Iron-II to iron-III change in complex formation We have an iron dust sample, which is treated with hydrogen chloride in aqueous solution. The solution enters in a reaction with a sodium hydroxide powder, from which we obtain a precipitate. Could we solubilize the precipitate with excess of ammonia? Note that, at the end of the reaction, qualitative chemical analysis identified the ion $\ce{Fe^{3+}}$, but not $\ce{Fe^{2+}}$.

First reaction is, obviously:
$$\ce{Fe + 2HCl -> FeCl2 + H2 ^}$$
Then, we have:
$$\ce{FeCl2 + 2NaOH -> 2NaCl + Fe(OH)2}$$
We observe that we have, now, only $\ce{Fe^{2+}}$ in the aqueous solution, not $\ce{Fe^{3+}}$. Precipitate solubilization means coordinative compounds formation. If for iron we have a similar reaction to the one with copper ion:
$$\ce{Cu(OH)2 + 4NH3 -> [Cu(NH3)](OH)2}$$
We would have only $\ce{Fe^{2+}}$. So, is there any redox process happening in background? Or a physical process (non-chemical)?

First iron(II) hydroxide $\ce{Fe(OH)2}$ is a precipitate made in aqueous solution, and it does not react with ammonia. Second, the iron(II) hydroxide $\ce{Fe(OH)2}$ is a green substance which is extremely sensitive to the oxygen of the air. In a couple of minutes, it gets brown, due to the formation of iron(III) hydroxide $\ce{Fe(OH)3}$ according to $$\ce{4 Fe(OH)2 + O2 + 2 H2O -> 4 Fe(OH)3}$$ If water is absent, the reaction may produce another compound of iron (III) $$\ce{4 Fe(OH)2 + O2 -> 4 FeOOH + 2 H2O}$$


Do L-sugars (enantiomers of natural sugars) have a sweet taste? I'm wondering about how L-sugars taste. They are obviously not digestible by humans, but what do they taste like? If they are sweet, too, why isn't sucrose with the unnatural chirality used as a zero-calorie table sugar substitute?

Take a look at tagatose. It is a 6-carbon sugar (a ketohexose like fructose) but is inefficiently metabolized and has about 30% the caloric content of sucrose. It is on the FDA GRAS (generally recognized as safe) list. Unlike common artificial sweeteners, you can cook with it like sugar. It has been industrially produced as a sugar alternative, but it is not clear if it commercially available to consumers at this time.


My confusion regarding Hydrogen peroxide In Hydrogen peroxide, the oxidation number of oxygen is "-1" instead of "-2".

But it seems to me that, the oxygen atoms have '-2' as their oxidation number as each oxygen atom here is connected to a hydrogen atom and an oxygen atom.

So, what's actually happening here?

If there's any problem in my question please inform me. Thanks!

Each oxygen atom is connected to a hydrogen atom(which develops -1 charge on oxygen and +1 on hydrogen) and to another $\textbf{oxygen}$ atom which contributes no charge to both(0 and 0). Similar for other oxygen atom. Hence oxygen state of each oxygen atom is -1 and for each hydrogen atom it is +1.

In general, in peroxide linkage, oxygen has -1 oxidation state


Reaction of acyl chlorides with diazomethane and silver(I) oxide $$\ce{R-COOH ->[SOCl2,CH2N2][Ag2O/CH3OH] X}$$

While solving Carboxylic derivatives recently, I came across this.
so following the reagents one by one-
$\ce{SOCl2}$ will form acid chloride.
$\ce{CH2N2}$ (diazomethane) on the other hand reacts with carboxylic acid to give a methyl ester. but we are currently on acid chloride, this part is a little confusing to me.
$\ce{Ag2O}$ in aqueous/alcoholic medium should give ($\ce{CH3O-}$) ions.
How to proceed? What is wrong with my reaction mechanism?

Your first reaction is correct. You do get an acid chloride. The reaction would proceed with the first step being what you have proposed:
$$\ce{R-COOH ->[SOCl2] R-COCl}$$
Now, the next step is a reaction known as Arndt-Eistert Reaction:
$$\ce{R-COCl ->[CH2N2][Ag2O/CH3OH] R-CH2-COOCH3}$$
The mechanism for this reaction is as follows:

Note: Mechanism is from Organic-Chemistry.org


Relating Gibbs Free Energy to Equilibrium Constants I am having some trouble relating these two concepts. I know that if K=1 then delta g is zero, and that we can interpret signs and magnitudes. However, i am having trouble determining required temparature based on given aimed K, and standart Gibbs free energy of a reaction.
Here is the kind of problem that i have trouble with:

To solve this problem, you suppose that $\Delta$$H°$  and $\Delta$$S°$ do not change with temperature. You calculate $\Delta$$G°$ by two approaches : $1)$ - $RT$ ln $K_p$, and $2)$ $\Delta$$H°$ - $T\Delta$$S°$. Then $T$ can be easily obtained.


Halogen bonds in QuteMol I've been using QuteMol for generating 3D animated molecular images and I've encountered a problem; for some reason that's a complete mystery to me when I create a pdb file for etravirine (using MarvinSketch to create a sdf file and then using OpenBabel to convert the sdf to pdb and add implicit hydrogens) and load it I found the bromine atom was all by itself with no bond connecting the two and I would like a bond in the image. Any ideas? Thanks in advance.

The following image is the result I'm getting from QuteMol.

I haven't used MarvinSketch in a while, but isn't it a 2D editor?

Generation of 3D coordinates from 2D files often means trouble. I tried that recently with a 2D BKChem drawing of a trypticene-based ligand for another answer and the result was useless.

I definitely recommend the use of Avogadro as a free 3D editor. Please have a look at their website or the article in Journal of Cheminformatics.

My typical workflow here is:

switch off auto-adding of hydrogen
draw the skeleton with the right multiplicity of bonds, but using carbon only
change the "wrong" carbons to heteroatoms
let the program add the hydogen atoms
run a force-field optimization
manually adjust some dihedral angles
run the force-field optimization again
(optional) align a bond with one of the principal axis
save as CML or MOL

I put a MOL file of etravirine created that way on pastebin. Please give QuteMol a try at it.

I generated a PDB from the molfile mentioned above and ran it through QuteMol. The image below shows that the $\ce{C-Br}$ bond does exist in the PDB file.

Admittedly, checking the PDB file for the respective CONNECT x y line would have been a (less flamboyant) proof too ;-)


How to assign overlapping multiplets in 1H NMR spectra? In the above $\ce{^1H}$ NMR spectrum ($\pu{400 MHz}$, $\ce{CDCl3}$) the methyl, alkene, and $\ce{CCl3CH2}$ protons are fairly easy to assign however, the protons in the cyclohexene ring show overlap and complex splitting (which I assume is due to the axial and equatorial arrangements). How would one go about differentiating between and assigning the remaining protons to the overlapping signals (if possible)?
I've actually been trying to find more generalised information to help with the assignment of a different (but similar) cyclohexene compound but couldn't really find anything, so took a suitable example from a paper I found.

Referring to your comment on Buttonwood's answer:

Unfortunately, in this case I only have 1D HNMR and IR available to analyse the product

If this is really the case, then you are essentially out of luck. It is not possible to extract much information from a series of overlapping multiplets, which may well be coupled to each other (these strong coupling effects are likely the cause of the "complex splitting" you mention). It may be the case that by going to a higher field you can alleviate some of this, as multiplets will be more dispersed along the frequency axis. However, you will quickly run into a hard limit this way, since our maximum field (as of the time of writing) is ~1.2 GHz and even that is likely not enough to fully resolve your overlapped multiplets.

With that out of the way, let's talk about some other NMR techniques which will be useful in this situation. I want to preface this with a few points:

These are general suggestions which are broadly applicable to a variety of different problems; they are not tailored to the specific molecule which you asked about, and indeed you may find that some of them are less suitable for your specific compound. For the specific case at hand I would suggest talking to a NMR specialist at your institution, if this is at all possible.

There is a caveat in that many of these techniques have slightly poorer performance in situations where there is strong coupling, although it by no means makes them useless.

You may find that you need more than one of these to get a complete understanding of the $\ce{^1H}$ spectrum (especially in light of the previous caveat).

I assume here that the overlapping multiplets belong to the same species, as is the case in the question. In the situation where you are studying a mixture of different species which may have peaks that overlap with each other, the following techniques will still remain very useful, but there are more options which will help to separate the different subspectra based on molecular properties (e.g. relaxation rates or diffusion coefficients). These are outside the scope of the current answer.

Finally, or further reading about all of these techniques I recommend Claridge, T. D. W. High-Resolution NMR Techniques in Organic Chemistry, 3rd ed.; Elsevier: Amsterdam, 2016..

(1) 2D correlation NMR
This has already been mentioned by Buttonwood, and is the standard answer to issues of resolution in 1D spectra. The downside is that 2D spectra are very, very rarely run with high resolution, partly because of time constraints, but especially so if heteronuclear decoupling is used (e.g. in a HSQC). Thus, instead of measuring well-resolved multiplet structure (which would be transferable to the 1D spectrum, thus allowing you to assign the 1D), you are more likely to get broad blobs that are centred on a certain frequency. This tells you nothing about the multiplets actual proton spectrum, it only tells you that at (say) 1.46 ppm you have some kind of multiplet that corresponds to a certain proton in your structure.
Despite this, one should not neglect the amount of information that is provided by standard 2D spectra (COSY, HSQC, HMBC, NOESY). Often even having a blob is enough to figure out which proton is which, even if it may not tell you the exact shape of the underlying multiplet.
(2) Pure shift NMR
This is a series of techniques which remove all coupling information, thereby collapsing multiplets into singlets and giving a 1D spectrum with exactly one singlet per (chemically inequivalent) proton. I would consider the TSE-PSYCHE method[1,2] to be the "state of the art" in this regard, especially with regard to strong coupling. This is very powerful but suffers from the same drawback as 2D NMR does: it removes all multiplet structure, so on its own does not lend any insight into the original $\ce{^1H}$ spectrum. Nevertheless, I would say that it is really worth having a pure shift spectrum in addition to some of the other methods, as it can greatly aid in assigning peaks.
Here's an example of a pure shift spectrum, versus the original proton spectrum of the same compound (personal data).

(3) 2D J NMR
A 2D J spectrum shows proton chemical shifts along one axis and proton multiplet structure along another axis. What this means is that every multiplet in a $\ce{^1H}$ spectrum is separated along a secondary axis according to its chemical shift. This allows you to not only figure out the position of each multiplet (which both of the above already provide you) but also the shape of each individual multiplet.
In particular, pure shift J spectra[3] are really the best versions in this regard as these provide much higher resolution and do not take a lot of time. To be technical, these don't provide "pure shift" characteristics because the 2D J $f_2$ axis (the horizontal axis) is already "pure shift", with the coupling information already separated out onto the $f_1$ axis. What they do is to convert the lineshapes in the spectrum from a "phase twist", which has long "tails" in both dimensions, to a pure absorption Lorentzian which is a single sharp peak. This is an example of how it looks (personal data again):

You can extract vertical traces from this spectrum to view each individual multiplet, with its chemical shift given simply by the value along the $f_2$ axis (horizontal axis).
(4) Selective excitation experiments
This refers to a family of 1D experiments where you selectively excite one multiplet and then perform some kind of mixing in order to look at spins that are coupled / near to the spin that you excited. A selective 1D NOE experiment (for example) would tell you which spins are in close proximity to a given spin that you chose to excite; it is analogous to acquiring only one row of a 2D NOESY. You could also run selective 1D TOCSY experiments, which would tell you which spins belong to the same spin system as the one you excited. By choosing an appropriate TOCSY mixing time you can limit the "spin diffusion" to one or two coupled spins away.
The benefit of this over the original 2D experiment is twofold: first, it takes much less time; and secondly, you can acquire much higher-resolution data. This allows you to disentangle different overlapping multiplets based on their coupling partners or spatial proximity to other spins. For example let's say you have two multiplets, corresponding to protons A and B, that are overlapping, but only proton A is in close proximity to proton C (whose multiplet is well-resolved). Then you can run a selective 1D NOE experiment where you excite proton C, and the resulting spectrum should include only multiplet A and not B. The same can be applied for scalar couplings via 1D TOCSY.
(5) Chemical shift-selective filters
If you find that you need to excite a peak inside an overlapped region, then the idea of a chemical shift selective filter may be useful. This seeks to excite one multiplet out of a series of overlapping ones, and traditionally involves recording multiple spectra and adding them up, allowing the multiplet of interest to add up while averaging all other multiplets to zero by virtue of them having a different phase. However, there is a recent implementation called GEMSTONE[4] which does this very cleanly in just a single experiment without needing to sum anything up. In my limited experience it works quite well although some of the experimental parameters need to be optimised for each peak to obtain best performance. This particular demonstration that I have (personal data, again) isn't so impressive because the overlap isn't as severe, but the paper has slightly more interesting examples.

References

Foroozandeh, M.; Adams, R. W.; Meharry, N. J.; Jeannerat, D.; Nilsson, M.; Morris, G. A. Ultrahigh-Resolution NMR Spectroscopy. Angew. Chem. Int. Ed. 2014, 53 (27), 6990–6992. DOI: 10.1002/anie.201404111.
Foroozandeh, M.; Morris, G. A.; Nilsson, M. PSYCHE Pure Shift NMR Spectroscopy. Chem. Eur. J. 2018, 24 (53), 13988–14000. DOI: 10.1002/chem.201800524
Foroozandeh, M.; Adams, R. W.; Kiraly, P.; Nilsson, M.; Morris, G. A. Measuring couplings in crowded NMR spectra: pure shift NMR with multiplet analysis. Chem. Commun. 2015, 51 (84), 15410–15413. DOI: 10.1039/c5cc06293d.
Kiraly, P.; Kern, N.; Plesniak, M. P.; Nilsson, M.; Procter, D. J.; Morris, G. A.; Adams, R. W. Single‐Scan Selective Excitation of Individual NMR Signals in Overlapping Multiplets. Angew. Chem. Int. Ed. 2021, 60 (2), 666–669. DOI: 10.1002/anie.202011642


Can these changes be reversed by changing temperature? An egg turning hard when boiled.
A spoon full of batter cooking on a hot griddle.

I said that they both cannot be reversed, because I think if you throw both the egg and batter into the freezer, it will freeze, but it will not be in its original form before you cooked both of them. Then again, I guess I could try this for real on my own but I don't know if that would be a good idea...

Both processes involved the denaturation of proteins and complex reactions so is impossible reverse them only with a change of temperature.

Boiling an Egg
Although in most case denaturation is not reversible there are some cases where it is.
However in the case of Egg white denaturation is irreversible (see here) and so the whole process is by the way irreversible.
Cooking batter
The same for the batter even if is unclear which flours are used, and the composition of the batter  you can be sure that after cooking batter on a hot griddle surely some irreversible reactions occur: denaturation, Maillard reaction, Amadori rearrangment.


Why don't water molecules react alongside aqueous reactions? To explain my question I'll use an example: We all know that sodium displaces copper in a solution of its sulfate in the reaction:

$$\ce{2Na (s) + CuSO4 (aq) \rightarrow Na2SO4 (aq) + Cu (s)}$$

My question is why don't some of the water molecules present in the aqueous solution of $\ce{CuSO4}$ also react with the sodium (Na) to form sodium hydroxide and hydrogen ($\ce{NaOH + H2}$) along side this displacement reaction. Shouldn't at least a bit of NaOH be formed?

I would appreciate a comprehensive response that a high school sophomore (me) would understand. Thanks in advance.

why don't some of the water molecules present in the aqueous solution of $\ce{CuSO4}$ also react with the sodium (Na) to form sodium hydroxide and hydrogen ($\ce{NaOH + H2}$)

I agree that sodium metal would react with the water.  Where/how did you come up with the equation implying otherwise?

I think you'll enjoy this video: https://www.youtube.com/watch?v=X5yLtqlopTM


Cis trans isomer and cyclic compounds 8 Calcitriol is a steroid hormone found in human blood.

Calcitriol shows geometrical isomerism.
Give the number of geometrical isomers of calcitriol, including calcitriol.
I’m confused I think it doesn’t have a single geometric isomer because rotation is restricted in cyclic compounds! What is the concept behind it!

for calcitriol:

Assuming only Z and E isomer 64 isomers are possible .For rest of isomers (3 ,ZZ,EZ,EE)
each will have 64 isomers.

For Z and E isomer 64 isomers are :

enter image description here

Quoting
MARCH’S ADVANCED ORGANIC CHEMISTRY REACTIONS, MECHANISMS,AND STRUCTURE SIXTH EDITION
Michael B. Smith ,Jerry March


How can decreasing in temperature indicate an endothermic reaction? I’m really wondering how can decreasing the temperature refer to an endothermic reaction.

I faced this when I was calculating the heat of a reaction Q in a constant-pressure calorimeter.

What I believe is that when something becomes cooler, then it has released some amount of heat.

Let’s say that you have a beaker of water. You measure the water’s temperature using a thermometer, and you determine that it is 25 °C. Now you add two reactants to the beaker, and mix them together well until they are completely reacted. If the reaction between these two reactants is endothermic, then the chemicals will absorb heat from the water. If you were to now remeasure the water’s temperature, you might find that it was something like 23 °C. This is because the heat that was absorbed by the chemicals was lost by the water.


What is the pKa of 2-methyl-2,4-pentanediol? What is the pKa of 2-methyl-2,4-pentanediol? Is there a reference that contains a list of pKa values for various molecules that I can consult or is there a way to determine/estimate the pKa of this molecule if such a reference does not exist?

Do you mean 2-methyl-2,4-pentanediol?

The $pK_a$ values of 2-propanol (isopropanol) and 2-methylpropan-2-ol (tert. butanol) are 16.5 and 17, respectively and I'd expect your diol to be in this range range, unless deprotonation is featured by intramolecular hydrogen bonding. In that case, the $pK_a$ might be somewhat smaller, around 15.


Why is the rate of reaction calculated as d[A]/dt? For example, for a simple chemical reaction like

$$\ce{A -> C}$$

the rate of the reaction could be calculated as the rate of appearance of $\ce{C}$ or the disappearance of $\ce{A}$, which makes a logical sense. But if the reaction is

$$\ce{A + B -> C}$$

as I read somewhere the rate could be calculated the same way as the first case either as the disappearance of $\ce{A}$ or $\ce{B}$, or the appearance of $\ce{C}$. Is it logical to calculate the rate as the disappearance of $\ce{A}$ or $\ce{B}$ as the rate?

Think about it, isn't the appearance of $\ce{C}$, the combined effect of the disappearance of both $\ce{A}$ and $\ce{B}$, which mean if the rate of the equation can be calculated as the change in concentration of $\ce{C}$ using the product, shouldn't the rate be calculated as the change in concentration of $\ce{A}$ plus the change in concentration of $\ce{B}$ through some time change $t$?

Although the appearance of C is due to the combined disappearance of A and B, the relation is still 1:1 between A and C. Therefore, the rate that A disappears = the rate that B disappears = the rate that C appears.

Let's look at an analogy in stoichiometry. Say I have 1 mole of A and 1 mole of B. Will I make 1 or 2 moles of C? By your logic, I should make 2 moles of C. But I don't.


Free automatic organic synthesis Linux command line program I'm trying to find a free automatic organic synthesis Linux command line program which take mol or smile files as input. I have random molecules and I'd like to know how difficult above a baseline they will be to synthesize. An implementation of synthetic accessibility and not a full step wise synthesis program would also be acceptable. Thanks for the help.

The SAscore is now implemented in RDkit and available for download.


How does a reference electrode work in a three electrode system? This is what I understand about three electrode electrochemistry (correct me if I am wrong):

The working electrode is where the reaction of interest happens, e.g. copper ions reduced to solid copper.
The auxiliary/counter electrode supplies the current to the working electrode, i.e. it is the source of the electrons (maybe not in all cases?) e.g. zinc metal is oxidised to zinc ions.
The reference electrode lets you control the potential difference so you don't apply too much driving force to the working electrode and end up reducing water etc. No current passes through it.

I don't understand how the reference electrode lets you control the potential and how it works if no current passes through it. How is it connected to the other electrodes?

The reference electrode lets you control the potential difference so you don't apply too much driving force to the working electrode and end up reducing water etc. No current passes through it.

You have the basic idea but if absolutely no current passed through the reference electrode then you couldn't measure any voltage!

The gist is that you can't measure the "voltage" of a single electrode. You need a complete cell to get a voltage. The reference electrode is used with a very very low current to measure the voltage of the electrolytic cell formed with the working electrode.


Why can't cyclohexylbenzene be called benzylcyclohexane? The following structure represents the compound cyclohexylbenzene:

Why can't the benzene ring be treated as a substituent and the compound be called benzylcyclohexane?

The relevant rule for the seniority order of parent structures in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows.

P-44.4 SENIORITY CRITERIA APPLICABLE TO RINGS, RING SYSTEMS, OR ACYCLIC CHAINS
P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.
The senior ring, ring system, or principal chain:
(a) has the greater number of multiple bonds (P-44.4.1.1);
(…)

P-44.4.1.1 The senior ring, ring system or principal chain has the greater number of multiple bonds; for the purpose of this criterion, mancude rings or ring systems are considered as consisting of noncumulative double bonds [criterion (a) in P-44.4.1].

Therefore, the name of the compound that is given in the question is cyclohexylbenzene rather than “phenylcyclohexane” since benzene has more multiple bonds than cyclohexane.
Note that this rule is not to be confused with the rules for systems composed of rings and chains, which are less strict. For systems composed of rings and chains, the context may favor the ring or the chain; nevertheless, the ring is selected as the parent structure for the preferred IUPAC name. In case of cyclohexylbenzene, however, benzene is always the parent structure – not only for the preferred IUPAC name.


What is the mechanism for the oxidative cleavage of diketones via permanganate under basic/hot conditions? The products of the reaction are carboxylates.
What exactly would cause the alkyne to cleave?

In J. Org. Chem. 1979, 44 (15), 2726–2730, the authors Lee and Chang have investigated the permanganate oxidation of alkynes to α-diketones. In addition, they investigated the oxidative cleavage of several α-diketones to carboxylic acids. In the case of hexadecane-8,9-dione, permanganate oxidation under at least three sets of conditions provided n-octanoic acid 8 (major) and n-heptanoic acid 7 (minor). They suggested that the oxidation of the dione 1 leads only to octanoic acid 8 via intermediate 4 while oxidation of enol 2 affords 2-oxononanoic acid 6 and ultimately octanoic acid 8 via oxidative decarboxylation. In addition, oxidation of heptanal 5 affords heptanoic acid 7. The three solvent systems employed were acetone/aqueous HOAc, aqueous H2SO4 and water. Starting material was recovered in each experiment. Tetradecane-7,8-dione showed similar behavior.
For the mechanism of permanganate oxidation of alkynes to α-diketones, see the accompanying post by @Raul Luciano.


Beer's law: UV-Vis absorbance values exceeding 2 In classical analytical chemistry, absorbance values in Beer's law can theoretically range from 0 to infinity.

As general rule of thumb neither high, nor very low absorbance are used for constructing calibration curves. The reason for avoiding high absorbance is that when absorbance is 2, only 1% light reaches the detector and with 3, only 0.1% light reaches the detector. Analytical chemists were taught to avoid absorbance > 1.5.  Low absorbance values are avoided because it hard to distinguish the light beam with and without the cuvet in place. The question then arises how much is the real absorbance and how much is contributed by noise.

Nowadays (= within the last decade), I am seeing UV-Vis spectrophotometers have absorbance ranges from 0 to 3, and some have even more. One day a PhD student showed an absorbance > 5 for a spectra, apparently without even thinking twice on the meaning of absorbance. Similarly, researchers showed an absorbance spectrum with the maximum at absorbance of 3.

My question is this a fundamental flaw in Beer's law that high absorbance values are erroneous or it is limitation in detector technology and the intensity of light sources? Please ignore the chemical reasons for deviations in Beer's law.

Imagine, if we had very high intensity light source, such as xenon arc lamp, instead of a typical deuterium or a tungsten lamp, would we get less error in the absorbance measurements of concentrated solutions? Absorbance is a ratio technique, but the number of photons reaching the detector will be high, even with conc. solutions, with intense light sources.

Can any spectroscopist shed some light on this issue (no pun intended) whether high intensity sources would beneficial in UV-Vis spectrophotometry of concentrated solutions?

Beer's law limitation can be more easily understood if the coefficient $\epsilon$ is considered as a cross-section effect. Let's start with a broad beam or a bundle of rays having a section $A$ (in $m^2$). Its measured intensity is $I_o$. Now suppose there exist plenty of tiny obstacles suspended in the bundle of rays, and that each such particle behave like a dark disk perpendicular to the bean, with a surface or a cross-section $\sigma$. $\sigma$/A is the fraction of $I_o$ that is blocked off by one such particule. If there are $n$ particules in the whole cylindrical bundle, the fraction of $I_o$ that is captured is $\Delta$$I$/$I_o$ = $n·\sigma$/$A$.

Now let l be the length of the bundle where the particules are present. The volume of this bundle filled with dark particles is $V=A·l$. So the fraction of $I_o$ absorbed by the particules is $\Delta$$I$/$I_o$ = $n·\sigma ·l/A·l$ = $(n/V)·\sigma ·l$ = $c·\sigma ·l$, if $c = n/V$ is the concentration of the particules in the volume crossed by the beam.

Going to the limit, this expression becomes : $dI/I_o$ = - $\sigma ·l·dc$.
Integration gives : $ln(I/I_o)$ = - $\sigma·c·l$

This last expression is similar to Beer's law :  $log(I/I_o) = - \epsilon ·c'·l$. I have written $c'$ because in the beginning of my text $c$ is the number of particules per volume unit. And the concentration in Beer's law is in mol/L. $c' = c/N_o$, where $N_o$ is Avogadro's number.

In the end we obtain the final formula : $\epsilon = N_o·\sigma /2.3$

Now if a particule is a disk of $0.2$ $nm$ diameter (dimension of a small molecule), the corresponding $\epsilon$ is about $10^5 L mol^{-1} cm^{-1}$. This is the order of magnitude of $\epsilon$ for the usual dye.

But if $\epsilon$ = $N_o·\sigma /2.3  = 10'400 m^2/mol$. And this is the point where I would like to go today. It means that if one mole of the dye molecules were all to be touched by the beam, they would have to cover a surface of $10'400 m^2$. This is as large as a football ground. It also means that when the concentration becomes too high, Beer's law does not apply any more. The majority of the molecules of the dye are hidden behind others.

You may even calculate that when $c > 1/ (\epsilon · cm), $ then $  : I/I_o < 0.01$, the absorbance is $> 2$, and this is the limit of application of Beer's law, because most of the molecules are hidden behind some other ones at higher values of $c$.

Sorry for the length of the demonstration.


Demonstrating orthonormality for wave functions for ground state and excited state of He The electron configuration of the ground state $\ce{He}$ atom is $\ce{1s^2}$, and the excited state configuration under study is $\ce{1s^2 2s^1}$. Show that these states are orthonormal, i.e.,

  $$\langle\Psi_0 \vert\Psi_0\rangle = 1$$

  $$\langle\Psi_1 \vert\Psi_1\rangle = 1$$

  $$\langle\Psi_0 \vert\Psi_1\rangle = 0$$

I start off by defining the wave functions as Slater determinants,

$$
\Psi_0 = \frac{1}{\sqrt{2}}
\begin{vmatrix}
\phi_1(1)\alpha(1) & \phi_1(1)\beta(1) \\
\phi_1(2)\alpha(2) & \phi_1(2)\beta(2)
\end{vmatrix}
$$

$$
\Psi_1 = \frac{1}{\sqrt{2}}
\begin{vmatrix}
\phi_1(1)\alpha(1) & \phi_2(1)\beta(1) \\
\phi_1(2)\alpha(2) & \phi_2(2)\beta(2)
\end{vmatrix}
$$

I am able to show that the wave functions are normalized, but I have trouble showing that they are also orthogonal. I get that $\langle\Psi_0 \vert\Psi_1\rangle = 1$. I set up the integral over all coordinates, carry out the multiplications, and add together the terms that are equivalent due to indistinguishable electrons (identical when all coordinates are interchanged). I end up with the expression

$$
\int d\tau \left[  \phi_1^*(1)\alpha(1)\phi_2^*(2)\beta(2)\phi_1(1)\alpha(1)\phi_2(2)\beta(2) - \phi_1^*(1)\alpha(1)\phi_2^*(2)\beta(2)\phi_1(2)\alpha(2)\phi_2(1)\beta(1)   \right]
$$

where $d\tau$ represents all coordinates. Separating the spin coordinates from the spatial coordinates, I get that

$$
\langle\Psi_0 \vert\Psi_1\rangle = \\
\sum_{m_{s1}=-1/2}^{1/2} \sum_{m_{s2}=-1/2}^{1/2} \alpha(1)\alpha(1)\beta(2)\beta(2) \cdot \langle\phi_1(1) \vert\phi_1(1)\rangle \cdot \langle\phi_2(2) \vert\phi_2(2)\rangle \\
- \sum_{m_{s1}=-1/2}^{1/2} \sum_{m_{s2}=-1/2}^{1/2} \alpha(1)\alpha(2)\beta(2)\beta(1) \cdot \langle\phi_1(1) \vert\phi_2(1)\rangle \cdot \langle\phi_2(2) \vert\phi_1(2)\rangle \\
= 1 - 0 = 1
$$

The first term reduces to $1$, I find, while the second term reduces to $0$. To get $\langle\Psi_0 \vert\Psi_1\rangle = 0$, either both terms need to be $1$, or both terms need to be $0$. Something must be incorrect. Is there some obvious flaw in my derivations?

First prove a well-known fact that any two spin orbitals $\psi_i$ and $\psi_j$ are orthogonal if the corresponding spatial orbitals from which they are constructed are themselves orthogonal.

For two spin orbitals constructed from two different spatial orbitals orthogonality follows from orthogonality of spatial orbitals regardless of spin components.
For two spin orbitals constructed from the same spatial orbital orthogonality follows from orthogonality of spin functions.

Then you can work with spin orbitals to (re)establish a well-known result that for two Slater determinants differ in just one spin orbital, the overlap integral $\langle \Psi_0 \vert \Psi_1 \rangle$ is zero.
$$
\Psi_0 = \frac{1}{\sqrt{2}}
\begin{vmatrix}
\psi_1(1) & \psi_2(1) \\
\psi_1(2) & \psi_2(2)
\end{vmatrix} \, ,
\quad
\Psi_1 = \frac{1}{\sqrt{2}}
\begin{vmatrix}
\psi_1(1) & \psi_3(1) \\
\psi_1(2) & \psi_3(2)
\end{vmatrix} \, ,
$$
where
$$
\psi_1(1) = \phi_1(1) \alpha(1) \, ,
\quad
\psi_2(1) = \phi_1(1) \beta(1) \, ,
\quad
\psi_3(1) = \phi_2(1) \beta(1) \, .
$$
Algebraically, it is very easy to show that,
$$
  \langle \Psi_0 \vert \Psi_1 \rangle =
  \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} \langle \psi_1(1) \psi_2(2) - \psi_2(1) \psi_1(2) \vert \psi_1(1) \psi_3(2) - \psi_3(1) \psi_1(2) \rangle \\
  = \frac{1}{2} \Big(
    \langle \psi_1(1) \psi_2(2) \vert \psi_1(1) \psi_3(2) \rangle
  - \langle \psi_1(1) \psi_2(2) \vert \psi_3(1) \psi_1(2) \rangle \\
  - \langle \psi_2(1) \psi_1(2) \vert \psi_1(1) \psi_3(2) \rangle
  + \langle \psi_2(1) \psi_1(2) \vert \psi_3(1) \psi_1(2) \rangle \Big) \\
  = \frac{1}{2} \Big(
    \langle \psi_1(1) \vert \psi_1(1) \rangle
    \langle \psi_2(2) \vert \psi_3(2) \rangle
  - \langle \psi_1(1) \vert \psi_3(1) \rangle
    \langle \psi_2(2) \vert \psi_1(2) \rangle \\
  - \langle \psi_2(1) \vert \psi_1(1) \rangle
    \langle \psi_1(2) \vert \psi_3(2) \rangle
  + \langle \psi_2(1) \vert \psi_3(1) \rangle
    \langle \psi_1(2) \vert \psi_1(2) \rangle \Big) \\
  = \frac{1}{2} \Big(c_1 \cdot 0 - 0 \cdot 0 - 0 \cdot 0 + 0 \cdot c_1 \Big) = 0 \, .
$$
In the case when spin orbitals are normalized (to one) $c_1 = 1$, but it is not required here: the overlap integral will vanish for spin orbitals that are just orthogonal and not necessarily normalized (to one) as well.


Order of Reactivity of Halogens in Electrophilic Addition What is the order of reactivity of halogens in electrophilic addition reaction via formation of cyclohalonium intermediate?
(A) $\ce{F2}$> $\ce{Cl2}$> $\ce{Br2}$> $\ce{I2}$
(B) $\ce{I2}$> $\ce{Br2}$> $\ce{Cl2}$> $\ce{F2}$
(C) $\ce{I2}$> $\ce{Br2}$> $\ce{F2}$> $\ce{Cl2}$
(D) $\ce{Br2}$> $\ce{Cl2}$> $\ce{I2}$> $\ce{F2}$

I tried to predict it using the mechanism, but couldn't reach a conclusion. Is this based on thermodynamic and kinetic data?
Here's the mechanism involving bromine, for reference:

Could someone provide a detailed explanation for the above problem, with data if possible?

The rate-determining step of halogenation of alkenes is the formation of cyclic intermediate.
The cyclohalonium intermediate of bromine will be more stable than that of chlorine owing to its lower electronegativity.
The following paragraph is taken from Peter Sykes [$1$, p.$181$-$182$]:

It is not normally possible to add fluorine directly to alkenes as the reaction is so exothermic that bond fission occurs. Many alkenes will not add iodine directly either and when the reaction does occur it is usually readily reversible.

As is clear $\ce{I2}$ will be more reactive than $\ce{F2}$.
Hence, the correct answer is D.
References:

Peter Sykes; A Guidebook to Mechanism in Organic Chemistry 6th ed.; 2003 ISBN-10:8177584332


Why is the bond dissociation energy of C-H bond higher than that of a N-H bond? Looking at a chart of BDEs (bond dissociation energies) regarding elements bonded with hydrogen, the general trend seems to be that the BDE increases as we go to the top and to the right. This can be explained by the fact that the elements to the right have stronger attraction for the covalent bonds, making homolytic cleavage harder. Likewise, those elements near the top have shorter radii, which means that their attraction is larger too.

However, there is a single anomaly in the $\ce{N-H}$ and $\ce{C-H}$ bonds, where the BDE of $\ce{C-H}$ is higher than $\ce{N-H}$. Why is this?

If you look at an orbital diagram of nitrogen you will see that each p-orbital has one electron in it, if you want to add a second electron to any one of them, it makes the orbital higher energy. Atoms would generally like to be in the lowest possible energy state, and half or completely filled shells are far more low energy, and thus, more stable, this is most clearly seen with the electron configuration of copper and chromium.

Cu: [Ar] 4s1 3d10

Cr: [Ar] 4s1 3d5

Nitrogen has a half filled 2p sub level, so it has to move to a substantially higher energy state to add a fourth electron. If it gets one in a bond with Hydrogen, Nitrogen finds it energetically cheaper to dissolve the bond and return to its base configuration. The reverse applies to Carbon. Because it now is able to get another electron from Hydrogen, it now has a half filled 2p sub-level and is in a fairly stable configuration that it would like to keep.

Most of this comes from Chemistry by Maciel, Traficante, and Lavallee. The book was published in 1978, so I am sorry if anything is dated.


How are the molar mass and molecular mass of any compound numerically the same? This observation is really annoying me, and the internet isn't providing me with any solid answers. Either their definitions of molar mass completely differ, or they don't stick to their own definition while actually calculating the molar mass of a certain element or compound.

Now, according to Wikipedia,

  The molar mass M is a physical property defined as the mass of a given
  substance (chemical element or chemical compound) divided by the
  amount of substance.

and it's unit is

  g/mol

My understanding of this is that we're calculating the total mass of all the atoms(or molecules) present in 1 mole of that element(or compund). This means that we would have to take the atomic mass of that element and multiply it by the number of atoms present in 1 mole of that element, which is 6.022 × 10^23 atoms.

So for example, the relative molecular mass of water is 18.02 u, and it's almost the same as the molar mass of water, which is 18.015 g/mol.

So my question is, how is 18.02 u × (6.022 × 10^23) = 18.015 g/mol?

So for example, the relative molecular mass of water is 18.02 u, and it's almost the same as the molar mass of water, which is 18.015 g/mol.

  So my question is, how is 18.02 u × (6.022 × 10^23) = 18.015 g/mol?

That is not quite correct.

The relative molecular mass $M_\mathrm r$ of water is a dimensionless quantity, i.e. the unit for relative molecular mass is the unit one and not $\mathrm u$:

$$M_\mathrm r=18.015\,28$$

It's the average mass per molecule $m_\mathrm f$ that has the unit $\mathrm u$:

$$m_\mathrm f=18.015\,28\ \mathrm u$$

Note that

$$1\ \mathrm{u}=1.660\,538\,921(73)\times10^{-27}\ \mathrm{kg}$$

Also note that you are missing the unit of the Avogadro constant:

$$N_\mathrm A=6.022\,140\,857(74)\times10^{23}\ \mathrm{mol^{-1}}$$

Thus

$$\begin{align}
M&=m_\mathrm f\cdot N_\mathrm A\\[6pt]
&=18.015\,28\ \mathrm u\times N_\mathrm A\\[6pt]
&=18.015\,28\times1.660\,538\,921\times10^{-27}\ \mathrm{kg}\times N_\mathrm A\\[6pt]
&=18.015\,28\times1.660\,538\,921\times10^{-27}\ \mathrm{kg}\times6.022\,140\,857\times10^{23}\ \mathrm{mol^{-1}}\\[6pt]
&=0.018\,015\,28\ \mathrm{kg\ mol^{-1}}\\[6pt]
&=18.015\,28\ \mathrm{g\ mol^{-1}}
\end{align}$$


Does an increase in pressure necessarily lead to an increase in temperature? Ideal gas law says that $pV = nRT$. So with compression, one way to add pressure which makes the volume smaller, can you actually increase the temperature?

It does not make sense to me that you can increase temperature, that is, add to the kinetic energy of the molecules, simply by increasing pressure, pressure meaning the force exerted on the gas from its surroundings. Can someone please explain this contradiction?

I think what you are asking is "why can the temperature of the gas increase when you compress it, even if the cylinder is adiabatic so that no heat can enter the gas?"  When you move the piston to compress the gas, you are doing work on the gas at the interface with the piston.  The piston is moving toward the gas, and the molecules of gas that collide with the piston leave with a greater average velocity than when they arrived.  So their average kinetic energy is increasing.  If expansion were occurring, such that the piston were moving away from the gas, the colliding molecules would leave with lower average kinetic energy.


Reaction between a vinegar/salt/hydrogen peroxide mixture as an etching solution and copper I am looking for the less harmful chemical solution to serve as an etching solution. After many searches, my interest went to a vinegar/salt/hydrogen peroxide solution.

I have to ask then if the residue of this solution is harmful and if it is ok to pour it in the regular water waste system.

The etching solution itself is fine to dispose of down the drain. However, if you use it to etch copper, the dissolved copper ions are quite toxic, and should not be disposed of through the sewer system.

You can bypass the problem by collecting the etch solution, and submerge aluminum foil. $$\ce{3Cu(CH3COO)2(aq) + 2Al(s)->2Al(CH3COO)3(aq) +3Cu(s)}$$React with the aluminum until the bluish color disappears, then collect the solid copper + remaining aluminum to throw in the trash. You can now flush the solution down the drain.


What is the preferred IUPAC name of o-toluidine? Why does o-toluidine have 2-aminomethylbenzene as its IUPAC name?

Why isn't it named as a derivative of amine (like 2-methylaniline)? Aren't amines given a higher priority than alkanes in nomenclature?

See the Priority Chart given on Master Organic Chemistry for reference.

In principle, your rationale is correct.

Generally, the selection of a preferred parent structure is based on the seniority of classes, which gives priority first to characteristic groups expressed as suffixes. Since o-toluidine (note that this name is no longer recommended) contains only one characteristic group $(\ce{{}-NH2})$, it is named as amine.

According to Subsection P-62.2.1.1.1 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the retained name “aniline” is the preferred IUPAC name (PIN) for the functional parent compound. The systematic name “benzenamine” may be used in general nomenclature.

For the retained name “aniline”, substitution is permitted at any position. Substitution of such functional parent compounds is limited to substituent groups having a lower seniority than that denoted by the (explicitly expressed or implied) suffix of the functional parent compound. Anyway, in this case, this limitation is not relevant since o-toluidine contains only one characteristic group and carbon compounds (rings or chains) have a lower seniority than amines.

Therefore, the PIN for o-toluidine is “2-methylaniline”. The systematic name “2-methylbenzenamine” may be used in general nomenclature.


Hydroboration on Limonene I saw that the hydroboration of Limonene (1-Methyl-4-(1-methylethenyl)-cyclohexene) does only happen on the methylethenyl group. Up to now, I wasn't able however to find out why. Generally speaking, a hydroboration can happen on any double bond, so why doesn't it react – at least a bit – with the cyclohexene?

I could argue, that the stereochemistry of the cyclohexene is not as favorable as the one of the methylethenyl group, but I don't see this being the case. On the cyclohexene the methyl group is in the way, where as on the methylethenyl group it is the cyclohexene.

The borane reagent in question, disiamylborane, is one of the many bulky hydroborating agents which preferentially add to less hindered double bonds.

In this case, the 1,1-disubstituted double bond outside the ring is more prone to hydroboration than the trisubstituted double bond within the ring, simply due to sterics.

There are many more examples of hydroborations with this reagent which are dictated by sterics: for further examples see its corresponding entry in the Encyclopedia of Reagents for Organic Synthesis. In all cases, disiamylborane is selective for terminal alkenes (i.e. either monosubstituted alkenes, or 1,1-disubstituted alkenes).


How to create different levels of hydrophobicity on glass beads surface? I have some glass beads and I need to create different levels of hydrophobicity on their surfaces (various contact angles). I know that this is possible by 'silanization'. However, I need to know about the following questions:

Whether there is any other reasonable method to do that.
In silanization, how can I change the level of hydrophobicity that my glass beads achieve? (e.g. by changing coating time, using different solutions or concentrations, etc.)

PS. I have the following solutions for silan. Please let me know if anybody has any comments on them or knows a more suitable solution for this purpose.
- CHLOROTRIMETHYLSILANE
- DICHLORODIMETHYLSILANE
- TETRAETHYL ORTHOSILICATE
- OCTADECYLTRICHLOROSILANE

Lets assume first that the surface $\ce{Si-O}$ bonds of teh glass are terminated by $\ce{Si-OH}$ groups. Hydrophilicity will depend greatly on the ration of $\ce{Si-OH}$ to $\ce{Si-R}$ groups on the surface of the glass. Chlorotrimethylsilane is great for a completely hydrophobic surface as it will cap all of the $\ce{Si-OH}$ groups to $\ce{Si-O-Si(CH3)3}$ groups allowing no polar interaction of the surface with the water. dichlorodimethylsilane is not as good leaving you with $\ce{-Si-O-(Si(CH3)2)-O-Si{-}}$ and $\ce{-Si-O-Si(CH3)2OH}$ groups at the surface which affords a bit more interaction.  Octadecyltrichlorosilane does provide even more polar groups than the last two, but because of the large octadecyl group this will likely make a more hydrophobic surface than the chlortrimethylsilane.  The tetraethyl orthosilicate or rather tetraethoxy silane (TEOS) will hydrolze to $\ce{Si(OH)4}$ in water which you can use to help make the surface more hydrophilic.

If I had to do this I would make various binary mixtures of TEOS, chlorotrimetyhlsilane, and dichlorodimethylsilane and treat the surface with them under constant conditions and see what spectrum of hydrophilicity you achieve. For the most hydrophobic surfaces I would use mixtures of octadecyltrichlorosilane and chlorotrimetyhlsilane.


What are simple applications of Monte Carlo simulations in chemistry, that can be done in Matlab? I am trying to learn simulation myself and I have installed Matlab on my machine for that purpose. In web I have gone through some of the basics of Monte Carlo simulation, but I am unable to find a simple Monte Carlo simulation experiment for chemistry. I am trying to simulate growth of a nano particle during a wet synthesis. I have tried searching research articles, but as I am new to simulations most of them I didn't understand. If possible can you explain a simple Monte Carlo experiment in chemistry, which can be done in Matlab. Is it possible to simulate the growth of nano particles using the Monte Carlo method?

I've done a little simulation as a senior year elective, though not in Matlab. Can you be a little more specific about how/what you're trying to model with nanoparticle growth? Perhaps direct us to an abstract or two.

The first Monte Carlo simulation you should attempt is the drunken sailor problem. The basic idea is that you have a drunken sailor who has equal probability of taking a step forwards or backwards, and you let him stagger around on a one dimensional line for some number of steps (1000) and repeat the experiment a large number of times. You then get some summary statistics and plots around your experiment.

After you're comfortable with that concept, you can look at increasing complexity into 2 or 3 dimensional random walks, changing the weighting function (not 50-50 for each direction), or some specific algorithm such as Metropolis-Hastings Monte Carlo.

I hope this helps. If you're looking for specific programming help, posting on stack exchange would be a good idea.


Zwitterions/IEP of glycine at pH 6? (Paradox?) I am trying to understand an experiment: Some crystalline, solid glycine was solved in water and a pH=6 was measured (also calculable with $\mathrm{pH} =\frac{1}{2}(\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a2} = \frac{1}{2}(2.34 + 9.6) = 5.97)$

I have tried to explain the acidic pH-value: Solid glycine is in a crystalline structure — all molecules are in a zwitterionic state. If they get solved in neutral water, the crystal-structure is lost and $100~\%$ of zwitterions are dissolved in the neutral solution. The carboxylic and amino group now act as an acid and base in aqueous solution. The acid reaction of the carboxylic group predominates ($\mathrm{p}K_\mathrm{a1} < \mathrm{p}K_\mathrm{a2}$). Some amino acids are now negative ions, some are still zwitterions.

But now is my question (the contradiction?): pH = 6 is also the isoelectric point IP with the definition, that all amino acids are in zwitterionic state. But according to my theory above, some amino acids have already reacted "away" from the zwitterionic state.

Are there flaws in my reasoning — is my explained "paradox" not true?

You need to separate the two concepts.

When dissolving a neutral amino acid in water, a buffered acid-base reaction will take place as you highlighted, giving a certain pH value. In this case, the following equilibrium:

$$\ce{HOOC-CH2-NH3+ <=> ^{-}OOC-CH2-NH3+ <=> ^{-}OOC-CH2-NH2}$$

will be somewhere between the pure second and the pure third species. (The neutral, non-zwitterionic species is neglectable in aquaeous solution.) Superfluous protons liberated into the solution will end up as $\ce{H3O+}$ explaining the overall reduction in pH value when compared to a neutral solution. You won’t notice it unless you perform really accurate experiments, but the solution’s pH value will be a tad higher than the isoelectric point.

If you go on and add external protons, i.e. by carefully adding an acid, you can adjust the overall solution’s pH. Since the glycinide anions are the strongest base present, they will take up the protons first, followed by the zwitterions. At a certain pH value (corresponding to a certain external proton concentration added), the concentrations of glycinide anions and glycinium cations will be identical and we have reached the isoelectric point of glycine. It is, and has to be, slightly different from the pH value obtained by dissolving glycine.


Percentages of radical constitutional isomers Is it possible to use the radical selectivity factors for chlorination based on primary, secondary, and tertiary substitutions to calculate the expected percentage of each constitutional isomer formed in the chlorination reaction? If so, how?

Your question is a bit hard to understand, you should change the wording or add more explanation. But I believe you want to know how to calculate the percentage of a particular isomer from the relative rates of chlorination. Well the calculation is as follows:

First, derive the ratio of different isomers. To find the terms of ratio, multiply the number of hydrogens which can be substitued to form the isomer, by the rate of halogenation of that type of hydrogen.

Next, calculate percentage from ratio. This is very easy arithmetic; you can do this by creating a fraction with the value of that particular isomer on the numerator and sum of all values from the ratio on the denominator. Then multiply the fraction by 100.

The ratio of reactivity of tertiary($\ce{3^\circ}$), secondary($\ce{2^\circ}$) and primary($\ce{1^\circ}$) H-atoms for chlorination is $\ce{5 : 3.8 : 1}$  at 298K. For bromination of $\ce{3^\circ}$, $\ce{2^\circ}$, $\ce{1^\circ}$ H, the ratio of reactivity is $\ce{1600:82:1}$ at 400K

Example: Monochlorination of n-butane. Here, two isomers are formed, 1-chlorobutane, and 2-chlorobutane.

Now, substitution of six $\ce{1^\circ}$ H  (shown by blue box) results in 1-chlorobutane

and substitution of four $\ce{2^\circ}$ H (shown by red box) results in 2-chlorobutane

So, the ratio is 1-chlorobutane : 2-chlorobutane  $\ce{=6\times1 : 4\times3.8= 6:15.2}$

Calculate percentage: 1-chlorobutane is $\ce{ \frac{6}{6+15.2}\times 100\%= 28.3\%}$

and similarly, for 2-chlorobutane, percentage is $\ce{71.7\%}$


Stability of CN molecule The $\ce{CN}$ molecule has been observed spectroscopically in comets where pressure has been estimated to be as low as $10^{-10}\ \mathrm{mmHg}$. Draw a Lewis structure for $\ce{CN}$. Suggest an explanation as to why this species can persist in a comet but is unstable under normal terrestrial conditions?

Having drawn the Lewis structure, the molecule is a free radical, hence it is very unstable. Would the species be unstable under terrestrial conditions due to a higher temperature (hence pressure) there? Or more specifically, how does a change in pressure effect a free radical molecule?

$\ce{NC-CN <=> 2CN}$

Which side is favored at low pressure?


IUPAC name for [CoSO4(NH3)4]NO3 What is the correct IUPAC nomenclature for the coordination compound $\ce{[CoSO4(NH3)4]NO3}$?

The answer I came up with is tetraamminesulphatocobalt(III) nitrate .


Why is the antibonding pi bond higher in energy than the normal pi bond? In my book, Organic Stereochemistry, the following picture is shown:

In the lowest diagram, in Figure 1.3, which shows a π* bond, the positive lobe of the p orbital of atom A is nearest to the negative lobe of the p orbital of atom B. By contrast, in the diagram above it, which shows a π bond, the positive lobe of the p orbital of atom A is nearest to the positive lobe of the p orbital of atom B. It is said that the π* bond is higher in energy, less stable, than the π bond. I would have thought it to be the other way around, due to electrostatic repulsion when two lobes of p orbitals which have the same charge overlap, and the electrostatic attraction when two lobes of p orbitals which have opposite charges overlap.

Why am I wrong? Why is it not this way around?

Firstly, the + and – signs you see do not indicate charge. They mark the sign of the electron’s wave function. In an atom, the electrons always have a negative charge, and the nucleus always has a positive one.

When a bonding orbital is formed, the electron density is concentrated in the area between the nuclei. The electrostatic attraction between the electrons and two (instead of one) nuclei, lowers the energy of the molecule. Furthermore, the electron density between the nuclei shields the electrostatic repulsion between the two positive nuclei to a certain degree.

In an antibonding orbital, the electron density is moved away from the internuclear area: the electron-nucleus attractions are minimized and the internuclear repulsion maximized. Thus, the molecule is destabilized.


lowest energy conformation of butan-2,4-diol Do the OH groups want to be furthest apart? or the CH3 groups?

I'm trying to determine the lowest energy conformation - and I don't know which group has the bigger steric effect?

I presume it is in the gauche form that its energy is minimun because of possible hydrogen bonding between the hydroxy groups


How do I balance the combustion reaction of butene? I have butene being burned in a fuel rich engine and I can't seem to balance this reaction:

$$\ce{C4H8 + $a$(O2 + 3.76 N2)
-> $b$ CO2 + $c$ H2O + $d$ C4H8 + $e$ (3.76 N2)}$$

When I try write equations for the elements, I get 5 equations and 4 unknowns which can't be solved. Here are my equations:

\begin{align}
\text{Carbon}   &:& 4     &= b  + 4d  \\
\text{Oxygen}   &:& 2a    &= 2b + c   \\
\text{Nitrogen} &:& 3.76a &= 3.76e    &  a &= e \\
\text{Hydrogen} &:& 2a    &= 2c + 8d  \\
\end{align}

I'm not sure if this makes a different but it says that water in the combustion products is removed before the dry analysis of the exhaust. The dry analysis has a composition of $14.95\% \ce{CO2}$, $0.75\% \ce{C4H8}$, $0\% \ce{CO}$, $0\% \ce{H2}$, $0\% \ce{O2}$, and balance $\ce{N2}$ (I don't know what that means).

The balanced equation for the combustion of butene is
$$\ce{C4H8 + 6O2 -> 4CO2 + 4H2O}.$$
Nitrogen is included only to indicate that it is present in the medium with some specific ratio to the oxygen amount – well technically it is included to make question look complicated.

$\ce{C4H8}$ in the products side is included to indicate that not all of it can be combusted but some escapes. And it really is arbitrary, depends on the the engineering of combustion room, etc. – which is why you have 4 equations, 5 unknowns: d is free to vary.  But you are given the dry analysis to fix its value.

Looking at the equations given, you have

\begin{align}
  b &= c = 4(1-d)\\
  a &= e = b + c/2 = 6(1-d) = 3b/2
\end{align}

And dry analysis says ratio of $\ce{CO2}$:$\ce{C4H8}$ is $14.95:0.75 = 19.93 (b/d)$ after the combustion. And this is the last equation you need.


Is iron (III) oxide-hydroxide the same as iron (III) hydroxide? Firstly, the Wikipedia article on iron(III) oxide-hydroxide says that $\ce{FeO(OH).H2O}$ is sometimes described as iron(III) hydroxide ($\ce{Fe(OH)3}$). If that is so, does iron(III) hydroxide not exist?

Secondly, the hydrated iron oxides section of the Wikipedia iron(III) oxide page says that on treatment of $\ce{Fe^3+}$ salts with alkali we get $\ce{Fe2O3·H2O}$ (also written as $\ce{Fe(O)OH}$) and not $\ce{Fe(OH)3}$. Is it the same as $\ce{FeO(OH).H2O}$ mentioned in the Iron(III) Oxide-Hydroxide page? If not, how are the two compounds different?

Thirdly, the iron(III) oxide-hydroxide page gives the following reaction:
$$\ce{FeCl3 + 3 NaOH → Fe(OH)3 + 3 NaCl}$$
Doesn't this directly contradict what's given in the iron(III) oxide page? And does $\ce{Fe(OH)3}$ here mean $\ce{FeO(OH).H2O}$.

Fourthly, the Wikipedia page on rust says that rust is composed of hydrated iron(III) oxides $\ce{Fe2O3·nH2O}$ and iron(III) oxide-hydroxide ($\ce{FeO(OH)}$, $\ce{Fe(OH)3}$). What are these hydrated iron oxides and how do they differ from hydrated iron(III) oxide-hydroxides $\ce{FeO(OH)·nH2O}$ described in the iron(III) oxide-hydroxide page?

Fifthly, the page on iron(III) oxide-hydroxide identifies $\ce{FeO(OH).H2O}$ as hydrated iron oxide, while the hydrated iron(III) oxide and rust page have me believing that $\ce{Fe2O3·nH2O}$ is hydrated iron(III) oxide. Are they wrong or is $\ce{FeO(OH).H2O}$ also a hydrated iron oxide?

I would also be interested in knowing the structure of these compounds, whether the water is just hydrogen bonded or if it takes part in some complex formation and where I can learn more about such unusual compounds.

Start with my answer here: What is chemical formula of Bauxite? On oxides and hydroxides of aluminium - that's a quick explainer on notation.

Now, to your questions:

  where I can learn more about such unusual compounds

Easy. Here: The Iron Oxides: Structure, Properties, Reactions, Occurrences, and Uses by R. M. Cornell and U. Schwertmann. ISBN: 3-527-30274-3. PDF link

The problem with iron hydroxides is that there are several polymorphs (same composition, different crystal structures), and they are often mixed together in one sample. Add to that that not all iron is fully oxidised, that these compounds are hygroscopic (they adsorb water moisture on their surfaces), and they can contain impurities of whatever was alloyed with the iron (nickel, carbon, chromium, etc etc) forming even more compounds.

A single polymorph rare exists by itself (but when they do, they are spectacular; examples: 1 2 3). Pure iron hydroxide is usually yellow, and pure iron oxide is bright red. Anything in between is going to be brown. When found in nature it's called "limonite", and when formed by rusting of iron, it's called "rust".


Synthesis of Potassium Trioxalatoaluminate (III) Trihydrate We mixed $\ce{Al}$ powder and hot water. Then we added $\ce{KOH}$ to the $\ce{Al}$ solution and it bubbled heavily (producing $\ce{H2}$). When that was over we added oxalic acid dihydrate to the boiling aluminate solution. We filtered and added cold ethanol. How do write these reactions?

I was thinking:

$$\ce{Al + KOH + 2H2O -> K[Al(OH)3] + H2}$$

and when oxalic acid comes in:

$$\ce{K[Al(OH)3] + C2H2O4.2H2O ->K3[Al(COO)3].3H2O}$$

The problem is this site says that the reaction is wrong and What is the reaction between oxalic acid and potassium permanganate?  is a similar question but has a completely different looking answer. I'm lost.

You have already concluded that the bubbles indicate that hydrogen gas is formed. Let's assume that two protons are reduced. Each receives an electron and they form a molecule of $\ce{H2}$:

$$\ce{2H+ + 2e- -> H2}$$

Where did the electrons come from? Did the aluminium powder dissolve?

$$\ce{Al -> Al^3+ + 3e-}$$

We are now having two equations, one for a reduction, one for an oxidation (of aluminium).

Let's add these two equations in a way that the electrons cancel out. For that, we multiply the first equation by 3, and the second equation by 2.

$$\ce{2Al +6H+-> 2Al^3+ + 3H2 ^}$$

Something's still wrong with the equation: You didn't throw the aluminium powder into a sea of protons, but suspended it in hot water. Let's add some $\ce{OH-}$ on both sides:

$$\ce{2Al +6H2O -> 2Al(OH)3 + 3H2 ^}$$

This is much better, but still not perfect: You did not describe a cloudy precipitate of a hydroxide being formed, and your first step took place in  the presence of potassium hydroxide:

$$\ce{2Al +6H2O +2KOH-> 2K[Al(OH)4] + 3H2 ^}$$

So far, we have explained the reaction of aluminium metal in aqueous alkaline solution: tetrahydroxoaluminate $\ce{[Al(OH)4]-}$ is formed.

Can you figure out the reaction with the oxalic acid on your own? Have a look at the structure of oxalate (Hint: bidentate ligand) and forget about the permanganate ;)

UPDATE

Let's add some oxalic acid, $\ce{HOOC-COOH}$ now. For the reaction, it doesn't make any difference whether it's a dihydrate, $\ce{H2C2O4*2H2O}$, or not. The reaction is performed in aqueous solution anyway. However, you might want to consider the difference in molecular weight between the anhydrous form and the dihydrate when you're planning to add a stoichiometric amount of oxalic acid.

$$\ce{[Al(OH)4]- + 3HOOC-COOH -> [Al(OOC-COO)3]^3- +2 H+ + 4H2O}$$


Can I predict the preferred resonance structure of carbon monoxide from its molecular orbital scheme? Because of the  8-valence electron rule I assume in CO the left resonance structure to be favored .

$$\ce{:\!\overset{\ominus}{C}#\overset{\oplus}{O}\!: ~<->~ :\!C=O\!:: ~<->~:\!\overset{\oplus}{C}-\overset{\ominus}{O}\!:::}$$

(from What is the Lewis Structure for CO?)

The molecular orbital scheme of CO looks like this:

Because the HOMO of CO is closer to the atomic orbital of O I think that the HOMO of CO is mainly localized on the oxygen atom, whereas the LUMO of CO is mainly localized on the carbon atom. Therefore I would have expected that the coefficient of the HOMO is higher on the O than on the C atom and therefore the top right resonance structure with the negative charge on the oxygen to be favored. However, in reality the top left resonance structure is favored. What is my error?

From another post, calculations have shown that the coefficient is higher on the carbon atom (How can the dipole moment of carbon monoxide be rationalised by molecular orbital theory?)
I guess my mistakes comes from the assumption that I say "HOMO closer to the AO of oxygen" means higher coefficient on oxygen. Is this assumption wrong and if yes why?

On the favoured Lewis structures

Well yes, you have correctly drawn a minus sign next to the oxygen in the rightmost structure and similarly a plus sign next to the oxygen in the leftmost structure. However, these are not true charges! These are formal charges. They can be representative of real charges (cf. $\ce{SO4^2-}$ but they do not have to be. They derive from the way Lewis structures are drawn and not from any physical significance.

Choosing a good Lewis structure requires balancing a lot of effects. Main group atoms should get electron octets (hydrogen dublets) is one of the main tipping points to take care of. In the leftmost structure, all atoms have electron octets. The central structure has an octet (on oxygen) and a sextet (on carbon) — less favoured. The rightmost structure only gives cabon a quartet, plus a positive charge. This combination is so strongly disfavoured that not even the negative charge on oxygen can save anything.

Minimising the total formal charges is a second balancing effect which would prefer the central structure over the other two; however, it is less strong than the octet idea. And finally, structures that have formal charges distributed according to electronegativity are slightly favoured over those that don’t, but as I said, the former rules are more important.

Remember that Lewis structures are just a representation; and not a particularly good one at that! You can sort-of derive things like bond orders or free electrons from them, but for most uses, they are inferior to the molecular orbital model.

On the MO diagram

  Because the HOMO of CO is closer to the atomic orbital of O

Stop right there. That’s not how MO diagrams work.

The general gist of these diagrams is that orbitals with a lower energy will be closer to the electronegative atom(s) while those with higher energy will have stronger contributions by the electropositive partner. This is the HOMO we are talking about in a diatomic molecule. It has to have a larger contribution by carbon.

Reasoning: Carbons atomic orbitals are higher per se. The higher the energy of a hybrid orbital, the higher the contribution of higher energy atomic orbitals must be. Remember that you need to distribute one entire atomic orbital across all the molecular orbitals that have the same (correct) symmetry. The contributions from each atom (squared) need to add up to one. It’s a lot easier for the atom to contribute to orbitals closer to the original energy.

Of course, the negative pole of the dipolar molecule is closer to oxygen. Why? All populated low-energy molecular orbitals have a stronger contribution by oxygen while the high-energy unpopulated ones have a stronger carbon contribution.

A side note, that was mentioned in the question’s comments: You have 5 $\sigma$ orbitals and 2 $\pi$ orbitals populated in the MO drawing. 1$\sigma$ and 2$\sigma$ can be thought of the 1s orbitals of carbon and oxygen, leaving five orbitals for bonding. One of them happens to have an antibonding structure (nodal plane through the bond axis) so you get $4 - 1 = 3$ a triple bond between carbon and oxygen. (This only works this easily because there are only two atoms.) Please also check out the pictures in the other thread

I’m not sure if I touched all the points you wanted answered concerning the molecular orbitals in here. If not, comment and I’ll be back to try to explain what I missed. There’s a lot more that you can read out of it if you want to give it a try.


What is the difference between radioactive equilibrium and transient equilibrium? Consider the following radioactive decay:
$\ce{A\rightarrow B\rightarrow C}$
In my textbook, it is written that radioactive equilibrium is when rate of formation of $\ce{B}$ is equal to rate of disappearence of $\ce{B}$.  In my textbook, it is also written that when the net formation of $\ce{B}$ is zero, we say that the system is in transient equilibrium i.e. $\frac{dB_{N}}{dt}=0$ where $B_{N}$ is the number of atoms of $\ce{B}$ at time $t$. Now, $\frac{dB_{N}}{dt}=0$ means that the rate of formation of $\ce{B}$ is equal to the rate of disappearance of
$\ce{B}$ which is exactly what radioactive equilibrium is. Then, what is the difference between radioactive equilibrium and transient equilibrium? Is there even any difference between them? Pleas explain. I am confused.

I guess you have been caught in a semantic trap. Adjectives radioactive and transient describe  two different aspects of the same thing.
It could be as well written as "radioactive transient equilibrium". Or, even better as "radioactive steady state", as it is not true equilibrium.
Radioactive describes its nature . It may not be radioactive,by can be thermal, chemical, radiative, mechanical etc .
Transient describes its dynamics. It need not to be transient ( temporary, also called  a steady state ), but it can be a permanent, timeless, true  equilibrium.


Software for chemical reaction simulation I wonder, what software packs / environments are available for chemical reaction simulation. The goal is to learn or just rehearse some basic knowledge in chemistry. I suppose there should be some to buy but are there any open source projects dedicated to this?

If you mean finding saddle point, then you should consider Gamess and Quantum Espresso packages: free, stable, mature packages, useful for molecule simulations and estimating reaction barriers. They, however, have to be manually compiled and require some skill to setup, as they are meant to be used in POSIX environment (various UNIX clones), not windows, and does not have user friendly result viewer or job builder/scheduler. It is not a problem for experienced linux user, though...

I'm currently using quantum espresso and I found it useful for my cases.


How to achieve clean dishes and glasses? I heard that there is a standard about cleaned glass, namely

  Chemically clean glass supports a uniform film of water, with no hanging droplets visible.

I tried to clean my dishes at home using detergent. After several efforts, the ceramic-made dish looks like a uniform film of water just after finishing the cleaning. If I hold the dish for a while, the water starts to concentrate at regions, forming a droplet.

In addition, in this video
http://www.youtube.com/watch?v=m7liHlmv5Kc
at 2 min, there is water droplet in the flask.

My questions are:

(i) Does the clean standard have any time range? Does it apply to arbitary time scale?

(ii) If the answer to question (i) is yes, how should I do to make the dishes so clean that there is absolutely no water concentrating at certain regions? I do not have a dishwasher.

P.S. Video demonstration is highly appreciated.

There are several laboratory techniques to rigorously clean glass. Here are a few of the more common ones, to add to the other answers. There are probably more procedures, and it's also common to perform more than one treatment to maximize efficiency.

Practically none of these have any use at home, because they are too dangerous, too expensive, and just outright unnecessary. A soapy scrub will do just fine cleaning anything in 99% of cases, even if water still beads.

1 - Ultrasonication in pure solvents or detergent solutions.

Often, the main impurities on the surface of glass are organic substances which are present at all times in air outside of strictly clean environments, and which strongly adhere to the glass when they come into contact. By dipping them into a solution with some affinity for the organic substances, it is possible to strip them off. Common options are regular dishwashing liquid (usually sulphonic acids with large alkyl chains), solutions of quaternary ammonium salts, aqueous ethanolanime, or pure deionized water, ethanol, isopropanol, acetone, among many others.

It is much more effective, however, to couple the aforementioned compounds with ultrasonication rather than simply dipping or scrubbing. In an ultrasonicator, the liquid is vibrated at frequencies higher than $10\ kHz$. Though macroscopically the liquid doesn't seem to change much, on the microscopic level there is chaos. The high frequency vibrations locally produce very fast solvent jets and cavitation bubbles, which effectively sandblast anything closeby. Glass manages to stay mostly unaffected due to its stiffness and strong bonding, but anything not very firmly attached to its surface will be flung away.

2 - Prolonged exposure to high temperatures

Common glass is thermally stable and can be heated strongly before it softens, so it can take temperatures at which many other substance decompose, evaporate or burn (if exposed to oxygen). Different types of glass have different maximum temperatures depending on their composition. Also, the thermal expansion while heating and contraction while cooling can be problematic for delicate glassware.

3 - Soak in base bath

Strong bases react with many substances and help dissolve them off the surface of the glass. However, while very inert, glass itself will also slowly react with concentrated basic solutions at room temperature. This also helps clean because if very persistent stains manage to survive the basic conditions, the glass surrounding them will begin dissolving slightly, and after a while the stain ends up detaching from the substrate as the glass behind it will have dissolved away. The regular options are concentrated $\ce{NaOH}$ or $\ce{KOH}$ in water or ethanol.

4 - Soak in highly oxidizing solutions

Glass is composted mostly of silicon already in its most oxidized form ($\ce{SiO2}$). Hence exposure to very oxidizing conditions will not damage the surface of the glass, while anything else on it is likely to dissolve, be it organic or inorganic matter. Classic options for dipping the glass are acidic piranha solutions (a mixture of pure $\ce{H2SO4}$ and concentrated $\ce{H2O2}$), basic piranha solutions (concentrated aqueous $\ce{NH4OH}$ and concentrated $\ce{H2O2}$), sulphocromic solution ($\ce{H2SO4}$ + $\ce{H2Cr2O7}$, in disuse due to the toxicity of $\ce{Cr^{6+}}$) and sulphonitric solution ($\ce{H2SO4}$ + $\ce{HNO3}$). These are all pretty heavy-duty mixtures. For example, here's why piranha solutions earn their name.

5 - Treatment with ozone/oxygen plasma

This is a step up from even the piranha solutions. If all you want to do is oxidize organic matter on the surface of the glass, then why not cut out the middleman and directly bathe the glass in reactive oxygen? The glass sample is first put into a closed chamber containing oxygen (air works, but pure oxygen is better). If ozone is being used, then the chamber is lit from the inside by a mercury lamp, which produces lots of ultraviolet light. The light is capable of dissociating oxygen molecules and producing ozone, which then is directly exposed to whatever is on the surface of the glass. In the case of oxygen plasma, the chamber is pumped down to a low pressure of pure oxygen gas before a coil produces strongly varying electric and magnetic fields inside the chamber, generating all sorts of oxygen species such as singlet oxygen (an electronically excited form of standard, triplet oxygen), free oxygen atoms, and several ions and radicals. Here's a nice video on how this process works in practice.

After all the work you put into cleaning a glass surface, it will immediately start adsorbing impurities if outside a vacuum, and it may take as little as a few minutes before it is too dirty to use in very sensitive and specialized applications. I have found that water already beads significantly again less than 10 minutes after the final treatment.


Formation of carbonic acid from breath? Is it possible that blowing on water (hot/cold) can cause a reaction between the $\ce{CO2}$ from expiration and the water, to form $\ce{H2CO3}$? Is $\ce{H2CO3}$ dangerous?

It is possible that carbonic acid will form; however, only in small amounts because the equilibrium of the hydration of $\ce{CO2}$ to $\ce{H2CO3}$ lies mainly on the educt side. As more $\ce{CO2}$ is soluble in cold water, more carbonic acid will be present in the solution at lower temperatures.

$$\ce{CO2 + H2O \rightleftharpoons H2CO3}$$

At $25~\mathrm{^\circ C}$ and in pure water, the quotient of the concentrations is $[\ce{H2CO3}]/[\ce{CO2}] \approx 1.7 \times 10^{-3}$, so most carbon dioxide remains physically dissolved in water. This makes carbonic acid - despite a $\mathrm{p}K_\mathrm{a}$ of $3.6$ for the first dissociation step - a fairly weak acid. It is not a dangerous compound, otherwise it would not be used in carbonated drinks. It is also a natural constituent of blood, where it acts as a pH buffer. In red blood cells, it is produced as an intermediate by the enzyme carbonic anhydrase, which catalyzes the hydration of carbon dioxide.


Why does pool chlorine smell only when wet? I'm in the swimming pool a lot since I'm on the swim team and I have noticed that if I don't shower after getting out of the pool and drying off, the typical "chlorinated pool" smell will mostly go away. But, if I decide to shower later that night, the chlorine smell will come back very strong when my skin and hair gets wet again from the water in the shower. My swim team friends and I even found this same thing to happen just from licking your arm. Why does water seem to activate the smell?

According to americanchemistry.com:

  Pool smell is due, not to chlorine, but to chloramines, chemical
  compounds that build up in pool water when it is improperly treated.
  Chloramines result from the combination of two ingredients:  (a)
  chlorine disinfectants and (b) perspiration, oils and urine that enter
  pools on the bodies of swimmers. Chlorine disinfectants are added to
  pool water to destroy germs that can give swimmers diarrhea, ear aches
  and athlete's foot. Perspiration, oils and urine, however, are
  unwanted additions to pool water. By showering before entering the
  pool, and washing these substances from the skin, swimmers can help
  minimize pool smell.   When chlorine disinfectants are added to water,
  two chemicals are unleashed that destroy waterborne germs:
  hypochlorous acid, $\ce{HOCl}$, and hypochlorite ion, $\ce{OCl-}$.

Some chemical reactions that then lead to the formation of chloramines include:

$$\ce{HOCl + NH3 -> NH2Cl + H2O}$$
$$\ce{HOCl + NH2Cl -> NHCl2 + H2O}$$
$$\ce{HOCl + NHCl2 -> NCl3 + H2O}$$

These are fairly volatile compounds, but can also, to different degrees, "stick" to your skin.  When you take a hot shower, these are then liberated and you can smell it again.


Why might two different chemical structures smell the same? From what I understand, it is the functional group attached to a structure that determines the odor it will give off, for the most part. Which explains why two similar molecules, except for their functional groups, would smell different. So, would that make it so that two different molecules with the same functional groups would smell the same?

The molecular determinants of odor perception is a disputed area in fragrance biochemistry.  There are two basic theories: the "shape" hypothesis and the "vibrations" hypothesis.  Luca Turin is the foremost exponent of the more controversial and less widespread "vibrations" hypothesis, but he does a good job of explaining both theories in his TED talk.

According to the vibrational theory, since different functional groups have different vibrational modes, they should smell quite different.  An example from Luca's talk is a complex alcohol vs. the corresponding thiol.

According to the shape theory, molecules which have similar but not identical functional groups should smell similarly, unless the protein receptors in our nose that sense odors have very exquisite shape sensitivity, enough to perceive the very slight difference in the shape of thiol vs. an alcohol in a large otherwise identical molecule.

  Why might two different chemical structures smell the same?

According to one theory, they would smell the same if they had similar vibrational (i.e. infrared) spectra.  In 2011, PNAS published is a very intriguing result that showed this was true for fruit flies in the case of some odorants.  They changed the vibrations by isotopic substitutions, a method that should leave essentially unchanged molecule's shape and its chemical reactivity.

  So, would that make it so that two different molecules with the same functional groups would smell the same?

In my mind this is true if you believe the vibrational theory as well.  But it is harder to understand why it is true if you believe the shape theory.

Of course, the real answer may be a combination of both theories.  I'm a tepid supporter of the vibration theory myself, but I would be very interested in a chem.se answer from another expert who is less impressed with Dr. Turin's theories than me.


Manually changing symmetry I have two crystal structures, one in R3m and one in Cm symmetry. Both structures also differ in stoichiometric ratios, $ABC_4$ and $A_2B_2C_8$, respectively. Both structures look nigh on identical, with the exception of double the amount of atoms in one than the other.
How would I go about manually changing the symmetry of the Cm structure to R3m symmetry? I normally use VESTA as a visualization tool, and Avogadro as a tool for imposing symmetry.

Your question is hard to answer because it lacks a lot of information.  $Cm$ (IT #8) belongs to the monoclinic which does not say anything about the threefold symmetry axes present in $R3m$ (IT #160).  But it is not unheard to transfer a monoclinic  model into one that belongs to the  trigonal / rhombohedral system.
Sadly, however, your question does not include a hint about the data and programs other than VESTA (for which I lack working knowledge) with you, nor your previous exposure to crystallography.  Of course the best were to start all over with reducing the diffraction data to write a new .hkl file, now with the hypothesis of higher symmetry present, to solve and refine the model again.  However, if you have a) a modern .cif file (those now including the hkl data and even better a good portion of the ShelX .res file), or b) one of the older .cif file plus the structure factors than you could give Platon's addsyminstruction a try. This routine was just written for the purpose of identifying missed symmetries.  It will write you a new .hkl, for lattice constants already adjusted, and an .ins file (with Laue symmetry updated accordingly) to start over with ShelX or Olex2.
By today's standards, Platon neither truly is CLI, nor GUI (mainly a large FORTRAN 77 executable + X-window; executables available Linux, Mac and Windows), but well documented in publications by Spek himself including many worked test data.  For addsym the page in question is here.  Don't forget to check the model if it still is chemically sensible (bond lengths / angles, ADP ellipsoids, void volumes) and significantly better, e.g. by Hamilton's R-factor ratio test.


Nernst equation: half cell vs overall reaction Can the Nernst equation always be applied to the overall redox reaction as opposed to only half cell reactions?

For example, in the following silver-silver chloride electrode: $$\ce{Ag (s)|AgCl (s)|Cl- (aq)}$$ rather than applying the Nernst eqution to the half cell reaction $$\ce{AgCl + e- -> Ag + Cl-}$$ can I apply it to $$\ce{AgCl -> Ag+ + Cl-}$$

The answer is the same for the question I was trying to solve (whether I use the overall reaction or the half-cell reaction), but I was wondering if this was just a coincidence or if it was the same.

If they are the same, are there any other factors I should consider such as whether or not the electrolyte fully dissociates in the given conditions?

It is coincidence, since
$$\ce{AgCl -> Ag+ + Cl-}$$
is not a half cell reaction.

The silver chloride electrode functions as a redox electrode and the reaction is between the silver metal (Ag) and its salt — silver chloride (AgCl). There are two ways to represent the processes (half reactions) happening on the electrode:
$$\ce{AgCl + e- -> Ag + Cl-}$$
or
$$\ce{Ag+ + e- -> Ag}$$
$$\ce{AgCl -> Ag+ + Cl-}$$

And you should be applying Nernst equation to one of the two half cell reactions shown above.

The reason why you got the same result/answer is because the Nernst equation for silver chloride electrode has the dependence on the activity or effective concentration of only chloride-ions.

$$E = E^0 - \frac{\mathrm RT}{\mathrm F}\ln(a(\ce{Cl^-}))$$


Why does the equilibrium constant depend on the temperature but not on pressure and concentration? According to my textbook (and intuitively) certain changes when the aforementioned 3 variables are altered occur in accordance with Le Chatelier's Principle. However, what I don't understand is what makes temperature change alter the equilibrium constant while the other two do not. All three cause changes in the equilibrium position (the state of concentrations at which equilibrium is achieved with lowest possible stress in the system), so why don't they all also change $K_{eq}$?

Please have a look at this answer of mine. It contains the derivation of the defining formula of the equilibrium constant.

\begin{equation}
\log \underbrace{\prod_i [a_{i}]^{\nu_{i}}}_{= \, K} = -\frac{\Delta G^{0}}{RT} \qquad \Rightarrow \qquad \log K = -\frac{\Delta G^{0}}{RT} \ .
\end{equation}

This formula shows the temperature dependence explicitely, so it is clear that $K$ must depend on $T$.

Apart from the temperature this formula also relates $K$ to the standard Gibbs free energy of reaction $\Delta G^{0}$.
This quantity is defined for specific standard conditions involving standard pressure and equilibrium concentrations.
Since there aren't any other quantities contained in the formula (apart from the constant $R$) it follows that $K$ is defined for this specific pressure and these specific concentrations too and so it has no pressure or concentration dependence.


Infrared spectroscopy of substances on a OHP film How can we record a IR spectrum of a substance coated on a Over Head Projector (OHP) sheet?
instrument has both transmission and ATR modes. IR spectrum is to be recorded for both the OHP sheet as well as the material that is coated on it

Your are going to analyze both, the OHP foil alone, and the part of the OHP foil that is covered by the coating and calculate the difference of the later to get the data of your coating material.
If you do want to do the classical way (transmission IR), just cut-off a little of the sample (a hole puncher as in the office may do the job, too), attempt to grind it with KBr and press the pellet as normal.

If you want to stick to ATR, place the foil on the dish without prior sample preparation.  This is quicker, and to some degree allows a space-resolved analysis, too -- the window (center dish of the table) of a single reflection geometry instrument (like the one below, Nicolet iS10 with smart-ATR-IR module) is of about 3 mm in diameter).  The window material of the ATR module may be Germanium (as here), yet there are instruments with a window of Diamond, too.


The science behind fertilizers Do you know some books or articles that elaborate on the chemistry behind fertilizers?

I have taken organic chemistry so far, and  inorganic chemistry as well. I am also familiar with the Haber-Bosch process.

Asides from a general search on Google Scholar, where, unfortunately, many resources are behind paywalls, here are some book references that may be of use for your broad-ranging request:

Methods in Agricultural Chemical Analysis: A Practical Handbook (Google eBook) (Faithfull, 2002) - some of this is available online through Google Books

Fertilizer Manual (UNIDO, 1998) - this, according to its contents has some information about the chemistry of frtilizers.

Chemistry For Agriculture And Ecology (Mido and Satake, 1994), A lot of this book goes back to basics, but chapter 18 is specifically about the chemistry of fertilizers.

Agricultural Chemicals and the Environment (Hester and Harrison, 1996) - once again, has some information about the chemical processes of fertilizers, with a focus on environmental concerns.

I hope these are useful.


Paramagnetic relaxation in 13C nmr I'm trying to make my 13C{1H} (hydrogen decoupled) nmr more quantitative. From the literature I've noticed that researchers either

increase the delay time in the experiment (which I can't do, as it takes too long to collect the spectra)
run the decoupling pulse only during data acquisition (to reduce the nuclear Overhauser effect), and/or
add a paramagnetic relaxation agent to the sample.

I want to try a paramagnetic relaxation agent (I'm working with $\ce{CDCl3}$ as a solvent).  The most common one seems to be $\ce{Cr(acac)_3}$.  Are there better relaxation agents out there (i.e. more effective at lower concentration)?  Also, the literature I've read shows all sorts of concentrations of the relaxation agent... would a low (0.01 M or less) concentration work, in your experience? Do these agents lead to too much spread in the peaks?

Also, will using a relaxation agent ruin the nmr tube for further use?

Paramagnetic relaxation in $\mathrm {~^{13}C}$ NMR:

Quantitative $\mathrm {~^{13}C}$ NMR requires full relaxation of all carbons at each scan.  Adding paramagnetic relaxation reagents could speed up the relaxation.  The most common reagent you can find online is chromium acetylacetonate, $\ce{Cr(acac)3}$. However, the amount you need to add to your solution is very critical. It is tricky that if your added amount to your solution is too little, $\mathrm T_1$ of the $\mathrm {~^{13}C}$ will not be short enough. I you added too much of $\ce{Cr(acac)3}$, the signal will be too much broadened due to too much shortening of $\mathrm T_2$, thus you will lose resolution. Typically used recommended amount by experts is Between $\mathrm{0.1wt\%}$ and $\mathrm{0.5wt\%}$ (https://blogs.umass.edu/weiguoh/?p=71). With that amount, the solution will have a light purple color [1] (courtesy of https://blogs.umass.edu/weiguoh/?p=71).

Above picture is a picture of a standard sample with $\mathrm{0.5wt\%}$ $\ce{Cr(acac)3}$, which is an excellent reference for you. It'd provide you the easiest way to avoid the mess of weighing $\ce{Cr(acac)3}$. What you have to do is to look at the color of your solution.  Following the color of the picture, you can add $\ce{Cr(acac)3}$ a little by little so that your sample has a similar or slightly lighter color than this.

To read about othe reagents download this thesis (available online): https://www.diva-portal.org/smash/get/diva2:793434/FULLTEXT01.pdf


Dichloromethane solubility in water I heard dichloromethane is soluble in water 1:50, but I also heard it is not because it can't form hydrogen bonds. I'm confused. Is there a way to extract it? Would putting the water to $40~\mathrm{^\circ C}$ make it evaporate trough water?

The solubility of dichloromethane in water (as well as the solubility of water in dichloromethane) depends on temperature.
At a temperature of $T=20\ \mathrm{^\circ C}$, the solubility amounts to $17.6\ \mathrm g$ dichloromethane per $1\ \mathrm{kg}$ water; i.e. the mass fraction of dichloromethane is $w_{\ce{C2H2Cl2}}=0.0173$ or $w_{\ce{C2H2Cl2}}=1.73\,\%$.
[source: “Aqueous Solubility and Henry’s Law Constants of Organic Compounds”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.]
(See here for values at other temperatures.)

You cannot simply evaporate pure dichloromethane (boiling point: $40\ \mathrm{^\circ C}$) from such mixtures since dichloromethane and water form an azeotrope, which boils at $38.8\ \mathrm{^\circ C}$. The mass fraction of water in this azeotrope is about $w_{\ce{H2O}}=1\,\%$.


Why does camphor sublimate? I was doing a laboratory experiment the other day and they asked us to heat a beaker that contained camphor and to cover it with a watch glass with ice on it.

I understand the crystallization process and I also know that the melting point of camphor is 149 C, so it is never reached during the experiment, but I do not get the sublimation process. How can the vapor pressure increase until it reaches a value close enough to the atmospheric pressure to sublimate without reaching its melting point? I think I am missing out an important fact, but I don't know which.

How can the vapor pressure increase until it reaches a value close enough to the atmospheric pressure to sublimate without reaching its melting point?

Compounds can sublimate even if the vapor pressure is much lower than the atmospheric pressure. This is how we can smell solids. If you put the solid substance in a closed container, the partial pressure of the substance in the gas phase will increase until it reaches the vapor pressure, and you will be at equilibrium. In an open system, it will continue to sublimate, with a concentration gradient in the gas phase (that is how the sense of smell is used to locate smelly objects).

  they asked us to heat a beaker that contained camphor and to cover it with a watch glass with ice on it.

In your experimental setup, you never reach an equilibrium. The vapor pressure of the camphor in the beaker is higher than the vapor pressure of the camphor deposited on the cold watch glass. Accordingly, camphor will continue to sublimate from the solid on the bottom and continue to deposit on the camphor on the watch glass.

  This seems weird. Does this happen to other substances?

Yes, you can see the snow sublimate in the winter when temperatures are below the freezing point of water. A dusting of snow on a drive way (or ice on a windshield) will "disappear" (sublimate) without liquid water present at any time. When food is freeze-dried, the same process is used. The key for this to work is that the air has to be dry (i.e. the partial pressure of water in the air has to be lower than the vapor pressure of ice).


Law of Definite Proportion: Do the numbers indicating moles count? I need to find the ratio of a substance as it converts from a bicarbonate to a carbonate:
ex:

$\ce{Fe(HCO3)2 -> FeCO3 + CO2 + H2O}$
$\ce{Fe(HCO3)2} / \ce{FeCO3}$ = $3/2$ *using molar masses

But what if the substances aren't  the same number of moles?
ex:

$\ce{2Fe(HCO3)2 -> FeCO3 + CO2 + H2O}$

Would the ratio be:

$\ce{2Fe(HCO3)2 / FeCO3} = x$

or still:

$\ce{Fe(HCO3)2} / \ce{FeCO3} = 3/2$

How do moles factor into this equation and why?

The stoichiometric coefficients are important. They make sure you have a valid relationship between reactants and products that obeys the Law of Conservation of Mass.

As Jerry hints at in his comment, there is another fundamental law that you need to consider: the Law of Conservation of Mass. Matter cannot be created or destroyed (outside of nuclear reactions and relativist physics). In chemistry this means you need the same number of atoms of each element on both sides of the equation. You cannot create additional atoms from nothing, nor can extra atoms of something vanish.

Consider your first equation:
$$\ce{Fe(HCO3)2 -> FeCO3 + CO2 + H2O}$$

This equation is balanced - it obeys the law of conservation of mass. Each side of the equation has one $\ce{Fe}$, two $\ce{H}$, two $\ce{C}$, and six $\ce{O}$ atoms. This is a valid chemical equation representing a real reaction relationship.

Your second equation is not balanced.

$$\ce{2Fe(HCO3)2 -> FeCO3 + CO2 + H2O}$$

There are now twice as many atoms of each element on the reactant side of the equation as on the product side. What happens to this extra atoms? They cannot vanish! This is not a valid chemical relationship. Balancing this equation generates a scalar multiple of your first equation, which will yield the same ratio of iron (ii) bicarbonate to iron (ii) carbonate.

$$\ce{2Fe(HCO3)2 -> 2FeCO3 + 2CO2 + 2H2O}$$

The Law of Definite Proportions tells you that chemical compounds always contain the same elemental compositions by mass. I am not sure how that is relevant to the ratio of product to reactant in your stoichiometry problem. In other words, it tells you that iron (ii) carbonate is always $\ce{FeCO3}$ and never $\ce{Fe3(CO3)4}$ or $\ce{Fe2(CO3)3}$ or $\ce{FeCO4}$ or anything else. Those other formulas may refer to real compounds (the first two are real), but they are not iron (ii) carbonate.


Are all carbohydrates a subset of alcohols? According to the definition,

  an alcohol is any organic compound in which the hydroxyl functional
  group (-OH) is bound to a saturated carbon atom.

Monosaccharides such as glucose contain several hydroxyl groups. Does that mean all carbohydrates are a subset of alcohols?

Whenever you see -OH is bound to a saturated carbon atom in organic chemistry it is Alcohol group.

So answer is YES. All carbohydrates are  subset of alcohols.

Though in linear forms they also have aldehyde or ketone group.