text:
Interpretation of 1H-NMR spectra for Z-ochtodenol

I'm struggling to understand the 1H-NMR data for this compound. I managed to find the compound using the data I had, but at the end I can't assign the peaks properly.
My confusions are:

Shouldn't the $1.96~\mathrm{ppm}$ (2H, s) protons couple with the proton from the blue carbon and give a (2H, d, $J~2~\mathrm{Hz}$) by allylic coupling? [Same issue with the protons from the blue carbon.]
What's up with the broad multiplet?
Counting the protons from the data gives us 15 instead of 18; why's that?

Let’s get the obvious out of the way first: Yes, the light blue protons should probably be the allylic ones since their shift is so similar to the other allylic protons’ shifts. (The green ones, that is; the red ones also experience deshielding from their neighbour oxygen.)
Then also, the report of the spectrum has been done badly. It should have included the field used. It makes a big difference if a spectrum is measured at $90~\mathrm{MHz}$ or at $700~\mathrm{MHz}$ — it may well be that weaker couplings are unobservable in lower field spectra but clear in higher ones. If I had to guess, I would assume a rather weak magnet since the smallest coupling reported is $6~\mathrm{Hz}$ — and they don’t even include a further significant figure.
Therefore, concerning your first bullet point: It is entirely possible that those protons do couple only the spectrum had too low a resolution to show it. However, remember that coupling should always be considered a possibility. If you check out the Karplus curve, you will see that for an angle of $\approx 90^\circ$ vicinial coupling effectively becomes zero. Same things may happen with long-range couplings: They may be absent with no apparant reason to the organic chemist since we cannot see the molecule’s exact orientation.
A ‘broad multiplet’ makes no sense to me whatsoever. In my opinion, multiplet alone should be enough. And I also strongly suspect an error in the book here: That is the shift where the four missing question-marked hydrogens should appear at. They should have badly overlapping, possibly even complex spins since each one is magnetically unequivalent and we are dealing with a cyclohexane system. This line should be, in my opinion:

$1.35-1.70$ (4H, m)

And that also explains why three protons are missing. The pink ones are the only ones that would allow more protons in the same spot.
All in all that last bit is an error that really shouldn’t have happened in a textbook exercise. I do hope there is an erratum somewhere.

Thanks for explaining! Now, knowing those facts, how should I approach structures that allow couplings with small coupling constant? Is it possible that a spectra can show only some long-range couplings and not all? (i.e. two allylic systems in a compound, from which we get long-range couplings only for one of those)

@L3ul Yes, that is definitely possible. Sometimes, you may even be able to use the presence or absence of a coupling to determine the stereochemistry. (i.e: If that coupling is missing but that is there then it is more likely that these residues be syn rather than anti or whatever.) How to approach them: Assume the absence of any coupling less than $5~\mathrm{Hz}$ initially unless you know the spectrum has been recorded at at least $400~\mathrm{MHz}$. If it’s there, celebrate and use as further confirmation ;)

Small couplings are afaik easier to see with lower field strength, not higher - I explained it a bit more in my answer.

Based on an old comment from NotEvans, it turns out that the reported NMR data is from the end-of-chapter problems of Organic Chemistry 2ed, Clayden et al. (Problem 11, Chapter 18). These can be accessed online using the username and password provided in the front cover of the textbook.

δH (ppm) 0.89 (6H, s), 1.35-1.70 (1H, broad m), 1.41* (1H, s), 1.96 (2H, s), 2.06 (2H, t, J 6 Hz), 4.11 (2H, d, J 7 Hz), and 5.48 (1H, t, J 7 Hz).

There is an erratum in this data; the "broad multiplet" should integrate to four protons, not one. This is not officially corrected in the published list of errata, but is sufficiently obvious that I am willing to go on record to say it is a typo.
In my experience, allylic couplings can sometimes be difficult to resolve, unless you manipulate the spectrum (e.g. with a suitable apodisation / window function). So, the lack of allylic coupling in the data shown above is not entirely surprising.
With this knowledge in hand, the spectral data can be reasonably easily assigned, and is handled admirably by Jan's answer.
However, there is something that I wanted to comment on, and it is that counter to intuition, small couplings (e.g. allylic coupling, ~1.5 to 2 Hz) are typically easier to see with lower field strength, not higher. It's because if you want to cover a spectral width of 20 ppm at 600 MHz, that corresponds to 1200 Hz (vs 400 Hz at 200 MHz). If all the experiment parameters are kept the same, the digital resolution (in terms of Hz per point) at 600 MHz is 3 times poorer than at 200 MHz.
This might not be the most instructive example, but I am not particularly bothered to find a better one. This is a methyl group which is displaying one large coupling ($^3\!J$) and two small couplings (one $^4\!J$ and one $^5\!J$). You can see that the resolution is ever-so-slightly better in the 400 MHz spectrum. Both spectra were acquired with the same spectral width (20 ppm), the same number of points in the FID (64k), and no zero filling has been applied. As a result, the digital resolution in the 500 MHz spectrum is 1.25 times worse (0.32 Hz vs 0.24 Hz for the 400 MHz).


text:
Coupling patter for a methyl-butyl ether
What would the high resolution H-NMR spectroscopy coupling pattern look like for the hydrogen attached to the carbon the arrow is pointing at? I am confused because there is a CH2 connected to either side of the carbon, but the hydrogens on each carbon are in different chemical environments since the one on the left is next to the oxygen in the ketone, so would it coupling pattern be a doublet of doublets? or just a doublet?

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Solubility of carbon dioxide in 1M sodium hydroxide solution?
I know that carbon dioxide is soluble and forms carbonic acid, but exactly how is solubility measured?
How soluble is carbon dioxide in a 1M solution of sodium hydroxide?

Difficult question, because there is the physical solubility of the gas itself which is in equilibrium with all the carbonate species.

The question is really somewhat out of whack. In sodium hydroxide any  carbon dioxide is converted to the carbonate anion. In a strong sodium +1 solution sodium carbonate has limited solubility, and will ppt out. So until the NaOH is substantially reacted there really isn't any "dissolved" carbon dioxide.

If you are still interested in this, do you mean the analytical methodology for quantifying carbon dioxide or the experimental procedure before the carbon dioxide analysis part?

This is a useful recent article for measuring solubility: https://pubs.acs.org/doi/10.1021/je200991x. Note that fundamentally solubility is a physical equilibrium but in many instances, it involves chemical reactions.

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text:
Why is it easier to determine the radius of inert/noble gases?
My textbook says it is possible to determine the atomic radius of the inert gases like He, Ne, Ar etc because they roam freely and are incapable of forming bonds. However it is very difficult to find out the atomic radius of elements that are chemically active because the position of electron in an atom is uncertain (according to Heisenberg's principle I guess). I am just curious that the principle is true for the electrons of inert gases too, then why did they say it was possible for those gases but very difficult for the chemically active atoms?

Not quite the right focus. The inert gases, in the gaseous state, have one "pure" electronic configuration, and one "pure form. For an ion like Sodium +1, it forms numerous compounds. The "pure" ionic charge of +1 is an abstraction. So sodium salts are in a variety of crystal lattices with various "partial" charges (something  between a pure ionic bond and a pure covalent bond). So for every salt sodium would have a slightly different radius for sodium.

There are several definitions of atomic radius. Guessing from context of your answer, I assume you are talking about  van der Waals atomic radius. By definition, it is a radius of atom in respect of equilibrium position with other atoms when only van der Waals forces are acting between the atoms. Since atoms of inert gases do not tend to make covalent bonds, the equilibrium distance between two in solid equals double atomic radius - plain and easy. Not so easy for other elements.
Since most metals and some non-metals form closely packed lattice in elemental form, there is no simple way to obtain their vdW radius. It is required to find a compound with the atom sterically unhindered and forming a crystal lattice containing close contact with said atom involved. Since most metals are electron-deficient and tend to interact with almost anything, no such experimental data are available for metals in a strict sense (though I can come with few examples when it may make sense). As such, vdW radii can be measured (and, actually, are meaningful) only for atoms of top-right corner of periodic table plus hydrogen. Furthermore, even for them vdW radius may fluctuate from compound to compound depending on the oxidation state and atomic neighborhood. Furthermore, the radius is pressure-dependent and, for chemically bound atom, may be unequal in different directions.
For covalent radii the problem is even larger, as the radius is dependent on near neighborhood and valence state in significant way (N-N bond legnth may vary from 1.10 Angstrom in dinitrogen to 1.45 in hydrazine)
As such, even if people talk about larger and smaller atoms, larger and smaller substituents, there is no strong definition for such relations.
As for atomic radius no related to vdW and/or covalent interaction, it depends on the definition of atomic border, which does not exist. A free atoms does not have a clear border, its electronic density fades exponentially with distance, but is non-zero for all finite distances.


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Green kerosene in copper container goes colorless
I recently got hold of an old blowlamp that had not been touched for more than 30 years.
The fuel tank of the blowlamp is made of brass.
I poured the old kerosene from the blowlamp into a clean polyethene flask, the kerosene was clear and quite green in color. The flask got about half full.
The next day the kerosene had turned colorless and there was a brown  precipitate. The flask was not shaken or subjected to strong light.
What happened here? Something that had been stable for years suddenly reacting. If it is a reaction with oxygen, does oxygen diffuse that easily through kerosene?
Was the green substance an inorganic copper salt dissolved in the kerosene or was it some kind of organic "copper kerosenate"?

You probably aerated the kerosene pretty well when you poured it from one container to another. The actual oxidization reaction may have taken a while, which is why you didn't observe any color change immediately.

The brass is a metal mixture of cooper and zinc. Copper compounds are known to form blue and green salts.
Examples: Copper Chloride

I believe that green color it's a copper compound and the copper reacted with oxygen (air) and the preciptate is copper oxide.

Also, the oxygen solubility in hidrocarbons it's around 60/70 ppm¹.
¹Corrosion for everybody - Alec Groissman


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What is the velocity of an ejected electron in lithium when subjected to UV radiation?
If UV radiation strikes an atom of lithium with one electron in its lowest allowed energy state with enough energy to eject the electron from orbit, what is the electron's velocity upon ejection? The wavelength is 7.1 * 10^-9 meters.
If this was hydrogen, I would use the fact that the energy of the radiation is the sum of the binding energy and kinetic energy of the electron, but the conditions are for lithium. How would I solve this problem given that it's lithium and not hydrogen?

The energy is still the sum of the kinetic and the binding energy. The only difference is how you calculate the binding energy, but you could just look it up


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How to calculate concentrations when dissolving two salts?

Some solid silver bromate was dissolved together with solid silver acetate what would be the highest achievable pH?
  \begin{align}
K_\mathrm{sp}(\ce{AgBrO3}) &= 5.38 \times 10^{-5}\\
K_\mathrm{sp}(\ce{AgCH3COO}) &= 1.94 \times 10^{-3}\\
\mathrm{p}K_\mathrm{a}(\ce{CH3COOH}) &= 4.76
\end{align}

Now I know this is just a question about the common ion effect. I know how to do the question if silver ions are already present in the solution then silver acetate is added. Also given the concentration of acetate ions, I can easily calculate the pH.
The real problem with this question is that you are dissolving 2 salts which are both in equilibrium. I don't know how to calculate the concentration of acetate ions in this scenario.
This is what I did:
$$5.38 \times 10^{-5} = [\ce{Ag+}][\ce{BrO3-}] = x^2$$
$$x = \ce{[Ag+]} = 7.33 \times 10^{-3}$$
$$1.94 \times 10^{-3} = [\ce{Ag+}][\ce{CH3COO}] = (7.33 \times 10^{-3} + y)(y)$$
I then used quadratic formula to find y.
I am pretty sure that this method is wrong because wouldn't the dissolution of silver acetate cause the equilibrium of silver bromate to shift backwards, causing the value of x to decrease. Or can we assume that this shift is negligible.
Also, by highest achievable pH doesn't that imply that the final pH of solution is variable. However, shouldn't it have the same pH every single time, making the words highest achievable pH redundant? Or am I missing something?

Different amounts of silver bromate and silver acetate would have different pH values. So "highest achievable pH" means to find the amounts of silver bromate and silver acetate which would give the highest pH.

Here you can find my answer to a similar question http://chemistry.stackexchange.com/a/40433/15235 :)

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How to distinguish diastereomers of unsaturated ketones by NMR?
I am synthesising a natural product that contains an α-methylated α,β-unsaturated ketone (see image below). The natural product’s stereochemistry is an (E) configured double bond. I selected this double bond as one of the retrosynthetic cuts of the overall molecule and I want to be sure that when I synthesise it that I get the correct stereochemistry. Is there any easy way to distinguish the (E) from the (Z) isomer?

Generic structure of an (E)-α-methylenone.

I do wonder why I got a downvote for this question. If the person who did so could leave a comment, I would be thankful.

Probably didn't like your hat-seeking behaviour. I mean, hat-seeking or not, it's still a good Q&A and we can do with more content like this. (flag as obsolete or too chatty if you feel evil)

According to Noack and Jones, unsaturated ketones with an α substituent — even if it is only methyl — generally adopt an s-trans conformation due to steric reasons. Only if the α substituent is hydrogen can an equilibrium between s-cis and s-trans be observed.[1] Therefore, we should better draw the structure this way:
$\hspace{30ex}$
$\hspace{28ex}$s-trans configuration of α-methylenones
I already explicitly added the unsaturated hydrogen atom to the image. As we can see, it is either rather close to the keto group or pointing away from it. If the alkene’s configuration is (E) and the hydrogen is thus pointing ‘inward’ it experiences a larger deshielding due to the close carbonyl group. Therefore, its signal in 1H-NMR is shifted to lower field when compared to the (Z)-configured double bond.
You can also compare experimental shifts. Unfortunately I didn’t find a pair of identical molecules differing only in the double bond orientation, and not even the solvent is the same, but I think the data still shows a clear trend, namely that for (E) enones the chemical shift is in the high alkene range while for the (Z) enones it is in the low range:
$\hspace{33ex}$
(E)-3-methylhex-3-en-2-one. 1H-NMR ($500~\mathrm{MHz}$, $\ce{CDCl3}$) $\delta =$ $\mathbf{6.59}$ (tq, $J = 7.2,1.3~\mathrm{Hz}$, 1 H), $2.28$ (s, 3 H), $2.24$ (app. qnt, $J = 7.4~\mathrm{Hz}$, 2 H), $1.73$ (s, 3 H), $1.05$ (t, $J = 7.6~\mathrm{Hz}$, 3 H).[2]
$\hspace{33ex}$
(Z)-3-methylnona-3,8-dien-2-one. 1H-NMR ($300~\mathrm{MHz}$, $\ce{C6D6}$) $\delta =$ $5.71$ (ddt, $J = 17.1, 10.5, 6.6~\mathrm{Hz}$, 1H), $\mathbf{5.38}$ (tt, $J = 7.5,  1.5~\mathrm{Hz}$, 1 H), $4.92$–$5.02$ (m, 2 H), $2.29$ (q, $J = 7.5~\mathrm{Hz}$, 2 H), $1.92$ (q, $J = 7.2~\mathrm{Hz}$, 2 H), $1.85$ (s, 3 H), $1.60$ (t, $J = 2.7~\mathrm{Hz}$, 3 H), $1.35$ (qnt, $J = 7.5~\mathrm{Hz}$, 2 H).[3]

References:
[1]: K. Noack, R. N. Jones, Can. J. Chem. 1961, 39, 2225. DOI: 10.1139/v61-294.
[2]: W. F. Austin, Y. Zhang, R. L. Danheiser, Org. Lett. 2005, 7, 3905. DOI: 10.1021/ol051307b.
[3]: L. Jiao, C. Yuan, Z.-X. Yu, J. Am. Chem. Soc. 2008, 130, 4421. DOI: 10.1021/ja7100449.

Wouldn't a more accurate way of determining it be getting an NOESY?


text:
Requirements for electroplating
I understand that electroplating involves using an anode (copper in this case) and an electrolytic solution containing Cu(II) ions with some sulfuric acid. Why exactly does the anode need to be copper if you want to plate something with copper; couldn't something like a graphite electrode be used instead, as the $\ce{Cu^2+}$ ions can be provided by the solution?

Yes, the anode can be a different metal. It probably is shown as the same metal so as the metal from solution would "bind" with the anode material, and so that the reaction would be reversible.

You are right in saying that the electroplating could equally well be carried out with any electrode. Just offhand though, I could think of a couple of reasons why you would want the anode to be made of copper.

Choosing copper as the anode makes the overall cell potential equal to zero. The reaction at the anode is $\ce{Cu -> Cu^2+ + 2e-}$ and the reaction at the cathode is $\ce{Cu^2+ + 2e- -> Cu}$ - they are identical but one is flipped, so $E^\circ_{\mathrm{cell}} = 0~\mathrm{V}$. Compare this to if you used graphite - the reaction at the anode would then be $\ce{2H2O -> O2 + 4H+ + 4e-}$ and you would have $E^\circ_{\mathrm{cell}} = -0.892~\mathrm{V}$, meaning you would need more energy to run the reaction.
More importantly, it also ensures that the concentration of $\ce{Cu^2+}$ ions in solution remains constant. If you use a graphite electrode, the solution can initially provide the $\ce{Cu^2+}$ ions required for electroplating, but as the concentration of $\ce{Cu^2+}$ decreases, $E_{\ce{Cu^2+}/\ce{Cu}}$ becomes smaller (by the Nernst equation) and it becomes harder and harder to reduce $\ce{Cu^2+}$ at the cathode.


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Reaction of cyclohexane-1,2,4-triol with acetone

What is the product formed when cyclohexane-1,2,4-triol reacts with 1 equivalent of acetone in an acidic medium?

There are 3 possible hemiacetal products. The major one is formed via nucleophilic attack by a less hindered hydroxyl group:

There is one possible cyclic acetal formed by the two adjacent $\ce{- OH}$ groups:

But which of the two products will be formed in a larger quantity? I have heard that cyclic acetals are more stable than acyclic ones but don't know the reason (because of entropy change or something?).

related Why are 1,3-diols protected with aldehydes but 1,2-diols are protected with ketones?

As long as you have a proton source in your solution (hint: all the time) these acetals will equilibrate back and forth until they find the thermodynamic hole. So you will never observe a hemiketal like the one you drew; rather, it will always move forward with the elimination of water (there is only little water around if acetone is the solvent, so liberation of water is entropically favourable) to the formation of the $\ce{O{,}O}$-ketal.

Note that you can form two different ketals if the stereochemistry of the hydroxy groups is correct: If they are 1,3-syn, the six-membered cyclic ketal can also be formed.

Which one you end up with will likely depend solely on the reaction conditions. Kinetically, the five-membered ring is favoured. Thermodynamically, I assume the six-membered ring to be more stable, if it is able to find a conformation with little steric stress.


text:
EMF measured while verification of Nernst equation keeps on increasing
I was doing this experiment to verify Nernst equation in my school's lab. I had two beakers one with $\ce{ZnSO4}$ and the other with $\ce{CuSO4}$ both having a concentration of $1~\mathrm{M}$. I had a saltbridge having $4~\%$ Agar in $1~\mathrm{M}\ \ce{KCl}$. The temperature was $21.3\ \mathrm{^\circ C}$ Now the problem was that when I connected a multimeter across the electrodes it showed an EMF of $0.8\ \mathrm{V}$ initially and it went on increasing to $1.001~\mathrm{V}$. The increase was rapid initially and became slower and slower. The EMF still kept on increasing but with a very slow pace.
By what I have studied the EMF I get initially should be the maximum one and then go on decreasing as the equilibrium is attained.
So why was the EMF in my experiment increasing? Is it because I had not used wool at the ends of the salt bridge? But the Agar was already a thick gel and was not flowing out of the U-tube.
I asked my teacher and she said that there might be some problem with the multimeter. Does any one of you know the reason for the increasing EMF?

Hi and welcome to chemistry.stackexchange.com. Thanks for already taking the [tour] =). I edited your post amoung others to include MathJax for maths and chemistry formatting. For more information on how to do so yourself, check out the [help], this meta-post or this one. Nice first question, btw!

Do you know the materials for which the electrodes are made?

Comment #1 re: "By what I have studied the EMF I get initially should be the maximum one and then go on decreasing as the equilibrium is attained. " // Not so much "equilibrium" but "steady state."  A drop in potential would indicate some sort of polarization at the electrodes. Comment #2 - Based on #1, there is a difference between the "cell potential" and the "working voltage" as a "lot" of current is flowing through the cell.

Comment # 3 - I'd expect the multimeter to have a high input impedance and hence that little current would flow in the system.

My guess - I'd guess that the observation is due to the salt bridge. I'm thinking that it would act like a capacitor in the system. So it takes some current flow through it to reach equilibrium. // Need someone with real knowledge of salt bridges to comment on this aspect.

You should arrive to a plateau value, the EMF can't increase perpetually. The ideal potential difference is around 1.1 V given the experimental arrange, this is also known as the Daniell Cell. I think the low initial value you get is because of the time response of the system, until it reaches a steady state.
There are basically two types of connections:
i) Electrode/Solution: Some specie B transfers an electron from the solution to electrode surface (by receiving or by giving). The electron transfer rate is very high, specially on batteries, so this can not limit the whole process.
ii) Ion migration through solution: This step is the limiting process. The ions are the charge carriers in bulk solution, which move much slower compared with electrons in a conductive metal. A way to overcome this is using strong acid solutions, as sulfuric acid in the Lead-Acid Battery used on cars.
The response you have will depend on the time of measure (immediately after closing the connection, 5 seconds after, etc.) and the time interval of such measure. It will depend also on the initial concentrations of the solutions (high or low). If the concentration is low, the ionic strength of solutions is weak and few ions are to carry the charge; adding table salt (NaCl) is a possible solution. The ionic bridge may be highly conductive but, if the water permeation is also limited, you have an extra limitation there, which will be higher as the phase thickness increases (ie using too much agar, or long bridge). A solution may be, using an U-Tube with NaCl 4 M and covered in the ends with commercial water-purifier filters well sealed to the glass.
A delay on response could be due to the hydration time of the agar gel, which can take a while. Once water has permeated the agar, the ions begin to move faster, until steady state is reached.


text:
Mechanism of protein precipitation with TCA
In practice, TCA (trichloroacetic acid) is widely used for quantitative protein precipitation, but so far I have failed to discover the exact mechanism of TCA action. I highly doubt it's just the effect of pH that precipitates proteins.
Any better guesses?

The paper "Trichloroacetic acid-induced protein precipitation involves the reversible association of a stable partially structured intermediate" (Rajalingam et al. 2009 Protein Science 18(5), 980–993.) may be relevant.
In the introduction, they mention the conventional explanation is that "TCA forces protein to precipitate by sequestering the protein-bound water" - that is, it may force protein precipitation in a related fashion to chaotropic agents or antichaotropic agents like ammonium sulfate.
Their experiments indicate a more complex explanation. They hypothesize  that the negatively charged TCA disrupts electrostatic interactions in the protein, resulting in a partially folded state which is prone to aggregation. The fact that TCA precipitation is less effective for intrinsically disordered proteins lends support to this hypothesis (as they don't have an aggregation-prone partially unfolded state).
The acidic nature of the TCA (the pH effect) helps with the rate of precipitation, but does not appear to be necessary, as the sodium salt of TCA is able to precipitate proteins in neutral solutions, although slower than acidic TCA.


text:
Hydrolysis of epoxide

There are two possibilities:

Oxonium ion formation
Cleaving of epoxide

Firstly, I don't know if epoxide is more reactive than carbonyl group.
If epoxide reacts with the proton,

But the given answer is:

In the above mechanism, an unstable carbocation is formed because its near a carbonyl group which is electron withdrawing:

The carbocation formation don't occurs. It's a concerted step.

Why? How do you know it has to be concerted?

You are correct in your last structure. This carbocation is very unstable, because this the mechanism is concerted.
See the two products, whats more stable?

Both are equally stable? Are you referring steric repulsion? But the product in my answer is stabilized by conjugation which is not there in the answer given.

In your answer the second carbocation seems unstable and the final product in your answer is more unstable in relationship the correct answer.
I think that the product that you propound are a subproduct.

Are you sure the answer given in your type script photocopy is correct?  A Reaxys search of your starting material gives this paper, http://www.nrcresearchpress.com/doi/pdf/10.1139/v75-242 which labels your starting material as cpd 12 and says that it gives the spiro diketone cpd 25  which has smiles string O=C1CCCC(C21CCCC2)=O and suggests an acyl migration.  In your diagram, the first protonated structure, the arrow would come from the bond between the carbonyl carbon and the hydroxyl carbon to the carbocation with subsequent loss of a proton.  Bear in mind the paper cited is from 1976.

I don't know. I will post the given options.

I would go with your approach but your final product is unstable due to high repulsion's by $\pi$ electrons.

Hence your final product will react further to form a more thermodynamically feasible product.

NOTES:
In second structure two anti-periplanar migrations are possible and the one which leads to the more stable carbonation is favored (or consider the better migrating group based on electronegativity). Third structure rotates about spiral carbon due to repulsion's.

Is any kind of solvent required for such a rearrangement?


text:
The logic for hybridisation formula
Hybridisation for a molecule is given by:
$$\frac{1}{2}(V + H - C + A)$$
Where,
V = Number of valance electrons in central atom
H = Number of surrounding monovalent atoms
C = Cationic charge
A = Anionic charge

If this expression is 2 then its $sp$ hybridised, 3 then $sp^2$ hybridised and so on.

However, I am not able to understand why this works.

duplicate of http://chemistry.stackexchange.com/questions/4399/

@MaxW No, it’s not.

@MaxW I've already read that question before asking.

Since you understand why it works for monovalent surrounding atoms, I will assume   that we have a molecule $\ce{MA_{$m$} B_{$n$}}$ where $\ce{M}$ is the central atom, $\ce{A}$ represents the monovalent atoms which may or may not be identical, and $\ce{B}$ represents not necessarily identical atoms which are polyvalent. First up, there are some assumptions to be made:

All "surrounding atoms" are directly bonded to the central atom $\ce{M}$, and with no other atom;
All atoms form as many covalent bonds as is indicated by valency;

Now say the $n\ \ce{B}$ atoms are absent. Let the steric number be $x$,
Then, $x=0.5\times (V+m-C+A)$
Now add the $\ce{B}$ atoms. Now the central atom forms $n$ more sigma bonds, but if the sigma bonds were not present, these would exist as lone pairs, and hence already counted in $V$. Take for example, $\ce{PCl_3}$ and $\ce{POCl_3}$. For $\ce{PCl_3}$ you would get $0.5\times (5+3)=4$ and for $\ce{POCl_3}$ you would get $0.5\times (5+3)=4$. The reason the $\ce{O}$ made no difference is because phosphorus uses two of its electrons to bond with both of which are nothing but valence electrons of phosphorus, only present as a lone pair in one case and a sigma bond in the other. Also, this formula does not work when you have triple bonds, say for $\ce{HCN}$.
As to how it works for mono valent atoms being the only surrounding atoms, it is simple.The number of sigma bonds is the number of mono valent atoms, and thus by adding that to the number of lone pairs we get the hybridization. Say you have $ n $ lone pairs and $s$ sigma bonds.You can see that $V=2n+s$, and that $s=H$. So $0.5*(H+V)=n+s$, as required.

Please format chemical expressions using the \ce{...} syntax. Also please remember to put a space behind a full stop.

I think that I didn't (or incorrectly) understand why hybridisation formula works for monovalent atoms too.  I am sorry for that, I'll change the question too.

@xXLemonProgrammerXx I've edited my answer.


text:
Electron withdrawing groups and acidity
I know that a strong electron withdrawing group reduces electron density and makes a molecule more acidic. Also, a meta substituent will have less effect than ortho and para substituence and the para isomer will be more acidic than the ortho one.
The point that I wonder: Does the acidity always increase with the number of withdrawing group as well?
Some group, let's say, phenol has one strong electron withdrawing group like nitro, carboxyl, chloro etc. If phenol would have two more of that same withdrawing group attached, would it be simply more acidic? Or, should another factors be considered to determine the acidity? I thought that maybe it would be more reliable to look $\mathrm{p}K_\mathrm{a}$ values but couldn't find any molecule that fits that situation.

Let’s look at some data:

phenol: $\mathrm{p}K_\mathrm{a} = 9.86$
3-nitrophenol: $\mathrm{p}K_\mathrm{a} = 8.34$
4-nitrophenol: $\mathrm{p}K_\mathrm{a} = 7.23$
3,4-dinitrophenol: $\mathrm{p}K_\mathrm{a} =5.42$

$\Delta \mathrm{p}K_\mathrm{a}(\text{P/3-NP}) = 1.52$
$\Delta \mathrm{p}K_\mathrm{a}(\text{4-NP/3,4-dNP}) = 1.81$
$\Delta \mathrm{p}K_\mathrm{a}(\text{P/4-NP}) = 2.63$
$\Delta \mathrm{p}K_\mathrm{a}(\text{3-NP/3,4-dNP}) = 2.92$
As you can see, the differences in $\mathrm{p}K_\mathrm{a}$ are not strictly additive but enfore each other; one could almost say in a cooperative manner.

Yes and no, with more electron withdrawing groups at ortho and para positions would certainly increase acidity but other factors do play a role. For example,
In trinitrobenzoic acid

here the two $\ce{NO2}$ at ortho positions are very bulky and due to steric repulsion rotate the $\ce{COOH}$ group thus making the compound non planar due to which there will be no more resonance thus, reducing acidity drastically. This is the most important factor that plays role in reducing the acidity.

@Mithoron edit made! I made the same mistake here that I made in the exam. But, can you tell why this doesn't happen in picric acid? I have seen that methyl group can rotate the OH group so why not NO2?

Compare their anions stability


text:
How to determine optical activity of a compound without asymmetric carbons theoretically?
I know that a compound doesn't need to have a chiral carbon to be optically active. So how can I find out whether a compound drawn on paper is chiral without access to any experiments? e.g. 1,2,2-tribromopropane
The question is expected to be solved using only pen and paper

But a sample of the compound in a polarimeter and try to measure optical activity.

@MaxW The problem is the question like these show up in MCQ based college entrance exams with no excess to lab.

@Jan I am expected to answer such questions without access to lab. I want to know if there is no chiral carbon in a compound how would I know that it has a non-superimposable mirror  image?

Draw it's mirror image, if the two images are not superimposable then the molecule is chiral.

It still looks like homework - in such simple cases checking for chiral atom is ok.

A substance will be optically active if it is not identical to its mirror image. Check out this great answer to see why. So the dullest and brute-force-est way to find out whether a molecule is chiral is to draw it, to mirror it, and then to attempt to overlay image and mirror image using only rotation. (Remember to draw molecules three-dimensionally!)
But there are a few shortcuts. You can look at the molecule as a whole and check out what symmetry features it has — determine its point group. If a molecule has a centre of symmetry (i.e. inversion results in the identical molecule), a plane of symmetry or improper rotation (which is a rotation followed by mirroring along a plane orthogonal to the rotation axis) it is achiral. If you cannot find any of these symmetry elements, the molecule is chiral.


text:
Calculating volume ethanol in an fermented aqueous solution of sucrose, water, and ethanol
I've been working on a biology experiment where I calculate the fermentation yield of yeast under various light induced stress responses inhibiting the translation of various fermentation enzymes. My solution initially contained 12.5g of sucrose, and 50g of water, resulting in 20% by mass sucrose.
Initially, I expected to measure this value before and after using a refractometer that measures in brix, which is a measure of percent mass of sucrose in a solution. I figured that the difference between these two values would represent the percent by mass ethanol in the final solution.
I failed to consider though that the yeast produced ethanol as the experiment progressed changing the mass of the solvent and therefore the percent mass of sucrose of the solution.
Primarily, my goal is to calculate the volume of ethanol in the solution. In order to do this though, I would need to be able to calculate the volume of the whole solution based on the change in the percent by mass sucrose in the solution, and the volume of the sucrose in the solution. With these values I could subtract the volume sucrose and the constant volume water from the solution and obtain the volume of ethanol.
Based on some internet searching, I've gathered that this is possible but it involves a knowledge of intermolecular interactions paired with a knowledge of the molarity of particular molecules. This is beyond my scope of knowledge for chemestry. Overall, I'm trying to develop a generalized equation for determining volume ethanol based on change in brix of the solution. This has been baffling me for a while now and I figured there may be other amateur researchers with the same questions I have so I figured I'd pose the question to the community at large. Thanks for the help!

It seems to me that your experimental design fails on several points. First yeast is inhibited by alcohol. I think about 16% (vol %) is the concentration that kills yeast. So I think your sugar content is beyond that. Second I'd expect the conversion to be non-linear. At first yeast is multiplying rapidly and hence converting rapidly. But as the alcohol % increases I think conversion would slow. Third I'd wonder about conversion of sucrose into glucose and fructose.  I know that brix is used in wine and beer industries so this may not be a problem.

@MaxW I can get the yield with an online calculator so I know it's possible but since I'm writing a paper on this I have to demonstrate the underlying calculations. Also I don't think it should matter that the yeast dies because it's a constant strain of yeast so it would all be inhibited at the same levels of ethanol exposure

I'm not sure if the data in the link can be adjusted to your sugar content, but boiling the solution to remove ethanol, then diluting to some constant volume would be a way to generate your own calibration curve. http://www.ithacoin.com/brewing/EtOH_to_Brix.htm#Measure_ABV

Normally a hydrometer is used to analyze alcohol content by measuring the specific gravity before and after fermentation. The problem here is that a reading needs to be taken prior to fermentation also to compare the final hydrometer value with. Maybe you could make up an identical solution of the original mixture , take a hydrometer reading then compare that with the hydrometer reading of the subject mixture that has already undergone fermentation.

I took a reading before hand. The refractometer measured in both brix and sg so I figured there would be a conversation. It was 20 brix initial

Some of the fundamental calculations are detailed here. http://web2.airmail.net/sgross/fermcalc/fermcalc_alcohol.html#honneyman Google search for additional references: https://www.google.com/search?as_q=sucrose+ethanol+brix&as_epq=&as_oq=&as_eq=&as_nlo=&as_nhi=&lr=&cr=&as_qdr=all&as_sitesearch=&as_occt=any&safe=images&as_filetype=&as_rights=

Before posting this, I wasn't aware of the homebrew stack exchange . I think I've got an answer based on http://homebrew.stackexchange.com/questions/2924/calculating-alcohol-by-volume so thanks.

@user2985955 Would you mind summarizing the chemically relevant portion here in an answer?  I think this might be useful to future readers over here as well.


text:
Is methanol more acidic than water?
Methanol is slightly more acidic than water. Their $\mathrm{p}K_\mathrm{a}$ values, in water, are $15.5$ and $15.7$, respectively. All other aliphatic alcohols, however, are less acidic than water.
Is the following reasoning correct? This is my best rationalization; is there anything better or anything that can be added?
In both hydroxide and methoxide, we have an $\ce{O}$ bearing negative charge. In the former, we have a hydrogen attached to the $\ce{O}$. In the latter, we have a carbon attached to the $\ce{O}$. Carbon is more electronegative than hydrogen, therefore, carbon should be able to better withdraw electron density via induction from the $\ce{O}$. This is a stabilizing interaction.
We often think of methyl groups as inductively donating, but that's molecular profiling. They can inductively donate to a carbocation, partly because a carbocation is highly electronegative, so the carbocation pulls the electrons toward itself pretty well (so there's both pushing and pulling of electrons going on). When you have a methyl group attached to an $\ce{O}$, however, the methyl group is often inductively withdrawing. NMR data supports this.
Now the question is why aren't other aliphatic alcohols more acidic than water? In longer-chain aliphatic alkoxides, you don't just have a methyl group attached to the $\ce{O}$ bearing the negative charge - you have a bunch of methyl groups strung together. These lessen the amount of inductive withdrawal that the alpha carbon can do. Each$\ce{-CH2 -}$ unit attached to the alpha $\ce{-CH2 -}$ is somewhat inductive donating to the alpha $\ce{-CH2 -}$.

The $pK_a$ of water is 14.0 so it is more acidic than methanol.  http://chemwiki.ucdavis.edu/Organic_Chemistry/Fundamentals/What_is_the_pKa_of_water%3F

@MaxW - well, that opens up a whole new can of worms because I do see the 15.7 value cited a lot. ;) Time for another question?

14.0 is $-\log(\ce{[H+][OH^{-}]})$; 15.7 is $-\log{\ce{[H+][OH^{-}]\over[H2O]}}$.

@IvanNeretin but can we say that the activity of water is its concentration? That is, we are determining the pka of water in water-so shouldn't we use the activity of water?

Some light in the darkness can be obtained from J. Chem. Educ. 2017, 94, 690−695 (DOI: 10.1021/acs.jchemed.6b00623) where the difference between the "rational" and the "conventional" pka (H2O) value is elegantly discussed. https://pubs.acs.org/doi/10.1021/acs.jchemed.6b00623

The relative acidities of these molecules in the aqueous solution$^{[1][2]}$ is:
$$\text{methanol > water > ethanol > isopropanol > t-butanol}$$
The relative acidities of these molecules in DMSO$^{[3]}$ is:
$$\text{methanol > ethanol > isopropanol > water > t-butanol}$$
The problem with only considering the acidities of these molecules in an aqueous solution is that it assumes only one explanation for acidity. The inversion of the order of acidity in DMSO indicates that there are multiple factors at play.
The acidity trend in DMSO is explained by polarizability and the anionic hyperconjugation effect$^{[4]}$ of the conjugate bases of the molecules. The larger size of the alkyl substituents in the conjugate bases allows for increased distribution of negative charge over a larger volume. This reduces the charge density, and thereby the Coulombic repulsion between the conjugate base and the solvent. In DMSO, water is much less acidic than even methanol, which is also consistent with the relative polarizability of a methyl group and a hydrogen atom, but fails to explain the relative acidity of t-butanol and methanol.
The trend in alcohol acidity is explained by anionic hyperconjugation, which is weakened in larger alkyl substituents.

For methanol and the methoxide anion, the primary interaction occurs between a filled lone-pair ($n$) orbital on oxygen and an unoccupied $\pi^*_\ce{Me}$ orbital of the methyl group, resulting in a two-electron stabilizing effect.

$\hspace{3.8cm}$

Deprotonation of the alcohol leads to an increase in the energy of the n orbitals on oxygen, and a decreased energy separation between the interacting orbitals $n$ and $\pi^*_\ce{Me}$, and hence a greater stabilizing interaction$^{[5]}$.

Although DMSO does solvate the solute to an extent, because is an aprotic solvent, it is incapable of hydrogen bonding with the solute. This allows DMSO acidities better reflect the intrinsic acidities of the molecules.
In aqueous solution, however, the conjugate bases are stabilized by hydrogen bonding. Because smaller ions are better solvated by water, water's acidity is increased more than methanol's, but because methanol is intrinsically more acidic due to its polarizability, it is still slightly more acidic than water, even in an aqueous solution. As the size of the alkyl substituent increases, the alcohol becomes weaker due to the decreasing ability of water to solvate the conjugate base. This has a greater effect on the alcohol's acidity than the dispersion of the negative charge, and overrides the intrinsic acidity, inverting the order of acidity.

$^{[1]}$ [Evans $\mathrm{p}K_\mathrm{a}$ Table, Alcohols][2]
$^{[2]}$ [Wikipedia, Ethanol][3]
$^{[3]}$ [Bordwell $\mathrm{p}K_\mathrm{a}$ Table, Alcohols][4]
$^{[4]}$ [ACS, Acidities of Water and Methanol in Aqueous Solution and DMSO][5]
$^{[5]}$ [ACS, Does a Methyl Substituent Stabilize or Destabilize Anions?][6]

Water is a complicated substance. Water molecules solvated by water are anomalously acidic due to unusual stabilization of the conjugate base ($\ce{OH^{-}}$) by very strong hydrogen bonds. Water in a non-aqueous solvent (especially aprotic solvents) suffers a decrease in acidity by many orders of magnitude more than one might expect. Here is a relevant past answer of mine, with a nice source.

But I just read this PDF article by Helvetica Chimica Acta that very convincingly explains why the pKa of water (the acid) in water (the solvent) is 14.0 and not 15.7! http://onlinelibrary.wiley.com/doi/10.1002/hlca.201300321/pdf and http://onlinelibrary.wiley.com/doi/10.1002/hlca.201300321/pdf

If we compare the +I effect of a methyl group and more substituted alkyl groups, then the methyl group has less +I effect as compared to other alkyl groups because oxygen is more electronegative than both carbon and hydrogen.
A methyl group cannot fulfill the electron needs of oxygen which is why oxygen takes electrons from hydrogen and a hydrogen ion is released easily.
If a more electron donating group was present it could have fulfilled the electron needs of oxygen and hydrogen ions wouldn't have been released easily thus making the compound less acidic

Methanol is more acidic than water because its conjugate base that is methoxide is weaket than waters conjugate base that is hydroxide and now we can say that weaker is the conjugate base more is its acidic character . or we can say that due to the presence of methyl group in CH3 bond between O and H become weak and it can easily release hydrogen where is in H2O bond between O and H is stronger so it will not easily release hydrogen and become a weak acid than methanol

Your answer is not very insightful. How does the presence of the methyl group cause the O-H bond in methanol to be weakened?

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


text:
Boiling of water does not involve temperature change, so why is there a change in internal energy during the process?
in an isothermal process, the value ΔT = 0 and therefore ΔU = 0 but during boiling if water at constant atmospheric pressure temperature doesn't change so why internal energy changes

The issue here is that this is a process that is isothermal, but involves a phase change. Heat is transferred and work is done, so $\Delta U \ne 0$. See my answer for more detail.

Change in internal energy in isothermal process is zero when no phase change or chemical changes are involved (and then only when changes in pressure are small). Since in boiling the phase changes from liquid to gas, there is also change in internal energy. The energy is required to break bonds between water molecules and is regained when the gas condenses back to liquid.
The answer in linked question provides a very thorough explanation.

Edited. Maybe it's now more accurate even though not complete.

Welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. Visit the [help] to learn more about how it works.

The change in internal energy $U$ is dependent on the heat transferred $q$ and the work done $w$.
$$\Delta U = q + w$$
$\Delta U = 0$ in two cases: when $q=-w$ and when $q=w=0$.
The heat transferred is not necessarily zero if there is no temperature change. The most common expression of heat transfer is $q=C\Delta T$, where $C$ is the heat capacity of the system and $\Delta T$ is a temperature change. We can also write $q=ms\Delta T$, where $m$ is the mass of the system and $s$ is its specific heat capacity (per unit mass). However, heat is being transferred during boiling (else how does the liquid boil?). We need a new equation for heat that does not require temperature.
In a constant pressure system (most boiling happens in a constant pressure,  open to the atmosphere, system), change in enthalpy is equal to $q$ since all of the work and other pressure-volume terms cancel. The enthalpy change of vaporization is usually given in $\mathrm{kJ}$ per mole or unit mass. Now we have a new equation for heat:
$$q_{vap}=m\Delta_{vap}H$$
Let's say we are boiling 100.0 g of water. Water's enthalpy of vaoprization is $2257\ \mathrm{kJ/kg}$. Thus the heat transfer is:
$$q=(0.1000\ \mathrm{kg})(2257\ \mathrm{kJ/kg})=225.7\ \mathrm{kJ}$$
Note that in this case, the work done is also not zero. There is a volume change of the system. Let's take our 100 grams of water.
$$V=0.1000\ \mathrm{kg}\times \dfrac{\mathrm{m^3}}{958.4\ \mathrm{kg}}=1.040\times 10^{-4}\ \mathrm{m^3}$$
At the boiling point, the density is $958.4\ \mathrm{kg/m^3}=0.9584\ \mathrm{kg/L}$. Using the ideal gas law (water vapor is reasonably ideal at atmospheric pressure and its boiling point), we can determine the volume of 100.0 g of water vapor, given that
$$100.0\ \mathrm{g}\ \ce{H2O}\times \dfrac{1\ \mathrm{mol}\ \ce{H2O}}{18.02\ \mathrm{g}\ \ce{H2O}}=5.549\ \mathrm{mol}\ \ce{H2O}$$
$$V=\dfrac{nRT}{P}=\dfrac{(5.549\ \mathrm{mol})(0.080257\ \mathrm{\frac{L\cdot atm}{K\cdot mol}})(373.15\ \mathrm{K})}{1\ \mathrm{atm}}=169.9\ \mathrm{L}$$
We can now calculate work. So that we get joules out, I am going to convert pressure to pascals and volume to $\mathrm{m^3}$.
$$P=1\ \mathrm{atm}=101,325\ \mathrm{Pa}\\
V_{gas} = 169.9\mathrm{\ L} =0.1699\ \mathrm{m^3}\\
\Delta V= 0.1699\ \mathrm{m^3} - 1.040\times 10^{-4}\ \mathrm{m^3} =0.1698\ \mathrm{m^3}\\
w=-P\Delta V=-101325\ \mathrm{Pa}\times 0.1698\ \mathrm{m^3}=-17.20\ \mathrm{kJ}$$
Now we calculated $\Delta U$:
$$\Delta U = q +w = 225.7\ \mathrm{kJ} - 17.20\ \mathrm{kJ} = 208.5\ \mathrm{kJ}$$


text:
To what degree an N99 active coal filter can protect against products of burning paraffin oil?
I sometimes do firedance and a part of it is waving a stick with burning endings made of Kevlar soaked in paraffin oil. I started to be a bit concerned about how healthy is it to inhale lot of paraffin oil fumes.
I was wondering if I can protect somehow against it and my idea is to wear a mask with N99 active coal filter. The vendor claims that it blocks dust of size 0,1 mm and more. What noxious products of burning paraffin oil can it block? Which of them it can't?

Do you smoke? If yes, ignore the troubles. If not, than simply don't inhale the smoke.

@permeakra How nasty a comment xDD But it’s true, most things one does in labs that are highly regulated (‘Wear dust masks! Do not inhale! Keep away from eyes!’) are much less dangerous than cigarette smoke ^^'

Activated carbon's forte is vapor adsorption of hydrocarbons due to the activated carbon's high surface area, and this mask is probably very good at adsorbing fumes from hot parafin.
However, activated carbon does not tend to make a good material for filtering out the most harmful components of that kind of smoke, namely particulates smaller than 2.5 microns. In fact, the specification on the mask for particulate removal is that "that it blocks dust of size 0.1 mm and more" (meaning 100 microns and larger).  The World Health Organization has classified particulate matter of 10 micron (one tenth of the size claimed to be removed by the mask) to be a Group 1 carcinogen $\mathrm{^{(1)}}$.
Your filters are likely very good at filtering the particulate size most relevant to a particular application. However, for your application (production of very fine smoke particulates), I suggest a mask that is rated to remove 95% of $\pu PM_{2.5}$.
1) Outdoor Particulate Matter Exposure and Lung Cancer: A Systematic Review and Meta-Analysis, Environ Health Perspect; DOI:10.1289/ehp.1408092


text:
What is the problem with forming 2,3-dimethylcyclohex-2-en-1-one?

I have suggested a mechanism for the formation of 3-ethyl-2-methylcyclohex-2-en-1-one A

but the question says problems would be encountered if a similar route was used to make 2,3-dimethylcyclohex-2-en-1-one B. I think perhaps a decarboxylation step is involved? Can someone please explain how to get to B and the difficulties in the mechanism?

It looks like some sort of trick question.  If you count the carbon atoms B has one less but it isn't clear how to remove it under the given conditions.  I think the answer is that B is a typographical error.  Hope this helps.

A similar route to B sounds to me like just removing one carbon atom from the chain to hope to arrive at B. I actually see no problem with this as long as you are in thermodynamic control … But I could be missing something obvious …

@jan how can i remove one carbon from the chain?

As in using a different reactant.

@Jan oh i see. As in changing the starting material completely? That was not clear from the question.. I will try it! Thanks!

@Jan i dont see what the problems are in this reaction?

@justbehappy Dito. That’s why I’m confused, too ;)

Could it be due to conjugation with the carbonyl so OH- would attack once more to give the enol compound. However, the ethyl group sterically hinders the approach of OH- @Jan

Your starting product for A is symmetrical, but your starting product for B will be asymmetrical - one carbon less on one side.  This is going to mean you may get 2 or more different products; B may not be your major product, and you'll have to separate them afterwards.  I don't believe there's inherently a tricky difficultly in the mechanism for B.

Actually no. Under thermodynamic control (i.e. given enough time) the product B is inherently more stable than the potential side product and it will form. Reasoning: Tetrasubstituted versus trisubstituted double bond.

@Jan My two cents: you are right, but the question may not necessarily be. I think this is the answer the question wants. If OP can edit and make a note of this piece of information I'd +1.

Wouldn't the more favoured enolate in the starting material needed to form B (6-oxo-2-octanone, probably not the IUPAC name) actually mean B as a product is not favoured?

@Beerhunter As I wrote above, under thermodynamic control B is more stable than the side product 2-ethylcyclohex-2-en-1-one and will thus be formed preferentially.


text:
When is a reaction feasible at room temperature?
Gibbs free energy determines whether a reaction will occur spontaneously or not. Still, the activation barrier of a reaction determines if a reaction will occur at a given temperature. At any non-zero temperature every energy level will be populated somehow, but when is population high enough to say that a reaction will take place?
So when is a reaction called feasible at room temperature? A source on this with an energy range for the barrier that can be overcome at a specific temperature (kJ or kcal/mol) so that one considers the reaction actually taking place would be much appreciated.

Related: How to relate a reaction barrier to the time the reaction needs to proceed?


text:
Acids generating nascent gas atoms
When sulfuric acid  reacts with copper, then $\ce{H2SO4}$ gives nascent oxygen $\ce{[O]}$. The reaction is:
$$\ce{H2SO4 -> H2O + SO2 + [O]}$$
This nascent oxygen reacts with copper to create $\ce{CuO}$.
Therefore my question is that when copper reacts with $\ce{HCl}$ then why doesn’t $\ce{HCl}$ give nascent chlorine $\ce{[Cl]}$ so that it could react with copper and create copper(II) chloride $\ce{CuCl2}$?

Honourable Sir, please take care to use proper capitalisation in further posts and ensure that your caps lock key does not get stuck halfway through the post. Ye may wish to take a [tour] of this establishment. Furthermore, I have taken the liberty to improve your correspondence by including MathJax syntax for chemical formulae; ye may learn more about this extraordinary feature in the [help] or in the storey us regulars call meta. I hope you enjoy your visit here and may your contributions be qualitative and plentiful. Yours sincerely,

This may be due to standard redox potentials between copper cations and chlorine anions, which can't give the driving force to generate nascent gas $Cl_{2}$, and the high solubility of $CuCl_{2}$ that doesn't allow this salt precipitation.

When concentrated sulfuric acid is heated, then it decomposes via the reaction
$$\ce{H2SO4 -> H2O + SO3}$$
I tried to determine under what circumstances the nascent oxygen (a free single atom of oxygen) is formed which would react with copper to create $\ce{CuO}$ but couldn't find the reaction in any modern source. I did find the proposed reaction in Essentials of Modern Chemistry
 By Charles Elwood Dull (1918) but I'm guessing that the reaction to give nascent oxygen is just wrong.
The gist here is that you can't piece reactions together from balanced equations but you must also define the conditions under which the reaction takes place. At a deeper level not only can the chemical equation be specified but also the mechanism of the reaction can be detailed.
Hydrochloric acid, $\ce{HCl}$, is typically in aqueous solution. Although copper won't react with pure hydrochloric acid, it can react with dissolved oxygen in hydrochloric acid.
$$\ce{Cu (s) + 2HCl (aq) + 1/2O2 (aq) -> CuCl2 (aq) + H2O}$$
Oxygen can also be formed in chlorine water:
$$\begin{align}
  \ce{Cl2 (aq) &-> HCl + HOCl}\\
  \ce{HOCl &-> HCl + 1/2O2}
\end{align}$$
But the concept of nacent oxygen from this reaction is wrong.

FYI: States of aggregation should not be subscripted, it is not wrong, but the recommendations (Sec. 2.1.) are different. I also adjusted the choice of maths environment and removed the blockquoting.


text:
Mechanism of pseudo buffer pH maintenance
Some solutions, like concentrated HCl, act as buffers even though they lack the buffering component. What is the cause for such a behaviour?

I do wonder what your question is, but I have a hunch. Assuming that what you’re asking is essentially:

In a concentrated $\ce{HCl}$ solution, the $\mathrm{pH}$ will more or less be the same value even if we add external $\ce{H+}$ or $\ce{OH-}$. Why?

Then there are multiple effects at work:

Aquaeous $\ce{HCl}$ contains a buffering component — the $\ce{H3O+}$ (or closest relative) ion. It is the most acidic ion that can be present is solution, so it ‘buffers down’ the acidity of said solution to what the corresponding $\ce{H3O+}$ acidity would be.
$\ce{HCl}$ in itself is an acid, but at the same time $\ce{Cl-}$ is also a (very weak) base. For high total $\ce{HCl}$ concentrations, the ratio of $\ce{HCl}$ to $\ce{Cl-}$ is at equilibrium. If we add additional $\ce{H+}$ to the mixture, we are effectively shifting that equilibrium back to the $\ce{HCl}$ side, reducing the proton count and thus buffering the $\mathrm{pH}$. Therefore also, if we add a proton scavenger (colloquially also known as ‘base’), then that will shift the equilibrium to the $\ce{Cl-}$ side, and as long as there is still enough $\ce{HCl}$ around, the total proton concentration won’t change notably.

In essence, everything that can act as a Brønsted acid or base will buffer at a specific $\mathrm{pH}$ value, even if that $\mathrm{pH}$ value is far from what one would typically call ‘buffered’.


text:
Which step below is rate determining
What is the rate determining step in the following energy profile? To clarify, the reaction is:
A -> B -> C
The energy of A is greater than B which in turn is also greater than C. The intermediate A-B is higher energy that the B-C intermediate but the energy gap between B and the intermediate B-C is greater than from A to the A-B intermediate. I hope this is clear.

you do realise that you've answered your question.

yes i was hoping to get some other opinion because the article on wikipedia has no reference and i have read some understandably more confused answers elsewhere.

Try this ref if you have access to a library, J. J. Murdoch:
What is the Rate-Limiting Step of a Multi-Step Reaction?
J. Chem. Ed., 1981, vol. 58, p. 32.

Here's a good starting point https://thewinnower.com/papers/mechanism-of-the-reduction-of-a-carboxylic-acid-by-borane-revisited-and-revised to get into the nuts and bolts of this thinking.

The second step is rate-determining. According to Wikipedia:

Given a reaction coordinate (energy diagram), the rate determining step can be determined by taking the largest energy difference between any starting material or intermediate on the diagram and any transition state that comes after it. That transition state will then be the rate-determining step of a given reaction. The transition state with highest absolute energy may not necessarily correspond to the rate determining step.

The second step is the rate determining step in the case that this is not an equilibrium reaction; however, if B can go back to A(therefore it is an equilibrium rxn) then it is easier for B to proceed to C instead of going back to A, right? The graph can go in both ways and then the first step will be the determining. It has the highest point of energy needed to be overcame.

This doesn't seem right. As drawn, species A and B should not be in equilibrium given the lower barrier to C. A and B would be in a pre-equilibrium if the transition state for B->C were higher in energy than that from B->A, but in that case, we've firmly established that B->C is rate determining, which can be confirmed via the definition in the accepted answer.

Well I never said that they are on equilibrium I was just saying that if this reaction was reversible then what I said above in the first part would apply

I still do not think that is correct. If it were reversible, it could not have the reaction coordinate diagram shown in the OP's question based on the arguments in my first comment.

I respect your opinion on what you think but I did confirm my answer with two different professors that I currently have just because of your skepticism... Thanks anyway for the feedback.

Sigh... I'm trying not be offended here. This is not my first foray into physical organic chemistry, and it's not like I've never taught this material before...


text:
What does "less hindered hydrogen" mean?
In a book they have given a statement,

In the process of pyrolysis of esters, esters having  beta hydrogen are heated at high temperature and then alkenes are formed along with carboxylic acid.
  If there are two beta hydrogens then beta hydrogen having less hindrance will participate in the reaction.

In the last line, what do they mean by less hindered hydrogen?

Consider these following two structures from the same compound:

The two Newman projections are along the horizontal $\ce{C-C}$ bond which is, of course, freely rotatable. The two structures represent two possible syn-elimination transition states. Note that in the left-hand structure there is a steric clash between the tert-butyl group and the methyl group. The right-hand structure has no such clash. It is therefore the preferred transition state for this reaction.
‘Less hindered hydrogen’ basically means that hydrogen that leads to the more stable transition state.

I didn't get it . We will remove hydrogen from methyl or from 3rd carbon?

Hm, we probably could remove the methyl one, too, since that is technically also beta. Let me think about this …

@VinayakAgarwal I just realised a major blunder I had in my images which has now been fixed. You could also draw a transition state towards the terminal methyl group rather than along the chain. That would work equally. However, the double bond you create would be terminal. Therefore, under thermodynamic control (evidently, it’s a pyrolysis after all) the higher-substituted double bond should be created as shown here. I mean, I could turn that methyl into another tBu group but I’m too lazy now.

By less hindered hydrogen do you mean hydrogen to be removed, leads to the more stable transition state.

Although I like this analysis, the wording of 'less hindered hydrogen' makes it seem that the hydrogen from the beta-methyl would be participating, at least according to the book in question. Do you have any evidence that the more substituted alkene is formed?

@jerepierre I admit, I just drew a methyl group there because it’s the easiest way to introduce stereochemistry on the double bond. But yes, you are correct; that can be done better.

The beta-hydrogen with less steric hindrance refers to the beta-hydrogen on the least substituted carbon. Your book is suggesting that the major product is the less-substituted alkene, which is not the most stable product (as pointed out by Jan).
Jerry March's book gives the example of the pyrolysis of sec-butyl acetate. The 'major' product is the less substituted alkene, although it appears to be statistically driven (there are three beta-hydrogens and only two-beta'-hydrogens), indicating that there is little energy difference in the transition states leading to the two products.

All else being equal, the less substituted alkene product (Hofmann's rule) should be predicted to be major. However, it appears that mixtures are typical and other factors must be considered for more complex substrates. The most important of these is that the reaction is a syn-elimination through a cyclic transition state, so the compound must be able to adopt the appropriate conformation. Other steric interactions may also influence the regioselectivity of the product away from the less substituted.
The Wikipedia page is a good summary of these intramolecular elimination reactions.

Does that mean that methyl group will have less hindered hydrogen than the carbon connected to tertiary butyl in the example given by Jan

Yes, in general the less substituted carbon will be less hindered. However, as I say on the answer, the effect is more likely statistical than steric in nature.

So if we want to say what do we mean by less hindered hydrogen in one line,we can say it as hydrogen on the least substituted carbon and it should lead to the more stable transition state.


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Why don't free radicals undergo rearrangement like carbocation/carbonium ions do?
Why don't free radicals and carbanions undergo rearrangement like carbocation/carbonium ions undergo in a reaction?

Short answer, they do.  They are significantly rarer, especially for radicals but a quick google of radical rearrangement will reveal several examples.  There are also several reactions (Wittig rearrangement being one) which can be drawn as either radical or polar and the one which operates is dependant on substrate.

First off, your observation is correct, rearrangements involving free radicals and carbanions are noticeably less common than rearrangements involving carbocations.
Many common carbocation rearrangements involve 1,2 shifts.  If we look at the MO diagram for a 1,2 shift and compare shifting a substituent to an empty p orbital (carbocation, case A in diagram) or to a filled p orbital (carbanion, case B in diagram), we see that the carbocation process involves a 2-electron system whereas the carbanion process is a 4-electron system.  The 2-electron process is allowed to proceed themally according to orbital symmetry, but the 4-electron process would need to proceed with inversion of configuration (bonding to the other lobe of the p orbital as we transform from reactant to product) in order to be thermally allowed - and this would be geometrically difficult.

(Source - V8rik at the English Wikipedia via Wikimedia Commons)
If we consider hydrogen shifts and replace the p orbital in our digram with an s orbital, we see that a 1,2 hydrogen shift in a carbocation is thermally allowed, but the corresponding shift in a carbanion is thermally forbidden (with an s orbital, there is no "other" lobe and proceeding with "inversion" is not possible)
Since thermal 1,2 shifts with retention at the migrating center are not allowed in carbanions by orbital symmetry, and a 1,2 shift with inversion at the migrating center is geometrically difficult, such processes would involve higher energy transition states.  As a consequence of the higher energy, 1,2 shifts in carbanion systems are less common than orbital symmetry allowed, thermal 1,2  shifts involving carbocations. One might reasonably expect that a radical process would be somewhere in between these two cases; and again the transition state would be of higher energy (since it is not orbital symmetry allowed) making such rearrangements less common.
While radical and carbanion shifts are not as common as carbocation shifts, they do occur.  In the case of radicals it requires high temperatures (hydrocarbon cracking) or weak bonds (peroxides for example). Carbanionic rearangements are more common when the carbanion is stabilized.  This can be achieved by having an electronegative heteroatom attached to the carbanionic center (as in the Wittig rearrangement) or by placing the anion on an $\ce{sp^2}$ or $\ce{sp}$ hybridized carbon, rather than an $\ce{sp^3}$ hybidized carbon (the Fritsch rearrangement provides an example here).

The short rationale I can provide is that you should consider the transition state for the rearrangement. In the case of hydride shifts in a carbocation, the transition state involves a hydrogen bonded to two carbons with 2 electrons. In a radical, this hydrogen would have 3 electrons. The analogy holds with carbon rearrangements possessing 8 and 9 electrons, respectively. So, we're breaking the octet rule on the migrating group. It is certainly possible to have such a rearrangement, particularly if the migrating group can accommodate an extra electron. For example, phenyl groups can delocalize the extra charge via resonance into the ring, and radical shifts of phenyl rings are well-known.


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Why was Lawrencium's symbol changed?
The initial proposal by Lawrence Berkeley National Laboratory scientists for naming element 103 was "Lawrencium" with the symbol "Lw", However, while IUPAC accepted the name "Lawrencium" for the element, it changed the symbol to "Lr".
Was there a stated (or unstated) rationale for this alteration? Was it in anyway connected with the competing claims from the Joint Institute for Nuclear Research in Dubna regarding who discovered the element first?

It was probably some political reason. Somewhat related: http://chemistry.stackexchange.com/q/16029/4945

Some interesting discussion here https://en.wikipedia.org/wiki/Talk%3ALawrencium#Why_change_from_Lw_to_Lr.3F Evidently the first letters of the first two syllables are the norm for new elements.

Ghiorso, Sikkeland, Larsh, and Latimer of the Lawrence Berkeley Laboratory reported the discovery of element 103 in 1961. It is correct that the discoverers suggested the name lawrencium, symbol Lw, for the new element in honor of E. O. Lawrence, inventor of the cyclotron and founder of the Lawrence Berkeley Laboratory. The original proposal can be found in Ghiorso, A.; Sikkeland, T., Larsh, A. E.; Latimer, R. M. New Element, Lawrencium, Atomic Number 103. Phys. Rev. Lett. 1961, 6 (9), 473–475:

In honor of the late Ernest O. Lawrence, we respectfully suggest that the new element be named lawrencium with the symbol Lw.

At the XXII Conference of the IUPAC in London 1963, the name was accepted but the symbol was changed to Lr. According to Comptes rendus / XXII Conference, London 5th to 9th July, 1963, the reason was simply the difficulty with the letter “w”:

Following the observation by H. Remy that “w” is an uncommon letter in many languages and difficult to pronounce it was proposed to alter the spelling to Laurentium. The Chairman emphasized that the Commission had no right to modify the spelling of the name, but that it could be pronounced according to the custom of the language. To make the name more acceptable it was decided to recommend the chemical symbol Lr and not Lw for element 103.

The case of Greek as one such language is especially relevant.  Greek lost the "w" sound as pronunciation evolved, changing it to "u" (upsilon).  So "u" and "w" became intertwined.  In English they were eventually separated, but how could (e.g.) Greek tell them apart?  Because of this confounding we could not always  distinguish clearly between Lw for lawrencium and Lu for lutetium.


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How to explain this acid-catalysed acetal rearrangement mechanistically?

I have spotted a protecting acetal group and thought this would undergo hydrolysis … My working is such:

But I cannot get to the desired isomer. Could someone please help? (Diagrams would be very much apprecited thanks!)

The two acetal oxygen's are different. Try protonating the "other" acetal oxygen.

Use the 1- and 3- hydroxys (where the carbonyl would be on the carbon at position 6). The original molecule has the acetal formed from the hydroxys at positions 1 and 2.

You have managed to draw the correct acyclic hydrolized compound with a decent mechanism.
This is a problem that can be solved by numbering the carbon atoms. Number the carbon atoms sequentially along the chain, starting with the acetal carbon (which has to be the carbonyl group). Likewise number the unpacked acyclic molecule. Use what you get to figure out which of the three hydroxy groups are used to make the second acetal.

In general, numbering your atoms using common structural features as anchors can help unravel complicated mechanism problems.

Upvote for numbering atoms.

I read a book called The Art of Writing Reasonable Organic Reaction Mechanisms and the first chapter was all about numbering atoms and mapping bonds broken and bonds formed.

That book is on my reading list too! Here's an Amazon link for anyone interested.


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Does a reaction have to have a rate determining step?
I am a bit confused about the concept of the rate determining step. From what I understand, a step in a reaction is the RDS if it meets the following requirements:

It is the slowest step in a reaction
It causes a bottleneck in the progress of the reaction

Therefore, if there is a catalyst, it means the slowest step in the reaction will no longer be the slowest step because the reaction can take place faster with a different mechanism involving the catalyst. Would it therefore be of any significance to say that a particular step in this reaction, say for example the step with the highest activation energy is the RDS, or is there no point in saying there is a rate determining step, ie, that it does not exist? Is it possible to say that the concept of the RDS has no significance in some reactions?

Here's a related question.

There are so many chemical reactions that any scenario is possible. Consider mixing two liquids, or dissolving a solid in a liquid. If the reaction is complete within that sort of time scale then the RDS is pretty insignificant. In aqueous solutions acid-base neutralizations are such reactions.

Any kind of chemical reaction can be modelled as a (complex) network of elementary reaction steps. For instance, if we model the conversion of R to P via the intermediates $\ce{I^1}$, $\ce{I^2}$, and $\ce{I^3}$, we have the following kinetic network:
$$\ce{R <=> I^1 <=> I^2 <=> I^3 <=> P}$$
Note that on the basis of microscopic reversibility, all these reactions can go both ways and the boundary conditions of the system (pressure, temperature and chemical composition) determine in which way and at what rate the reactions occur.
To answer your question: there exists a formal way of identifying the rate-determining step. If we look at the overall reaction rate $r$; then we can define a concept called degree of rate control:
$$
\chi_{i} = \left( \frac{\partial \ln r}{\partial \ln k} \right)_{k_{i \ne j}, K_{i}}
$$
Here, $\chi_{i}$ is the degree of rate control coefficient that indicates the extend to which an elementary reaction steps controls the overall rate and $k$ is the rate constant of the particular elementary reaction step you want to calculate the DRC coefficient for. Importantly, in this calculation you keep the rate constants for all other reaction steps and for the equilibrium constant of the elementary reaction step you are looking at constant. (in other words, you only change the height of the reaction barrier). When you find that this coefficient is 1 and all other elementary reaction steps have a coefficient of 0; in that case this elementary reaction step is the rate-determining step.
Interestingly, when investigating a lot of chemical systems, it turns out that only roughly half of these systems have a rate-determining step. A lot of the systems (especially in catalysis) have multiple rate-controlling steps and even rate-inhibiting steps (where the DRC coefficient is less than zero).
I will not go too much in-depth, but there exists a complete mathematical treatment and derivation of this theory that shows that the sum of all coefficients has to be unity. This means that a chemical system should always have either a single rate controlling step (the RDS) or multiple rate-controlling steps and possible one or more rate-inhibiting steps.
To conclude: Not every overall chemical reaction has an RDS, but they should have one or more rate-determining steps.
Notable references:
The above theory was proposed by Campbell and co-authors in this research paper: J. Am. Chem. Soc. 2009, 131 (23), 8077–8082. DOI:10.1021/ja9000097.

Simple answer: No, there is no requirement for a reaction to have a rate determining step, though largely we are able to show through kinetic experiments that the vast majority of reactions do have a rate determining step.
Longer answer: Most reactions have a reaction pathway that consists of one of more intermediates (one or more 'steps'). In these cases, one of the steps is likely (though not necessarily) slower than the others, leading to the concept of a rate determining step (RDS).
The RDS is, as you say, a bottleneck to the reaction. If you imagine the reaction where...
$$\ce{SM -> I -> P}$$
...and $\ce{I -> P}$ is the rate determining step, then no matter how fast you convert $\ce{SM -> I}$, the $\ce{I}$ will just be accumulating until it reacts to form $\ce{P}$ and hence the second 'step' is said to be rate limiting. This can be derived mathematically, but as you are asking the question I assume you're yet to reach this topic yet.
In essence, the $\ce{I -> P}$ rate constant ($k$) dominates the kinetics, making the rate constant for $\ce{SM -> I}$  irrelevant. This simplification is incredibly useful in physical organic chemistry as if we know the rate limiting step, we can think of ways to increase the overall reaction rate by considering how to speed up the slowest step on the reaction pathway.
Catalysts lower the activation energy of a step, allowing it to proceed faster, and in some cases this might mean that the rate determining step is no longer rate determining and some other step along the reaction pathway is.  They may do this in a variety of ways, one of which is a change in mechanism (i.e. providing an alternate reaction pathway), when this happens, its not really valid to keep using the same kinetic argument developed for a non-catalytic variant of the reaction, and as such a new rate equation (and hence the rate limiting step) should be developed (if this is important).

Of course, it is extremely time consuming to measure these kinetic parameters, and even when measured it can be difficult to interpret the data for complex reactions. When we discuss the 'rate determining step' as practicing organic chemists, we are often generalising based on similar systems for which the kinetics have been studied.
There are, however, reactions that occur in a single, concerted step. These are known as pericyclic reactions. All of the bond making and breaking occurs via a single transition state, with no intermediates formed and as such there is no RDS (everything occurs in one step).
A famous example of this is the Diels–Alder cycloaddition:

Nice answer; I’ll upvote tomorrow (used up all my votes already ^^') But technically, if a reaction is single-step than that step is rate-determining by definition, isn’t it? Yes, I’m being nitpicky ;)

I did originally write 'yes or no, depending on how you look at it', but I thought I was being more philosophical at that point. But yes, I do agree with you, if theres one step, it is, by definition, rate limiting. ^_^

In addition to contribution of other people.
Usually, there is at least one bottleneck on the reaction path even in case of catalysis. However, it is possible for reactions to form a network with cross-influence. A well known, widely recognized case is Belousov–Zhabotinsky reaction, where intermediates accumulate and, once critical concentration is reached, fall further the path together in a big bunch, following with new accumulation period.
In case of heterogeneous catalysts a reaction still can have a simple path with no weird behavior, but the reaction rate may be determined not by any chemical interaction, but diffusion of reagents towards the surfaces. This is the case of steam conversion of methane.


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Coordinates for molecules except cartesian and zmatrix
Every program in theoretical chemistry that I know uses different ways of writing their input. The same holds true for writing the geometry of the molecules of interest. But they can all be reduced to be either cartesian or zmatrix representation.
Intuitively these two are the natural answers to the problem of specifying a molecule geometry.
Nevertheless:
Are there programs in use that use anything different e.g. zylindrical, or elliptical coordinates?

What would be the point? It is natural to use cylindrical coordinates in a problem with cylindrical symmetry - that is, in a linear molecule. But such molecules look perfectly fine in the standard cartesian form as well, so why bother. Come to think of it, there are certain molecules (fullerenes, for instance) which might look kinda natural in spherical coordinates, but again, this is a field too narrow to justify the necessary adaptations.

Besides Cartesian and z-matrix, some codes specify coordinates in reciprocal space, particularly for solid-state codes (i.e., you give the full unit cell in fractional XYZ).

That's not quite what we call reciprocal space. Good point, though.

I had the same thoughts as @IvanNeretin in mind. But since I write a python module myself for converting between xyz and zmat files. I wanted to make sure that I did not forget another important kind of coordinate specifications.
(I know that babel exists but it is too inaccurate for my purpose)

@mcocdawc If you think Babel is "inaccurate," I'd really like to get an e-mail explaining your problems.

@Geoff Hutchison Thank you for your interest. I made yesterday a bug report on openbabel on GitHub regarding this problem (same user name).


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Is the reaction between powdered soap and water exothermic?
I have done a fair share of hand washing using powdered soaps and in the first few seconds after mixing the soap with water, I have in most cases felt a subtle warmth in my hands. Is it that the reaction (if there's any) between soap and water is exothermic? Or there is another  explanation for this phenomenon?

Not having the exact sequence, I'd guess that the result is that the soap reduces the surface tension of the water. Thus there is better contact between your hand and the water. So I'm guessing that you start moving your hand in the "pure" water first, then you keep moving your hand as the soap is added. You feel the extra warmth as you add the soap.

@MaxW So this is more of a physics than chemistry related occurence?


text:
How can a sample that is in two different states be at a uniform temperature?
In my chemistry course, the professor has discussed how a sample that contains both solid and liquid water would be at 0 degrees Celsius no matter what the proportion of solid to liquid water is. I find this concept to be confusing, because wouldn't the areas of the sample that had already changed from solid to liquid begin increasing in temperature even while the other parts of the sample continue to be in the process of fusion?
The other part of this idea that confuses me is that if the sample was at a uniform temperature, then why do bodies of water (ex. lakes that freeze in the wintertime) continue to have different temperatures of water within them?

And in addition to @DrMoishePippik's answer below on how, in equilibrium, the liquid and solid are at the same temperature, note that a lake tends not to be in equilibrium - the bottom will not be at the same temperature as the top.

What is confusing you is the thermodynamic concept of equilibrium. A system at equilibrium has no external source of energy flowing onto it and is perfectly mixed internally. These are not common circumstances in the real world.
In a system that is in equilibrium, a solid and its corresponding liquid can exist at one temperature. Any extra energy added will, when equilibrium is restored, serve to convert the solid to liquid (this takes energy). Since, at equilibrium, we have perfect mixing this will not result in a temperature change. For water it takes a lot of energy to convert ice to liquid water.
Of course, in the real world, nothing is at equilibrium and there is always a flow of energy in or out of the system. And for an isolated system at equilibrium, when energy is added there will be a period where parts of the system will be hotter than others. But, since being at equilibrium means the system will be perfectly mixed, this is temporary and the system will regain equilibrium with a constant temperature but a different mix of water and ice.
So you professor is right if he specifies that the system is in equilibrium but you are right in the real world for most systems. The idea of equilibrium is a useful idea for understanding the limiting behaviour systems if a little unrealistic as a description of the real world.

I'll throw in that many times in chemistry we'll assume an adiabatic process - in other words the system doesn't gain or lose heat to the rest of the universe. In reality such perfect isolation of a system is impossible. So it is a great theoretical concept even through it is impractical in an absolute sense.

For a system in equilibrium: Think of it as a closed system, with a certain amount of energy shared between the solid and liquid components (discounting the container).  At any instant, some of the faster-moving molecules of the liquid could hit the ice, melting a bit, and thereby losing some of their own energy and freezing.
It takes ~80 cal/g to change ice at 0 °C to water at 0 °C. This enthalpy of fusion, aka latent heat of fusion , "absorbs" any additional heat applied until all the ice has melted. So a gram of water at 0 °C has ~80 calories more heat than an equal mass of ice at 0 °C. If you have a 1 gram chunk of ice and apply 40 calories as heat, then about half will melt.
BTW, pure water does not necessarily freeze at 0 °C... it may supercool, which will change the calculations a bit. See this video or this one for a supercool experiment!
As @MaxW states, in real life most systems are not in equilibrium (they need to be shaken or stirred, despite Ian Fleming's direction). To view a clearly non-equilibrium demo, where there is water in three states at once, see Red Hot Nickel Ball on Ice.

Ice/water at 0C assumes that the ice and water are at an equilibrium temperature. As the OP questioned, it is possible to have a lake where the bottom water is warmer than the surface water (with ice perhaps).

True... I'm amending my answer, which had applied only to a system in equilibrium. Thanks for that correction.


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Diastereoisomer of quinic acid and its proton NMR?
Question:

Quinic acid, A, has the gross structure shown below:

(i) Draw all the possible diastereoisomers of A and indicate which are chiral.
  (ii) Quinic acid is optically active and the marked hydroxyl groups are cis to one another; which of your structures is consistent with this data?
Using the $\ce{^{1}H}$-NMR data below, draw a likely conformation for quinic acid, making sure that you fully assign all these data.
$\ce{^{1}H}$-NMR for quinic acid in $\ce{D2O}$: δ 4.09 (1H, q, 3 Hz), 3.99 (1H, ddd, 11 Hz, 9 Hz, 5 Hz), 3.39 (1H, dd, 3 Hz, 9 Hz), 2.12 (1H, dd, 13 Hz, 5 Hz), 2.05 (2H, d, 3 Hz), 1.86 (1H, dd, 13 Hz, 11 Hz).

My working so far:
$\hspace{29ex}$
Firstly, have I missed any diastereoisomers? And what does it mean by 'conformation' of quinic acid? Does it mean a chair conformer?
But I'm not sure about the shift at 4.09 ppm since I made the assumption that the quartet is actually a doublet of doublets. Also surely $\ce{H_{f}}$ would vicinal couple with $\ce{OH}$ on the same carbon?

Also, you write that you only need the deduction for the last proton … so how did you deduce the placement of the other six? (This is a homework question which means you should show as much work as possible ;).)

Here’s a hint for the diastereomers: Grab one group and bind it in place. I suggest the carbon attached to $\ce{COOH}$. Then, wobble the remaining ones back and forth. Since one stereocentre will not budge, you can be sure to not generate any enantiomers/identical structures.

The last comment was supposed to mean: the structure which solves it is missing. If the quartet really had been a dd, they would have reported it as such; and if the dd had fallen together due to identical coupling constants, it would have fallen together into a triplet, not a quartet. I’ll add that to my answer.

Confirming that if I can see it correctly in that minitiature image of yours, you got the diastereomers correct ^^

I see one major issue in your working immediately: The NMR has been recorded in $\ce{D2O}$, hence there will be no more $\ce{OH}$ groups present in your molecule, only $\ce{OD}$ groups. Since deuterium is not hydrogen, you cannot observe any coupling to the $\ce{OD}$ groups as if they contained hydrogen.
Also, you said:

I made the assumption that the quartet is actually a doublet of doublets.

This is an incorrect assumption. They have multiple dd’s that they reported as dd which gives us less reason to suspect an incorrect report. Note that typical dd signals have two different coupling constants, e.g. the proton at 2.12. Now you could imagine that the two doublets fall together. But in that case, they would fall together to become a triplet, not a quartet. You can try it with splitting trees: Either a dd stays dd, or it collapses into a t. And finally, the quartet cannot be a set of two doublet signals from distinct hydrogens since it only has an integral of 1.
That said, the attribution of different hydrogens to their position in the ring is remarkably simple. Let’s label the quaternary carbon $\mathrm{C1}$ and then go around the ring in any direction. Our first stop is $\mathrm{C2}$.

 Since this is a $\ce{CH2}$ group, its chemical shift must be lower than $3~\mathrm{ppm}$. We have two choices. Either we take the two hydrogens that exhibit a $^2J$ coupling or the other two. Since we can easily identify the $^2J$ coupling, let’s take those.

With the hydrogens of $\mathrm{C2}$ in hand, we can check, which hydrogen must be on $\mathrm{C3}$.

 Again it’s pretty simple. The two coupling constants that are not $^2J$ constants must all be present in the hydrogen’s signal. There is only one in question.

Once we have gotten that sorted, we can go back and determine the stereochemistry that must be present here.

 We have one very large $^3J$ coupling and one mediocre one. The very large coupling is, of course, due to an axial-axial relationship between the two hydrogens.

We now have three hydrogens firmly in place: Those on $\mathrm{C2}$ and the one on $\mathrm{C3}$. Let’s move on to $\mathrm{C4}$.

 There is one coupling constant on the hydrogen at $\mathrm{C3}$ which is not yet assigned and which thus must belong to the coupling to this hydrogen. It’s magnitude again tells us pretty clearly which dihedral angle we expect and thus what the relationship between the two must be.

With $\mathrm{C4}$’s stereochemistry sorted out, $\mathrm{C5}$ is next.

 The coupling constant is rather different this time, indicating the opposite relative stereochemistry. Note that the hydrogen at $\mathrm{C5}$ is reported as a quartet of the same coupling. We’ll get back to that in a second. For now, we know what must be substantially different here.

And now we can move on to $\mathrm{C6}$, the one carbon we are still missing.

 As I just mentioned, we have an apparant quartet for the hydrogen on $\mathrm{C5}$. But that’s okay. It just means that there are three rather similar coupling constants to neighbouring hydrogens. One is taken from the coupling to $\mathrm{C4}$’s hydrogen, meaning the two of $\mathrm{C6}$ must be the other two. Thankfully, there is one signal with an integral of two left, which also has the same coupling constant. And also thankfully, if you did everything correctly, there should be no questions as to why this stereochemistry along the $\ce{C{5}-C{6}}$ bond be correct.

Finally, we have a hint in the question as to the overall structure:

 Namely that the two hydrogens on $\mathrm{C1}$ and $\mathrm{C4}$ are syn. Thus, the carboxylic acid must be in equatorial position and the hydroxy group axial.

A note on finding the diastereomers: Enantiomers are only enantiomers if all stereocentres are reversed. Thus, if you take one centre and keep it the same all the time and only exchange the direction of the others, you are bound to find diastereomers only. I suggest immobilising the $\mathrm{C1}$ carbon since it is rather prominent up there all alone; it gives the entire thing less hassle. Also, any achiral isomers will always have a plane of symmetry through it which helps, too.


text:
van't hoff factor
Basically I was doing a freezing point depression question where enough information was given to calculate the van't Hoff factor(i) which was 1.86. Anyways the question was asking to calculate the percent dissociation based on the above value. Note: the equation given was AB3 -> A + 3B so assuming no dissociation i would have equaled 4. Everyone I asked is conflicted whether the answer is 45.5 or 28 percent. Which one is correct?


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Galvanic corrosion: Copper, iron and aluminum in an automotive cooling system
I am building a small copper heat exchanger I am going to plumb into my coolant system. I'm afraid that the copper will corrode my aluminum radiator or iron engine through galvanic action via the coolant. How was this prevented on older vehicles with iron engines and copper radiators? Or is this less of a problem then I expect it will be?

Interesting question. The answer seems to be the inhibitors in the engine coolant. See for example http://www.dtic.mil/dtic/tr/fulltext/u2/a289588.pdf

If the copper components are not in direct contact with the aluminum or iron parts, then there should be no electrolytic corrosion. This would require that non-conductive rubber hoses are used for coolant lines (avoid those with steel spring cores to prevent collapsing on the vacuum side, lest the rubber crack through and allow metal-to-metal contact). Also mount the copper parts on rubber shock mounts, rather than directly to the chassis.
Actually, corrosion was not well prevented in the past. For example, some Renault engines had an aluminum block with steel cylinder liners (shrunk with liquid nitrogen and dropped in place to expand for a very tight metal-to-metal seal, if I remember correctly). This short-circuited cell corroded rapidly. Modern antifreeze contains corrosion inhibitors such as silicate/phosphate mixtures that form an insulating film on metal and should help prevent any issue with the copper parts. Use a good quality ethylene glycol antifreeze with inhibitors.