is easy to convert to the other without extreme conditions. Which is why you can wear diamond jewellery without constant anxiety.

In short the stable allotropes of carbon are a function of the way they were made. And there are a variety of ways to join carbon atoms up to give different structures. Once formed they don't easily interconvert. Hence we find a variety of allotropes depending on the variety of conditions that led to their formation.

NB carbon isn't the only element with multiple allotropes. Tin, for example, changes crystal structure and properties depending on the temperature (and, unlike carbon, this interconversion is fast enough to have been a problem in the era when tin cans were actually made of tin and stored in very cold conditions).


Is the correct configuration assigned to the given pair of geometrical isomers of 1,2-dichlorocyclopropane? I suppose that the configurations assigned are opposite as in the left isomer of 1,2-dichlorocyclopropane, the left chlorine atom is coming out of the plane of the paper while the other is going into the plane of paper. So it must be trans. Am I going wrong wrong somewhere?

Those drawings are rather confusing.  In the following drawing I've reproduced your "cis" drawing and then below, I've redrawn it more clearly.  We now see that your molecule is actually the "trans" isomer.

The dark wedge bonds should be either drawn both "up" or both "down".  The same applies to the dashed bonds, they too are either drawn both "up" or both "down".  Your drawing is different, it has one wedge bond "up" and the other "down".  This is confusing and makes interpretation somewhat difficult.  In my redrawn version both dark wedge bonds are "up" and both dashed bonds are "down" making interpretation easier.


Using cryptands to make a negatively-charged solution? In our discussion group, my friend mentioned that metal cations can be separated from other cations in solution using cryptands. After addition of the cryptands to the solution of the cations, chelation of the cations takes place and since the complexes formed are more soluble in organic solvent, addition of an organic solvent to this resultant solution would allow these complexes to be migrate to the organic layer. The organic layer can then be decanted and the metal ions have now been separated successfully from the aqueous solution. This was my friend's proposal.

From hearing this, I quickly thought of the possibility of creating a negatively-charged solution. By applying the method of using cryptands which my friend has proposed to separating sodium ions from a solution of sodium chloride, wouldn't the remaining chloride ions in solution give the resultant solution an overall negative charge? This led me to think that there must be something wrong with his idea as this should not be able to occur, based on my instinct. However, I do not know exactly how to refute my friend, to say his proposal does not work. Thus, I would like to ask if anyone knows the reason why my friend's method would not work.

For some reason this question reminds me of Maxwell's demon paradox. You don't really separate anions and cations here; in reality the resulting metal ($\ce{M+}$) cryptand ($\ce{L}$) complex is going to be associated with anionic part ($\ce{X-}$) in organic solvent ($\ce{s}$) after it's being extracted from aqueous phase ($\ce{w}$). For the detailed process, refer to [1], for example.

There are two main processes to consider:

  $$
\begin{align}
\ce{L_\mathrm{s} + M^+_\mathrm{s} &<=> LM^+_\mathrm{s}}\label{rxn:1}\tag{1}\\
\ce{\overline{\ce{L}} + M^+_\mathrm{w} + X^-_\mathrm{w} &<=> \overline{\ce{LMX}}}\tag{2}
\end{align}$$

You would be correct if only \eqref{rxn:1} occurs, however, this is not the case.

Reference

Fyles, T. M. Can. J. Chem. 1987, 65 (4), 884–891. DOI 10.1139/v87-149 (Open Access).


Can I convert from an emission amount (in tonnes per area) to a concentration (in ppbv)? Hi, I was wondering if it is possible to estimate the concentration of SO2 given the total mass of SO2 emissions?

Europa's EDGAR has a resource called "Global Emissions EDGAR v4.2" (see http://edgar.jrc.ec.europa.eu/overview.php?v=42). A database of SO2 emissions in Metric Tons per Area is provided, but I need concentration instead.

Currently I am determining the volume of the gas by assuming it's at STP, then I am dividing this by the estimated total containing volume, which I tried to estimate as follows.

  Total Containing Volume = (11120 m*11120 m)*altitude

or

  Total Containing Volume = (apprx. area of 1° latitude by 1° longitude "square")*altitude

I calculated an altitude that makes the calculation match another database of SO2 concentrations that I have. Unfortunately the other database of SO2 concentrations is limited in the area that it covers, so it is not suitable.

Is this estimation attempt misguided or is there a better way to do it? If there is another database that contains global or continental SO2 concentration grids that would also fix the problem.

What you are doing seems reasonable but there are many hidden assumptions in your technique:

The data source you are using actually provides a rate, i.e. metric tons per area per year.
Thus your calculation is also implicitly a rate: the results will be in tons per volume per year.
You are assuming that the only source of atmospheric $\ce{SO2}$ is the emissions quantified by your data source.  It may well be that there are other, natural sources of $\ce{SO2}$, or more likely, natural sinks of $\ce{SO2}$.
If you want to know actual concentrations, you will need to quantify all the other sources and sinks of $\ce{SO2}$, and integrate all sources and sinks over time.


Determining identity of unknown compound with NMR data I have the following IR spec, C NMR and H NMR data. The boiling points determined for the unknown compound were 116.8 and 119.1°C (two trials). Given the possible compounds, the unknown can be one of:
3-methyl-2-butanol
1-butanol
3-pentanol
2-pentanol
2-methyl-1-butanol. (These are based on boiling point)
I have eliminated some of these choices based on the number of different types of carbons (4, as shown on C NMR) but am not sure where the H of the OH is indicated in H NMR. There is no singlet present among the 3 groups of peaks, so where would it be?

Peaks of acidic protons in an NMR spectrum can change (in position, form and intensity) due to exchange with other volatile protons.

A very fast exchange makes the peaks coalesce, you get one singlet between the two "original" (expected) chemical shifts, the exact position depends on the respective amount of both.
A slower exchange, one "on the NMR timescale", broadens that peak considerably, typical case is residual water and a COH proton. The peak might in some cases be so broad it becomes invisible.
Even slower, this peak splits into two, which become sharper and further removed from each other.
A very slow exchange makes no change to the positions, but an exchange with deuterium (i.e. from the solvent) changes the relative intensity of the respective peaks over time.


How many significant figures are appropriate when a new digit is created? In my chem class the other day, we were working through a question about average atomic mass. We were given the following information about two isotopes of boron:

Boron-10: Atomic Mass = 10.013, % in nature: 19.8%

Boron-11: Atomic Mass = 11.009, % in nature: 80.2%

For the weighted average (before sig fig rounding), then, Boron 10 contributes 10.013*.198 = 1.982574 AMU, and Boron 11 contributes 11.009*.802 = 8.829218 AMU.

Now, if we were just finding one of these values, we would report 1.98 or 8.93 AMU (to 3 sig figs). Adding these together, we get 10.81 AMU. However, if we follow "convention" and do all rounding at the end, we get 10.8 AMU (3 sig figs).

My question is, which of these is correct? On the one hand, 10.8 is correct since it has the same number of sig figs as the least accurate given value (19.8% or 80.2%). On the other, why should we go from hundredths to tenths precision just because we got a new digit in the least precise position? One could argue that the "10" is essentially its own digit.

Personally, I agree with the response here (Significant Figures Interpretation) that says sig figs don't make a whole lot of sense. But for the purposes of this question, which is the proper way to express the answer?

Thanks in advance!

Actually, you are correct - when adding significant figures together, you keep the largest last significant figure between the two numbers. For example, $15.31+2.2=17.5$. On the other hand, when you are multiplying numbers, you keep the same number of significant figures. Using the same numbers, $15.31*2.2=34$. While this decision might appear arbitrary, there is a good reason behind it.

Since we do not know anything past the tenths place about $2.2$, lets say this number is actually $2.2x$ where $x$ is a digit that would cause this number to round down (value between 0 and 4). $15.31+2.2x = 17.5[x+1]$ (where a number in brackets is a digit that is the result of a calculation. It can carry to the next place, but for reasons that I will explain later, we assume that it does not). Since we now know nothing about the hundredths place after adding these two numbers, we cannot include it in our result.

Likewise for multiplication, $15.31 * 2.2x = 33.[6+x][8+5x][2+3x][x].$ Since we know nothing about the value of $x$, we know nothing after the decimal point. Because our next digit is guaranteed to round up, regardless of the value of $x$, we round up to get $15.31*2.2x=34$. This is not a special case. In fact - the number of digits that will not depend on the value of $x$ is always equal to the number of significant figures in that number (this is - again - because we assume that $x$ will not cause a digit to carry), thus giving us our multiplication rule for significant figures.

Now for why it is okay to assume that $x$ will not cause a digit to carry. When you record an experimental value, you are supposed to record one more digit than your device can differentiate (ex. guessing the location of a meniscus between two numbers in a graduated cylinder). As such, the final digit that you use in your calculations is inaccurate anyways. Because of this, it is safe to assume that $x$ did not cause any additional overflow. If we were to assume that x overflowed, then it could cause your doubtful digit to also carry, which would result in numbers that disagree with the measured value.

I hope this helps (I know I went into a lot more detail than you were looking for).


What am I doing wrong with the fuel and packing I am using for a cannon I made? Recently I have been using flash powder in a cannon that I made, for a chemistry project, however, it is not causing the rubber ball I put in it to propel out of the tube. First I put in 30g of flash powder, then pack a small wad of paper on top, and lastly insert the ball in an aluminum foil bowl. The fuse is then inserted into the bottom and lit on fire, but the only thing that has been happening is fire flaming from the end of the tube and melting parts of the rubber ball. Do you have any small suggestions, as I am running out of time to figure this out? I attached a picture of the cannon. (Also I do not have time to make black powder).

Flash powder and gun powder are two different mixtures, designed to do two different things. Gun powder is designed to produce a lot of gas very quickly (Carbon dioxide, Sulfur dioxide, Nitrogen) which propels the projectile out of the barrel.

Flash powder - typically potassium chlorate/perchlorate and aluminium powder, examples here and here, produce a lot of heat and light very quickly but not much gas - so no propulsion.