text:
Need to find series of equation products from given reagents
I can balance equations but only if I already know both the sides. I am trying to find the following products:
$$\begin{aligned}
\ce{Na2S2O5 + O2 ->}\ ?\\
\ce{Na2S2O5 + KNO3 ->}\ ?\\
\ce{Na2S2O5 + KNO3 + Fe2O3 ->}\ ?\\
\ce{Na2S2O5 + KNO3 + C12H22O11 ->}\ ?\\
\ce{KNO3 + C12H22O11 ->}\ ?\\
\ce{KNO3 + C12H22O11 + Fe2O3 ->}\ ?
\end{aligned}$$
I'm guessing that the sodium metabisulfate will produce sulfur dioxide and sodium oxide, but that would just be plain decomposition, so I'm not sure where the oxygen would fit. I know that $\ce{KNO3 + C12H22O11}$ produces carbon dioxide and water, but I'm not sure where the nitrogen and potassium go.
Of course once I know the products, balancing and calculating specific energy is simple, if a touch laborious.

Welcome to Chemistry.SE. For homework-type questions, we want to see what your efforts have been so that we can better help you.

This is the first step of a rather lengthy project/problem. I'll need to balance the equations, calculate bond energies, solve for specific energy, and go on from there. I just need the reaction products as a first step.

Then share what you can in your question.

Most (all?) of these seem like redox reactions.

Yes, they are all redox reactions. I'm just very, very bad at chemistry.

I will assume that all of these reactions are happening in solution. Since we've established that all these reactions are redox reactions, we know the general structure of each reaction--- $$\text{reducing agent} + \text{oxidizing agent} \to \text{oxidized thing} + \text{reduced thing}$$ ---and we generally won't have to worry about other things (like what you've mentioned in your edit) happening. I will work out one reaction, and leave you with the rest. Let's look at reaction 4.
We can ignore $\ce{Na+}$ and $\ce{K+}$, because we know from prior experience that these are relatively inert spectator ions that won't participate in most reactions. Then we have the three species $\ce{S2O5^2-}$, $\ce{NO3^-}$, and $\ce{C12H22O11}$.

$\ce{N}$ has its maximum oxidation state of $+5$ in $\ce{NO3^-}$, so we expect that $\ce{NO3^-}$ will be a good oxidizing agent. $\ce{NO3^-}$ is commonly reduced to $\ce{NO2^-}$ in solution.
On the other hand,  $\ce{S}$ has an oxidation state of $+4$ in $\ce{S2O5^2-}$ and a maximum oxidation state of $+6$, so we might expect $\ce{S2O5^2-}$ to be a reducing agent. Fully oxidized $\ce{S}$ is usually present as $\ce{SO4^2-}$ in solution.
Finally, $\ce{C}$ also has its maximum oxidation state of $\ce{+4}$ in $\ce{C12H22O11}$, so $\ce{C12H22O11}$ is likely to also be a reducing agent. Fully oxidized $\ce{C}$ is commonly present as $\ce{CO2}$.

This gets us the three half-reactions
$$\begin{aligned} \ce{NO3^- + H2O + 2e- &-> NO2^- + 2OH-}\\ \ce{S2O5^2- + 6OH- &-> 2SO4^2- + 3H2O + 4e^-}\\ \ce{C12H22O11 + 48OH- &-> 12CO2 + 35H2O + 48e-} \end{aligned}$$ where we've also assumed basic conditions and hence added $\ce{H2O}$ and $\ce{OH-}$ as required to balance out charges and atoms on both sides of each equation. Now we simply need to take linear combinations of these half-reactions to cancel out electrons on both sides, and we're done.

In response to your comment, everyone starts out bad at chemistry. You'll get better.

the result really depends on reaction conditions. These products are believable:
Na2S2O5 + O2 -> Na2SO4 + SO2
Na2S2O5 + KNO3 -> Na2SO4 + KNO2 +NO2
Na2S2O5 + KNO3 + Fe2O3 -> Na2SO4 + KNO2 + [Fe(NO3)3 + Fe(NO2)3]
This one makes little sense. Two reducing agents and one oxydizer. Try making this:
Na2S2O5 + KNO3 + C12H22O11 -> Na2SO4 + K2SO4 + CO2 + H2O
KNO3 + C12H22O11 -> K2CO3 + N2 + CO2 + H2O
KNO3 + C12H22O11 + Fe2O3-> K2CO3 + N2 + CO2 + H2O + FeO


text:
Does >2e contribution of an atomic orbital (AO) to a hybrid AO of a molecule violate Pauli exclusion?
This question is inspired by On the Role of d Orbital Hybridization in the Chemistry Curriculum J. Chem. Educ., 2007, 84 (5), p 783.
The article calculates 6 bonding sulfur atomic orbitals in SF6 as each having a 0.768s term (s meaning 3s). This implies $6\times0.768^2 = 3.5$ of the 6 valence electrons corresponding to 3s character.
Is it valid to say there is that much 3s character?

I would agree that this looks strange, but without details of the calculations it is difficult to understand what all this means. Suddenly, "Readers requiring further details concerning theoretical methods are asked to contact the author directly." Without details we could only speculate...

Just a random thought: IIRC, a bonding orbital does not necessarily implies a filled one. So, while there might exist 6 such hybrid orbitals, they should not necessarily be all filled (with 6 electrons).

@Wildcat later the article does say "Thus, the six bonds
in SF6 can be considered to be made up of four bonding orbitals
and two nonbonding orbitals located primarily on the
fluorines resulting in a bond order of 2/3 for each SF bond", although even 2/3 of 3.5 is greater than 2.

I also agree that this look incredibly strange, but I need to look at the article which I cannot currently do. It would be interesting to see what the rest of the orbitals look like. What happened to the p orbitals? And how did they calculate the value.

@Martin-マーチン The 6 sulfur hybrid atomic orbitals (disregarding the small 3d terms) are: 0.768s +0.636pz, 0.768s -0.636pz, 0.768s +0.636py, 0.768s -0.636py, 0.768s +0.636px, 0.768s -0.636px.

The answer to your question is yes, such a contribution violates the Pauli principle, and is hence not a meaningful wave function.
The demonstration of this calculation (paper) aimed at a different statement and is unfortunately very thin on the calculation details. Furthermore it is a bit difficult to distinguish between the different models they use.
However, I believe the key is actually given in the first sentence of the "Results and Discussion" section (see emphasis below). As far as I understand it, the formula presented in equations (2) are not normalised. Furthermore it neglects any ionic contribution and/or occupation numbers. This further manifests itself that the structure is about 600 kcal/mol higher in energy than the resonance hybrid.

When the $\ce{3s}$, $\ce{3_{$px,y,z$}}$, and $\ce{3d_{$x^2−y^2$}}$ sulfur AOs are free to mix in any proportion to form six bonding orbitals on the central sulfur, the resulting energy is 1305.8 kcal mol1 (entry 1,
  Table 1) with sulfur orbital mixing as described by [..., equation (2)].
  John Morrison Galbraith, J. Chem. Educ. 2007, 84 (5), 783-787. (emphasis mine)

While the wave function constructed in equation (3) seems to obey the criteria of a physically meaningful wave function,  but it produces an energy even higher.
After that they suddenly jump into the MO picture of the argument, but not really explaining that. In this description one can clearly see in figure 3, that the s orbital only contributes to one molecular orbital.

While I completely agree with the basic premise of the paper, from my point of view it completely fails to present a compelling argument. For example: how did they come up with an energy of >300 kcal/mol of $\ce{SF6}$ relative to the $\ce{S + 6F}$ fragmentation? (And who uses Gaussian98 in 2007?)
I personally would rather recommend our own discussion:

Why do compounds like SF6 and SF4 exist but SH6 and SH4 don't?

Last but not least I would like to recommend some educational papers that I find more enlightening on the matter (no order):

Ronald J Gillespie, Bernard Silvi, Coord. Chem. Rev. 2002, 233-234, 53-62. The octet rule and hypervalence: two misunderstood concepts;
Ronald J. Gillespie, Edward A. Robinson, Inorg. Chem. 1995, 34 (4), 978–979. Hypervalence and the Octet Rule;
Thomas A. Halgren, Leo D. Brown, Daniel A. Kleier, William N. Lipscomb, J. Am. Chem. Soc. 1977, 99 (21), 6793–6806. Apparent octet rule violations, fractionality, and the interpretation of localized molecular orbital structures. Polarization and hybridization functions in chemical bonding.

Martin, what do you mean by "formula presented in equations (2) are not normalized".  (0.768)^2 + (0.635)^2 + (0.079)^2 = 0.999.  The sum of the squares of the coefficients are 1 for each of the 6 equations.  Do you mean something else by normalized?

@DavePhD at the time I wrote it, it made perfect sense to me. I can't puzzle it together on my phone now and I'm going on vacation soon. I have another go at it soon.

Thanks Martin, enjoy your vacation.  Maybe you could eventually add more about in principle whether or not a linear combination of atomic orbitals (as an approximation of the actual best orbital approximation of the wavefunction) is strictly limited to a 2 electron contribution from a single atomic orbital.  Does the fact that the quantum numbers for the atomic orbitals are not actual quantum numbers of the molecule loosen the requirement at all?

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text:
Number of CO2 in pyrolysis
In the following pyrolysis reaction, we have to find how many $\ce{CO2}$ is formed.
According to me it should be 5 as there are five carboxylic acid.
But the answer is 4, how?

Tip: the presence of a carbonyl group in beta position to a carboxyl facilitates decarboxylation. Try to remove all carboxyl groups (one at a time) which fulfill that condition.

@vapid so there would be three carboxyl group

A carboxyl group also includes carbonyl.

@vapid yeah i know that

@vapid Please elaborate your answer . Am I coorect that there are 3 carboxyl group for decarboxylation

While it would be possible to decarboxylate almost any carboxylic acid with appropriate conditions, simple aliphatic acids are not readily decarboxylated. Carbonyl group in beta position facilitate decarboxylation due to formation of a cyclic transition state:
Now, your example has five carboxyl group, and each has a carbonyl in beta position (carboxyl also counts as a carbonyl compound). Let's look at the picture:

First we can remove the three top carboxyls marked with arrows (A). There is a carbonyl in the ring that will aid in the decarboxylation, so we will arrive to a compound at B) (I'm not saying that the reaction will proceed that way, it is just an explanation). As I said before, one carboxyl group can facilitate the removal of the other carboxyl. But once we remove one group, we get compound C). That compound does not have any groups that would help to get rid of the last carboxyl. So there is your answer.

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text:
Why is the value of "a" is positive instead of negative in Van der Waals equation?
Higher value of a means higher atttraction force between molecules. Attraction force causes the pressure to be less than expected. Now the actual pressure becomes $P + \frac{n^2a}{V^2}$ .
Which means if a increases the neat pressure increases too. But isn't it the other way around?
(I assume here P is the pressure exerted on the wall of the gas container considering there is no intermolecular force)

You understand the chemistry: intermolecular attraction forces cause the pressure to be less than expected. The problem here is with the math. Though unintuitive at first sight, a positive $a$ leads to a negative correction to pressure. Let's look at a numerical example.

Assume that the pressure inside a certain container is $50$ atm with the ideal gas approximation and $45$ atm with the van der Waals equation of state. Then $P_\text{ideal} = P_\text{vdW} + 5$.

In other words, the corrected pressure is $P_\text{vdW}$, not $P_\text{vdW} + a\frac{n^2}{V^2}$.

The formula is good because it is correct in most of the regions and the parameters make sense. It is attractive because it both describes behavior of gas and approximate behavior of a liquid.
This formula has a problem in a region that corresponds to phase transition (boiling of a liquid). To account for that region correctly you need a function that is constant within some range and changes within other ranges.
Complex analysis (a branch of mathematical analysis) has a theorem that proves that you cannot have an analytical function that is constant on one interval and changing over the other intervals.
This sort of functions exist, but they are not analytic and behave poorly (hard to work with).
As a free bonus it turns out that the "strange" region describes behavior of overheated liquid and overcooled gas. Those two have different behavior at the same temperature.

I downvoted this answer because it is irrelevant to the posed question.


text:
Why does the pH of soft drinks get lower as they go flat?
I am doing an investigation about the chemistry of soft drinks and as part of this I observed the effect of time on the pH of soft drinks fully expecting the pH to increase as the carbonic acid, a weak acid, decomposed and CO2 escaped. Instead, time after time, the pH decreased, not by much though in fact it hardly made a difference to the acidity of the drinks but it the pH still decreased every time.
So I did a control experiment with soda water and found that the pH did increase when the CO2 escaped the drink. I originally did it with Coke and Pepsi, which aside from carbonic acid also have phosphoric acid (Pepsi has citric acid as well) so the increase in acidity probably has to do with an interaction between the the weak carbonic acid and the strong phosphoric/citric acid. Can anyone help me to explain this?

What sort of soft drinks and what sort of soda water have you measured ?

I think you said it... phosphoric acid. Carbonic acid is "buffering" phosphoric acid... test seltzer water (composed of water and carbon dioxide with no other additives).
Another way of thinking about this is to consider carbonic acid (nominally a weak acid) to actually be a base. Looking at the pKa (the pH at which half of the acid is dissociated) is the best way to describe this.
The pKa of Carbonic acid is 6.35
The pKa Phosphoric acid is 2.16
pH is basically a measure of free (dissociated) H+ in water... so as CO2 is removed, it is not available to act like a buffer and bind with H+... so the free H+ increases.

Thanks a lot. I'll have to read up on buffers I guess. We are in the middle of an acids and bases unit at the time of our summer break and are going to learn about buffers in the first few weeks of the new year, which isn't very practical as the draft for this paper is due after the first week.

I am a little fuzzy about this.  While the pKa of CO2 seems high enough, for it to act as a base either CO2 or H2CO3 would have to accept a proton, so what counts is if the pKa of the resulting cation exceeds the soda pH.  Do we know that?

@OscarLanzi : you're right, the pH cannot possibly decrease if an acid ($CO_2$) is removed (and nothing else is being added or removed, e.g. the solution is not evaporating).

The observation is interesting. If you have two acids (in your case H2CO3 and H3PO4) and one of them (H2CO3 as CO2) leaves solution pH must increase.
If you just mix Na2HPO4/NaH2PO4/CO2/water and let CO2 evaporate pH will increase or (if solution is very acidic) stay unchanged.
Can you please write the method of measurement ("pH paper" or "pH electrode"), conditions ("wait for 1 h at 60C" or "shake vigorously at 20 C for 5 minutes") and the results ("pH dropped from 5 to 4.5 then to 4.0" or "pH was 6, didn't change for an hour, then dropped to 4.2 in 1 minute"). Is this diet or normal coke? Do you see the same effect in plastic bottle and in a glass cup?
Oxidation of sugar is a plausible cause, but I need more info to come up with a better hypothesis.

pH electrode. I did this at a range of temperatures ranging from ice cold to 40°C (10°C intervals) with two trials for each temperature for both Coke and Pepsi, leaving the drinks alone for 50 minutes whilst controlling the temperature of the ice/water baths. To make sure these results were legitimate and not just faulty apparatus I did an informal experiment where I used a magnetic stirrer to shake the drinks and the pH still went down. Basically, over the course of 50 minutes, I typically saw falls from 3.0 pH to ~2.75 pH, numerically slight falls but significant. I used normal Coke+Pepsi.

Continuing on from last comment (reached character max) --- Coke and Pepsi both aluminium canned. If it also helps, I live in South East Asia so the sugar in the drink would most likely have been glucose and not high fructose corn syrup, they use the cheapest ingredient regionally. Also when doing research I came across a science fair paper saying that it didn't matter whether the drink was bottled, canned, etc... Not the most reliable source but...

And when I did the magnetic stirrer experiment it was at room temp. With the pH electrodes, they did observe distilled water to be a little bit alkali (6.15) and slowly drifting towards a little acidic (~6.9) but my chem teacher said it should've been around around pH 6. They were kept in the same beaker for weeks and I only bothered to change the water when mould started developing. Same calibration solution used for weeks and weeks as well. Even if the apparatus were not well kept, they aren't fully responsible for this observation.

And the sugar is probably sucrose, not glucose.

@1111 " (6.15) and slowly drifting towards a little acidic (~6.9)" Wait... 6.1 is slightly acidic. 6.9 is almost neutral. 8 is slightly alkali. But earlier you give correct numbers... from 3.0 pH to ~2.75 pH... Hmm, I can't explain this behavior and it looks strange to me... Good luck searching for the reason.

Sorry, I meant 7.15 drifting to 6.9. But thanks anyway

You can just use the three reversible reactions involved in the amount of carbon dioxide dissolved in a fizzy drink, use Le Chatelier's principle and solve for the changes in pH:
\begin{align}
\ce{CO2(g) &<=> CO2 (aq)}\tag{1}\\
\ce{CO2(aq) + H2O(l) &<=> H2CO3(aq)}\tag{2}\\
\ce{H2CO3(aq) &<=> H+(aq) + HCO3-(aq)}\tag{3}
\end{align}


text:
"Limit test" vs. "Titration" in a pharmaceutical guide
From the "Technical guide for the elaboration of monographs" by EDQM, edition 7, 2015, page 35:

Where the introduction of tests for foreign anions in organic substances is considered then a single one, either for chlorides, sulfates or − less commonly − nitrates, will usually suffice even when several could theoretically be present. The test is then to be carried out on the most abundant anion. When a test for chlorides is considered, up to 1000 ppm (0.10 %), a limit test should be used rather than titration.

I don't understand this. Isn't titration one of the possible ways to perform a limit test (to check whether the chloride content is within the specified limits)?
Maybe my understanding of "limit test" is flawed?

From what I know, rules for making a monograph are not about giving an exhaustive list of all the analyses possible. Guides describe international agreements about what is recommended and sufficient to identify a substance and its impurities.
A titration is considerably more tedious to perform and validate (see ICH Guideline). If you use titration in a monograph, people referring to it will also have to use this method. Unless you also included a limit test but, if you did so, why bothering carrying out titration? Overall, you would waste your time and, most importantly, others' time.
So the sentence "limit test should be used..." should be read "we agreed that a limit test gives enough information in this context and we prefer it over titration".
HTH

Thank you! So "titration" could not be called "a limit test"? Technically speaking, it could be used as a limit test, but it's too cumbersome?

Titration gives you a concentration (or a quantity) while the limit test gives a yes/no answer so they are different. Yes, you can use titration but only if you have proven that it is suitable for such use, i.e. you have determined the detection limit, quantitation limit, linearity... A limit test is much easier to validate.


text:
Why are the absorption and emission spectra mirror images of each other?
In one of my chemistry courses we learned that the absorption and emission spectra are mirror images of each other, but I can't rationalize why.
I know that the electronic transitions in emission spectra is not very different from that of absorption spectra, however the spectra is shifted slightly to the right due to the presence of non-radiative decays releasing some of the energy absorbed. I can see how similar energy transitions can account for a similar shape, but why is it a mirror image in most cases?

In so many words, "what goes up must come down, and what is down can go up, and by the same distance too".

Have a look at the Wikipedia page on the Franck-Condon principle here.

The sketch below explains how the mirror image spectrum comes about. (the figure is not at all to scale, its just illustrative, but does show the salient features.)

There are some conditions attached.
(a) mirror image spectrum is only observed in the condensed phase. This is because solvent collisions cause vibrational relaxation from the vibrational level excited and removes$^*$ energy to the solvent much faster than fluorescence can occur, thus all emission occurs from v=0. This vibrational relaxation occurs with a rate constant of $\approx 10^{13}$ s$^{-1}$.(This is not true in the vapour phase where if the pressure is low enough no collisions may occur within the excited state lifetime and fluorescence occurs from the level excited) The intensity of each transition (not wavelength) in absorption or emission is governed by the Frack-Condon Principle. The strength of a transition is proportional to the squared modulus of the two wavefunctions. i.e. $|\psi _{s0} \psi _{s1}  |^2$. In practice this means that overlap between vibrational levels in the two states is only large at the turning points of the wavefunctions, so transitions appear to be vertical. Classically this makes sense as momentum has to be conserved and this can only be achieved from vertical transitions starting and ending at the potential curve where kinetic energy is zero. In the quantum world the uncertainty principle limits change in position and momentum and so Franck -Condon Factors.
$^*$(The collisions preferentially remove energy as generally the excited state is at a higher 'temperature' than the surrounding solvent, i.e. it has more vibrational energy that the average $3k_BT/2$ of the solvent)
(b)The geometry of the excited state has to be similar to that of the ground state, thus mirror image spectra are observed for 'rigid'$^{**}$ molecules such as anthracene (3 benzene rings attached side by side) and not for flexible molecules such as cis or trans stilbene (phenyl=phenyl, where = is a double bond and this molecule isomerizes in the excited singlet state).

Relative displacement of potential energy surfaces (for the normal modes mode involved in the transitions) is not the only consideration, for the shape of the potential has to be similar in both states, thus the vibrational frequencies are similar and so gaps between vibrational peaks in absorption and fluorescence are similar. (Gaps between peaks in absorption give excited state vibrational frequencies and vice versa)
If the lowest triplet state has a similar potential to the singlet, which is usual if it derives from that singlet state (and not another higher singlet state) the the phosphorescence spectrum is very similar to the fluorescence, but shifter to longer wavelengths
$^{**}$ By 'rigid' is meant that the molecule has only its normal vibrational modes and no large torsional motion as is the case in stilbenes.


text:
Decarboxylation of Sodium Pyruvate
Can Sodium Pyruvate decarboxylate to acetic acid + CO2 in the presence of just moisture and heat?  If so, what kind of temperature is required?

I would expect acetaldehyde, not acetic acid.

As pointed out in the comments above, the expected product would be acetaldehyde, not acetic acid.

If you stop and think about the mechanism of any thermal decarboxylation, you immediately stumble upon the issue. The decarboxylation would lead to a d1 type synthon, which from organic chemistry you should realise is the wrong polarity for a carbonyl group (the natural polarity of the C=O bond is to have a partial positive charge on carbon and partial negative charge on oxygen).


text:
Confusions about oxidation and reduction
My teacher told me that elements get oxidised when:
1)There is a loss of electron
2) There is an addition of oxygen
3)There is a loss of hydrogen
And that they get reduced when:
1)There is a gain of electron
2) There is a loss of oxygen
3)There is an addition of hydrogen
Now I don't understand these things:
1) Most elements will lose an electron (and get oxidised) on the addition of oxygen due to its high electronegativity. But what about OF2? In this case shouldn't oxygen get oxidised and fluorine get reduced due to the higher electronegativity of fluorine? (I know its a covalent bond but I am talking in terms of on which side the electron will get attracted)
2) How does loss of hydrogen (or addition of hydrogen) oxidise (or reduce) an element? Like there are elements that have higher electronegativity than hydrogen then how will they get oxidised on the loss of hydrogen? They should get reduced because of the gain of an electron from hydrogen, right?

Fluorine is an exception, since it is more electronegative than oxygen, it gets reduced.

The formal definition of an oxidation/reduction is linked to the loss/gain of electrons. Considering the addition or loss of hydrogen and oxygen is not a global rule, but rather a trick generally used in organic chemistry. If you consider organic molecules, thus mainly composed of carbon, "adding" oxygen atoms suggests an oxidation of the carbon, as oxygen is more electronegative than carbon, while "adding" hydrogen atoms suggests a reduction, as hydrogen is less electronegative than carbon.
An example would be the successive oxidation of ethanol to ethanal, then acetic acid : $$\ce{CH3CH2OH ->[-2 e^-][``-2\text{ H}"] CH3CHO ->[-2e^-][``+1\text{ O}"] CH3COOH}$$

Your teacher is correct in these definitions. However, in many reactions there is no oxygen or hydrogen involved as such, so the loss/gain of electrons is the most general definition. It is important to understand that in one reaction a species may be oxidised but reduced in another, which is what you suggest from your question.
For example in the reaction  2Fe$^{3+}$+Zn= 2Fe$^{2+}$+Zn$^{2+}$ the zinc is oxidised to Zn$^{2+}$ and so reduces the Fe$^{3+}$ to Fe$^{2+}$
In the reaction I$_2$ + H$_2$SO$_3$ +H$_2$O= 2HI + H$_2$SO$_4$  the iodine acts as an oxidiser (so is reduced) to the strong reducer H$_2$SO$_3$, (which is oxidised) but in the reaction
I$_2$ +5Cl$_2$+6H$_2$O=HIO$_3$+10HCl  iodine acts as a reducer as chlorine is a much stronger oxidiser.
In complicated reactions like these it is useful to use oxidation numbers to work out what is oxidised and reduced.
If you find a textbook with a table of redox reactions you will see that species have different potentials which is how we know which are the strongest oxidisers (most oxidising Fluorine positive potential) and reducers (most reducing, lithium, negative potential). There are standard ways of working out whether a reaction will occur between oxidising and reducing species.
There is a description of calculating oxidation numbers in my answer to this question
Oxidation Number Of Phosphorus In PH3

But your answer doesn't actually answer my question. Does the addition of oxygen always oxidise an element? Then what about compounds like OF2  where oxygen is reducing fluorine. And how does the loss of hydrogen oxidise an element which is in a compound containing hydrogen?

I made some general comments which I thought would help you understand that some atoms can be either oxidising or reducing depending on their reaction partner. As fluorine is the most oxidising species it oxidises oxygen and is itself reduced. (It may help to think of oxygen as  some species X then statements like oxidising  oxygen seem less absurd). If you look at oxidation numbers, oxygen goes from 0 to +2, oxygen acts as reducer, and fluorine from 0 to -1,  acts as oxidiser.

So my teacher's statements are not always correct?

They are correct as far as I can see within the scope of the present knowledge of those that are taking your course. You will realise that as you progress things become more complicated, and hence more interesting, but in teaching its usually necessary to give a simplified version of things to get one started on a new subject.

But how does loss of hydrogen oxidise an element? Can you please give me an example?

Water to oxygen & hydrogen.

i don't understand


text:
Why can't we use orbital correlation diagrams for sigmatropic rearrangements?
In Ian Fleming's Molecular Orbitals and Organic Chemical Reactions book, he describes the use of correlation diagrams as a method for predicting whether pericyclic reactions will be allowed/forbidden on a symmetry argument (he calls this a basis for the Woodward–Hoffmann rules).

Correlation diagrams have given us a convincing sense of where the barriers come from for those reactions that we have been calling forbidden.

Whilst there are given examples of cycloadditions and electrocyclic reactions there are no examples of sigmatropic rearrangements, and on the internet a few sources say the diagram is not appropriate for these sigmatropic rearrangements.
I am curious why a correlation diagram does not work for sigmatropic rearrangements, since the the Woodward–Hoffmann rules are able predict which are allowed and forbidden.

Please go through the following reference, which deals with the use of correlation diagram (dynamic correlation diagram) to explain the sigmatropic rearrangements.
J. Theor. Comput. Chem., 2017, 16, 1750055. (DOI: 10.1142/S0219633617500559) - Dr. C. R. Ramanathan

The reason that comes up again and again is the one below, taken from Orbital Interaction Theory of Organic Chemistry.

Orbital correlation diagrams are useful for cycloadditions and electrocyclic reactions but not for sigmatropic rearrangements since no element of symmetry is preserved.
Source:   Orbital Interaction Theory of Organic Chemistry. Arvi Rauk.

In order to (easily) draw a correlation diagram you need to find some symmetry element that is preserved throughout the reaction. A correlation diagram is shown below for an electrocyclic ring closing reaction, note how the C2 axis is preserved during the process, allowing the various correlations to be made.

Source: Dr Wallace's web notes on pericyclic reactions

Since you mention Fleming's book, I'll quote him on this matter:

Step 3. Identify any symmetry elements maintained throughout the course of the reaction.
Source: Molecular Orbitals and Organic Chemical Reactions: Reference ed. Ian Fleming.

Essentially, you have to be able to look at the orbitals directly involved in the reaction (i.e. we don't care about random groups hanging off the side that play no role).
In a sigmatropic rearrangement there is rarely (possibly never but I'm sure there are exceptions to the rule) a preserved symmetry element, which makes it impossible to draw a correlation diagram.
There are, of course other approaches, each of which have advantages and limitations depending on the reaction in question:

Just apply the Woodward–Hoffmann rules
Use frontier orbital theory to look at the HOMO–LUMO interactions
Use a Zimmerman approach and look if the transition state is aromatic, or anti-aromatic.

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text:
Which MO approximation best fits with lab results and observation?
I understand that there are many different methods that can be used to approximate the set of MOs for an arbitrary molecule, as the actual quantum mechanical description is taxing to calculate. I just wanted to know which computational method comes closest to reality. B3LYP/6-31G(d) seems to be a popular choice, but I realize that there many others such as RHF/STO-3G, MP2/MINI, etc.

If there were one single method that would come closest to reality, who would bother to use (or remember, for that matter) the rest of them?

@IvanNeretin, those people without access to the computer time they need. Inferior methods are constantly used to avoid using years of CPU time for trivial things, no?

Full CI would be about the most exact thing you can do as it is only limited by the basis set employed. But it is too time consuming to be used for most practical problems.

I'm voting to close this - it's incredibly broad and opinion-based (e.g., what exact property you mean). Different methods are good for different properties. Consider, e.g., http://chemistry.stackexchange.com/questions/27302/dft-functional-selection-criteria


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Imaginary Bonding Interactions
Usually in chemistry, we deal with bonding interactions. That is, if I have the diatomic A-A molecule or A-B molecule, there's a favorable interaction (i.e., a bond) and a prototypical MO diagram like this:

As I'd say in a lecture, one orbital goes down in energy, one orbital goes up in energy. I can use Hückel theory to define the energies of the bonding and anti-bonding interactions:
$$e_{ab} = e_a + \frac{H_{ab}^2}{e_a - e_b}$$
$$e^*_{ab} = e_b + \frac{H_{ab}^2}{e_b - e_a}$$
Now, consider if $H_{ab}$ is imaginary.. we'd use the same equations to get a very unusual MO diagram:

How would you describe such an interaction? It's clearly not the traditional favorable bonding. It's not a non-bonding interaction. And it's not the traditional anti-bonding case like $\ce{He2}$, because if I put 2 electrons into this system, the anti-bonding orbital isn't filled.
I ask the question because in a survey we've performed using DFT, we find many such combinations.
I'm leaning towards calling it an imaginary bond, but if there are known discussions, I'd be curious to read references.

If you put two electrons into this system, together they will be better off than $(e_a+e_b)/2$, which means this is still a bonding interaction.

@IvanNeretin Exactly. My question is more that it's a very unusual bond, so I'm trying to come up with a description suiting the interaction diagram.

I'm not sure it's all that unusual.

@IvanNeretin Based on our survey, you might be right. Yet, I can't find much in the way of discussion. Pick up a book on MO theory, and you'll see exclusively real-valued $H_{ab}$ pictures.. the "traditional" kind. I have yet to find any discussion of the imaginary case.

I guess one aspect is whether the matrix representation of Hamiltonian is Hermitian. If it is, $H_{ab} = H_{ba}^* $. Solving the secular equation will lead to $H_{ab} H_{ab}^*$, which is positive-semi definite (different than  than $H_{ab}^2$). On the other hand, if such Hamiltonian is used to describe open-system, it is in-general non-Hermitian, the imaginary bond will be possible

IIRC, in Hückel theory the Hamiltonian matrix is assumed to be real from the get go: $H_{ab} = H_{ba} = \beta$ and the MO energies formulas are valid only within this assumption. Thus, this formulas simply could not be used if  $H_{ab}$ is not necessarily real. One have to apply the variational principle for this different case to get the MO energies formulas first.

This is very interesting, but I'm an experimentalist not theorist, so I naturally ask a naive question; as interactions are of the form $\phi _a \phi _b /r $  e.g. between pairs of electrons or pairs of nuclei, and between electrons and nuclei leading to Overlap, Coulomb and Exchange integrals etc, what form of interaction will lead to complex matrix elements? Also, the orbital energies in your diagram would lead to observable transitions, presumably you can calculate their properties and see if this can be observed.

Yes, as you surmised, for closed systems, the Hamiltonian should be Hermitian, so the integral should give real-valued numbers. @Wildcat - while Hückel theory has real-valued parameters, it's often helpful to use simpler theories like this to understand more complex systems.

ok, but I'm still curious, what sort of 'system' is not 'closed', what would the molecule's potential look like?

@user26143: What do you mean by "open system"?

I'm no professional and I could be completely wrong about this, but my first thought on seeing that diagram is that both electrons would end up on the left side and any interaction would be purely ionic.

@GeoffHutchison , can you give any reference about the equations that you wrote please. I can not realize how to get them from the Hückel theory. Thank you.

Just want to add that in accordance with the IUPAC definition of the bonding MO the above mentioned case is, in fact, "traditional favorable bonding". I quote from the definition: "Generally, the energy level of a bonding MO lies lower than the average of the valence orbitals of the atoms constituting the molecule." Note that bonding MO is not even required to lie lower than the valence orbitals, just lower than the average of them.

The short answer is that I need to come up with better wording to describe the interaction and that Hückel is surprisingly bad for these systems.
I'll give a slightly longer answer for now, and add a reference as we submit the paper.
First, an example system:

So "A" is a strongly electron-deficient dinitrothiophene ring (an electron acceptor) and "B" is a strongly electron-rich diaminothiophene ring. The figure in the question is a subset of the entire MO diagram, focusing on the HOMOs of the monomers and the HOMO and HOMO-1 of the dimer.
Why do I think this is a strange system? Well, the particle-in-a-box and Hückel models are usually pretty good at explaining conjugated organic molecules. If you make a longer conjugated system, the HOMO should go up in energy, and the HOMO-LUMO gap goes down (delocalization).

The alternative, localization, would have the orbitals remain at the same energy level:

Both of those "limits" can be explained in the Hückel model by changing the $H_{ab}$ or $\beta$ parameter (depending on how you write the equation, the symbol is different). In the localized case, $H_{ab} = 0$.
What's strange about these strong donor-acceptor cases is that as I said, it's not really delocalized, or localized.
Instead, as one comment correctly deduced, the DFT calculation (B3LYP/6-31G*) shows a large degree of charge transfer, but not necessarily localization. The dipole moment computed to be >7 Debye.
If one attempts to use Hückel to analyze these results, you must infer that $H_{ab}$ is imaginary. This indicates an "open system" because electrons are indeed transferred - from monomer "B" to monomer "A" creating the large dipole moment.
My overly-simplified question was probably a bit confusing, but:

It's interesting that no simple model captures charge transfer unless you allow Hückel to have imaginary (non-physical) $H_{ab}$ parameters. (NB, my colleagues aren't too happy with this concept, suggesting it simply indicates Hückel is a poor choice.)
It's surprising that in prototypical conjugated systems, I can't (yet) find a discussion about this alternative to delocalization and localization. We find it commonly as the difference in energy level increases.


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Why is 1,1-dimethylbenzene impossible?
It seems possible, as each carbon needs four bonds, so the first carbon could be bonded to two adjacent carbons and two methyl groups.

The carbons in the benzene ring are connected by alternating double and single bonds.

@MaxLi You are thinking of it the wrong way. Each methyl is a substituent, which means that it substitutes something (at least nominally). That is, your methyl is there instead of something (usually hydrogen). Are there two hydrogens on that carbon in benzene? Why or why not?

or put another way if you want to place two methyl groups onto one carbon atom then you would loose one double bond completely and leave the adjacent carbon with one spare electron and one H atom. In this way you would destroy the aromaticity.

Have you drawn the structure of this compound?

One of the determining features of benzene is a delocalised 'cloud' of electrons, which overall counts for 3 double bonds. In other words, Each carbon atom in the ring has a single bond to one neighbour, and a double bond to the other - this leaves one remaining bonding orbital - so only one more substituent can bond to each carbon on the ring. Adding a second substituent to a single carbon on the ring would break one of the C=C double bonds in the benzene ring in order to limit the number of bonds to the central carbon to 4.
This would be a very unfavourable reaction because benzene is an aromatic molecule, meaning that it is more stable than its hypothetical non-aromatic analogue (a cyclohexatriene) such that almost all of benzene's reactions will lead to a product which retains the aromaticity.
Once the double bond is broken, the ring is no longer aromatic, so is no longer called a benzene ring, instead it would be called a 1,3-diene. As you can see from the image below, adding two substituents to one carbon atom in the ring is possible, but the product would be less stable than the reactant.

It would be radical nor diene until hydrogen atom would be added

@Mithoron my radical chemistry is pretty rusty, but I doubt you would be able to isolate the radical species due to it being incredibly unstable. To be honest I'm not sure which reagents you could use in order to obtain the diene from benzene in the first place.


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What is the energy of the London dispersion force holding together two atoms of the same element?
I read somewhere that the energy is around $U=-\frac{3\alpha I}{4r^6}$ where $\alpha$ is the polarizability and $I$ is the ionization energy but I'm obtaining way too large energies for a lithium crystal.
The shortest distance between two lithium atoms in the crystal is $304 pm$
The first ionization energy of lithium is $520.2 kJ/mol$
I've found various polarizability constants online in papers but I can't really figure them out or the units they are using.
Also, I suspect that I want the dispersion force between $\ce{Li^+}$ ions and not neutral $\ce{Li}$ atoms.

Polarisability as used here has dimensions of volume. The London formula is $$U \approx -\frac{3}2\frac{I_aI_b}{I_a+I_b}\frac{\alpha _a \alpha _b}{r^6}$$  for two species a and b so you just need to square $\alpha$.

So, for a polarizability constant of $24\times 10^{-30}$ which I'm not sure is correct I got a total energy of 290kJ/mol which still seems too high. Is this correct?

still seems very high

having looked at some values the polarisability of 24 seems very large, atoms seem to have values not bigger than 10 although I have not done a thorough check on this.

The source at https://arxiv.org/abs/quant-ph/0506106 says about 24 and the source says other sources say 22 and 24. Maybe they mean a different kind of polarizability?

Volume polarisability is correct. I had a look again and found values for benzene of 25 .10$^{-24}$ ml so that is consistent with your values  in m$^3$. Carbon tetrachloride has a value 10.5 for example. ( I had only looked at nitrogen (1.73) and water (1.44),  before, apologies). An interaction energy of approx 24000 cm$^{-1}$  is still huge, its as large as an electronic excitation, but I can't see what wrong  :(   The ionisation energy and bond length seem ok .

Thinking again, in the crystal one calculates the Madelung energy $\alpha$ to get at the binding energy $U=-\alpha (1-\rho /R_0)$ where $\rho$ is about 0.1$R_0$ the near neighbour distance and accounts for repulsion. (see the classic C. Kittel  ' Introduction to Solid State Physics') The reason that your previous calculation gives such a large value is the small separation you use.  Normally the centre to centre separation of molecules is not as small  as a bond length.

The polarizability of ions would be very small right? So in the sea of electron model where we are dealing with $\ce{Li^+}$ the energy would be much smaller right?

yes; the values I quoted are for molecules, notions, a value similar to He would seem to be reasonable  approx $0.2.10^{-30}$ m$^3$.

yes; the values I quoted are for molecules, a value similar to He would seem to be reasonable ; value approx $0.2.10^{-30}$ m$^3$ and this would reduce your value very considerably. Atoms and tri-atomics with small atomic number atoms seem to have values in the range 0.2 to 2. H$_2$O = 1.45, H$_2$S = 3.6  as heavy atoms are more polarisable than light ones.


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Can toluene be used as a solvent for nucleophilic substitution?
This is a continuation to a previous question of mine regarding a change to a one-pot procedure for the preparation of Wellbutrin that I performed in college last semester. Could toluene have been used as the extraction solvent for the a-bromoketone and for the nucleophilic substitution? The procedure we were given used DCM and then removed it under vacuum for a solvent switch to NMP. Also, what was the reason for this? Could a mixture of NMP and DCM have been used?
Previous question: Suggestions for solvents capable of efficiently performing alpha-bromination and nucleophilic substitutions

Please provide a link to your previous question.

I added the link in an edit.


text:
2-bromobutane absolute configuration from structural formula to 3D
For the 2-bromobutane drawn below, given the 3d image I can tell it is R configuration

Annotated from source:
However, given the structural formula I can't figure out how to get this 3d shape correctly.
Structural formula:

When I try to visualise it I get two possibilities:

My sticking point is: when moving the H (the lowest priority group here) to the back, which way (to the left or right) does the CH3 go? Depending on which I draw the answer changes.

This is a Fischer projection. By convention, the horizontal groups come towards you and the vertical ones go away.

Oh I see. Thanks. By the way, how do I close the question or select an answer if  the answer is in the comments?

Well, that's the thing, you cannot select an answer from the comments. You have basically two options now. Hope that @f'' is considering to write a small answer (or maybe someone else), or when that does not happen, write one yourself how you figured out the solution. Ideally it should contain a little more information than just a comment.

@f'' How can you be sure it is a Fischer projection? The page you link says, The use of Fischer projections in non-carbohydrates is discouraged, as such drawings are ambiguous when confused with other types of drawing. If it was a Fischer projection, would it not be more natural to have Br at one side of the "cross", and then have the C-C bonds vertical? Maybe the answer to this is, if you know it is a Fischer projection, the exact configuration can be inferred from that. If you do not know if it is a Fischer projection, the precise stereoisomerism cannot be inferred from the "flat" diagram.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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Why doesn't 1,3-dichloro-2,4-dimethylcyclobutane have a plane of symmetry?
In (1R,2R,3S,4S)-1,3-dichloro-2,4-dimethylcyclobutane, are the blue lines not planes of symmetry?

My book says it has a centre of symmetry, but not a plane of symmetry.  But if I took a cut through any of the blue lines wouldn't both sides be symmetrical?

One of your suggested planes transforms upward-facing chlorine to downward-facing chlorine and vice versa (which is not the behavior expected of a vertical plane of symmetry). Another does the same to methyl groups.

The point group is C$_i$ with only a centre of inversion (and the identity). This can also be described as an S$_2$ operation which is rotation by 180$^o$ and reflection in the plane perpendicular to the rotation axis. This axis points out of the plane of the carbons.

Ivan already stated quite briefly in the comments, that the planes you drew would transform an upward facing chlorine to a downward facing chlorine and vice versa, and that the same holds for the methyl groups. This is of course true and the first and most obvious point, why there is no symmetry plane.
A more close investigation will, however, point out another important feature of the compound itself. The cyclobutane ring itself is not planar. In the (electronically) lowest conformation of the molecule it has no symmetry whatsoever (technically $C_1$ symmetry), not even inversion symmetry like your book states. This is due to the fact, that the four-membered ring is puckered. This is best visualised by looking at the 3D structure of the compound. There are multiple programs that will allow you to view such a structure and I will xyz coordinates at the end of this posting.
The following image will hopefully already show you what I am talking about. It was optimised on the DF-BP86/def2-SVP level of theory.

In order for the the molecule to have inversion symmetry, the shown angle needs to be exactly $0^\circ$. The following animation shows the molecule rotating hopefully further clarifying this point.

You can clearly see the puckering of the cyclobutane ring.
One final comment: For a molecule like this it really is necessary to specify the stereochemistry with the name. There are other stereoisomers, which do indeed have a symmetry plane. I leave it to you to figure out which of them do.

 Except the last one, all of them have at least one plane of symmetry.

Addendum: Transition state structure
The symmetric ($C_\mathrm{i}$) transition state is on the DF-BP86/def2-SVP level of theory only about $\Delta G (298.15~\mathrm{K}) = 8~\mathrm{kJ/mol}$ higher in energy than the puckered asymmetric forms. It is therefore worthwhile to have a quick look at its structure. In the following graphics the red ball indicates the inversion centre.
Below that is an animation of the transitional mode. It briefly pauses at the puckered structures (these are extrapolated, not optimised structures) and the transition state to make it easier to see the movement in between them.
I would expect this process to be readily available at room temperature and therefore almost impossible to distinguish the different lower lying structures.

Coordinates of the lowest structure:
18
G09D.01; E(RB-P86/def2-SVP/W06) = -1154.66760674 A.U.
C        0.051334000      1.246599000      0.089432000
C       -0.768087000      0.117382000      0.789824000
C        0.002884000     -0.950933000     -0.020424000
C        1.086640000      0.142749000     -0.293350000
H        0.449081000      2.008047000      0.790972000
H       -0.535742000      0.060683000      1.871003000
H       -0.532992000     -1.105843000     -0.982729000
H        1.517474000      0.183811000     -1.312785000
C        0.343625000     -2.291820000      0.610089000
H        0.939420000     -2.156389000      1.535300000
H       -0.586821000     -2.839917000      0.866159000
H        0.932752000     -2.926935000     -0.083624000
C       -0.606103000      1.926898000     -1.110977000
H       -1.475939000      2.532182000     -0.784546000
H        0.116584000      2.600709000     -1.616245000
H       -0.977177000      1.195914000     -1.858831000
Cl       2.502295000      0.083512000      0.845211000
Cl      -2.570729000      0.126790000      0.662589000

Coordinates of the transition state (X is a dummy atom indicating the inversion centre)
19
G09D.01; E(RB-P86/def2-SVP/W06) = -1154.6659061600 A.U.
C       -0.188954000      1.025076554      0.415795307
C       -0.873066584     -0.383572154      0.501227861
C        0.188954000     -0.991627446     -0.457606693
C        0.898153416      0.404477846     -0.509590139
H        0.242239584      1.289691385      1.404629307
H       -0.864241753     -0.820445015      1.518694431
H       -0.267326416     -1.206068615     -1.446440693
H        0.939502247      0.849712985     -1.522875569
C        0.953473154     -2.212350139      0.042720768
H        1.426842139     -2.018382400      1.026791199
H        0.267219307     -3.079692431      0.145048491
H        1.760843738     -2.494108569     -0.663022216
C       -0.974378846      2.203987861     -0.147249232
H       -1.806836262      2.489927431      0.529225784
H       -0.309030693      3.083873569     -0.266301509
H       -1.418479861      1.964027600     -1.135500801
Cl       2.591747000      0.489781723      0.126797539
Cl      -2.591747000     -0.506506277     -0.064080461
X        0.006271708      0.013588700     -0.012543416

The model seen in your animation is even chiral. One notices that the two Cl-C bonds (green/black sticks) have different (non-oriented) angles to the axis of rotation you have chosen. If for example, I "hold" the animation at the "stop" where the upper methyl group is closest to my eyes, I can see that the Cl-C bond then to the right is more "vertical" (smaller angle with the axis you chose for the rotation), and the Cl-C bond to the left is more "horizontal". If you had given us the mirror image, that would be opposite. So the model you present, is chiral.

@Jeppe I was trying to imply that with stating that it has no symmetry (C1). The transformation of one of the forms into the other at room temperature will be very fast at room temperature and I would not be surprised if you find no chirality since the transition state does have an inversion centre and is therefore achiral.


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Is it possible to detect the component of an unknown substances using FTIR?
Currently I'm doing research involving photo-polymer which contains unknown substances. Is it possible to detect each component of the substance by using FTIR? Will it be able to separate the photoinitator component, the monomer, and the oligomer?

Not really... FTIR is more like a fingerprinting device. FT basically means identification is done with your computer. But you just have one line/graph that represents all three together. It doesn't separate the spectra; but with practice you can "see" things that are mixed, although some chemical tend to dominate the spectrum. Also, the FTIR device that you get will probably have a program that can help interpret peaks (which I have found to be utterly useless). The real way it (the spectrometer) works is by searching through the library. Again, the device you get will come with a library, but it's usually not worth anything. You will need to create your own library including all of the possible combinations (identify/name each sample that you make).
I would say that with enough practice and repeated sampling (building a library of known mixtures), that you could might be able to identify an unknown sample as a mixture of the components; and possibly the amounts of mixed components (depending on what they are, exactly).
In other words, let's suppose you have decane, benzene, and acetone. You will need to create a library like:

pure decane
pure benzene
pure acetone
5% Benzene in decane
25% Benzene in decane
50% Benzene in decane
75% Benzene in decane
95% Benzene in decane
5% acetone in decane
25% acetone in decane
50% acetone in decane
...
35% acetone 55% decane 10% benzene
...

And you should run each sample 3 times. Then, after you have exhausted the list of known culprits, you should be able to nearly identify an unknown sample which is known to be mixture of these three chemicals (supposing that the amounts of each are what is unknown)... but be sure to run/analyze the unknown sample at least three times (adding it to the library each time) for best results.
Now with regards to photoinitiators... if using metals (like silver), it might be more tricky. IR Spec is most useful for organic chemistry.

‘IR spec is most useful for organic chemistry’? I argue it’s not useful at all …

You will clearly be able to produce a difference spectrum, light on vs light-off, and possibly identify some groups therein, but as to identifying species, I would not hold my breath ! You should try something like MALDI, I'm sure it would be more fruitful and is already used in polymer chemistry.

FTIR wouldn't help you in find out the exact structure of a compound, but rather gave you information about the functional group present in a molecule. If you want to find out the exact structure of a compound you need Nuclear Magnetic Resonance( NMR) spectroscopy.


text:
Cheletropic carbene addition to ethene under Woodward-Hoffmann rules
Woodward and Hoffmann state in their seminal paper [1], that a ground-state pericyclic reaction is thermally allowed when $p+q=\textrm{odd}$, with $p=(4n)_{antara}$ and $q = (4m+2)_{supra}$. When I look at a cheletropic carbene addition to an ethene (see Figure, from [2]), this rule gets a bit shady for me.

Essentially, there are two ways to look at it:
1) The HOMO is the two-electron sp2 orbital on the carbene and the LUMO the the $\pi^*$ LUMO on the olefin.
2) The HOMO is the $\pi$ orbital on the olefin and the LUMO the pz orbital on the carbene.

Case 1) results in a $p=1$ from the sp2 and $q=1$ from the $\pi^*$, so $p+q=\textrm{even}$. The same can be verified for case 2). How can this be, since these kinds of insertion are usually thermally allowed? Even when I would add together both situations, I'd get an even number. Does the Woodward-Hoffmann system work here or am I missing something?
[1] http://onlinelibrary.wiley.com/doi/10.1002/anie.196907811/abstract
[2] http://chemistry.caltech.edu/courses/ch242/H1.pdf

Carbene additions are a little finicky under Woodward-Hoffmann rules - if you have library access, I would refer you to Ian Fleming's Pericyclic Reactions, pp 47-48. He also has lots to say in his other text Molecular Orbitals and Organic Chemical Reactions - there is a huge section on W-H rules and the part on carbenes begins at p281 of the Reference Edition.

Molecular Orbitals and Organic Chemical Reactions is indeed an excellent suggestion. This [https://www.scripps.edu/baran/images/grpmtgpdf/Martinez_13.pdf] PDF from a Baran group meeting might also be useful, it discusses the same kind of cheloptropic additions you're interested in


text:
MO diagram for allyl cation
Given below is an image from google

In $\pi_2$, isn't there two possibilities? On the third carbon, The green coloured lobe can be either on top or on bottom relative to the lobes of the first carbon.
Does $\pi_3$ mean that there are no double bonds? Because it would result in an out of phase combination. What type of bonding will be present for $\pi_2$? Will it be same as $\pi_3$?

Are you missing the fact that $\pi_3$ is not filled? The $\pi$ bond arises from the filled $\pi_1$ orbital

(1). No, the two orbitals have to be of opposite phase relative to each other. Recall that the MO wavefunctions for $\pi$-delocalization are very much like the wavefunctions for the particle-in-a-box, which requires that the phase alternate. Furthermore, a general rule is that molecular orbitals higher up in energy have more nodes, and the phases of the orbitals in $\pi_2$ must alternate for the middle lobe to be a node.
(2). We can speak of double bonds only for the molecule, not for molecular orbitals. Molecular orbitals do, however, have bonding character, and we characterize $\pi_2$ as non-bonding (little interaction between orbitals) and $\pi_3$ as anti-bonding (unfavorable interaction between orbitals).

where can I read more about (1)? I have never heard something like that

I like Clayden's Organic Chemistry. In the first edition, I think this is chapter 7.


text:
What compound does this chemical formula represent?
I recently saw this:

What chemical does this formula represent?

It's adrenaline (epinephrine) https://en.wikipedia.org/wiki/Epinephrine

The structure represents adrenaline or 'epinephrine' (en.wikipedia.org/wiki/Epinephrine). As in:

Go out and have an adrenaline rush on your birthday!!

The image on wikipedia has a slightly rotated ring, so doesn't look identical.

You can easily search for chemical structures online. Just one example is ChemSpider. It has an online editor in which you can draw the structure. You can also upload images.
Always make sure to compare the resulting molecule with your original structure to see if it's exactly the same.
But yes, it's adrenaline/epinephrine.

Nice advice, however that last one about exact comparison might be misleading to chemistry beginners.

Good point. I just want to warn against overseeing for instance an additional OH group, or miscounting the number of C atoms; mistakes easily made and leading to a completely different molecule. However, as gilleain already mentioned in his answer, one molecule can indeed be displayed in several ways. This depends on the viewing angle, and the orientation of the atoms in space (the "conformation") that can change by rotating around single bonds. Basically just check if all the atoms are linked in the same way.


text:
How to dissolve bronze at room temperature
Is there any way to dissolve about 10 kg of bronze at room temperature?
If yes, which substance would be the best to use and how would the reaction look like?

Any hints at what exactly you'd consider simple, and also how much bronze you need to dissolve, might be of great help to us. Nitric acid, for example, would certainly do the job; it would also poison you if treated without due respect. With that in mind, welcome to Chem.SE.

@IvanNeretin Thanks for remarks! I've edited the question.

Good grief! Why do you want to do that?

Wow. 10 kg is quite a lot. Still, nitric acid would do the job, but you'll need a lot of it, and you'll get a great deal of toxic fumes, too. I'd definitely not try it at home. I'd probably not try it at the lab either, because why would I, really?

@JonCuster Actually, I write a Sherlock Holmes chemical story about disappearance of bronze statues :D There will be great revelation mentioning the process used!

Letting a glass dagger vanish in  a refractive-index matched fluid seems far more impressive to me.

Supposing that the bronze is only 78% copper and 12% tin (and there are no other metals). Nitric acid will dissolve the copper... but this reaction is exothermic, which means it will heat up.
At boiling temperature. Nitric acid will also react with tin to make tin oxide which is soluble. However, if you kept it cool/room temperature with an ice bath, then you would have leftover tin which did not react.
Tin will react with aqua regia to form tin (IV) chloride, but copper is slower to react with aqua regia... so it would be better to use nitric acid first (to dissolve the copper) and then add hydrochloric acid to react with and dissolve the tin. Once again, the reaction between tin and hydrochloric acid (or aqua regia) will proceed much faster with heat, but it will happen at room temperature.
Aqua regia = 1 mole of Nitric acid and 3 moles of hydrochloric acid).

Any guess on just now much nitric and aqua regia that will take? Disposal will be a problem to consider before starting, since it will be hazardous waste.

@JonCuster Since its 10kg of bronze... sure; 7.8 kg / the molar weight of copper = recycle the nitrate. Here's a youtube video about making nitric acid.

Thanks for an excellent answer! Could you make an estimate about how long would it take? (considering 10 kg of bronze)

@MatějKripner It depends on factors like temperature and surface area (powdered bronze might go boom). This is kind of a dangerous reaction, so I would recommend going slowly with lots of ventillation (like outside with fans), and fire extinguishers and perhaps the fire deparment (this could be of some educational importance to them too). Don't breathe any red gas. If possible, start small (grams) and work your way up (to ounces). I could see this reaction taking any where from 1-6 hours to complete (depending on how quickly you feed it and the temperature and thickness/shape of the bronze).

@MatějKripner I just saw that you are a novelist... with practice and good equipment, I suppose you could dissolve a statue pretty quickly, but if you are considering something like spraying a bronze statue, it's not going to work in a practical way. You would need to soak the statue in a vat of nitric acid.


text:
Use of water in IR spectroscopy
I am familiar with the process of IR spectroscopy and the preparation of IR samples.
However, I am curious as to why you cannot use hygroscopic or protic solvents for IR spectroscopy. Is it just because it dissolves the cell (KBr/NaCl), or does it give an OH peak, or are there other reasons?
And are there any protocols that allow you to use water as a solvent for obtaining an IR spectrum?

I applaud this question. I have not really tried to find any solution, but I definitely recognize the trouble with water... nonsense/worthless reading. As far as why... easy; water absorbs like, all of the IR spectrum. Below is the IR spectrum of liquid water (from NIST's Chemistry Webbook).

I actually have an ATR IR Spec, which is (supposedly) meant for reading aqueous solutions...

...but it doesn't really work... it can't be used for figuring out dilutions (well maybe, give or take 20%). In other words, 50% water + sample looks a lot like 95% water + sample... which looks a lot like any sample + water. Water basically drowns the reading- so it all looks alike.
Don't get me wrong... it's super nice that I don't need to oven dry my sample or deal with KBr. Basically, I just put the sample on the top and I get a decent reading, but if the sample has a lot of water in it (like greater than 10%), it loses a lot of definition.

@benNorris That's what a sample of say anything plus 50% water looks like

Its a continual problem. Usually its due interfering absorptions (often total absorption) from water so try using D$_2$O instead, the frequency shift may be enough for you to be able to use it.  Also change to using ATR (if possible), we use this in the teaching  lab for several experiments on liquid samples without the need for any cell. Signal to noise is not so good though. You can also try subtraction via a water only spectrum against water plus sample. It will be messy but better than so data at all.


text:
Could compressed refrigerant be stored feasibly in residential setting?
In Vapor-compression refrigeration, a refrigerant is compressed releasing heat (typically outside a building) and then decompressed in another location, absorbing heat.
A residential air conditioner works on this principle with a compressor outside the home and a blower over the decompression coils to circulate cold air.
Would it be feasible to compress ahead of cooling demand, storing the compressed refrigerant in a high pressure container for use at a later time of the day, thus shifting the power consumption to times of the day where there is less electrical demand?
By feasibility, I'm thinking of questions like:

How much storage volume would one need for a typical household usage over an afternoon of running air conditioning given a container type that would be available cheaply, ex: propane canisters?
What would be the sources of efficiency loss added by something like this?
etc.

Storage of the liquid refrigerant would take a fairly large tank. The much bigger problem is how to store the decompressed gas afterwards. sCO2 might work though!

I happen to have a bottle of refrigerant in my house and another in my car (besides the stuff already being stored in my AC units).
While this is definitely more of a physics question, or perhaps a question for the HVAC guys on the Home Improvement SE... to generally answer your question, there are many factors for cooling a home, like size and location, for example. But let's say your average 2-ton window-unit AC pumps 24,000 BTU's per hour... and 4000 BTUs requires 1 gallon of refrigerant per minute.... you need 6 gallons per minute (being released through an evaporator coil) to match a window unit AC.

So 6 gal/min * lets say 2hrs of cooling = 720 gallons. Could that be reduced further by additional compression? For example, a quick search reveals that a typical A/C compressor is about 200-400 PSI, whereas, for example, the hydrogen tanks in a toyota mirai can store hydrogen at 10,000 PSI.
Does pressure/volume work linearly like this such that going from 720 gallons at 400PSI would give you 28.8 gallons at 10,000 PSI?

That's liquid refridgerant... which is just run around in circles. Liquids are not really compressible. At 10,000 psi you might get 1% or 2% compression. So at 10,000 psi you need ~712 gallons. Or at 100 psi you would need 720 gallons of liquid refrigerant (for 2 hours use).

Ah, bummer. Just curious, but this doesn't sound very feasible. Thanks!

A BTU is a unit of heat, just like joules or calories. An air conditioner's capabilities are measured in cooling/time and $\pu{24,000 BTU}$ is colloquial used to mean $\pu{24,000 BTU/hour}$. The circulation requirements would be defined by

X gallons of refrigerant circulating per hour would yield Y BTU/h of cooling capacity

Apparently (I'm not an AC guy) that translates to ~$\pu{66,7 BTU/minute/gallon}$, or $\pu{4,000 BTU/hour/gallon}$.
If the main AC unit's cooling capacity is $\pu{24,000 BTU/hour}$, which comes down to about $\pu{7 kW}$ or $\pu{25 MJ/h}$ (that's astounding for a home appliance!), you need a certain amount of refrigerant in constant cycle.
If the assumption is $\pu{4,000 BTU/hour}$ per gallon of circulating refrigerant per minute, you would then need at least $6$ gallons of refrigerant to reach the 24,000 BTU/h. More is always a possibility, though.
P.S. I speak metric, so I rounded off enthusiastically in conversions.


text:
I cannot find the structural formula with a particular compound's IUPAC name
I currently need to read a paper. In that paper the following compound was mentioned:
1,2-benzo-6-(N-ethyl-N-isobutylamino)fluoran
I tried to find the corresponding structural formula using the OPSIN parser at: http://opsin.ch.cam.ac.uk/
but I got the error message:

1,2-benzo-6-(N-ethyl-N-isobutylamino)fluoran was uninterpretable due to the following section of the name: 1,2-benz

I also tried just Googling the compound's name but only found patents which weren't very useful.
I hope somebody can help me find the structural formula.

Do you have a link to the paper you found the compound in? Some context would be useful

@NotNicolaou Yes. See Table 1 example 6.  https://www.google.com/patents/US6494950

The name given is not standard.
Fluoran itself is a spiro compound with two components:

(it's strictly an isobenzofuranone, but I'm lazy to type that multiple times.)
Because of the nomenclature rules of spiro compounds (which I'm sure someone will come along and tell you all about), the atoms in the benzofuran system are labelled 1-7 and those in the xanthene system are labelled 1'-10', with an apostrophe.
Atoms 1 and 2 of the benzofuran system are both within the furan ring. If I am not wrong, atom 1 is the carbonyl carbon, and atom 2 is the ring oxygen. The given name, that starts with "1,2-benzo", therefore makes no sense as you cannot fuse a benzene ring to atoms 1 and 2 of the benzofuran system.
After a little bit of Googling, I came across this compound: 6'-(Diethylamino)-1',2'-benzofluoran. I think with the apostrophes, it makes much more sense.
The numbering within the xanthene system parallels that of anthracene:

Having "1',2'-benzo" means that you have an additional fused benzene ring to carbons 1' and 2'. The 6'-substituent also makes sense. Chances are if the paper you were reading labelled the xanthene atoms with 1-10 without apostrophes, the 6-amino group should be a 6'-amino instead. Your desired structural formula is probably

ChemDraw's name to structure generator gives this molecule when you input 6'-(N-ethyl-N-isobutylamino)-1',2'-benzofluoran. The amino group, which is just an ordinary substituent, should probably be a prefix to "benzofluoran" which is the parent compound.
I am sure in due time someone will come along and give you all the official IUPAC rules but I figured that if you just wanted the structural formula, my not-very-rigorous explanation would suffice.

There appears to be an (old) alternate convention where the benzofuran has the primes instead, which the source could be using. See page 560 in this proposal: fluoran (156) is listed at the bottom.

If that is the numbering that is being used, then the "6-amino" group would be on the carbon now numbered as 5'.

@orthocresol In view of similar structure formulas in Y. Hatano: The Chemistry of Fluoran Leuco Dyes and related patents, your interpretation of the name is probably correct.

By the way, the preferred systematic name of this compound is 9'-[ethyl(2-methylpropyl)amino]-3H-spiro[2-benzofuran-1,12'-benzo[a]xanthene]-3-one. However, a detailed explanation of this name would probably require at least one new question.


text:
Does the structure of water change when a drop of color is added?
If I add a drop of color to water, would it change its structure entirely?

Can you please clarify and add some specifics? How much water are we talking here? What is the drop comprised of (oil, dye, etc.)?

There are similar questions about dissolving non-polar molecules in water.

No it does not change the structure of water.
Yes the H-bonds between water molecules are broken to accomodate the dye molecules. However, liquid water does not have a set shape or fix molecular arrangement. So what happens is,
The dye molecules simply diffuse out all throught and among the water molecules. The dye molecules become hydrated to some extent, that is, the charged groups and areas on the dye molecule interact with water to form H-bonds or polar bonds which allow the molecule to remain suspended among the water.
Source, General chemistry course work.


text:
Why is the photon energy of lithium flame coloring lower than that of cesium?
Why does $\ce{Li}$ give a red flame color which has lower photon energy compared to $\ce{Cs}$ which gives a violet flame color according to $E=h\nu$?
This is my attempt: Since lithium has smaller size it has a high ionization energy, thus, if an electron jumps from a higher energy level to the ground state then it releases a large amount of energy compared to cesium which has a larger size and lower ionization energy. Therefore lithium's photon energy should be higher compared to cesium. But it is not. Why?
The problem is the photon energy of flame coloration should gradually decrease down the group of alkali metals but rather it's increasing.

The colours you get in flame tests / burning metals, comes from the electrons in one shell being 'knocked' (by the heat), into a higher shell; then shortly after the electron 'drops' back (in one or more steps). The electron which dropped down a shell has a lower energy state, so to conserve energy, a photon is emitted. The further it 'drops' the higher energy/frequency/ colour it has. (Lower energy Infrared -> ROYGBV -> UltraViolet -> X-rays Higher Energy)
An element with more protons has a 'deeper' bottom shell. More protons => more attraction to electrons => more energy required to 'knock' up to a higher shell => higher energy/frequency photons.
So in general, a larger element would give higher energy photons off; in a flame test.
However it is a little more complex than that.
Because elements have many shells, an electron may jump back in more than one hop; giving off several different colours. For example Calcium has an orange flame colour, but heaver Strontium has a Scarlet red colour. If you look at a Strontium flame through a prism, you get Red, Green, and Violet.
Note: You may get infrared and ultraviolet in the mix too.
The ionization energy (IE) is qualitatively defined as the amount of energy required to remove the most loosely bound electron. Normally an electron in its outermost (filled) electron shell.
Lithium only has electrons in its bottom 2 shells. Cesium has electrons in its bottom 6 shells. Its outer most one has a single electron and is relatively "further away" from the protons in the nucleus. So it is easier to liberate Cesium's outer most electron than Lithium's outer most. In a flame test, if the flame is hot enough, electrons from deeper shells may be 'knocked' out of place and so you get higher energy photons.

I don't mean to hijack the question, but I'm curious about something. If photons are only emitted when the electron drops down a shell, why is there color while they are in the flame? When they are in the flame wouldn't the electron have the energy to remain in the higher shell? @DarcyThomas

@Noel Good comment/question.  The metal atoms in the flame will constantly be having their electrons 'knocked up' as the hot atoms bang into each other, but almost immediately after (less than a nano second I think) the electron will 'drop' down; before being bumped into an excited state again shortly after that etc, etc.  Note also that the metal atoms emitting photons is one of the sources of light/photons. That gives you narrow specific colours, there are others too. Like hot bits of carbon soot which glow with a range of colour (but looks yellow/orange when all mixed together).

But the photon released should have energy somehow nearly equal to the ionization energy of the atom. I think this gives more specific information of the energy of photon. And since Lithium has higher I.E.(Ionization Energy) to Cesium, thus Li should have high photon energy in flame coloration. But it's not. Why?

@YashpalYadav Lithium only has electrons in its bottom 2 shells. Cesium has 6 shells. Its outer most one has one electron and is relatively "further away" from the protons in the nucleus. So it is easier to liberate  Cesium's outer most electron than Lithium's outer most. The outer electron shell is the one which normally is important in (basic) chemistry ( not the case with   transition metals ). However in a flame test, if the flame is hot enough, electrons from deeper shells may be 'knocked' out of place and so you get higher energy photons.

Can you explain this "An element with more protons has a 'deeper' bottom shell. More protons => more attraction to electrons => more energy required to 'knock' up to a higher shell => higher energy/frequency photons." further?

@chemsonraun What doesn't make sense?

Maybe it's a term I'm not familiar with, but what do you mean by "a 'deeper' bottom shell"? Also, why do you say that "more energy is required to 'knock' up to a higher shell" since the ionization energy decreases as we go down the group? (I know that ionization energy refers to an electron completely leaving the atom and not just going to a higher energy orbital, but I think it's the same thing in the sense that an electron goes to a higher energy orbital in this case too, only with n->∞)

@chemsonraun  1. It is from the metaphor 'energy well' Here we are talking about an electron well, a gravity well is similar. An atom with more protons is like a planet with more mass. A planet with more mass has a deeper gravity well, the height from its surface to the height you would need to throw something for it to escape orbit. See https://xkcd.com/681_large/  Note: you can have satellite orbits at a range of heights, but electron orbitals  have discrete levels (shells) with discrete number of electrons they each can hold.   2. see the bold part of last paragraph from my answer .

To estimate the approximate wavelength of electronic transitions is alkaline metals, the Rydberg formula $(1)$ may be used. Strictly speaking, it is only valid for hydrogen and hydrogen-like atoms ($\ce{He+, Li^2+, \dots}$), however, if we assume the core electrons to be of spherical symmetry we can assume the effective nuclear charge to be $1+$ and the entire core to resemble the proton of hydrogen.
$$\frac{1}{\lambda_\text{obs}} = R\left( \frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\tag{1}\\
\begin{array}{ll}
\text{where:}\\
&R = \pu{1.097e-7 m^{-1}}\\
& n_1 = \text{lower shell’s principal quantum number}\\
& n_2 = \text{higher shell’s principal quantum number}\end{array}$$
For lithium, we expect the lowest transition to be $\ce{3\bond{->}2}$ while for caesium we expect it to be $\ce{7\bond{->}6}$. Plugging these values into the formula gives us:
$$\begin{align}
\lambda_\ce{3\bond{->}2}(\ce{Li}) &= \pu{656nm}\\
\lambda_\ce{7\bond{->}6}(\ce{Cs}) &= \pu{12369nm}\end{align}$$
Thus, we expect the lowest-energy photon emitted by caesium indeed to have a lower energy than that of lithium. However, this transition is firmly within the infrared wavelength range. It is therefore not observable visually.
What you are seeing in caesium’s case are a number of different transitions that do not correspond to the valence electron being excited by one shell and dropping back down. You can see all, part or a combination of:

high-shell core electrons being displaced into the valence shell and subsequent relaxations
relaxation of the valence electron from higher outer shells
high-shell core electrons being displaced into the valence shell and another, higher-shell electron relaxing
…

There are a lot of possible transitions covering the infrared range, the visible range and probably even the UV range. Some of them will be intense enough for the colour to be visible in a spectroscope or as part of the flame colour.

This is an excellent answer. The fact that a photon emitted by caesium may have a higher wavelength than one emitted by lithium but is not within the visible range, is a key insight that the accepted answer lacks.

A question: why do we expect the lowest transition to be $7 \rightarrow 6$ for caesium?

@RayBradbury Because caesium is in the 6th period that means its valence electron is a 6s electron. Excite this to a higher shell (7p, for example) and let it relax and we get $\ce{7 -> 6}$.


text:
Why the optimization in gaussian shows no bond with some atoms in diborane?
I have optimized the structure of diborane with Gaussian at the B3LYP/6-31+g(d,p) level and when check it on Gaussview I found that there are two hydrogens with no bond to boron atoms (like free atoms) and the two B atoms are bonded. Frequency calculations gives zero imaginary frequencies and when I compute properties (like ionization potential) the values are good approximations.
I wonder if the problem is with the calculation or only the graphical interface of Gaussview doesn't recognize it as an actual bond.
Also, the bond distances are in accordance with the literature (~1.3 Angstroms for H shared between the borons and ~1.19 Angstroms for the others).

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I find it highly likely that the distance exceeds to preset parameter in Gview and hence it is not displayed as a bond. It would be helpful if you could add the optimised geometry (as xyz coordinates) and the absolute energy for members to reproduce the situation. In the meantime I recommend trying to view the compound with a different software, e.g. molden, avogadro, etc...

Roughly speaking, there are no such things as bonds. There is molecular geometry, and if it is realistic, then what's the problem? Use the visualization software of your choice to draw the sticks wherever you like them.

I wouldn't put too much faith in the representation of bonding in GaussView, it's simply determined by bond distances and often it doesn't show the bond at all for some unknown reason. If your geometry is good, then you can use your knowledge and intuition of bonding to recall the true structure.

As mentioned GaussView as do many other visualizers use distances and angles to determine bonds. If you want bonding information, you need to resort to atoms in molecules or natural bond order calculations. Assuming you are using Gaussian, the latter is available via the NBO keyword.

I'll update this with links when I'm not on a phone.


text:
How slow is a ester reaction without a catalyst? (in a whiskey barrel?)
The typical description of a high school esterification reaction of an alcohol and an organic acid uses a strong acid catalyst like $\ce{H2SO4}$; otherwise the reaction is said to be "slow".
I would like to know, how slow? For context I am interested in the esterification reaction which happen in the time frame of distillation of whiskey and whiskey maturing in a barrel.
New spirit (what goes into a barrel to age and become whiskey) is approximately 60% EtOH, 39% water and less than 1% (I think) other alcohols, and organic acids like acetic and lactic acid.
How long would this system take to come to equilibrium, without any catalyst? (Hours, days, months, years?)

Please specify the reaction you are interested in: esterification? Ester hydrolysis? Transesterification? Do you have any particular ester in mind? Depending on their chemical structure, esters reactivity can vary greatly.

@vapid I am interested to know the time frame, to form an equilibrium. So Esterfication would be the main thing but that would be offset by Ester hydrolysis naturally. The the main esters would be Ethyl Acetate and Ethyl Lactate

The question is built of several assumption, most important, there is no catalysts present. There are ALWAYS catalysts present! I also, I highly double that the esters are straightforward products of acids and the alcohol present. In most organic matter there are significant amount of esters presents, as well they can be formed during distillation at high temperature.  Since boiling point of esters are lower than acids, i would expect them to be actually enriched in any distilled alcohol.

@Greg There always caveats. However having some ball park of how long would be useful. Is it on the order of hrs, days, months, years?

Could be months. Go and check it yourself, after all. Chemistry is an experimental science.

@IvanNeretin I am - Kinda. Making some whiskey (legal where I am from). I don't have a lab to make quantitative tests though. I only have  subjective qualitative olfactory test equipment available at this time. Which I am making the best use I can of :P

Realistically it totally depends on what reaction you are thinking about. Esterification? Ester hydrolysis? Transesterification? Either way without a catalyst from my research it would probably take anywhere from one hour to eight hours

The best info I could come up with is from wikipedia (with no citations for the relevant sections)
https://en.wikipedia.org/wiki/Fischer–Speier_esterification
For reaction with an excess of one reactant and an acid catalysts, it says:

"Typical reaction times vary from 1–10 hours at temperatures of 60-110
  °C."

Later it talks about ester reactions in wine:

"Of course, when compared to sulfuric acid conditions, the acid
  conditions in a wine are mild, so yield is low (often in tenths or
  hundredths of a percentage point by volume) and take years for ester
  to accumulate."

Which is a bit of an Apple to Oranges comparison, and is not whiskey, but it gives an indication of 'Yes, slowwwwww'. Or something like that.


text:
Can we do column chromatography for the purification of metal complexs?
In case of organic ligands there are several techniques for purification, but in case of metal complex (using $\ce{Cr,Mn,Fe,Co}$ as central metals), I have not seen any literature for the purification of them except recrystallization.
Can we do column chromatography for the purification of metal complexes?

Depends on the specific complex. Some are ameanable to columning, others decompose.  I think generally if things are crystalline and will recrystallise its a whole lot easier to purify that way.       Ferrocene and acetyl ferrocene are two classic examples of iron complexes that can be columned

For some yes, for some no. Typically, the ‘more organic’ a complex gets, the easier it would be to purify it by column chromatography.
Example: I synthesised zincporphyrin complexes in my master’s thesis and used silica gel columns. When synthesising the porphyrin itself, if I used sodium carbonate to neutralise propionic acid used in the reaction, I would get a (green, assumed) sodium complex which I could separate from the free porphyrin again by column chromatography.
Remember though, that many complexes are very polar and thus difficult to separate on standard phase columns.

Of course you can, but it really depends on your complex stability, the stationary phase and the solvent.
The first test you should carry out is a "2D TLC": cut a TLC plate to be a square. Draw a line at 1cm of the border in two adjacent borders (so the 2 lines intersect at one point). Put your sample at that exact intersection point then elute the TLC along one axis all the way up. Dry it quickly but without too much heating. Then elute it along the other axis. Dry it and reveal it: stable compounds have the exactly same Rf in the 2 dimensions (because they do not change), so their final spot is on the diagonal; unstable compounds will decompose to create new molecules which are more polar or less polar than their precursor. This will lead to spots out of the diagonal.

thanku sir for ur advice, for more knowledge about this technique, can u send me any reference literature or video..

Have a look at http://chem.chem.rochester.edu/~nvd/pages/chromatography.php?page=tlc_troubleshooting


text:
I'm a layman, but I'd like to know to to use GAMESS for some specific tasks

I have Avogadro and GAMESS. I modeled the sulfuric acid molecule using Avogadro, then I asked it to generate an input for GAMMES to calculate the transitional state (I'm hoping this will get me somehow closer to my computational modeling of real chemistry). But GAMESS says the job finished with an error, and I don't know why.

It's worth noticing Avogadro seems to spell one of the commands wrong, and I have to manually correct it. Avogadro issues the command \$STATTP but GAMESS does not recognize this command, so I change it to \$STSTEP, and it ends with an error, and I try the other keyword possible \$STPT and I also get an error. Here's my 2 inputs and outputs:

http://pastebin.com/TabWC9wr

This is what happens when I do what you told me, @Mitradip Das:

INPUT

!   File created by the GAMESS Input Deck Generator Plugin for Avogadro
 $BASIS GBASIS=STO NGAUSS=3 $END
 $CONTRL SCFTYP=RHF RUNTYP=SADPOINT $END
 $STATPT OPTTOL=0,0001 NSTEP=20 $END

 $DATA
Title
C1
S    16.0    -3.61564     3.32108     0.04978
O     8.0    -4.51377     4.03518    -0.82228
O     8.0    -2.75016     3.96121     1.00661
O     8.0    -2.67033     2.39549    -0.85711
O     8.0    -4.49699     2.21900     0.81326
H     1.0    -5.20320     1.93219     0.19859
H     1.0    -1.81232     2.32072    -0.38885
 $END

OUTPUT (showing only the error part, full output can be viewed here (http://pastebin.com/HRh1fS4d)

 CONSTANT LIST TOO LONG
 **** ERROR READING INPUT GROUP $STATPT   *****
 THE PROBLEM IS WITH THIS INPUT LINE, NEAR THE X MARKER
 $STATPT OPTTOL=0,0001 NSTEP=20 $END
                       X
 THE ONLY KEYWORDS ACCEPTED IN THIS GROUP ARE:
  METHOD    UPHESS    NSTEP     NNEG      IFOLOW    DXMAX
  RMIN      RMAX      RLIM      RESTAR    HESS      OPTTOL
  IHREP     HSSEND    IFREEZ    NPRT      NPUN      PURIFY
  ITBMAT    MOVIE     TRUPD     TRMAX     TRMIN     STPT
  STSTEP    PROJCT    IFCART    NPRTHS    KDIAGH    IACTAT
  IHMCON    FHMCON    SHMCON
 TROUBLE INTERPRETING $STATPT
 OBTAINING INITIAL HESSIAN, HESS=READ
 ***** ERROR, YOU DID NOT SUPPLY A $HESS GROUP.
 EXECUTION OF GAMESS TERMINATED -ABNORMALLY- AT Wed Jul 27 12:23:49 2016
                   580000  WORDS OF DYNAMIC MEMORY USED
 STEP CPU TIME =     0.00 TOTAL CPU TIME =        0.0 (    0.0 MIN)
 TOTAL WALL CLOCK TIME=        0.0 SECONDS, CPU UTILIZATION IS 100.01%

job aborted:

In general, this is not the right place to ask these questions since they are entirely about the software you use. You should also try to work through at least the first few examples in the manual. Starting to learn a program with a saddle point optimisation is a very bad choice, since they really need some understanding of the program first. The error you get is problematic because it points you to the wrong position. Try opttol=0.001. You might also need to tell the program how to obtain the hessian.

If the input geometry is not close to a transition state (which one? what do you want to find?), you might not get any conversion. Try running a single point calculation first. Don't rely on the avogadro input generator, it makes mistakes. Understand the keywords you are using and structure your input file accordingly. Leave comments in your input explaining what you are actually trying to achieve. Gamess is difficult to learn and even if your calculation runs without error, you might not get any meaningful results.

I see. It is not a very intuitive program to use. even the command line to invoke it is verbose.

You're asking it to do a transition state calculation (RUNTYP=SADPOINT ) which doesn't make sense.  A transition state occurs between two different structures, such as between reactants and product or when ammonia goes through its umbrella.  Try an equilibrium run with RUNTYP=OPTIMIZE instead.  This should get you going.  If you are having difficulty I will generate an input file for you.  It is well worth persevering.

For GAMESS, the command is not $\$STATTP$, but it is $\$STATPT$. Also, please change the commas to dot, otherwise it considers it to be a part of the number.
Please refer to the GAMESS manual for better understanding. The link is: www.msg.ameslab.gov/gamess/GAMESS_Manual/input.pdf

I don't think I'm able to use that software. Oh well.

Man you have given OPTTOL as 0,0001. Make it 0.0001. It will work. Let me know if it works.


text:
Why is acetone so cold?
I have a summer job as a custodian and one cleaner I use is called Goof Off, which is pretty much just acetone. Another is a disinfectant which is mostly water.
Why is it that whenever I put the acetone on my rag, it is always colder than when I use another cleaner or plain water?
The acetone also seems to stay cold longer than water does. Both cleaners are left in the open so they are both the same temperature.

Acetone is significantly more volatile than water. When you pour it onto a rag, it begins evaporating, taking some of the heat with it as it does so hence why it appears slightly cool

Evaporative cooling is also the exact purpose of sweating. Too bad we don't sweat acetone, would be handy on a hot day (although we'd probably be re-painting a lot of things...)

Goof-off superglue remover is acetone. Their other adhesive remover is ethylene glycol; basically antifreeze.

Evaporation is an endothermic phenomenon, i.e. it absorbs heat in order to proceed. Acetone is a volatile solvent (it evaporates easily) so it absorbs much heat when evaporating, and your skin gets colder because of that. That is what you are feeling. So acetone is not "colder", it just cools down your skin more easily when evaporating.
Ether (aka diethyl ether) is even more powerful at this.

On another note, while acetone is a pretty safe solvent, it can damage your skin because it can take away skin's lipids and you get "dry hands". In France, your employer would be forced to provide you with protecting gloves, e.g. latex gloves.

On a Good News™ note: Acetone isn't so bad, compared to some cleaning solvents in active use.

In active use in countries that don't mandate basic security gear like rubber gloves, that is.

(With suitable fire precautions and ventilation) dip a thermometer wrapped in a rag into some room temperature acetone. I bet you will see the temperature drop as the acetone starts to evaporate. So while SteffX is correct that the cold feeling on your skin is because heat is drawn quickly from your skin, the acetone is actually colder. http://highschoolenergy.acs.org/content/hsef/en/how-can-energy-change/energy-of-evaporation.html

@dcorking But StellX states clearly that the "skin gets colder because of [the evaporation]".

@Crowley Yes, the answer is great, because the lower temperature, and the sensation of cold skin, are both caused by evaporation. My only concern is with this sentence, 'So acetone is not "colder", it just cools down your skin more easily when evaporating,' which is not precisely correct.

I think that it was meant that "if you pour your hand with water and acetone both with same temperatures, the acetone will cool your hand faster (easier)". Nevertheless, the sentence is  not written well.

Acetone is never let evaporating in a large piece of glassware in real life. It is most usually stored in a wash bottle and evaporation is very limited. Original question was about the difference between washing with water and acetone. Hence my answer.

@dcorking No, the acetone is NOT colder. The fact that the thermometer would suffer a drop in temperature is due to the evaporation and the fact that the thermometer suffers first from the endothermic process. It will then exchange with the rest of the air in a slower manner and be at equilibrium. If you do the experiment in conditions where evaporation is not possible (close to the fusion point -94.6°C for instance), then the thermometer will not suffer from any temperature drop.

@Martigan exactly: any liquid that is adiabatically evaporating will get colder.


text:
Are there any observable effects of casein binding to capsaicin?
I would like to complete an experiment on balancing capsaicin. Since both capsaicin and milk are bases on the pH scale, I prefer not to use the common term "neutralizing".
I will be extracting the capsaicin and I do not wish to taste-test it. Is the bound casein protein with the hydrophobic capsaicin oil visible? Will it curdle in order for me to obtain quantifiable amounts?

Are you sure that it is the casein, not the fat, in milk that neutralizes the heat from capsaicin?  Also, keep in mind that "neutralizing" capsaicin just means preventing it from binding to a G-protein coupled receptor in your mouth.  Since GCPRs are proteins, all that needs to happen for it to be "neutralized" is that it needs to bind to or stick to something more strongly than it sticks to GCPRs in your mouth.

Capsaicin is very oil-soluble. Most of the things that "neutralise" the perceived hotness are actually ways of moving the compound into the oil phase of a mixture (which is what I think happens in milk) so that less of it is stuck in the heat receptors where its effect is felt.

Also, the Oleic acid found in milk is an inhibitor of TRPV1 (aka, the capsaicin receptor).

Capsaicin doesn't dissociate in water and can't act as a base (like well-known bases such as Sodium Hydroxide or Ammonia) and won't react with the casein in milk. So there won't be any curdling/precipitation of casein protein. Casein will precipitate through pH adjustment if you lower its pH to the isoelectric point (from 6.6 to 4.6).
Notes:

Casein isn't a base, neither an acid, it has an amphoteric nature.
Bases generally don't react with each other.
Pure capsaicin is a powder, not an oil.