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Why is the entropy of deuterium greater than that of hydrogen gas
Why is the entropy of deuterium more than that of hydrogen gas?
I thought about it but couldn't figure out a reason for why that would happen. Why would the number of neutrons in the nucleus affect the entropy of the gas?

I guess this has something to do with nuclear spin.

@ Ivan Neretin ;  yes nuclear spin seems to be is the cause, D has a spin of 1 whereas H has spin 1/2. In $\ce{H2}$ the ortho-para ratio is 3:1 but 2:1 in $\ce{D2}$  there is some detail here http://chemistry.stackexchange.com/questions/67316/why-do-spin-isomers-of-hydrogen-ortho-and-para-hydrogen-change-their-nuclear-s/67329#67329 which you will have to modify for deuterium.

also there is the fact that the rotational constant for deuterium is almost half that for hydrogen, $60.85 \pu{cm^{-1}}$, so many more rotational levels are occupied at a given temperature.

The entropy can be split into three parts arising from translational, rotational and vibrational energies and using statistical mechanics each of there can be calculated from measure properties. The gases $\ce{H2}$ and $\ce{D2}$ differ in each of these three energies due to their differing mass and nuclear spin. The experimental values are hydrogen $130.7$ and deuterium $144.9 ~\pu{J mol^{-1}K^{-1}}$
The translational entropy is given by the Sakur-Tetrode equation,
$$S_{trans} = R\left(\frac{5}{2}+\ln \left(\frac{V}{N} \right)+ \frac{3}{2}\ln\left( \frac{2\pi mkT}{h^2} \right )\right)$$
where m is the mass, V volume, N Avogadro’s number, h the Planck constant and k the Boltzmann constant. Assuming that the gases behave ideally the difference in entropy is small as it depends only on the mass difference all other terms being the same, so $S_{trans}$ is greater for $\ce{D2}$ than $\ce{H2}$ by $(3R/2)\ln(2) \approx 8.64~ \pu{ J mol^{-1}K^{-1}}$.
The vibrational entropy can be ignored at low temperature, for example at room temperature $kT \approx 210~ \pu{cm^{-1}}$ whereas the vibrational quantum are $4395$ and $3118~ \pu{cm^{-1}}$ for $\ce{H2}$ and $\ce{D2}$ respectively.
The contribution from the rotational motion depends both on the mass, as the moment of inertia determines the rotational constant, and on the nuclear spin which via the symmetry properties of Fermions and Bosons affects the weighting of the rotational energy levels.
To calculate the entropy the partition function $Z$ is required. Once this is obtained the entropy is calculated as
$$ S = k\ln(Z)+\frac{U}{T}$$
where the internal energy $U$ is
$$U= kT^2\left ( \frac{\partial \ln(Z)}{\partial T} \right)_V$$
The partition function for odd mass number, assuming a rigid rotor model, is (see this answer for details Why do spin isomers of hydrogen (ortho and para hydrogen) change their nuclear spin with temperature variance? )
$$Z_{odd}= \frac{g_n(g_n-1)}{2}\sum _{J=0,2,..} (2J+1)\exp(-B_eJ(J+1)/(k_BT)) +\\ \frac{g_n(g_n+1)}{2}\sum_{J=1,3,..} (2J+1)\exp(-B_eJ(J+1)/(k_BT)) $$
where $g_n = 2s_n+1$ where $s_n$ is the nuclear spin, $1/2$ for hydrogen and $1$ for deuterium. The rotational quantum numbers are J and $B_e$ the rotational constant assumed to be the same for all J but which differs between hydrogen ($60.85 \pu{cm^{-1}}$) and deuterium ($30.44 \pu{cm^{-1}}$) by a factor of two. (These rotational constants are not typical, values for most molecules which are heavier and have longer bonds are far smaller, for example, $\ce{F2}, B_e =1 \pu{cm^{-1}}$)
The partition function for even mass number
$$Z_{even}= \frac{g_n(g_n+1)}{2}\sum _{J=0,2,..} (2J+1)\exp(-B_eJ(J+1)/(k_BT)) +\\ \frac{g_n(g_n-1)}{2}\sum_{J=1,3,..} (2J+1)\exp(-B_eJ(J+1)/(k_BT)) $$
where the only difference is that due to the weighting (those terms in $g_n$ ) in from of the summations which run over even and odd values of J. Hydrogen has weightings $1$ and $3$, deuterium $6$  and $2$ respectively in the equations above.
Thus Z for hydrogen with $g_n=2$ is
$$Z=1\cdot\Sigma (even~J) + 3\cdot\Sigma(odd~J)$$
and for deuterium, $g_n=3$
$$Z=6\cdot\Sigma (even~J) + 2\cdot\Sigma(odd~J)$$
At low temperatures the argument in the exponential is large meaning that the exponential is small. e.g. $exp(-large ~number) \rightarrow 0$  for all J terms except $J=0$ where the exponential is $1$. Thus in hydrogen the limiting partition function at low temperatures is $Z_{T\rightarrow 0}= 1 + 3(small~number) \approx 1$ and the limiting entropy is $S_{T\rightarrow 0 } R\ln(1) \rightarrow 0$; (and $U(0)=0$)
In deuterium the limit is $Z_{T\rightarrow 0 } = 6 + 2(small~ number) \rightarrow 6 $ and thus the limiting entropy is $S_{T\rightarrow 0 } R\ln(6) \rightarrow 14.9~ \pu{J mol^{-1}K^{-1}}$.
The change of rotational entropy with temperature is shown in the figure.


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Electrolysis of PbCl2 and use of Nernst Equation

The answer is -0.09V.
And the explanation is that the concentration of  $\ce{Pb^2+}$
will decrease.
My question is how will  [$\ce{Pb^2+}$] decrease?
In my opinion, [$\ce{Pb^2+}$] should remain constant. The cell is a voltaic cell meaning that $\ce{Pb}$ should oxidize to  $\ce{Pb^2+}$ to keep the cell running (since  $\ce{\frac{Pb^2+}{Pb}}$ = -0.13V). As a result, the concentration of $\ce{Pb^2+}$ will increase. But since the solution is saturated, no more lead can dissolve keeping the value of [$\ce{Pb^2+}$] constant.
Can somebody please point out the error in my explanation?

B and D are presumably Pt electrodes, and B is part of an SHE. What is the composition of C?

This was to be answered in the first part. The answer is any solution containing H+ ions.

I've finally read this question. I don't know what the source is, but I disagree with both the answer and the explanation.
Consider first the reduction half reaction for lead:
$$\ce{Pb^{2+} + 2e- -> Pb}, E^{\varnothing}=-0.13\ \mathrm{V}$$
Now, this is for the reduction at standard conditions, which means that the concentration of lead ion is $1\ \mathrm{M}$. If you decrease the concentration of the reactant ions (and you are, if you consider that the concentration of the dissolved lead chloride is less than 1), then the reaction should be less favorable. This means that the potential should be more negative, not less negative.
This is easy to check based on the form of the Nernst equation as the lead ion concentration enters into the denominator of $Q$. Since $\Delta Q<1$, $\Delta \left(\frac{RT}{nF}\ln Q\right) > 0$.

Can you please explain a little more on how the Pb2+ concentration is decreased?

And one more thing if Pb2+ is decreasing and it goes into denominator of Q then, decrease in Pb2+ will lead to increase in Q

The lead concentration is explicitly provided via the maximum solubility of lead chloride. for the second comment, $\frac{1}{x}$ gets bigger if $x$ is smaller.

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Extracting Copper from Copper Carbonate
In a crucible, we added copper carbonate and the total weight of both is 37.04g. After, we heated the crucible with the copper carbonate. It was easy to judge when the reaction had taken place as the colour of the substance had changed from the light green of copper carbonate to the black of copper oxide.After we observed this, we weighed the copper oxide and we notified the result which was 36.42g. We should have heated and weighed the crucible again, but we weren't able because it was broken(due to temperature change). But theoretically, which would be approximately the value of the second weighing?I'm curious because after searching on the Internet, I have seen both increase and reduction of the remaining mass.

Do you know the weight of the empty crucible? Also, I assume that you mean the total mass of the crucible plus the copper oxide was 36.42g?

Unless I'm missing something, the mass should not have changed with a second heating, assuming all of the copper carbonate was oxidized during the first heating. Were the conditions of the second heating supposed to be the same as the first? Are you supposed to calculate the total amount of copper or anything else?

@airhuff Yes, the total mass of the crucible plus the copper oxide was 36.42g. The weight of the empty crucible was 34.86g and the weight of the copper carbonate was 2.18g.

The instruction said that for the second time we should heat it for 1 minute and then we should weigh it again. It also said that if the two measurements differed more than 0.03g we should heat it for 1 minute again.  I'm supposed to think which should be the theoretical value of the second weight.

Cool, I get it now. The repeated heatings were just to ensure that you had oxidized all of the copper carbonate to copper oxide.  If you heated it a second time and the mass decreased, that would just mean that some of the copper carbonated had not been oxidized by the first heating. Then you would just keep heating until the mass quit decreasing, meaning that you had definitely oxidized all of it. So the answer to you question is simply that the mass theoretically should not have decreased at all with a second heating, unless some of the copper carbonate had not been oxidized the first time.

No problem. I'll write it into an answer real quick here, which you can accept if it was helpful to you. Thanks :)

Perhaps The carbonate used may have had a small amount of water in the product initially. Was it anhydrous before being loaded into the crucible? I agree with the above statement from Airhuff, the second heating was likely to just ensure everything was oxidized.

The repeated heatings were just to ensure that you had oxidized all of the copper carbonate to copper oxide. If you heated it a second time and the mass decreased, that would just mean that some of the copper carbonated had not been oxidized by the first heating. Then you would just keep heating until the mass quit decreasing, meaning that you had definitely oxidized all of it.
So the answer to you question is simply that the mass theoretically should not have decreased at all with a second heating, unless some of the copper carbonate had not been oxidized the first time.

I'm sorry for asking again, but is it right to say that the theoritical amount of Cu we extract will be the same as the mass of Copper Carbonate we added? if the whole copper carbonate oxidized to copper oxide?

There should be no loss of Cu, so the moles of CuO you made should be the same as the moles of CuCO3 that you started with. So, take the mass of CuCO3 you started with and divide that my it's molecular weight (123.6 g/mol) to get the moles of CuCO3 (which is the same as the moles of CuO and CuC3 in this experiment). Then multiply the moles of CuCO3 by the molecular weight of CuO, and this should give you the theoretical mass of CuO that you should have gotten if the experiment all worked correctly. Don't hesitate to ask if you have more questions.

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

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

MO symmetries should reflect the symmetry of the molecule. This is not a symmetry element for butadiene.

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

yes, thought about that but there must be 8 variants ( 2 for each ethene ,that's 2x2=4 and also their interactions ( node in middle or not ) that's 8 ) .also why pi* of one ethene can't interact with pi of another ( it would be the picture of mine )

The main reason you can't have an MO like that is, going back to the two ethenes example, the orbitals that overlap to form the MO have to be of the same symmetry (ie we couldn't form an MO from one orbital with $\pi*$ symmetry and another with $\pi$ symmetry). Also, we only form as many MOs for butadiene as we put in from the two ethenes, so we can only form 4 MOs total.


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How to determine acidity or basicity from pKa
I am trying to figure out if a drug that I have is acid or base and strong or weak. On drugbank database, I found the following pKa. I have been reading and understand that higher the pKa stronger the acid (and weaker the base) and vice versa, however, based on two reported pKa values, how do you determine if it is a base or acid and strong or weak?

The lower the pKa of an ionizable group the more likely it is to dissociate an H+ ion (proton) thus the more acidic the group is. pKa is nothing more than the negative logarithm of the Ka which is the acid dissociation constant. Determining whether a group is an acid or base is more difficult, because this depends on a couple of factors such as the inductive effect caused by other atoms in the molecule. There are quite some tables online of the most functional groups with their respective pKa's, but in general carboxyl groups are weak acids and amines weak bases.

Hello and welcome to Chemistry.SE. If you have any questions about the site a good place to start is with the short [tour].  You can also visit the [help]. I see you've already gotten some feedback on you question.  Good luck with it, and enjoy!

First value is "normal" pKa, second is pKa of conjugated acid.

Remember pKa is the -log of the Ka and if you can remember the relationship between Ka and how acidic something is then you'll be set.

thanks everyone for clarification. Your answers are very helpful.

Sorry, you have misunderstood the concept of acidity. Low $\mathrm{p}K_\mathrm{a}$ means stronger acids. High $\mathrm{p}K_\mathrm{a}$ means weaker acids.
I will give you two examples. Acetic acid has $\mathrm{p}K_\mathrm{a} = 4.76$. Hydrochloric acid has $\mathrm{p}K_\mathrm{a}$ of about $-7$. Surely, hydrochloric acid is a much stronger acid than acetic acid.

$K_\mathrm a$ refers to the degree of dissociation of the acid.
Consider:
$$\ce{HC2H3O2(aq)<=> H+(aq) + C2H3O2-(aq)}$$
$$K_\mathrm a = \frac{[\ce{C2H3O2-}][\ce{H+}]}{[\ce{HC2H3O2}]}$$
If the acid dissociates to a large extent, i.e. high $[\ce{H+}]$ ions, then its $K_\mathrm a$ will be high. When you convert a high $K_\mathrm a$ to $\mathrm p K_\mathrm a$, $\mathrm p K_\mathrm a$ will be low (just a mathematical relationship).
Thus, high $K_\mathrm a$ = low $\mathrm p K_\mathrm a$ = strong acid. High $K_\mathrm b$ = low $\mathrm p K_\mathrm b$ = strong base.


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What's the difference between RHF and UHF?
I am doing a course on electronic structure methods and am somewhat puzzled by the content (I have a physics background). From what I understand Hartree-Fock is a method where you only use the Slater determinant for the ground state of occupied spin orbitals. Then configuration interaction (CI) is an extension of this where you include unoccupied spin orbitals for your Slater determinants. But if Hartree-Fock considers the ground state energy, then how can you have unrestricted Hartree-Fock?

The names RHF and UHF refer to the invariance of Hartree-Fock wave functions with respect to certain symmetry operations. Fukutome has worked out the details. Also, Stuber gives the details in his PhD work. I.e., a rRHF ( = real Restricted Hartree-Fock) wave function is invariant under the $\hat{S}^2, \hat{S}_z, \hat{K}, \hat{\theta}$, where $\hat{S}$ is the spin operator, $\hat{K}$ the complex conjugation operator and $\hat{\theta}$ the time reversal operator. rUHF is invariant only w.r.t. the operators $\hat{S}_z$ and $\hat{K}$. So the operators form a group and the labels rRHF, rUHF, $\ldots$ are used to group wave functions.
Configuration Interaction (CI) is the interaction of a predefined set of Slater-Determinants (SD) or Configuration State functions (CSF) with each other. I.e., one forms a CI wave function
$$
\Psi_\mathrm{CI} = \Phi_0 + \sum\limits_{r}^{a} \Phi_{r}^{a} + \sum\limits_{rs}^{ab} \Phi_{rs}^{ab} + \ldots
$$
$\Phi_0$ is the reference wave function and $\Phi_r^a$ denotes the wave function constructed from the reference by exciting one electron from orbital $r$ to orbital $a$. The choice of $\Phi_0$, however, is up to you. Oftentimes a rUHF solution is chosen. But similarly, rRHF is possible.
Ground state in your nomenclature just refers to the lowest energy HF solution of a given symmetry. But in some cases, the wave function can be unstable w.r.t. the loss of some symmetries. E.g., you can calculate an rRHF wave function for the disscociated $\ce{H2}$ molecule and you will find it to be a (local) minimum. Using this wave function as a starting point for an rUHF calculation, you will quickly discover the local rRHF minimum to be a local rUHF maximum. Hence an optimization will lower the energy. This is closely connected to the terms singlet or triplet instability. For a more complete discussion please refer to books as, e.g., Molecular Electronic-Structure Theory by Helgaker, Jørgensen and Olsen.
In the mentioned case of $\ce{H2}$ dissociation it is also quite logical. As indicated already by others, RHF implies one set of spatial orbitals for all electrons of $\alpha$ and $\beta$ spin or, to put it differently, double occupation of each molecular orbital (MO). Triplet states cannot be represented with an RHF wave function! But while dissociating $\ce{H2}$ it is unreasonable to assume that the electron spins on the atoms stay paired forever. One would rather expect both atoms to be completely identical at infinite distance and thus to end up with two hydrogen atoms instead of a $\ce{H+H-}$ system. And this is what happens when you use UHF.
The complex versions of HF become interesting especially when dealing with relativistic effects. Due to the Pauli matrices the Hamiltonian becomes complex and hence the MOs become complex as well. By this the wave function is not symmetric w.r.t. to $\hat{K}$ anymore and hence belongs to a different group in the scheme of Fukutome and Stuber.

this is a good answer, but I think it would help if you elaborated on the applications a little bit. When and Why would someone choose to use RHF vs. UHF for example? What is a situation where one or the other would be unsuitable?

The restricted and unrestricted qualifiers refer to whether certain properties of the orbitals you use to make the HF Slater determinant are allowed to change. Specifically, RHF requires that every spatial orbital is used to form both an $\alpha$ and $\beta$ spin orbital, whereas UHF lifts this restriction and allows every spin orbital to be formed from a different spatial orbital.
A great resource that explains Hartee-Fock, from the very basics to the in-depth details of calculations, is Modern Quantum Chemistry by Szabo and Ostlund.
Edit: The main reason we do UHF calculations is that restricted determinants lead to poor results for systems where there are unpaired electrons. A good example of this is lithium. We could describe it by the restricted determinant $|\psi_{1s}\bar\psi_{1s}\psi_{2s}$>. There are two electrons in the same $\psi_{1s}$ spatial orbital, but each electron experiences a different effective potential (the $\alpha$ spin electron in $1s$ has an exchange interaction with the $\alpha$ spin electron in $2s$). So, we would expect that the $1s$ electrons would be better described by their own spatial wavefunctions which are consistent with the potential they feel. In this case, the new unrestricted determinant $|\psi_{1s}^{\alpha}\bar\psi_{1s}^{\beta}\psi_{2s}^{\alpha}$>, where $\psi_{1s}^{\alpha}$ and $\bar\psi_{1s}^{\beta}$ have different spatial components, gives a lower energy for the ground state of lithium.
This is detailed in chapter 2 of the book above.


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Equilibrium constant when given Ka and Kb of reactants

The first problem has completely stumped me on even how to begin. I just need some sort of direction.

It should help to look at the form of the constants:
$K_\mathrm{eq}=\ce{\frac{[NH4^+][CN^-]}{[HCN][NH3]}}$
For $\ce{HCN}$, $K_\mathrm{a}=\ce{\frac{[H^+][CN^-]}{[HCN]}}$
For $\ce{NH3}$, $K_\mathrm{b}=\ce{\frac{[OH^-][NH4^+]}{[NH3]}} = \frac{K_\mathrm{w} \cdot \ce{[NH4^+]}}{\ce{[NH3][H^+]}}$ where $K_\mathrm{w}=\ce{[H^+][OH^-]} = {1*10^{-14}}$
Is there anyway you can use some combination of $K_\mathrm{a}$ and $K_\mathrm{b}$ to get $K_\mathrm{eq}$?

So could I treat it like a Hess's law in a way where I add the Ka and the Kb and the reverse of Kw to get the overall equation then multiply them all?

You can treat it similar to Hess's law. However, I would just look at the product of $K_a$ and $K_b$. How does this product compare to $K_{eq}$?

Thanks for the help I ended up getting it how you could figure that out.

$K_\mathrm{eq}=\ce{\frac{[CN^-] [H+][NH4^+] [OH-]}{[HCN][NH3] [OH-][H+]}}$
$K_\mathrm{eq}=\ce{\frac{$K_\mathrm{a}$K_\mathrm{b}}{K_\mathrm{w}}}$
$K_\mathrm{eq}=\ce{\frac{(4.0*10^10-)(1.8*10^5-) }{10^14- }}$ = 0.72


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Mechanism of homologation of aldehyde to alkyne: Ohira–Bestmann reaction
This reaction was part of an organic scheme I was doing, and I was wondering what the mechanism of the reaction would be.

I tried first forming a carbene by letting $\ce{N2}$ leave from the diazocarbonyl. Then, I did a Wolff rearrangement, attacked the ketene with water, decarboxylated the COOH, then did a Wittig reaction with the phosphonate ester. However, I ended up with an alkene and not the alkyne.
I also tried first doing the Wittig rxn, but I didn't know where to go after that (since you can't form the carbene anymore).
Any help with the mechanism of this reaction would be appreciated.

This reaction is the Ohira–Bestmann modification1,2 of the Seyferth–Gilbert homologation.3,4
In the original Seyferth–Gilbert homologation, a diazomethane phosphonate is used to convert carbonyl compounds to alkynes with one extra carbon in the presence of potassium t-butoxide. I used acetophenone as the reactant here just to simplify things, but the same mechanism applies to your substrate.

In the last step, the diazoalkene may be equivalently thought of as a carbene. You could lose nitrogen to form a carbene, which undergoes a 1,2-alkyl shift to generate the alkyne. I have drawn it as being concerted but I am not entirely sure whether it is really concerted (as far as I know, there is still some degree of contention over the Wolff rearrangement — but feel free to correct me if I am wrong).
This procedure has some issues due to the strongly basic conditions. For example, in your substrate, the presence of t-butoxide can easily lead to racemisation α to the carbonyl group. The Ohira–Bestmann conditions are much milder and generally do not lead to racemisation.
As Waylander mentioned in a comment, you need to specify the solvent, but generally the Ohira–Bestmann version is carried out with potassium carbonate in methanol. The combination of carbonate and alcohol generates small amounts of alkoxide, which then attacks the carbonyl group in the diazophosphonate to produce the same reactive species:

However, I looked up your compound on Reaxys and it seems that the authors used caesium carbonate in t-butanol.5 I am very hesitant to suggest t-butoxide attacking a carbonyl group as a nucleophile, so I would go out on a limb and propose an alternative mechanism. The ketene generated can probably be intercepted by the t-butanol to generate t-butyl acetate. I don't have any evidence for this, however.

In general you had a lot of good ideas (Wittig-type chemistry and a Wolff rearrangement) and were mostly on the right track. The missing ingredient was probably the solvent, which is crucial for the generation of the reactive carbanion.
References

Ohira, S. Methanolysis of Dimethyl (1-Diazo-2-oxopropyl) Phosphonate: Generation of Dimethyl (Diazomethyl) Phosphonate and Reaction with Carbonyl Compounds. Synth. Commun. 1989, 19 (3–4), 561–564. DOI: 10.1080/00397918908050700.

Müller, S.; Liepold, B.; Roth, G. J.; Bestmann, H. J. An Improved One-pot Procedure for the Synthesis of Alkynes from Aldehydes. Synlett 1996, No. 6, 521–522. DOI: 10.1055/s-1996-5474.

Gilbert, J. C.; Weerasooriya, U. Elaboration of Aldehydes and Ketones to Alkynes: Improved Methodology. J. Org. Chem. 1979, 44 (26), 4997–4998. DOI: 10.1021/jo00394a063.

Kürti, L.; Czakó, B. Strategic Applications of Named Reactions in Organic Synthesis; Elsevier: Amsterdam, 2005; pp 402–403.

Brenneman, J. B.; Martin, S. F. Application of Intramolecular Enyne Metathesis to the Synthesis of Aza[4.2.1]bicyclics:  Enantiospecific Total Synthesis of (+)-Anatoxin-a. Org. Lett. 2004, 6 (8), 1329–1331. DOI: 10.1021/ol049631e.

Odd solvent choice. Would really have expected Pot. Carb. in tBuOH to racemise the proton next to the aldehyde.

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Remove labels from jars?
How can I remove labels from jars?
Sometimes I like to recycle jars for my own use, but normally I want to remove the label first. I find this extremely difficult to do. With ordinary stickers, like the stickers found as price tags on books, acetone works very reliably.
However, the labels on food jars are stuck on with a much more tenacious glue that does not respond to vegetable oil, acetone, methyl ethyl ketone, hot vinegar or ethanol. What is this stuff and how can I dissolve it?
Right now the only way I have found is to laboriously scrape them off with razor blades, but it is extremely time consuming to do this.
(Note that the tough labels I suspect are "hot melt" adhesives used to affix polypropylene labels. For example, traditional hot melt compositions are ethylene/vinyl acetate copolymer and ethylene/(meth)acrylate copolymers, but obviously there are others. I am hoping somebody with expertise can identify the range of typical compositions and corresponding solvents.)

Related question on Arts & Crafts: How do I remove labels from glass jars/bottles?

@Loong I have seen that. I am more interested in a scientific answer, not folklore from arts and crafts people.

Based on your deduction that the adhesive is of the types you mentioned, you could try isopropyl alcohol (or methanol if you can be safe about it). I am surprised acetone didn't work, my only reason to suggest an alcohol is that they are slightly more polar and that could work with the acetate/acyrlate portion of the glue.

I think this question is impossible to answer. There are dozens of chemical types for label adhesives. Most are thermoplastic, meaning they don't form new covalent bonds with the surface. But some are thermosets, meaning they react and cross-link.
In general, it is quite difficult to dissolve a crosslinked polymer since the dissolution process requires destruction (scission) of covalent bonds. The ways to do that are obvious: heat, oxidation, reduction, acids, or bases.
One thing that needs to be said here is consideration of both the opportunity cost as well as the environmental cost of your method generally makes all but the easiest methods inadvisable. When you say the adhesive "does not respond" to X, do you soak the label in X? (I typically use a slightly larger container (say part of a PET bottle) and a shallow layer of the solvent and allow it to soak overnight.
It's always a good idea, in my opinion, to start with water with a drop or two of surfactant in it. If hot water doesn't work, I move on to 70% IsoPrOH and give that a few hrs. Next, I use an alkaline (basic) detergent solution and give that another overnight soak. If it's worth my time, I end with an acetone soak (I have acetone which I reuse for this purpose in its own container.) One of these works most of the time. When none of these work, I toss (recycle) the jar. You may or may not know that not all polymers are soluble. So, your assumption that there is a solvent for the label you want to remove isn't necessarily true.
The most aggressive solvent I am aware of is mixtures of methylene chloride, dodecylbenzene sulfonic acid, and N-methyl pyrrolidone. That was used to de-pot Soviet microelectronic back in the Cold War days to reverse-engineer them. Anyway, glass is mostly Si-O-Si bonds and alkali bases should loosen any organic's adhesion to it.
Be sure to remove any surface layer on the label - abrasion is generally quickest. Oh, speaking of dangerous chemicals (methylene chloride is a carcinogen, and ddbsa is corrosive to the skin), I've also used aqueous HF to clean glass. The HF actually removes some of the silica, but if done carefully enough doesn't visibly damage the glass surface (never would try this with glass jars; why bother? but with expensive glass flasks, and other apparatuses, it was worth the hassle. The disposal of the waste solvent was a pain, though.)


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How does mCPBA convert an amine to an N-oxide?

Is this how the mechanism works?
Since it is a very stereoselective reagent I thought maybe it proceed via one step only, but how would that then work?

Firstly, any stereoselectivity in this reaction is not affected by the second step you drew (which is a simple proton transfer). This is because your second step neither creates nor destroys any stereochemistry present in the reactants.
However, mechanistically it is probably easier to deprotonate in the same step using the carbonyl oxygen of mCPBA.


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if the shared electrons in covalent bond revolves around their nucleus, how they hold the two molecules together?
Covalent bonds are formed by sharing electrons. The pictures depicting this bond make it seem like that the shared electrons stay in a fixed/limited place. What happens in reality? If they rotate around their nucleus how the shared electrons hold the two nuclei in an atom?

Electrons don't rotate, nor do they rotate around their associated nuclei if we are taking about atoms in a molecule.

Let's consider a simple case, $\ce{H2}$. When the two nuclei come closer to each other (starting from a long distance), the electrons benefit in terms of energy because they now enjoy being attracted to two nuclei instead of just one. This energy gain is (partly) responsible for the bond dissociation energy, which needs to be invested in order to break the bond again. For a discussion on why the nucleus-electron interaction outweighs the electron-electron repulsion, see this question.


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Electronic configuration of uranium
I read that the electronic configuration of uranium is [Rn] 5f³ 6d¹ 7s²  .
Given that the subshells fill in the order 5f --> 6d, why is the 5f subshell only partially filled? Why do electrons fill the 5f subshell partially then proceed to fill the 6d subshell?

Nobody really knows.

@OscarLanzi maybe when we will get a chance to simulate atoms using quantum computer we will realize why it is what it is.

@Volpina that will depend on how well developed is the theory with which we program the computers.

@OscarLanzi IBM's quantum computer is already at 500+ qubits so if each qubit represents 1 electron then we already are very capable of simulating such atoms but I am not a expert in quantum computing so I could be wrong.

@Volpina We aren't either. We need to effectively translate that information into what we, as Chemistry folks, could understand.

Nobody really knows. Our theory of how electrons are ordered in the ground states of atoms is a case of good news and bad news. And, as in human endeavors, the bad news ends up getting the headlines.
First, the good news. Electrons do occupy shells (numbered $n$ in typical nonenclature) in perfect order, just as Bohr imagined -- when we look only at each individual level of angular momentum ($l$). Among $s$ orbitals ($l=0$) we have $1s$ at the lowest level, then $2s$, then $3s$ and so on. With one unit of angular momentum, which we don't have until the second shell, the energies are $2p<3p<4p<...$. And so on with $d$ orbitals ($l=2$), $f$ orbitals ($l=3$), etc.
The bad news creeps in when we try to consider variation in $n$ as well as $l$; that is, when we try to consider more than one of the angular-monentum-based tiers of shells at the same time. We know that with one electron, the energy level depends on just the shell number $n$; thus both $2p$ as well as $2s$ will be at higher energy than $1s$, and $4s$ has to sit above $3p$ and $3d$ as well as $3s$. But with multiple electrons, the shielding effects of electron-electron repulsions causes higher angular momentum states of each shell to rise above lower angular momentum states; we have $3d>3p>3s$ as well as $4s>3s$, but these relationships may allow, let us say, $3d>4s$ even though $3<4$. We get neutral iron atoms with the $4s$ orbital filled while $3d$ orbitals are left short, or worse yet a situation where the clashing orbital levels are balanced in such a way that filled and empty orbitals are most stably distributed between both (chromium = $\ce{[Ar]}3d^54s^1$).
We might explain the configurations of chromium and copper in terms of exchange energy or spherical symmetry, but such arguments cannot begin to cope with the fifth-period elements, let alone subsequent periods when $f$ orbitals come into play. By the time we get to uranium we are guessing as much as knowing, even with computer outputs whose underlying programs are also based on educated guesses.
There is one way to simplify things — a little. If we add extra protons (and any needed neutrons for nuclear stability) to the nucleus and thus form a positive ion, we tend to push the electrons back towards the shell-by-shell order that the energy levels of a single electron would obey. This is because the extra nuclear charge breaks through the electron-electron repulsive shielding. Even a small ionic charge can have a large effect on valence electrons, whose effective nuclear charge is also relatively small in the neutral atoms of most metals. The case of $24$ electrons is described in both Jan's and my own answers to this question: adding just two extra positive charges to the nucleus converts the $\ce{[Ar]}3d^54s^1$ of $\ce{Cr^0}$ to $\ce{[Ar]}3d^6$ of $\ce{Fe^{2+}}$. When we put in the full electronic configuration of the argon core, we find that the extra protons in the ferrous ion have pushed all the electrons into shell-by-shell order.
Similarly, with $88$ electrons the configuration of $\ce{Ra^0}$ is $\ce{[Rn]}7s^2$ whereas that of $\ce{U^{4+}}$ (a common oxidation state of uranium) is $\ce{[Rn]}5f^2$. As with the ferrous ion, the valence electrons have dropped into a lower shell under the influence of the ionic charge on the uranium. However, with the heavier atoms we still have lower angular momentum states ($s,p$) occupied up to the sixth shell even as higher angular momentum states fail to fill the fifth shell. The radon core, containing electrons in the low-angular momentum $6s$ and $6p$ subshell but none in the higher angular-momentum $5g$, is itself not in shell-by-shell order so the idea that it holds up in seventh-period metal ions is also sonewhat of a guess. So the shell-by-shell ordering remains imperfect even for chemically accessible ions in this case and generally beyond the fourth period.

Uranium has the electronic configuration $\ce{[Rn] 5f^3 6d^1 7s^2}$. You might alternately ask as to why it doesn't have the configuration $\ce{[Rn] 5f^4 6d^0 7s^2}$ if it should follow Aufbau principle.
But remember one more fact that d orbitals are slightly better shielders than f orbitals. Now the problem with $\ce{[Rn] 5f^4 6d^0 7s^2}$ is that there are no d electrons which causes the effective nuclear charge $(Z_{eff})$ to increase on the $\ce {7s^2}$ electrons. But this increased $Z_{eff}$ would tend to draw $\ce{7s^2}$ electrons toward the nucleus and thus decreasing the stability of a large orbital of $\ce{7s^2}$ and the electrons in it.
On the other hand, $\ce{[Rn] 5f^3 6d^1 7s^2}$ contains one d orbital which will, with it's slightly better screening, would decrease $Z_{eff}$ on $\ce{7s^2}$ electrons and thus keeping it distant from the nucleus and hence stabilizing it.

I am sure you are familiar with the rules for assigning electron orbitals, I will briefly descibe them here:
Electrons fill orbitals in a way to minimize the energy of the atom. Therefore, the electrons in an atom fill the principal energy levels in order of increasing energy (the electrons are getting farther from the nucleus). The order of levels filled looks like this:

Pauli Exclusion Principle
The Pauli exclusion principle states that no two electrons can have the same four quantum numbers. The first three (n, l, and ml) may be the same, but the fourth quantum number must be different. A single orbital can hold a maximum of two electrons, which must have opposing spins; otherwise they would have the same four quantum numbers, which is forbidden.
Hund's Rule
When assigning electrons in orbitals, each electron will first fill all the orbitals with similar energy (also referred to as degenerate) before pairing with another electron in a half-filled orbital. Atoms at ground states tend to have as many unpaired electrons as possible. This explains the behaviour of Chromium: Z:24 [Ar] 3d54s1 (note here the the one electron in 4s orbital while the d orbitals are occupied with single electrons of one spin direction)
Exceptions
Although the Aufbau rule accurately predicts the electron configuration of most elements, there are notable exceptions among the transition metals and heavier elements. The reason these exceptions occur is that some elements are more stable with fewer electrons in some subshells and more electrons in others and a notable example is uranium, for it to acquire maximum stability is usually having this ground state: Uranium: Z:92 [Rn] 7s2 5f3 6d1
References

Rules for Assigning Electron Orbitals

Ah so uranium is an exception to this rule. What is it about this specific configuration that makes it so stable?

Its not only Uranium,  read again I have also mentioned Chromium. There are other elements for example Copper, Niobium, Palladium, Silver, Thorium etc which deviate from this trend. The reason like described is partly based on the combination of the rules. Remember that in a ground state of an element the electron configuration has its lowest energy. The lower the energy the more the stability. In some cases this kind of stability can only be obtained when there are fewer electrons in a particular orbital say Uranium configuration.

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


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Electron pairing and repulsion
If two electrons that get paired occupy the same orbital, then wouldn't there be a heavy amount of repulsion between the two?
As is, since electrostatic force ${\displaystyle \propto }$  $\ce{ 1/ (distance)^2}$   so since the distance between two electrons is way smaller than between the electrons and nucleus, I don't really assume it has any role here.
Does this have anything to do with the strong nuclear force? As far as I know its range is too short and exists only between protons and neutrons.
Summing up, what causes two paired electrons to stay in the same orbital rather than get thrown away due to "the force"?

No, it has nothing to do with the strong nuclear force, which operates, well, in the nucleus. The pairing of electrons in atomic wave functions is a natural consequence of quantum mechanics.

No strong nuclear force is involved. Instead, tt's the same force as always: electromagnetic interaction. By staying in the same low orbital as the other electron, the electron in question will continue to be subject to a stronger attractive force from the nucleus. Note that the MO diagrams (see e.g. wikipedia) typically already account for the intra-orbital electron-electron repulsion, and you still have HOMO-LUMO gap.
It is also helpful to remember that electrons are "spread out" over the orbital, so once you do the math, you realize that the average distance between electrons does not have to be small compared to the average electron-nucleus distance.
Let's get semi-quantitative: $\ce{H_2}$, 1.4 bohr, HF/STO-6G($\zeta = 1.20$). After the SCF is completed, I transformed the integrals from the AO basis to the MO basis, which yields the following, relevant integrals in Mulliken notation:
(1|h|1) -1.24964
(2|h|2) -0.49918
(11|11)  0.65720
(22|22)  0.68091
(11|22)  0.64686
(12|12)  0.17440

Nuclear repulsion is 0.71429, all values in hartree. We can then calculate the low-spin, ground-state energy as:
$$
E_\text{gs}^\text{RHF} = E_{nn} + 2\cdot(1|h|1) + (11|11) = -1.12778,
$$
wherein we add the nuclear repulsion energy, the one-electron energy of the lower MO (once for each electron) (one-electron meaning that this is the sum of the electron's kinetic energy and its nuclear attraction) and the relevant two-electron integral for the Coulomb repulsion. Since the electrons are of opposite spin, no exchange term (as is responsible for the spin-pairing energy, cf. Hund's rule) is needed.
The energy of the high-spin, "excited" state can be calculated according to:
$$
E_\text{triplet}^\text{RHF} = E_{nn} + (1|h|1) + (2|h|2) + (11|22) - (12|12) = -0.56207,
$$
where an additional exchange term enters the equation to account for partly-correlated electrons due to Pauli-repulsion.
This calculation is semi-quantitative because RHF in such a small basis set does not yield quantitative results. One would have to go to natural orbitals obtained through full CI in a larger basis set for fully quantitative results. However, for illustrating the relative size of the quantities involved, this level of calculation is sufficient.


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Can equivalent mass be more than the molecular mass? Can it be applied to radical reactions?
Consider a reaction involving radicals:
$$\ce{CH3^. + Cl^. -> CH3Cl}$$
Is the definition of equivalent mass applicable here? If yes, how?
Are there reactions where equivalent mass of a reagent is more than its molecular mass?

I think this Chem-guide gets at what you are asking about in the section about compounds, though I must admit I'm almost entirely unfamiliar with the concept of equivalent mass.

https://en.m.wikipedia.org/wiki/Equivalent_weight  radicals aren't included in the Wikipedia definition


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Do Li4 or Li8 molecules exist?
I've read that alkali metals form ionic bonds; $\ce{Li}$ is an exception which majorly forms covalent bonds. Wikipedia says dilithium exists. This makes me wonder why $\ce{Li_8}$ doesn't exist. It seems having all 8 electrons shared produces a low energy state (because the valence shell in each of the 8 atoms will be completely filled) ?

You're wrong, such clusters (and others) do exist http://www.sciencedirect.com/science/article/pii/0304510275800071 , http://iffwww.iff.kfa-juelich.de/~jones/jcp97jlh.pdf

The very word "exists", as it happens, can be profoundly confusing. You can't buy dilithium from a chemical supply store. Neither can you make it by yourself in any significant quantity, no matter how hard you try.

Yes, they do exist and were characterised spectroscopically, see here (and there is a note on similar clusters for sodium):
Blanc, J.; Bonačić‐Koutecký, V.; Broyer, M.; Chevaleyre, J.; Dugourd, P.; Koutecký, J.; Scheuch, C.; Wolf, J. P.; Wöste, L. Evolution of the electronic structure of lithium clusters between four and eight atoms. J. Chem. Phys. 1992, 96 (3), 1793–1809. DOI: 10.1063/1.462846.
They are not, however, "stable" enough to exist in solid state or even to obtain a pure gas, rather complicated arcane trickery is used to work with (very small amount of) them.
Lithium has rather complicated chemistry, easily forming polyhedral structures when bound to proper partner. For example, methyllithium is a tetramer. The simplest rationalization is that its s- and p-orbitals have very close energies and thus low-energy vacant p-orbitals participate in formation of multicentre electron-deficient bonds.

You may be applying the octet rule to a situation where there is no p orbitals involved, the electronic structure of the (ground state) Li atom. Li has the electronic configuration 1s²2s1. So, the covalent bond is between two 2s orbitals, each with 1 electron. Using the Lewis structure, Li2 is Li:Li and that is a full bonding orbital, there's no p atomic orbitals involved in the bonding which might increase valence to 8 (an octet).

Looking at the electron configuration of lithium ($\ce{(1s)^2(2s)^1}$), it's very difficult for the $\ce{1s}$ electrons to be involved in bond formation/sharing.
The reason is because of the extent of attraction of those electrons in the 1s orbital as they are very close to the nucleus (very stable) and rarely available for any bond formation. Again, say if we were to look at the ionisation energy required to extract those electrons in the $\ce{1s}$ orbital, you would agree with me that it's almost impossible.
Looking at the $\ce{2s}$ orbital, you notice it's not as stable since there is more room for electrons to come and occupy in the $\ce{2s}$ (or $\ce{2p}$) orbital to attain maximum stability.
In short, we can't have a $\ce{Li8}$ molecule.

@alphonse . Thanks for the correction, I was basing my explanation on most stable bonds/ compounds formed by Li. Some Li compounds are rare and very difficult to isolate so I had taken that angle. I apologise for this mistake and I do agree with permeakra's explanation


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Is glass amorphous solid or supercooled liquid?
I've read that glass is a supercooled liquid. But lately I've came to know in a science info post that glass is amorphous solid. How? Please someone explain me.

Is the glass half-empty, or half-full?


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(Single) Amino acid rotational degrees of freedom
When defining the rotational degrees of freedom for a single molecule, should we just include the psi and phi angles (red lines in the picture) or the rotation around every single bond (yellow lines in the picture) or both?

You should be concerned with bonds where the rotation is non-trivial. For example, rotation about a C-H or O-H bond doesn't do anything. But rotation around the serine side-chain C-O bound or the side-chain central carbon bond or the C-O carboxylate bond all matter.

Thank you! So, we would say that a single serine has 4 rotational degrees of freedom (C-N, 2 * C-O, C-C)?

Are you just talking about bonds? You also have overall rotation...

Also I count a second C-C.

Thank you again; really appreciate it. I forgot the 2nd C-C. Since there are these bonds of interest, I was thinking of calculating all the possible combinations of rotations around them to get the final number, taking into account that, additionally, each of the possible non-linear "final molecules" has 3 degrees of rotational freedom.


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Energy of an electron (why negative?)
Why is the energy of an electron in an orbit is always negative?
What does it mean?
Why isn't it positive?
Does it mean that energy is released by revolving electron?

This doesn't mean anything, because it is not a fact of nature. Rather, it is a matter of where you set your zero.

Let me explain you in the simplest manner.
For a two object system,
If they repel each other when you move them towards each other, that means it takes work to do this, and the potential energy is positive.
If they attract each other when you move them towards each other, then potential energy of the two object system goes on decreasing.
Potential energy of a two charged object system at a particular orientation is defined as work done by external agent to bring the second body from infinity to desired point in presence of the field created by the first body. Now think the first body to be nucleus and the second body to be electron.  Now to find the potential energy, according to potential energy defination I just said above, we need to being the electron from infinity to any desired location, lets say to a distance of "x" from the nucleus. As we have to bring electron from infinite that means initially electron is infinitely far away from nucleus and hence at zero potential.
Potential Energy of electron:
When we are bring the electron towards the nucleus, offcourse they are going to attract each other and as said earlier if two objects attract each other, then potential energy of the system goes on decreasing. So as we move electron towards the nucleus, energy goes on decreasing. Note that initially as electrons and nucleus were at infinite distance from one another, hence initially at zero potential and any further decrease in potential energy implies negative potential energy.
So potential energy of the electron is negative inside any atom.
Kinetic Energy of electron:
The electron also has kinetic energy. Kinetic energy can never be negative (because in K.E expression we have squared of velocity which is always positive) and it is smaller in magnitude than the potential energy.
Total Energy of electron:
So the total energy i.e kinetic energy  + potential energy is negative.


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Retrosynthesis of a cyclopropane type molecule
I'm trying to work out a retrosynthesis of the following molecule:

The issue I'm having is regarding the building up of the carbon skeleton. I know I can build the cyclopropane group using a carbene (diiodomethane, zinc) on the alkene. However, the alkene is proving difficult to build, I can't see why an obvious way around building this (Witting/from the equivalent alkyne) wouldn't work here. Is there something I am missing here like a rearrangement?

Alkyne seems workable.

@orthocresol I guess you could start with methyl propiolate, then could deprotonate the alkyne and get the other part on the alkyne, however the equivalent for that would be something like 2-bromo-2-methoxypropane which isn't readily available and can't really think of a synthetic equivalent for it

Count your carbons again - you should deprotonate acetylene and react with acetone, followed by methylation of the alkoxide. Then you deprotonate again and add methyl chloroformate or something. Lindlar hydrogenation and cyclopropanation would give you the desired compound.

@orthocresol Would the deprotonated alkyne react with acetone without any other reagent? The rest makes sense, to get the correct stereochemistry would just hydrogenate with Lindlar's catalyst and use the carbene to make the cyclopropane. Thanks for the guidance!

As far as I know alkynes (or their conjugate bases to be precise) generally add to carbonyls without issues, I think acetone should be no different.

You will need to protect the ester though.

How about this? Cyclopropane from carbene + maleic anhydride. Add 2 eq methyl Grignard and trap with methyl iodide.
Possibly, you could do the exhaustive addition of Grignard first, then cyclopropanate and methylate in one shot with diazomethane. Thanks to Beerhunter for this suggestion in comments.

very much nicer than making it from the alkyne!

Did you mean carbene rather than methyl carbene? This is a lovely proposal. I would rejig this slightly to use the double methyl addition first, then acidic work up to give the gamma hydroxyacid. From there, exhaustive reaction with diazomethane (done by Aerojet on plant scale), with manipulation of conditions, to methylate the hydroxy, the acid and make the ring in one go. I'm not posting this as an answer as the one above is the definitive starting point for a strategy.

Yes, I meant carbene. Sorry for the confusion. Fixed. I'm also going to include your suggestion to cyclopropanate afterwards.


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Chemical theory, formula(s) that predict final *volume* of a solution?
I asked this question a couple years back regarding the volume of a mixture being less than the volume of the original solute and solvent, but this question asks if there is theory and/or formulation that can predict what the final volume would be. And based on the answers from the last question I suspect there may be more than one answer depending on the nature of constituents (ionic vs molecular solution, etc.)
Furthermore in the case of ionic salts is there a theory/formulation that predict which saturated salt solution (e.g. a chloride, acetate, bromide, etc.) would result in the most efficient packing between water molecules (at the ame temperature).
Does such theory, formulation exist?

Look at: http://www.thermopedia.com/content/734/
Equations of state can be used to describe the behaviour of single and multi-phase systems to varying degrees of accuracy. I'm not familiar with any work addressing your question about the water packing in a salt solution.

https://chemistry.stackexchange.com/questions/44487/theory-of-volume-contraction-in-solution


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If an action increases the reaction rate in a particular equilibrium mixture, does it mean that both forward and backward rates are increased?
Say for example we increase the temperature in a closed system containing gaseous reactants. The forward reaction is exothermic so the yield of the product will decrease as we increase the temperature. But increasing the temperature increases the reaction rate so does it mean that the forward and backward rates are both increased ?

The forward and reverse reaction rates both increase, but the reverse rate increases by more. By Le Chatelier's principle, we know that $K_{eq}$ for an exothermic reaction will decrease when the temperature is raised. We also know that $K_{\text{eq}}=\frac{k_{\text{forward}}}{k_{\text{reverse}}}$ and that $k=Ae^{\frac{-E_a}{RT}}$ (The  Arrhenius equation). Since this is an exothermic reaction, $E_a(\text{forward})<E_a(\text{reverse})$, which means any increase in $T$ will increase $k_{\text{forward}}$, but it will increase $k_{\text{reverse}}$ by more. This lead to a decrease in $K_{\text{eq}}$, consistent with what was predicted by Le Chatelier's principle.

Another action you could take to speed up a reaction is to add a catalyst. In this case, if you keep the temperature the same, the proportion of the two species will remain unchanged, and both forward and reverse reactions are accelerated by the same factor.


text:
If the same atoms (cells) that now compose my body get arranged in N years away, am I that person?
Probably a very stupid question, even may deserve to be downvoted, but I have always wondered: Hypothetically, let's say somehow tomorrow I die in an accident. Now lets say that after a long time (e.g. centillion of years) for one reason or another the cells that made up my body -right before I died- are part of a new body arranged exactly the same doing the exact same functions. My questions are:
1) It is possible? -Possible in the sense that atoms that are cremated tomorrow or ate by worms, can they go back to their previous chemical state?-
2) Am I the same person? Not philosophically, I mean are there in my brain the same synapses? or are other variables to take into account? (I guess this one is more about neurochemistry than chemistry)
Let's not discuss the likelihood of happening, at least not much, because time is infinite one can argue that it may be bound to happen. But time being infinite or not and related questions may be more of an cosmology issue.

I would say this is outside the scope of chemistry. You might be able to stretch and say it is a biological question, but I think it is more suitable as a philosophy question. I don't think science has an answer on whether the same arrangement of the same atoms would constitute the same person. We have yet to achieve cloning or understand the brain on anywhere close to the level to say the collection of cells would be the same person.

Let me start from the bottom up: If we assume the earth as a closed system, i.e. exchange of energy with the surrounding space is possible, but exchange of matter impossible, than naturally our body contains atoms previously present in other form of matter.  Already our breakfast is matter we incorporate, subject to chemical reaction (digestion), and eventually excrete in one form, or an other.  Hence it is likely that we contain some atoms that once constituted a mammoth, or a dinosaur, or by your tomato in the back yard.
Digestion often splits the larger molecules present in our food into smaller ones.  This eases transport of matter across membranes, for example.  Think of proteins de-constructed into amino acids; our organism uses the smaller to build new proteins in need.
Beside chemical degradation as in aforementioned example of the proteins, digestion includes physical degradation, too -- chewing food, for example increases the surface area the chemicals responsible for digestion may access.  By  both means of action, physical and chemical degradation, entire cells face only a tiny chance to remain intact "as such".  It is more likely that they are dismantled, are split into their constituents, eventually -- as molecules -- harvested in the ways briefly mentioned earlier.
To sum up: so far there is no evidence a synapse of yours will be recycled, maintaining the original form and function, to find its way back as synapse in an other brain, similar to a spare tire.  Atoms of yours however, as easily they entered your body, eventually will leave you; at their pace.


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Rotational Constant for CO2 from an IR Vibrational-Rotational spectra
I have been trying to calculate the rotational constants ($B$) for $\ce{CO}$ and $\ce{CO2}$ from IR Vibrational-Rotational spectra.
I know that for $\ce{CO}$ the peak spacing is approximately $2B$ (ignoring effects due to centrifugal distortion) however for $\ce{CO2}$ it seems the peak spacing represents $4B$ instead when I compare my results to literature?
Does anybody have any explanation as to why this is, or could provide reading on this subject to explain why the peak spacing for $\ce{CO2}$ is $4B$?

I'm assuming you are measuring the rotationally well resolved band at $\approx 2349 \pu{cm^{-1}}$ and you are correct.
The rotational band structure for the antisymmetric vibration centred at $\approx 2349 \pu{cm^{-1}}$ (also called parallel $v_3$ band) has alternate lines missing, which are those having odd J. $\ce{CO2}$ is a linear symmetric molecule ($D_{\infty h}$ point group) but nuclear spins on $^{12}\ce{C}$ and $^{16}\ce{O}$ are both zero and means that antisymmetric rotational lines are missing. The ground vibrational state is $\Sigma^{+}_g $ resulting in the odd levels being absent.
The effect is due to the symmetry properties of the wavefunction, and  what happens on exchange of coordinates. For fermions ( multiples of spin 1/2 , eg. electrons, protons,) the wavefunction must be antisymmetric but symmetric for Bosons (integer spin 0,1,2 etc) which is the case for $\ce{CO2}$. Look for 'statistical weight', 'linear molecules', and 'nuclear spin' in a good spectroscopy book, Herzberg 'Infra Red and Raman Spectra' for example.


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What fraction of the vapour in electronic cigarettes is water?
Not a chemist but looking for an answer I can understand.
Electronic cigarettes use propylene glycol and/or vegetable glycerin as a base liquid. This is heated and vaporized at a low temperature (my e-cig is 7 watt). When exhaled the VG/PG absorbs water from the air producing the visible vapour cloud.
Assuming standard temperature ($\pu{20^\circ C}$) and humidity ($\pu{50\%}$), what ratio (or percentage) of the visible cloud is water?
It's not that I can't find an answer to this, just can't find an answer I believe. Most based on rumor and warm fuzzy feelings of what people want to believe.

The short answer to this is that there is no single answer to this problem as the vapor composition that is produced by any given atomizer is dependant upon a large number of components. To name a few:

The initial composition of the liquid, which will vary flavor to flavor and manufacturer to manufacturer. You can find out a bit more about this with the following articles

Characterisation of mainstream and passive vapours emitted by selected electronic cigarettes DOI 10.1016/j.ijheh.2014.10.001
Electronic cigarettes: overview of chemical composition and exposure estimation PMCID: PMC4304610
     A note to consider with the above is that many toxic organic solvents such as Formaldehyde, Acetaldehyde, Acetone, Acrolein, Propanal, Ethylene glycol, and    Diethylene glycol are found in eLiquids, in various concentrations.

The temperature that the eLiquid is heated to internally (this affects the vapor pressures of the various components in the liquid)
The concentrations of all of the, solvents, flavorings and such which are adsorbed by the person who is using the atomizer

Given all of these factors, and the large number that I am sure that I am missing, there is simply too many confounding variables to be able to give a straight answer to your question without directly measuring the concentration - not that I would know how you would make such a measurement.

Disclosure: I'm not a chemist.
First off you can't be sure what temperatures your atomizer reaches purely by wattage because it depends on coil resistance, material and if you run your vaporizer in TC mode (I assume you do not have that option available).
Humidity will only have a visual impact on cloud density after you exhale. Depending on atmospheric humidity glycerin molecules will effectively bond with water in the air. The more humidity it is, the more water will be picked up.
The ratio of glycerin to water in the cloud forming then will depend on:
- temperature
- humidity
- volume of air in which the cloud is expanding
What I'm more interested in is if this contributes to atomizers popping (especially open  air ones like RDA's). Say I leave it standing on my desk overnight soaked in juice. What will be the liquid water % be then.
Also how does this effect contribute to nicotine neutralisation? I red that it gets broken after contact with oxygen and I guess that it will get plenty of it while floating in an areosol particle.


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lowest energy conformation of butan-2,4-diol
Do the OH groups want to be furthest apart? or the CH3 groups?
I'm trying to determine the lowest energy conformation - and I don't know which group has the bigger steric effect?

Consider H-bonding.

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

In butan-2-ol you have one alcohol, which is indeed smaller than the methyl group.

sorry i've just realised that I meant a diol.


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Does water have a chemical name?
Does water have a chemical name?
If so, what is it?
P.S. I checked up the web and got all sorts of crazy answers like dihydrogenmonoxide, oxidane, hydrogendihydride etc. Please validate.

Related: Do chemists refer to water as “dihydrogen monoxide”?

What exactly are you expecting of this "chemical name"? In what way is "water" "unchemical"? In what context do you want to use it, and why is "water" insufficient?

@orthocresol Any reason why this is not considered a duplicate?

@Dmitry The question I linked asks about the usage of one specific name, "dihydrogen monoxide". This question is "what is water called". Close, but not quite the same, imo. As you can see, my answer didn't touch on the use of "dihydrogen monoxide".

TL;DR IUPAC hasn’t made up their mind, but plain old water appears to be an appropriate name. However, chemical derivatives of water may not be named using water.

In Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 it is stated (P-21.1.1.2) that

The common names water, ammonia, [...] are used in these recommendations but their use as preferred IUPAC names is deferred pending publication of recommendations for the selection of preferred inorganic names; thus, no PIN label will be assigned in names including them.

I’m not sure when the next edition of the Red Book (i.e. inorganic chemistry nomenclature recommendations) is coming out, but I can quote the relevant section from the most recent version.
Table IR-6.1 of Nomenclature of Inorganic Chemistry: IUPAC Recommendations 2005 lists "oxidane" as the parent hydride name for $\ce{H2O}$. However, it also adds the caveat that

The names ‘azane’ and ‘oxidane’ are only intended for use in naming derivatives of ammonia and water, respectively, by substitutive nomenclature, and they form the basis for naming polynuclear entities (e.g. triazane, dioxidane). Examples of such use may be found in Section IR-6.4 and Table IX.

Therefore the compound $\ce{ONONO} = \ce{(ON)2O}$, a nitrosylated derivative of water, is named "dinitrosooxidane" and not "dinitrosowater". More examples may be found in Table IX of the same publication.
Water itself is still called water. For example, tritiated water $\ce{H^3HO}$ is named (3H1)water (Section IR-2.2.3.2).

To make a short answer shorter, yes, dihydrogen monoxide.
To expand on it; the chemical name is based on the chemical equation of the elements, in this case $\ce{H2O}$, in which water is made up of two hydrogen atoms and an oxygen atom thus the chemical name is created to tell this so you get dihydrogen monoxide, "di" meaning two, "mo" being short for mono, meaning one, and "oxide" meaning oxygen. And it works the same for most elements.

Analogy with hydrogen peroxide ($\ce{H2O2}$) and hydrogen superoxide ($\ce{HO2}$) suggests that the number prefixes would be omitted: "hydrogen oxide".

@zwol That's not correct.  Hydrogen Peroxide is named weirdly for partly historic and partly chemical reasons.  See here, and  here. HO₂ is a superoxide, not hydrogen superoxide.  See here and here.  There are rules for naming covalent compounds.


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When would the presence of lone pair result in a polar molecule?
In what cases would the presence of a lone pair result in a polar molecule? For instance, O3 is polar despite only consisting of O atoms due to the presence of lone pair of electrons. However, O2 also has lone pair of electrons but is not polar, why? Also, for HAt, the electronegativity difference between H and At is 0, but is HAt polar or non-polar?

You misunderstand why ozone is polar. Draw the resonance structures and think about its shape.

Lone pairs have nothing to do with polarity. Also, the element At is better thought of as non-existent.

Then why is O3 polar given that it only has oxygen atoms and has no electronegativity difference?

A polar molecule must have a finite net dipole moment.

A dipole moment is a result of separation of charges.
The below explanation requires you to have a basic understanding of dipole moment, vectors and resonating structures.

Ozone is not polar because of the presence of lone pair on oxygen atoms. In ozone, all the oxygen atoms are not same. Some oxygen atom wants to hold on the electrons more than the others which leads to separation of charges. This separation of charges results in a dipole which makes a molecule polar.

Electronegativity difference in a bond causes one atom to have more affinity for the electrons. This makes that atom attract the pair of electrons more, hence developing a slight negative charge on itself(because of more electron density) and a slight positive charge on the other atom(because of less electron density). This leads to separation of charges and creates a dipole. In the figure, a dipole is shown from carbon atom to chlorine atom. A dipole is represented as a vector with its tail pointing towards the atom having more electron density.

It is not the dipole but the net dipole moment that makes a molecule polar. The net dipole moment is a vector sum of all the dipole vectors. Now, why need vectors for finding the net dipole moment?
Suppose you are being surrounded by a bunch of strong guys who are pulling you through a rope from all sides. They are applying heavy forces to pull you towards them but you won't move because the net force on you is zero. Similarly, in the second figure, the four chlorine atoms are pulling the bond pair of electrons toward itself from 4 different orientations. To account for those different orientations which we know will affect the net pull of electrons by chlorine atoms, we need vectors.

Consider this case of carbon tetrachloride. The carbon can be thought of being placed at the centre of a regular tetrahedron and the chlorine atoms are placed at the corner of the tetrahedron. All the chlorine atoms are at equal distance from carbon. Even though you might say there is electronegativity difference between carbon and chlorine, the net pull on the electrons of carbon is zero. Hence, the dipole moment is zero despite the electronegativity difference.

The non-polarity of oxygen atom has got to do nothing with the presence of lone pair on oxygen atom. The oxygen molecule is not completely non-polar. There is instantaneous dipole in oxygen molecule.

No molecule is completely non-polar at every instant of time. H-At is also polar, although its dipole moment is very less. H and At are different atoms. One atom must have a little more affinity toward the electron and hence there will be separation of charges. The electronegavity difference between H and At is very less, therefore the separation of charges is very less which makes its dipole moment near to zero, but still finite.


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Classification of Oxides
There is a way to classify oxides according to their behavior where we have the following categories: Acidic, Basic, Amphoteric and Neutral. But is there a way of classifying the oxides according to their chemical bond?

Related: http://chemistry.stackexchange.com/questions/22189/how-to-decide-acidic-or-basic-nature-of-oxides?rq=1

The classification we use to distinguish bonds in other chemical compounds will obviously be valid for the classification of oxides.
They can be classified as covalent or ionic.


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E, Z designation - what to do when substituents are identical on one side of the double bond?

Here, A has a greater atomic number than B, which has a greater atomic number than C. To assign E,Z designations you would divide the double bond into left and right portions and assign priorities. On the right side, A > C. However, on the left side, B = B. A (first priority on right side) is simultaneously on the same side and on opposite sides to B (first priority on left side). How would you class this compound then?

How would you classify an ethylene molecule without any substituents? Is it E or Z?

@IvanNeretin I take it that you mean E,Z nomenclature can permissibly fail to class a molecule?

How would you classify a methane molecule without any substituents and without any double bonds? Is it E or Z?

@IvanNeretin I thought that E,Z nomenclature was used for compounds with double/triple bonds.

That's right (that is, if you throw away the triple bonds), but there's a catch. First and foremost, any nomenclature (not just E,Z) exists to tell apart different things. Your molecule is identical, no matter which way you orient those A and C. There is nothing to tell apart.

When both the substituents are identical on one side of the double bond, the compound does not show geometrical isomerism.
There is no point in using E-Z nomenclature in that case.


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Can redox reactions be reversed?
Suppose I did this reaction by dipping a $\ce{Zn}$ rod in $\ce{CuSO_4}$ solution.
$\ce {CuSO4 + Zn -> ZnSO4 + Cu}$
Is there a way to reverse this reaction to
$\ce{Cu + ZnSO4 -> Zn + CuSO_4}$?
If so, how?

Short answer: The reverse reaction can occur using an electrolytic cell.
Long answer: There are two types of cell-electrochemical cell and electrolytic cell.

Electrochemical cell uses a spontaneous reaction to convert chemical energy to electrical energy.

$\ce{CuSO4 + Zn -> ZnSO4 + Cu}$
This reaction is spontaneous and the cell setup for this reaction is an electrochemical cell.

Electrolytic cell is just the opposite of electrochemical cell. Non spontaneous reactions can be made to happen using an electrolytic cell.

$\ce{Cu + ZnSO4 -> Zn + CuSO4}$
This reaction is non spontaneous. Energy from an external source will be used to make this reaction happen. If you want to know how these cells work, refer to this site.

Yes,It is possible to reverse such reactions, given that you supply the required electrode potential in a cell with proper electrolytes, you can get the reverse reaction to occur

Can you please explain how to do that for this example?


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Why in aromatic nucleophilic substitution the Nucleophile attacks carbon and not the halogen?
In nucleophilic aromatic substitution of halobenzene why does Nucleophile attack the carbon atom bonded to halide X not the halide.
The carbon atom atom has partial positive charge whereas the halide has positive charge due to resonance as its lone pair is delocalised in the benzene ring doesn't resonance always dominate so why does Nucleophile behave so?
I can think of because of lone pair repulsions between electron rich Nucleophile and halogen but I'm not sure. May someone help ?
(Ofcourse Here the benzene ring has electron withdrawing groups at o p position)

Your argument applies to every other nucleophilic substitution reaction, yet they all don't proceed that way. Therefore, what other factors are more important?

I dont know. What are you saying ?

First I'm saying that you're looking at the wrong factors because they also don't work to explain anything else. Second, I'm saying that should consider transition state stabilization because that really governs whether or not these reactions proceed at all.

Yeah I know that. Im asking why these factors are wrong and dont explain anything else ?

I'm not sure how to answer this question. These are statements, but I don't see what they have to do with understanding the mechanism of an ipso attack...


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DIY spectrometers limits as a water quality sensor
What pathogens or harmful stuff IN WATER can you detect with these homemade video spectrometers (webcam) + UV light?
https://www.youtube.com/watch?v=IA5BTD-aelo
I'm trying to do a cheap spectral analysis to analyze water quality, something electronic, so I can do continues monitoring remotely.

Unfortunately the answer is probably no. Water absorbs very little in the visible spectrum which is good for your application, but absorbs a huge amount in the infra-red and UV, which is not good as it absorbs all the light so none reaches detector.  Detection in visible will require visibly coloured pathogens and to get sensitivity, a very long cell of water to pass the light through; by 'long'  this could be metres. Detection in near UV (250-350 nm) requires UV transmitting windows and detector sensitive here.

@porphyrin so UV light is the problem, what if I use a different kind of light? Will it solve the problem? If so what kind of light should I choose?

no; its not the light that is the problem, in general you have to have absorbance from substances that you want top study in a wavelength range that you can detect. It is common to detect in the uv with commercial or special lab built instruments but hard at home unless you buy special uv optics and uv detectors.  And especially hard for trace amounts of substances if you don't know before hand at what wavelengths to look.  Also the change in transmitted light intensity you are looking for may be only 1 part in 1000 or less.

In general, with this setup, nothing. You would need a UV detector instead of a webcam. And, assuming you had a UV light source and the right detector, it would depend on the UV source spectrum. You could use a blacklight for mid 300 nm or a Steripen for 254. A broad spectrum lamp would be best. Water doesn't start to absorb UV strongly until you get below approx. 200 nm, so compounds that absorb above that will show up. This would be on a commercial unit like a Hach DR6000. On a DIY kit, I doubt you'll be able to detect much of anything unless it is either at high concentrations (very high) or your path length is long (ie Beers law). With all that being said, UV spec isn't a very good "fingerprint" analysis like ftir or mass spec. For example DNA and many organics all absorb at 254 nm, so you won't be able to differentiate them. Probably the easiest water quality parameter you could measure is called UV254, but that's really only useful for determining organic content and disinfection byproduct formation potential.
Okay, will all that being said, you might have a better option using visible light. Hach sells chemical kits for measuring certain constituents - they are coloromertric so they convert the constituent to a colored compounds that is detectable at a certain visible light wavelength. For example, you could test for nitrate or chloride. A used portable Hach meter will run you around 500, these are basically a fancy visible light spec. They work decent, but not great and even that would be worlds better than any of these DIY kits if that tells you anything.

what's the cheapest way to do it under $100 ?
What about this?   http://renaud.schleck.free.fr/spectrometre.php
It won't work for water analysis?

could you tell me more about the  UV254  ?
is this the one?  -> http://www.ebay.com/itm/UVP-Mineralight-UVG-11-UV-lamp-transilluminator-254-nm-/282391850250

Example of nitrate powder pillow, absorption wavelength is in the documents: https://www.hach.com/nitraver-5-nitrate-reagent-powder-pillows-10-ml-pk-100/product?id=7640208933&callback=qs

Yes, that lamp would give you the wavelength you want. It will be monochromatic though, so only 254 nm (well, actually with a bit of 185 nm too).

Again though, I think you need to be a little realistic here - what type of water are you trying to measure? Tap water? Surface water? Ground water? If tap water, you likely won't have concentrations high enough and if surface water there will likely be too much "stuff" to differentiate anything. So...you might be overestimating the ability of a homemade spectrometer.

You can find quite a few used UV/Vis specs on Ebay for cheap. They would need new lamps I assume. Example: http://www.ebay.com/itm/PERKIN-ELMER-UV-VIS-SPECTROMETER-LAMBDA-BIO-NICE-/250708682289?hash=item3a5f66e631:m:m5hL3J4kSv4LeI9YWR3w9fg

yesterday   mmm, so this project
http://blog.ted.com/waterwise-fellows-friday-with-sonaar-luthra/
might be ... "smoke" I'm trying to build something like that,
It seems they use a nano-spectrometer to detect some kind of bacteria, but there is no much information about it


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Very fast brass oxidation
I work for a company that manufactures pyrotechnics for the oil and gas industry, and we have been having an odd phenomenon occur.
One part, a brass tube with a copper clip filled with black powder, is placed into a steel receptacle. Another part, consisting of a blend of black powder, asphalt, epoxy, and a little ammonium nitrate (and possibly a few other materials) in a steel tube is placed into the same atmospheric chamber, but does not come in physical contact with the first part. The chamber is sealed up, and the brass tube is covered in blue corrosion within minutes. I don't think that the second part is offgasing, because it's in a sealed package prior to use which never builds up any pressure. What could be causing it to corrode so fast? I think the customers are using a caustic cleaner in the tool which isn't being fully removed that's reacting with the brass, but they seem to think it's an interaction between the two tools. Any thoughts?


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Is it possible for three carbon double bonds to form next to each other in a carbon chain?
In a typical carbon double bond $sp^2$ hybidrization is used to explain the resulting bond. When two carbon atoms A and B form a double bond one $sp^2$ orbital from A bonds to another $sp^2$ orbital from B and one $p$ orbital from A bonds to a $p$ orbital from B. This leaves two $sp$ orbitals sticking out the end of the ethene-like molecule. I think one could stick two ethene-like molecules end to end to stick together with two $sp^2$ orbitals from each end but then I think one would have an interfering $p$ orbital that would subtract and leave a bond order of 1. I think one might then have further hybridization and end up with three 1.67 bonds (instead of one single and two double bonds) but not three double bonds. The second possibility is sticking together two $sp^2$ orbitals and one $p$ orbital from A to two $sp^2$ orbitals and one $p$ orbital from B leaving behind one $sp^2$ bond sticking out on each end. I think one could stick together such pair with another pair rotated at an angle. After hybridization I think one ends up with three net 2 bonds together but this is confusing to me. A third possibility is sticking together two ethyne-like molecules together but I think that would be a hybrid of two 3-bonds and a -1 bond which would be 1.67 again.

https://en.wikipedia.org/wiki/Cumulene

There's two different ways to stick double bonds next to each other.
The first is to put the two double bonds right next to each other, the "cumulenes":

Image from Wikipedia Commons
In these compounds, the pi orbital of one double bond does NOT hybridize/delocalize with the pi orbital of the adjacent double bond. For example, in the structure on the far left, the three carbon propadiene, the two end carbons are hybridized $\ce{sp^2}$, and the center carbon is hybridized $\ce{sp}$. The remaining two $\ce{p}$ orbitals of the center carbon participate in the double bonds, one to each. But because the two $\ce{p}$ orbitals are perpendicular to each other, there's no communication between the two pi bonds - they're perpendicular to each other, too. One of them is in the plane of the screen, and the other is in the plane going in/out of the screen. (Or, it is in the simplest model of organic bonding that's taught. Once you get into things like molecular orbital theory things get more complicated, but the fact of two non-interacting orbitals remains.)
You can see this by looking at where the hydrogens are located in propadiene.

There's another way double bonds can be combined, and that with an interveneing single bond, the "polyenes":

Image from Wikipedia Commons
In a polyene, all the carbons are $\ce{sp^2}$ hybridized. Unlike a cumulene, the pi bonds of a polyene are all on the same "side" of the polyene. For example, in the polyene shown here, all the pi bonds are sticking in/out of the screen.
This means that you can draw resonance structures where the double bonds "travel" along the carbon chain, effectively swapping the double and single bonds. Like benzene, the distinction between double and single bond starts to break down, and it's more correct to think of a continous pi bond throughout the whole of the polyene chain.
But unlike benzene, the end effects for the bond-swap resonance form means it's less stable than the "normal" bond locations. This means it contributes less to the hybridized form. So while the polyene chain doesn't behave entirely like an alternating single/double bond form, it's not like it's really a "1.67" or "1.5" bond order, either.


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Can hydrocarbons undergo combustion without oxygen present?
Most of us are familiar with the standard reaction of methane gas combustion:
$$\ce{CH4 + 2O2 -> CO2 + 2 H2O}$$
In standard combustion, oxygen is generally used as the oxidant. Would it be possible for hydrocarbons to undergo combustion using alternative oxidants? If so, under what conditions could this be reproduced?
For example, could we achieve anoxic combustion by replacing oxygen in the atmosphere with another highly electronegative element such as chlorine or fluorine? Or perhaps compounds in high oxidation states such as permanganate or dichromate under different pressure/temperature conditions such that they were gaseous?

Note that permanganates and dichromates both contain oxygen.

This Q&A discusses, among other things, how a combustion engine could work in a Cl2/Ar atmosphere. Nothing about this is good for humans, but just from a perspective of chemical curiosity, sure you could do it. Since Earth doesn't have an appreciable amount of Cl2 in the air, you would have to carry around both oxidant (Cl2) and reductant (hydrocarbon). Then, in the simplest case of methane, I suppose your exhaust would be carbon tet and HCl. No thanks ;)

They would certainly combust if meeting a worse adversary than oxygen. $\ce{ClF3}$ would oxidize them, with extreme certainty, rapidity and violence.
Here is a short excerpt from the Wikipedia article of $\ce{ClF3}$ which amused me:

It is, of course, extremely toxic, but that's the least of the
  problem. It is hypergolic with every known fuel, and so rapidly
  hypergolic that no ignition delay has ever been measured. It is also
  hypergolic with such things as cloth, wood, and test engineers, not to
  mention asbestos, sand, and water — with which it reacts explosively.
  It can be kept in some of the ordinary structural metals — steel,
  copper, aluminum, etc. — because of the formation of a thin film of
  insoluble metal fluoride which protects the bulk of the metal, just as
  the invisible coat of oxide on aluminum keeps it from burning up in
  the atmosphere. If, however, this coat is melted or scrubbed off, and
  has no chance to reform, the operator is confronted with the problem
  of coping with a metal-fluorine fire. For dealing with this situation,
  I have always recommended a good pair of running shoes.

check out azidoazideazide,mabye that can perform better

@QuarkyLittleThing hah, that sounds like uncomfortably many nitrogens in one place. I don't want to be present when the experiment is being done, but I want the footage =D


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Calculating pKa's using Gaussian
I've been learning to use Gaussian recently and I found a question here that seemed like a good opportunity for me to practice. I know that I can obtain $K_a$ from $$\Delta G=-RT\ln(K_a)$$ using the general reaction $$\ce{HA(aq) + H_2O(aq)->A-(aq) +H_3O+(aq)}$$ and taking the sum of thermal and electronic free energy for each compound as their value of $G$ However, I'm not confident that the method I'm using in Gaussian is correct.
I performed an optimization/frequency calculations on all these molecules using the APFD functional, with basis set 6-311G+(2d,p) and SMD solvation using water.
After working out my calculation, I obtained $pK_a=26.10$. My main question comes down to: did I use a sufficient level of theory, at least approximately, to capture the effect of solvation? I've looked at the Gaussian white book for thermochemistry as well as a few journal articles (1 2) and couldn't get a clear sense of if I had the right level of theory or if my calculations were completely off-base.
Another finer point might be: are there any fairly similar experimental values to which I could compare my calculation to see if I'm in the right ballpark?

Such a small model for water protonation is generally way off, and you will not improve it by eg better basis set.

@Greg So would it require better theory or a better solvation model(like explicit solvation) to make this work? If it winds up being too burdensome a calculation I might just leave it, but I would be interested to see any resources that point to a sufficient model size for this type of problem.

In my experience, the solvation is the critical part that often fails when heterolytic dissociation is studied.

Are you just calculating the free energy change of the dissociation using implicit solvation? One thing that you could do is to use a more suitable thermodynamic cycle. You can optimise the structures at a lower level of theory and then calculate electronic energies in the gas phase and in implicit solvent at a higher level to get better solvation free energies while using the experimental value of the solvation energy of the proton (-265.9 kcal/mol).

The approach itself is flawed. There is no H3O+ in water.

Here's a useful reference ... http://nopr.niscair.res.in/bitstream/123456789/3900/1/IJCA%2048A(4)%20520-525.pdf

AMT I found the general equation I used in my 1st link. @Snakeysnakes Maybe I'm misunderstanding, but isn't the solvation free energy just the difference between the gas and aqueous free energy for a particular compound? Would I not get the same result as using just using the free energy obtained from the implicitly solvated molecules as I would from adding the solvated free energy to the gaseous molecules? I can see how using the experimental solvation energy for the proton could fix some of the error of modeling it implicitly.

The point of doing that is that you can do single point calculations at a higher level of theory than you used to optimise your structures. That way you can get a better approximation to the solvation energy. Look for the review by Alonghi, theoretical calculations of acid-dissociation constants. By the way, if you are looking to calculate the pKa of a series of structurally similar compounds, do you have an experimental pKa value for one of them? In that case there might be an even better way to get some reasonably reliable values.

@Snakeysnakes I was actually trying to see if I could give someone a reasonable estimate of the pKa since they were unable to obtain it experimentally, though how it is looking as though a good calculation would require at least some calibration with respect to similar experimental or well-validated theoretical calculations of pKa.

This is unfortunately not a real answer to your question as such, but I'd like to provide some pointers why this approach is a lot more complicated than it seems at first. It might very well be (currently) impossible.
That is probably also a reason why you don't find a lot of publications on the theory of it and so far no black-box method has been developed. Solvation is still one of the major problems in computational chemistry, hence the use of implicit solvents and counter-charges. Such methods naturally come with the implicit disclaimer that these are approximate corrections only.
The most elaborate method I know so far is COSMO-RS, which still has large error margins, at least that is the my current state of knowledge.
I think for calculations you would at least need to run molecular dynamics simulations with many explicit solvent molecules to describe the equilibrium. This is important, because the proton which dissociates will be caught in the bulk of water molecules and a highly complex network of hydrogen bonds. This is very well demonstrated by the Grotthuss mechanism, which shows how easily bond lengths can change. I would say it is impossible to describe this complex system with a single quantum chemical calculation.
Furthermore, DFT is notoriously bad at describing non-covalent interactions, and including dispersion corrections will only get you so far. I have not checked recently, but I believe there are quite a few publications discussing various sizes of water clusters on this level. I have neither used nor heard of AFPD before, so it would be wise to thoroughly search for literature (benchmarks) about its performance on these clusters. Suffice to say, that one particular functional will not give you a reliable result as you have nothing to compare it to. Especially in such complicated cases you have to calibrate your results.
Another problem here is that you are the describing heterolytic bond dissociation. More often than not this may lead to additional problems. I don't always have confidence in my results when I treat simple donor-acceptor complexes where there is no charge separation and all singlet species. A supermolecular approach would probably a better choice, but could heavily suffer from a BSSE. However, it might be worth considering $$\ce{H2O\bond{...}HA <=> H2OH\bond{...}A}.$$
I think the key for calculating acid dissociation constants is treating the whole system as one ensemble.
Recall the common definition of the equilibrium constant for the reaction
\begin{align}\ce{AH (aq) &<=> A- (aq) + H+ (aq)},&
K &= \frac{a(\ce{A-})\,a(\ce{H+})}{a(\ce{HA})}.\end{align}
We conveniently ignore water here, because it is abundant and we assume ideal solution. However, when your system is too small, you will get a large error for $K$. This is most likely due to your solvated proton, which will be described worst with solvation models. Explicitly you are writing
$$\ce{HA (aq) + H2O (aq) -> A− (aq) + H3O+ (aq)},$$
which is not really what you are calculating. Instead you are (more ore less) using already approximations to treat thermal and solvent corrections
$$\ce{HA (g, th, solv) + H2O (g, th, solv) -> A− (g, th, solv) + H3O+ (g, th, solv)}.$$
And this doesn't even consider that you are calculating two states in an equilibrium and you will get the equilibrium constant (using activities, not concentrations) for only these two states (which might not even be the ground sates),
$$K = \exp\left\{-\frac{\Delta G}{RT}\right\}.$$
You have also not accounted for the possible conformational space of $\ce{HA}$/$\ce{A-}$, which might become very important as soon as you have ethyl moieties, or even cyclohexyl moieties.
These are just some issues that immediately come to mind, there might very well be more underlying insufficiencies. I have looked into this a couple of years ago and came to the conclusion not to bother with it any more.
What can you do instead?
Assessing acid strengths does not need to be at the level of what experimental chemists understand. If they really need the $\ce{pKa}$ they should run the appropriate measurements and obtain reliable results instead of educated guessing. If they want you to classify whether an acid is stronger or weaker, use a rigorous approach to obtain the conformational space for all species and compare them with isodesmic equations. This will have the advantage that insufficiencies of the methodology will likely cancel each other.
Another purely theoretical approach would be calculating gas-phase proton affinities. From this you can judge if a compound behaves stronger or weaker. Keep in mind that this approach is better suited for bases or nucleophiles. If you have the resources, you can also evaluate Lewis acidities,
$$\ce{Base + BX3 <=> Base-BX3}.$$
This should give you at least a good idea how on a relative level your compounds behave.
You can also use a first-principle derived value for your proton, which will likely cause the largest error. A quick check turned up a few promising results. (For example: Chang-Guo Zhan and David A. Dixon. J. Phys. Chem. A 2001, 105 (51), 11534–11540.) You can then go ahead and calculate know compounds and try to find a correlation between $\ce{pKa}$ and $\Delta G$. This way of "calibration" might lead to some significant error compensation.
I would also recommend to look at COSMO-RS in more detail. There might be a lot of development that I have missed and they might have found a way to obtain more reliable results. As far as I remember, the foundation of this was solid and a good step into easier capturing solvent effects without the need to resort to full blown MD simulations.

Thank you! This is definitely helpful and gives me a nice range of ways that I could go forward with the calculation or look at the problem from a different angle. I'm might try using  something other than DFT (I defaulted to using it since it was what is used in general in book I'm using to learn Gaussview).  I also think looking at the supermolecular equilibrium could be promising.

@Tyberius In my opinion GaussView is not a programme worth learning as it is only a visualisation tool of the results, not even a good one. Learn how Gaussian operates, how an input needs to be structured, what different things get reported in the output, how you can analyse those results further. When you do this, it's easier to switch to other calculation packages and analysis tools. I have to reiterate, without a proper size assessment and MD treatment, you won't get reasonable results.

I misphrased that, it is mainly for Gaussian and includes the specific structure of the input/output files, though it gives instructions for how to view results or set up jobs within Gaussview. But yeah I can see that finding a pKa for an individual compound might be outside of the scope of strictly Gaussian calculations without supplementing with MD, experiment, or some calibration of with a series of similar compounds.

@Tyberius Recently the topic came up on CCL (http://www.ccl.net/cgi-bin/ccl/message-new?2020+04+22+001), and Marcel Swart shared this link: https://comp.chem.umn.edu/solvation/, which is a good resource I thought I share with you.


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Mechanism for cyclic enamine formation after N-Boc deprotection
In this reaction taken from Snyder's total syntheses of Coccinellid alkaloids,1 how does the Boc deprotection lead to ring closure? I understand that the $\ce{PBr3}$ step simply converts the alcohol to a bromide.

Sherwood, T. C.; Trotta, A. H.; Snyder, S. A. A strategy for complex dimer formation when biomimicry fails: total synthesis of ten coccinellid alkaloids. J. Am. Chem. Soc. 2014, 136 (27), 9743–9753. DOI: 10.1021/ja5045852.

Those conditions just give the amine. One you have an amine and an aldehyde or ketone, you should form an imine or iminium ion. Imines are relatively stable, but secondary amines, such as the one in this question, can only form the iminium ion. But these iminium ions frequently generate enamines instead by tautomerization.

You should consult your study materials on imine and iminium ion formation. The mechanisms there are very good to know in general because they reinforce the general reactivity pattern for carbonyl addition and elimination from the tetrahedral intermediate.