text:
HDEHP and complexes formed in extraction
I did an solvent extraction of $\ce{Eu(III)}$ using $\ce{HDEHP}$ from nitric and hydrochloric acid solutions to determine which acid, and which concentration, will yield the highest distribution coefficient.
My results were interesting, I got that the hydrochloric acid extractions had much lower distribution coefficients, and less dependence on acid concentration (the max distribution coefficient was at 0.3 M).  The nitric acid extractions were far more dependent on acid concentration with the global max at 0.01 M and another local max at 0.6 M, as well as being higher in general than the $\ce{HCl}$ distribution coefficients.
I did a literature review, and based on this paper, I think I know the structure of the $\ce{HDEPE/Eu}$ complex.  My original complexes were $\ce{EuCl3}$ and $\ce{Eu(NO3)3}$ dissolved in the acid solution, and extracted with 0.8 M HDEHP in dodecane.  I know the melting point of solid $\ce{EuCl3}$ is about $\sim\pu{600°C}$ which is far higher than $\ce{Eu(NO3)3}$ ($\sim\pu{85°C}$) so I'm wondering if the stronger bonds in the $\ce{EuCl3}$ complex are preventing the $\ce{HDEHP}$ complex from forming, or if the conjugate bases are somehow participating in the complex formation between $\ce{HDEHP}$ and $\ce{Eu(III)}$.
Does anyone have any insight into my results?

We need some more details such as what diluent you used. Maybe you were using pure DEHPA as the organic phase which is something which I would suggest that you do not do.
I would like to know if you did triplicate experiments, sorry if I offend you but I have no idea how good you are at the measurement of D values by solvent extraction. I like to think I am good at it, but I always use triplicate experiments. I will shake three different vials for each set of conditions. This allows me to see how reproduceable my work is, it also guards against me doing something stupid with one of the vials.
If you only do single extractions then it is possible that a random error causes the experiment to give misleading results. I hold the view that a beginner needs triplicate shaking experiments more than anyone. When you are new to solvent extraction you are more likely to make errors, also when you have done it for a while you might get careless and then you would need the triplicate experiments again to guard against mistakes.
The triplicate experiments will not prevent you making mistakes, but they will help to alert you to the fact that something has gone wrong.
I also worry about the use of very low acid concentrations, the Eu will tend to form hydroxides.
I think that you need to pay attention to the activity coefficents. I would be much more happy if you had used sodium nitrate to keep the ionic strength of the aqueous phase constant. I would want to do the experiment with a constant nitrate concentration to allow you to separate the effects of changing the nitrate concentration from that of the solvated protons.
I would like to point out that at very high nitrate concentrations the extraction mechanism for lanthanides with DEHPA can change from the normal ion exchange mechanism to a solvation mechanism which is similar to the extraction of plutonium from nitric acid by tributyl phospahte (PUREX process),


text:
On stretching a rubberband does the entropy increase or decrease?
It seems that on stretching a rubberband, entropy or the degree of disorder is likely to increase. Is this the correct answer?
My text says that on stretching, the arrangement of particles become more 'ordered' and hence entropy 'decreases'. How?

Think of it this way, when an elastic polymer is in its relaxed state, these molecules are all tangled up with each other and have no particular direction to them, but when you apply a force, i.e stretch the rubber band, you end up disentangling some of the polymer molecules.
We are not stretching the actual polymer molecules, just sort of changing the way they are organised. This disentanglement corresponds to a reduction in micro-states that the system occupies and thus, a decrease in entropy.
In simpler words, there's more ways for a polymer chains to jumbled up with each other (more microstates) than for them to be aligned.
More simply-- If the chain were totally stretched out, it would only have one possible conformation. By coiling itself, it increases the number of
conformations it can have, and therefore, its entropy.
A more rigorous treatment can be found if you read up about the ideal chain, which serves as a model for such systems.
Thus, assuming that no other energy exchanges takes place, the tendency of the elastic band to shrink back to its original conformation is due to something called an entropic force, i.e a tendency of the system to move to a configuration that maximises entropy.
Some Details
Note: I am not quite confident with this material and might've made some mistakes--if that be the case, I am sure someone here will correct me
Consider the following model system: The chain is made of $n$ rigid segments of size $b$; Thus, the length of the chain is $L = nb$. Also, assume that the chain has one fixed end, and one free end and the distance between the two is denoted as $r$.
Then, one can derive the probability of finding the chain ends a distance $r$  apart is given by the following Gaussian distribution:
$$ P(r,n)\mathrm dr = 4\pi r^2 \left(\frac{2nb^2\pi}{3}\right)^{\frac{-3}{2}}\exp\left(\frac{-3r^2}{2nb^2}\right)\mathrm dr$$
The classical entropy is approximately $ S = k_\mathrm b\ln(P(r,n))$
From hereon, one can go on to show that for an ideal chain, maximising its entropy means reducing the distance between its two free ends (i.e reducing $r$)

Nice job of articulating the concept of configurational entropy as it applies to polymer molecules.

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text:
How to quantize the shape of a molecule?
I am looking for numbers which tell you something about the 3D shape of the molecule, and is also easily calculated.
One of these is the acentricity factor. I am aware of this one. My problem is you calculate acentricit, from thermochemical data, which are usually hard to find, and can contain a great degree of experimental errors.
A more suitable quantity would be constructed from the groups of the molecule. Is there such a thing? Or what are other quantifiers of molecular shape other than acentricit, factor?

This sounds like it could be a bit of an XY problem. What are you actually trying to do?

I am planning to relate some physical-chemical properties of large sets of molecules to their shapes. For example, trying to find a qualitative picture of what is the effect of the molecular shape on the critical temperature. I am also planning to characterize the efficiency of currently used descriptor calculation methods.

So you already know a bunch of descriptors, and we should figure out ones you may not know... Wouldn't it be helpful if you mention which ones you know? Also, it is nice to have an argument behind using a descriptor, not just some random parameter you tried and seemed to wok ona limited number of molecules.

There are lots of descriptors of molecular size and shape that come to mind:

Moments of inertia (e.g., how different are they)
Radii (e.g., enclosing the molecule in a best-fit box)
Surface area or polar surface area (there are existing group-based methods)
Volume (e.g., of the van der Waals surface)

Some people have also used 3D shape descriptors like "spectrophores".

Both the absolute magnitudes and relative values of the moments of inertia contain useful information, I think.  Absolute values give overall size, whereas the relative values (ratios close to $1$, or very different) give symmetry/asymmetry. The moments can't capture convexity or concavity, though.

I haven't seen any good measures of convexity or concavity, but those would likely be nice too.

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Formation of thermodynamic product from 1,4-addition
Question:

The temperature is high enough for formation of the thermodynamic product. Now, there are two possibilities:
Starting with the top double bond
2
Or, starting from the bottom double bond,

The answer given specifies that the second product is formed. Why is this, why isn't the first product formed?

Both of your products have a tetrasubstituted double bond.  This makes me think that they have similar stabilities, therefore there is no thermodynamically preferred product.
In both the top and bottom lines of your drawing you show the formation of allyl carbocations.  You use an arrow like this "$\ce{->}$" to connect the second structure from the left to the third structure from the left. The allyl carbocation is a resonance stabilized carbocation.  The second and third structures are really resonance structures and should be connected with a resonance arrow "$\ce{<->}$".  Both resonance structures taken together describe the allyl carbocation.
If you compare the allyl carbocation on the top line to the one on the bottom line, you'll notice that the one on the bottom has the positive charge delocalized onto 2 tertiary carbons while the allyl carbocation on the top line has the positive charge delocalized onto 1 tertiary and 1 secondary carbon. Therefore, the bottom allyl carbocation is more stable and the product derived from it will be preferred.


text:
Placement of Electrodes for Electrochemistry?
I'm kind of curious as to the relationship between resistance and the yield of electrochemical methods. If I have my electrodes rather close together in a conductive solution the resistance will be smaller than if I had them further apart and more current can flow. Would there be any penalty for having the electrodes close together like this? Will I miss some ions in solution, or are they distributed equally in solution and so will always move towards the electrodes as some will be taken out of the solution as gases (such as electrolysis of (H2O -> 2H2 + O2))?

It really depends on the process and how
homogeneous electric filed is needed. The more are electrodes close together, the inhomogeneity will be greater.
The second problem will be with the homogenity of the solution. You can assume, that electrolyte is kinda homogeneous in a bulk. But more close you get to the electrode, the more is the electrolyte separated.
Ions, as a charged particles, are transferred to near electrode surface.


text:
What does ozone smell when treating with household water?
We live in a rural area and our well water comes from storage tanks that are continously being ozonated before passing through an infra red lamp.
The water in the house smells like it contains chlorine and so does the water in the storage tank upon climbing to the top cover and smelling.
There is no source of chlorine and want to know if the ozone can be the source of the bleach odor.

Do you also have water softener?

Although the primary products of chloride oxidation resulting from the ozonation of saline waters are chlorate salts ("Photochemical oxidation of chloride ion by ozone in acid aqueous solution", Environ Sci Pollut Res (2015) 22:16554 – 16569), a portion of the chloride is also known to oxidized to molecular chlorine, $\ce{Cl2}$.
Human olfactory senses can detect $\ce{Cl2}$ at levels as low as 0.2 ppm in air. This World Health Organization publication reports that some non-seawater contaminated surface waters in the UK contain greater than 40 mg/L chloride, and some seawater-contaminated aquifers used for drinking water in both the US and Philippines contain chloride concentrations on the order of a few hundred mg/L.
The degree of oxidation of $\ce{Cl-}$ to $\ce{Cl2}$ by ozone depends on several factors including temperature, lighting conditions, pH of the well water in question, and of course chloride concentration. However, based on the information given above, low-percent level conversion rates of $\ce{Cl-}$ to $\ce{Cl2}$ by ozone oxidation should be feasible, which under the right conditions, would result in the production of a noticeable chlorine odor. It indeed seems that $\ce{Cl-}$ to $\ce{Cl2}$ conversion represent the most likely scenario for the chlorine odor detected from the treated well water.
NOTE: OP mentioned an "infra red lamp" as part of the disinfectant process for the drinking water.  I suspect this really meant, or was in addition to, a UV lamp, which would further promote the oxidation of chloride per my first reference.


text:
Why do hydroxyl groups only produce a singlet in proton NMR?
Take, for example, methanol ($\ce{CH3OH}$). The hydroxyl proton ($\ce{-O$\color{red}{\ce{H}}$}$) is three bonds away from three more protons ($\ce{-C$\color{blue}{\ce{H}}$_3}$), which aren't equivalent to it, so why would this not produce a quartet using the $n+1$ rule?

You are both right and wrong.
At low temperatures, the NMR spectrum will indeed be as you predicted: the $\ce{H}$ from $\ce{OH}$ will produce a quartet and likewise the $\ce{H}$ from $\ce{CH_3}$ will produce a doublet. Note that at low temperatures the formation of $\ce{H}$ bonds is favored, hence leading to more stable components and significantly decreasing proton exchange.

At higher temperatures however, the following spectrum will be observed:

Due to the increase in temperature, the $\ce{H}$ bonds are weakened, and hence shields its signal. Most importantly, the intermolecular proton exchange, non negligible at those temperatures, will render the interactions between the two groups of $\ce{H}$ undetectable via NMR. Hence why you get two singlets.

Wouldn't the H-bonding cause shielding, not deshielding, based on chemical shift here? I would expect higher electron density on the hydoxyl proton during hydrogen bonding, too, so shielding makes sense to me both theoretically and experimentally here.

@DGS this and that source among others indicate that H bonding display resonance signals down-field of their usual signal. "This
downfield shift is a result of a number of not yet fully understood and partially
competing factors, including a decrease in the electron density around the hydrogen
nucleus and deshielding effects from the electronic currents of the acceptor atom."

But that contradicts the data you presesented. Unless there is actually increased hydrogen bonding at low temperature, which I would believe. But then that would negate the idea that H-bonding reduces J-coupling.

Could it be that at lower temperature, there is more H-bonding but less exchange, so you see an downfield shift but still see J-coupling?

@DGS You're totally right, I mixed up the two cases in my answer. Amended. Please do tell me if there's still something that seems wrong.

This is provided as a comment as much as anything - but a lengthy one, in order to address some concerns I have with the accepted answer - not that it is entirely incorrect. There is much too much emphasis on H-bonding, and not enough about exchange in order to justify whether one observes coupling or not.
The loss of coupling in this instance is a classic example of exchange decoupling. The reason that the coupling disappears is due to exchange processes. The important part in factoring whether coupling will be observed relates to whether there are any mechanisms that will promote or inhibit this exchange. H-bonding may be a mechanism to reduce exchange at low temperature, however it is also a well-established mechanism that promotes exchange especially at higher concentrations. The original paper by Kivelson and Kivelson in 1958 clearly demonstrated that the observation of coupling in methanol was dependent on reduced concentrations which lower the probability of H-bonding. The opposite could be argued for ethanol, and Shoolery (also in 1958) showed that higher concentrations of ethanol actually give rise to observation of coupling between alcohol and CH2. Neat ethanol clearly shows this coupling at room temperature. This is also explained by H-bonding mechanisms giving rise to stable dimers and higher order oligomers, which in fact undergo very fast signal averaging (exchange). So, H-bonding is not the answer for whether coupling is observed - it is purely a factor to consider in determining whether active exchange pathways exist for chemical exchange.
In a nutshell, a dilute sample of methanol in acetone will show splitting quite clearly at RT, but a concentrated sample does not. S.Burt quite rightly points out that using a dry solvent will improve your chances of observing coupling. Choice of solvent is critical, and aprotic and nonpolar solvents, removed of any acidic impurities help to promote observation of coupling. I use dry pyridine-d5 with a number of sugars to demonstrate coupling through to alcohol protons, and then add a drop of D2O to make them disappear. Dry CCl4 is another great solvent, and will easily allow the observation of couplings in a dilute sample of methanol. Methanol is used as a low temperature thermometer and low temperature calibration standard down to about 180K, and coupling is never observed in these samples.

What you propose actually happens if your solvent is very dry.  A colleague of mine works with sugar chemistry and when his students run very dry samples, they not only see every alcohol on the sugar, but the alcohol is split by the vicinal protons (and those protons are split by the OH) even in CDCl3!
Typically, there is enough water in typical samples (or the solvents themselves!) to ensure that the hydroxyl proton is rapidly exchanging and this effectively decouples the spin from its neighbors.
If you're in a position to make a sample yourself - get a stable organic solid with a hydroxyl that you can dry in a drying oven (or lyopholize it if you have access to a lyopholizer) and prepare a sample in dry DMSO-d6.  You almost always see the splitting to/from hydroxyl peaks in DMSO.

Good answer. Using the right solvent also is a great help for observing couplings to alcohol protons. Pyridine-d5 is quite an easy solvent to get dry, and I use this for running sugar samples to see couplings to the alcohol protons.

You don’t even necessarily need very dry solvents. I have had my fair share of couplings to hydroxy protons in 2-hydroxycarbonyl compounds where an intramolecular hydrogen bond makes hydrogen exchange less favourable.

have a look at the solution of this problem which has been published some decades ago:
D. Gerritzen, H. H. Limbach, Ber. Bunsenges. Phys. Chem., 1981, 85, 527-535. http://dx.doi.org/10.1002/bbpc.19810850702 Kinetic and Equilibrium Isotope Effects of Proton Exchange and Autoprotolysis of Pure Methanol Studied by Dynamic NMR Spectroscopy.
H. H. Limbach, J. Magn. Reson., 1979, 36, 287-300. http://dx.doi.org/10.1016/0022-2364(79)90106-9 NMR Lineshape Theory of Superimposed Intermolecular Spin Exchange Reactions and Its Application to the System Acetic Acid/Methanol/Tetrahydrofuran-d8.
H. H. Limbach, W. Seiffert, J. Am. Chem. Soc., 1980, 102, 538-542. http://dx.doi.org/ 1H NMR Spectroscopic Study of Cyclic Proton Exchange between Acetic Acid and Methanol in Tetrahydrofuran-d8.

Welcome to Chemistry SE! If you haven't yet, take the [tour] and visit the [help]. While we definitely want well sourced answers on the site, we generally prefer that the answer doesn't just consist of sources or links to material. Could you offer a brief summary of the key points from those papers that apply to this question? @Hans-HeinrichLimbach

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How is the pka related to the ph of the solution for an amine?
So was reading a magazine for competitive exams and this question is bugging me a lot.

How is pka values of a base associated with the pH of the solution. The solution for it is that the Nitrogens would remain this way and only the H from OH would be dissociated. That's obvious because the medium is basic and the acidic H would be dissociated, but what about the nitrogen atoms? How is the pka of a base represented? Is that the equilibrium constant for it's conjugate base?

The pKa's quoted for the amine groups are for the protonated forms of the amines. The protonated forms would add a +1 charge to the molecule in solution. Since amines are fundamentally basic, the protonated amine is a conjugate acid.

But we use pka for bases too right? And how does it explain which amine will be protonated and which would not?

No the pKb value is for a base to accept a proton. For aqueous solutions: $$\mathrm{pKa + pKb = 14}$$

Ok I get that but how would you explain which amine will protonated and which not at a particular pH?

That is a different question which I side stepped on purpose. The short answer is that I would look at the information provided in the question (image). // The longer and more complicated answer is that histidine has  a side chain imidazole group. The imidazole group is a planar  aromatic heterocycle which is amphoteric. see the wikipeda article https://en.wikipedia.org/wiki/Imidazole

Hey in that article its written pka for imidazole and pka for its conjugate acid given separately. But you told that pka is only for acids and conjugate acids??

https://en.m.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation

The neutral imidazole group (or molecule) can accept a proton or donate one. If both nitrogens are protonated then the cation formed has a +1 charge. If both nitrogens are deprotonated then the anion formed has a -1 charge.

A base accepting a proton won't represent the base dissociation constant, right? Because base dissociation constant involves donation of OH-

Sigh... $\ce{OH^{-}}$ is the base. $\ce{OH^{-}}$ accepts a proton from $\ce{H3O^{+}}$ to become $\ce{H2O}$.

Yeah that's the general equation for a base. The term Kb isn't used in this equation.


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Does the synthesis of beta-keto phosphonates from esters have a name?
Consider the Horner–Wadsworth–Emmons reaction; one of its reactants is a phosphonate, usually stabilised by a β-carbonyl group which is then deprotonated in α position (figure 1). These phosphonate esters can be synthesised in a Michaelis–Arbuzov reaction from a phosphorous acid triester and an alkyl halide (figure 2). Another viable way is to start from diethyl ethylphosphonate, deprotonate that (e.g. with n-BuLi) and let it attack a carboxylic ester much like a Grignard reagent would (figure 3).

Figure 1: Horner–Wadsworth–Emmons reaction. Image taken from Wikipedia, originally created by Wikipedia user ~K.

Figure 2: Michaelis–Arbuzov reaction. Image taken from Wikipedia, originally created by Wikipedia user ~K.

Figure 3: Alternative synthesis of HWE phosphonates. Image self-drawn.
Since about every reaction in organic chemistry has a name of some kind, is the reaction in figure 3 also a named reaction, and if so, what is the reaction’s name?

I might call this reaction in scheme 3 a Claisen reaction/condensation, but the purist might contend that the Claisen condensation is for ester nucleophiles and other carbonyl electrophiles and not for other carbon nucleophiles with ester electrophiles.

@BenNorris We'll just call it a phosphonate ester and be happy!

@Jan Does reaction 3 actually stop after the first addition? (Sorry, I don't have access to general chem lit searches any more.)

@Zhe Yes. I used 3 equivalents of diethyl ethylphosphonate and BuLi and yet good pretty good yields of the ketophosphonate.

My inclination is to call this reaction a "phosphono-Claisen condensation." A  Claisen condensation is the alpha-acylation of an ester with another ester. In this case, the typical carbon-centered ester is replaced with a phosphonate, so phosphono-Claisen condensation specifies that difference.  Avoiding the proper name, "alpha-acylation of a phosphonate" is a suitable description of the reaction.
Montchamp and coworkers published a paper describing the "phospha-Claisen condensation" reaction, which they limit to the condensation of two phosphonates to give a 1,1-diphosphono compound.
The website maintained by the venerable Hans Reich describes this reaction as "Claisen-like":

(MeO)2P(O)CH2Li will also react with carboxylic acid derivatives like
  esters, lactones and amides in a Claisen-like reaction to form keto
  phosphonates.

The Organic Chemistry Portal describes the reaction as "the condensation of esters and phosphonates."
A brief survey of the literature featuring this reaction type (Thanks SciFinder!) doesn't show any convention for naming this reaction beyond calling it the acylation of a phosphonate. In my SciFinder searching, the earliest usage of this reaction is from 1970, and that paper cites a Corey paper from 1966. The latter paper doesn't have a proper experimental on the reaction but contains the note:

Anions derived from 1 (X = H or alkyl) are of value in the synthesis
  of keto phosphonates 1 (X = RCO) by reaction with esters. The
  preparation of these intermediates, which are of value in olefin
  synthesis [see H. Takahashi, K. Fujiwara, and M. Ohta, Bull. Chem.
  Soc. Japan, 35, 1498 (1962)], is usually impractical by the
  Michaelis-Arbusov reaction.

It's a little nebulous who performed this chemistry first, but perhaps someone could have christened it the "Corey-Claisen condensation" or the "Corey variant of the Claisen condensation". Of course there is already a Corey-Claisen rearrangement so this would be really confusing.

Two separate papers I found containing this paritcular type of reaction had the following wording:

Protection of the ketoester as the dioxolan followed by reaction with lithium diethylmethylphosphonate and hydrolysis of the intermediate dioxolanphosphate led to the diketophosphonate.$^{[1]}$

and

"... treatment with lithiated ethylphosphonate furnished
the target phosphonate."$^{[2]}$

The lack of results online and the wording in these papers leads me to believe that there isn't any particular name for this reaction. You might say that this is a Claisen-type reaction, though instead of an ester enolate nucleophile, it is a lithium phosphonate nucleophile.

$[1]$ Begley, M. J.; Cooper, K.; Pattenden, G. Chemischer Informationsdienst 1982, 13 (18).
$[2]$ Mohan, P.; Koushik, K.; Fuertes, M. J. Tetrahedron Letters 2012, 53 (22), 2730–2732.

Yesterday I chanced upon a rather obscure name for it: the "Corey–Kwiatkowski condensation".

The reaction between lithium phosphonamide or phosphonate and ester to give β-keto phosphonamide or phosphonate is called the Corey–Kwiatkowski
  condensation.
Wang, Z. Corey-Kwiatkowski Reaction. Comprehensive Organic Name Reactions and Reagents [Online]; Wiley: 2010, 160, 730–733. DOI: 10.1002/9780470638859.conrr160.

This cites Ciufolini's synthesis of (+)-camptothecin:

Radical bromination and methanolysis advanced 13 to 14, which delivered phosphonate 4 upon Corey–Kwiatkowski reaction.

Ciufolini, M. A.; Roschangar, F. Total Synthesis of (+)-Camptothecin. Angew. Chem. Int. Ed. 1996, 35 (15), 1692–1694. DOI: 10.1002/anie.199616921.

In turn Ciufolini cites a 1966 paper by Corey and Kwiatkowski. Oddly enough, that paper makes no mention of any such reaction, except in footnote 6. This was already mentioned in jerepierre's answer, but it is reproduced here:

Anions derived from 1 (X = H or alkyl) are of value in the synthesis of keto phosphonates 1 (X = RCO) by reaction with esters. The preparation of these intermediates, which are of value in olefin synthesis [see H. Takahashi, K. Fujiwara, and M. Ohta, Bull. Chem. Soc. Japan, 35, 1498 (1962)], is usually impractical by the Michaelis-Arbusov reaction.

Corey, E. J.; Kwiatkowski, G. T. The Synthesis of Olefins from O,O'-Dialkyl β-Lithioalkylphosphonothioate Esters. J. Am. Chem. Soc. 1966, 88 (23), 5654–5656. DOI: 10.1021/ja00975a057.

Anyway, I wouldn't really recommend using this name, as (1) it is not well-known (2) there is potential for confusion.
The name "Corey–Kwiatkowski _____" is also sometimes used to describe the reaction of a lithium phosphonamide with an aldehyde to give an olefin directly (this is what Corey's 1966 paper was actually talking about). In the very first book cited, Comprehensive Organic Name Reactions and Reagents by Wang, the "Corey–Kwiatkowski reaction" refers to precisely that. Also, in Kürti and Czakó's section on HWE olefination,

The Corey–Kwiatkowski modification uses phosphoric acid bisamides to prepare (Z)-alkenes stereoselectively, $\ce{(Me2N)2P(O)CH2R}$, where R = aryl.
Kürti, L.; Czakó, B. Strategic Applications of Named Reactions in Organic Synthesis; Elsevier: Amsterdam, 2005; pp 212–213.

In turn they cite two papers by Corey and Kwiatkowski, different from the one already quoted above:

Corey, E. J.; Kwiatkowski, G. T. The Synthesis of Olefins from Carbonyl Compounds and Phosphonic Acid Bisamides. J. Am. Chem. Soc. 1966, 88 (23), 5652–5653. DOI: 10.1021/ja00975a055.
Corey, E. J.; Kwiatkowski, G. T. Synthesis of olefins from carbonyl compounds and phosphonic acid bis amides. J. Am. Chem. Soc. 1968, 90 (24), 6816–6821. DOI: 10.1021/ja01026a045.

If Kürti-Czakó says that the Corey-Kwiatkowski is that, it is that. We might still call it Corey-Kwiatkowski phophonylation or something … I have been calling it phosphono Claisen long enough ;)


text:
Does this NMR show the carboxy group?

This is the NMR for the molecule shown. I was told that carboxy group will be in the above 10ppm range. Does that tiny signal at about 12.9 ppm indicate the carboxy group though? I dont understand why the methoxy would show large integration while carboxy barely shows.
Also my general question is, how much absorption is necessary for it to indicate some group present?  Is there "noise" on NMR spectra or does every signal indicate a group?

Indeed, the very small signal at 12.9 ppm is due to the carboxyl group. The reason why it's so low is because, as explained here, at room temperature the proton exchange makes it nearly undetectable. At higher temperatures, you would probably not even see that spike.
The methoxy group, on the other hand, is not affected by such effects.
There will always be a tiny amount of "noise" on experimental NMR spectra, because a solution always contains unwanted elements; but usually anything that looks like a well defined spike is a significant signal.

The compound is 4-Chloro-2-methoxybenzoic acid and nmr is given here with assignments.  http://www.hanhonggroup.com/nmr/nmr_en/B35477.html

What you'd observe depends also heavily on solvent and concentration. If you'd use benzene-d6 as a solvent, you should get very clear O-H signal.

Don't mistake peak intensity for peak area.
The (somewhat oversimplified) rule in NMR is that the area under the peak is proportional to the number of equivalent hydrogens, not the peak height.
If you look carefully at your spectrum, the methyl singlet is very very narrow - this will lead to a very tall peak (in addition to the fact that it's a singlet).  Notice that it's considerably more than 3 times the height of the aromatic protons.  The splitting of these protons into multiplets (difficult to see in your image, but you can deduce what they should be) gives them a shorter peak height than they would have if they were also singlets.  The carboxylic acid proton is a singlet, but it's very broad and this leads to a dramatically lower peak height (you can see this in your image if you look carefully - but from experience I know that you often have to integrate over several ppm to get the integral to be even close to 1.
As @Hippalectryon mentions, this broadening is due to the exchangeable nature of this proton - how broad it is (and therefore, how tall it is) will depend strongly on temperature, concentration, the pH of your solution, how much water is in your sample, etc.
While probably not the case here, you can also get differences in peak areas when one proton has a significantly different relaxation time than other protons (due to motion, nearby quadrupolar nuclei, etc.) this will lead to differences in peak area even if they have the same linewidth assuming their T2* values are the same.


text:
Electrolysis of Water: Overvoltage?
I've heard from a few sources that over-voltage in a electrolysis cell for water will cause a greater heat buildup with no yield. Is this true, and if it is, why is it true? From my electronics background, voltage can be though of as electric potential, or thinking of it as water in a pipe for an analogy it is the strength of the substance flowing through the pipe, or as the strength of each electron, so why does it cause extra heating with no appreciable/no yield?

To first order, the power dissipated in an electrochemical cell is equal to that of an electrical circuit: $P=IV$. Thus, as the overpotential necessary to achieve a given current throughput increases, the power consumption increases also.
In the fluid flow analogy, think of an elevated overpotential as a throttling orifice inserted into one of the pipes.  In order to keep the water moving through the pipe at the same volume flow rate, you have to increase the pressure drop across the system.  A greater pressure drop to achieve the same flow rate requires a pump with a greater power rating.

Wow, that was really a duh type moment, I didn't even think to go to the power function.

@SarahSzabo  It wasn't obvious to me, when I first started working in earnest with electrochemistry. Why should it be that an electrochemical boundary layer behaves at all like a circuit component, in terms of power dissipation? But, it apparently does.

You regard water in pipe as electrons in conductor, it's well. No yield means water pressure is not enough to wash the barrier away, increasing the water pressure is needed, the increment of water pressure is overvoltage. The barrier usually comes from polarization of electrode reaction (electrolysis of water).

This looks like a bunch of random Markov chains. At least NAA

@IͶΔ No, the underlying thread here is pretty accurate, actually, though mostly duplicative of what I wrote. The last sentence, actually, is important for someone to have mentioned, that polarization of the interface is part of the physical explanation for the overvoltage. Definitely needs clarification.


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Single Point Energy in Gaussian software
I used Gaussian software to determine the single point energy (SPE) of 3 molecules namely ethane, ethene and phenol.
The SPE for the molecules are -79.83, -78.59 and for phenol it is quite large at -307. However, I am not able to explain the trend.
I am thinking does SPE have anything to do with the molecular weight of the molecule for example, larger mass = more negative SPE?? I cant think of a better explanation. Can someone please help me out?
thank you

The energies you're seeing are those resulting from solving for the eigenfunctions of the Coulomb Hamiltonian under the Born-Oppenheimer approximation, where the Hamiltonian can be broken down into the following 5 components:

The most important thing to take away from these expressions is that they all contain summations over nuclei, electrons, or both; combined with the presence of the set of nuclear charges $\{Z\}$, there is a non-linear dependence of the molecular energy on more than just the nuclear masses.
This is one reason why it usually isn't instructive to compare absolute energies of unrelated molecules.

Although you did not provided the method used to calculate the energies, it is most probably HF or DFT. These methods use the electronic Hamiltonian, as provided in answer by pentavalentcarbon. The energy corresponds to formation of the compound from bare nuclei and electrons. Such value is of little use on itself.
You probably expect the result to be Heat of Formation, i.e. formation from atoms at their standard state. To obtain such a number, you'd need to calculate the electronic energies of corresponding atoms and formulate the corresponding chemical equations.
Alternatively, semi-empirical methods (AM1, PM3, PM6 in Gaussian) give the Heat of Formation directly.
To overcome the difficulty, one often uses the isodesmic reaction to obtain the relative energies, which are of better use.


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Is there a difference between negative/positive controls and control/experimental groups?
First of all, just so I can fully understand, what is the correct definition of a control regarding science experiments?
Is there a difference between the definitions of a control/experiment group and negative/positive controls?
I know that the control group is the group that will not be tested on while the experimental group is the group that will be experimented on, but that seems very similar to negative and positive control.
Is it true that negative controls are the controls in the control group while positive controls are the controls in the experiment group?

It would seem so. https://en.wikipedia.org/wiki/Scientific_control#Types_of_control

Consider a chemical test, such as a test for the presence of lead in drinking water. You would use both experimental and control samples. Whenever you run the test on experimental samples, you also need to run it on both positive and negative control samples--positive samples are prepared deliberately with a certain known concentration of lead, and negative samples are known not to contain any lead (you have to be careful here!). If you get the incorrect results for the control samples, then you know that the results for the experimental samples may be off, so you have to repeat the test. You also might want to periodically have the analyst run a proficiency test with the standard positive and negative controls as well as samples containing various known amounts of lead.
Experimental and control groups might more typically refer to situations such as drug trials, where you are comparing a new drug to another drug that is already approved and used to treat an illness. The experimental group would be the group of subjects having the illness that are given the new drug and the control group would also have the illness but would be given the approved drug (preferably in a double-blind study design--where neither the subjects nor the people administering the drug know which treatment is which). Note that both the experimental group and control group are "experimented on" - otherwise they could tell whether they received the treatment or not, and the results could be erroneous due to the placebo effect.


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Formation of water vapour during combustion
When we burn fuel, water vapour is produced because hydrogen is burnt as well. But since the vapour is in gaseous state already, it shouldn't steal any heat from fuel gases, should it?

Water vapor produced during the burning of hydrogen-containing fuels should not steal any heat from the fuel, unless they condense on solid surfaces. In such cases, it may require a small amount of heat to evaporate them again. However, in my experiences, there is a stage during burning that there is a lot of water vapor on the solid object, and as the object is warmed up, there is no more vapor condensed.

thanx shadow.I have another question.does the condensed water present in the fuel before combustion makes the heat difference between gross calorific value and the net calorific value then?

See https://en.wikipedia.org/wiki/Heat_of_combustion. You should consider posting another question, rather than asking in the comments.


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Combustibility of gas/carbon dioxide mix
Via electrolysis I produced an ethane/hydrogen/carbon dioxide mix. Normally, there is 1 part ethane, 1 part hydrogen and 2 parts carbon dioxide. If I simply used a Bunsen burner with air hole closed to try to burn this gas, would it burn at all? Would the hydrogen burn properly but the ethane produce lots of soot and carbon monoxide?

I say this not entirely in jest, but..why not try it out and see for yourself?

Can't, because I don't have any pipes that won't melt right now, and am busy so can't really get any right now

Then try it when you can. Chemistry is an experimental science, after all. My guess is that it would burn quite readily.

@IvanNeretin  This makes me wonder.  Are many or most electrolytic gas mixtures self combustible?  Could the changes to the electrolyte result in some fuel or oxidiser component not being a gas leaving only one or the other type requiring external air or fuel for combustion?

Many (though not most) electrolytic gas mixtures contain hydrogen, hence yes, they are combustible. If by "self combustible" you mean "containing both oxidant and reducing agent", then the right word is "explosive". Some mixtures are like that, some aren't.


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Why does the temperature of the gas in a container moving with constant velocity does not change?

Systematic changes do not affect thermodynamic equilibrium.

What does this mean? And what kind of systematic changes are allowed?
The container with gas is stationary till some time then its given a constant velocity and the final temperature is asked , the answer says that systematic changes dont affect thermodynamic eqiulibrium and temperature remains constant .

What does your title have to do with your question? Maybe I just don't know what your title question is asking.

@DGS I suppose that the statement in the question is an example of a systematic change that does not affect thermodynamic equilibrium. In other words, the velocity of the container (systematic change) does not affect the temperature(thermodynamic equilibrium) of the gas .

I feel like there is an answer based on relative motion that I cannot quite grasp. Even though I voted to reopen this question, it may find a better home on Physics.SE.


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GCMS Peak Identification: Online or Open Source / Free Programs?
Suppose I have a GCMS peak graph for a component, what are possible options to figure out what the molecule may be?
Any tips about good software that will do this? I do know that for many compounds there is a hit in the NIST library, but how does one go about automatically checking the library?
Google came up with OpenChrom. Has anyone used this? The peak library in it is unfortunately not accessible in the free version.

This is what I use. For a formula, at least. Other methods then allow me to deduce structure. MS can give you many structural clues, depending on the ionization method.

@DGS Thanks. Correct me if I am wrong, but that tool only seems to use the MW based on the top peak, right? Isn't there a way to use the other lower peaks as well?

Type in the mass of a lower peak?

@DGS Thanks. Maybe I'm not getting the right way to use this. e.g. Say for an unknown sample I've a GCMS consisting of several MW / peak height pairs. What one would want to do is identify a hit from the library that approximates this distribution most closely, right? If I just enter a lower peak's MW in this tool it just produces a hit to a low MW compound but that's not what we want right? Its just a low MW fragment produced from the original high MW compound we are trying to identify, correct?

That's right. Maybe I wasn't completely clear on your intension. I don't think there is a library for MS data like that. I could be wrong. ChemDraw includes a MS prediction program, but not the other way around, as far as I know.

@DGS I know that most GCMS manufacturers include programs that will do such matching for you but those are expensive & propreitory. OTOH, NIST does have a library but not sure what tools to use to access it.

Oh, okay, my bad.

Not exactly what you are looking for but a good resource for spectroscopic identification is http://sdbs.db.aist.go.jp/sdbs/cgi-bin/cre_list.cgi

Have you tried the Golm Metabolome Database?
It is a collection of GCMS mass spectra. The focus is on metabolites.
However, on their webpage is the entry "Decision Trees". If you follow the link, there is the option to enter your own mass spectra information and receive a prediction of functional groups in the measured substance.
This might be of help to identify your molecule.
There are more details in the Wikipedia entry.

This answer would be much more useful if you elaborate on the good and bad sides of "Golm Metabolome Database".

Thanks for the hint, @IͶΔ. I have extended the answer. Is that more helpful?

Yes, much better. Thank you and welcome to the site!

OpenChrom is free of charge, but the NIST db is not. That is why you need to have the NIST db already installed on your system.
Have a look at this tutorial.

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


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Formation of Tropylium carbocation
I've read that Tropylium carbocation is formed by the ring expansion of benzyl carbocation. But in benzyl carbocation, the $\ce{C}$ atoms in the benzene ring are $sp^2$ hybridized i.e. $120^\circ$  is the required angle for stability. And in the benzene ring of the benzyl carbocation, the angle is already 120 degrees. Therefore benzyl ring need not expand for stability. Therefore why does benzyl rearrange to give tropylium?

Note that such rearrangement is typical for mass spectrometry, not "normal" reactions.

More stable products are lower in energy.  At first glance the bonding arrangements might seem equivalent in both cases in terms of number of $\ce{C-C}$ and $\ce{C-H}$ bonds from $\ce{sp^2}$ carbons, which would result in no net change.  However, the change in molecular structure leads to a change in the molecular orbitals and the resulting energies of the molecule for the aromatic/resonance component to the energy. Using the Frost diagram technique we want to compare the orbital energies, nicely pictured here:

The key point to notice is the filled orbitals for seven member ring are slightly lower in energy than the filled orbitals for the six-member ring.  This means there is an increase in resonance stabilization energy for making a fully aromatic system rather than just a cation in resonance with benzene ring.

It does not work that simply.  In benzyl cation the non-ring carbon is conjugated with the ring and that alters the Frost circle.

This comment is less helpful than one might assume at first glance.  Do you have, for example, some QM calculations to back this up?

if we are using Frost circles then we can generalize it to the Huckel model.  The pi electron stabilization I get from that model = $8.00\beta $ (vs nonbonding $p $ orbitals) for benzene but $8.72\beta $ for benzyl cation, versus $8.99\beta $ for tropylium.  The predicted driving force for rearrangement is still positive but much less than if you ignored the conjugation of the benzylic carbon.  So,  I would expect the benzylic conjugation to be fairly important in more advanced models too.

That gets a little tricky, since benzylic conjugation would disrupt the aromaticity, ya?

That's what the more general model is for.  For an all carbon system you just set up the connectivity matrix and get the eigenvalues in units of $\beta $, and those are your energy levels (positive= bonding, negative =antibonding).  The Frost circle is just the pattern of eigenvalues specific to a single ring with no pendant pi bonds.

Ah.  I believe you are saying the model I am employing is too crude.  I'm happy to concede it is crude.  But it is probably good enough.


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Stability Comparison between the Tropylium and Tricyclopropylcarbinyl Carbocation
Why is the Tropylium carbocation less stable than the tricyclopropylcarbinyl  carbocation?  The tricyclopropylcarbinyl carbocation undergoes a sigma-tropic rearrangement whereas tropylium is highly stable due to conjugated system, that being, it is resonance stabilized and the number of canonical forms of tropylium is more. Then, why is it less stable?

This is a difficutl question to answer, because of the somewhat contentious nature of the concept of hyperconjugation.

related: What is the reason for the exceptional stability of the cyclopropylmethyl carbocation?

It seems likely that the tricyclopropylcarbinyl carbocation would be more stable than the tropylium carbocation for the reasons I'll outline below, but if you have a reference proving this point, it would be nice to add it to your question.
Background
The cyclopropyl group is similar to an olefinic double bond in that it is very effective at stabilizing adjacent positive charge. The reason for this is that the $\ce{C-C}$ bonds in a cyclopropane ring are close to being $\mathrm{sp^4}$ hybridized (see this earlier answer for further elaboration of this point), said differently, there is a lot of p-character in the bond. In fact, the p-character is high enough that a cyclopropane compound will absorb bromine much like an olefin. If that cyclopropane high p-character bond (orbital) can align itself with the p-orbital on the adjacent carbocation center, it will stabilize the carbocation, just like a p-orbital on an adjacent double bond (the allyl system) stabilizes a carbocation.
Hart and Sandri [J. Am. Chem. Soc. 1959, 81 (2), 320–326] studied the solvolysis of the following 3 compounds (Ar = p-nitrophenyl) and found that the relative rates of solvolysis were 1 : 246 : 23,500.

The corresponding tricyclopropyl p-nitroester could not be synthesized, so to complete the series Hart and Law [J. Am. Chem. Soc. 1964, 86 (10), 1957–1959] prepared the analogous phenyl esters and found the relative rates of solvolysis to be 1 : 1,080.

Clearly, every time a cycloproyl group is added significant additional stabilization of the carbocation center occurs.
Answer
The number of substituents stabilizing an adjacent positive charge is critical in determining the stability of a carbocation. Just as the trityl carbocation is much more stable than the benzyl carbocation, we see above that the  tricyclopropylcarbinyl carbocation is much more stable than the cyclopropylcarbinyl carbocation. Three stabilizing substituents are better than two substituents which is better than one.
In the tropylium carbocation there are 2 double bonds and a hydrogen stabilizing the carbocation center.  In the tricyclopropylcarbinyl carbocation there are 3 cyclopropyl groups stabilizing the carbocation center. Three groups provide  better stabilization than two groups, hence we might expect the tricyclopropylcarbinyl carbocation to be more stable than the tropylium carbocation.
A Word about Resonance Structures
The number of significant resonance structures for the tropylium cation is 7, the positive charge can be delocalized to each of the seven carbons as shown below. However 7 resonance structures can also be drawn for the tricyclopropylcarbinyl carbocation. So actually both carbocations have the same number of resonance structures.

Isn’t a corresponding cyclobutyl cation also taking part in cyclopropylmethyl cation’s resonance?

@Jan  The nuclei are in different positions for those two cations;  they are in equilibrium with each other rather than resonance structures.


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Why is copper(II) carbonate not selected as an unknown compound for back titration? (Whereas another unknown carbonate is used instead?)
Why is copper(II) carbonate not selected as an unknown compound for back titration? (Whereas another carbonate is used instead)
I know that the carbonate to be provided on the day (not Copper (II) obviously) will be reacting with excess HCl. The excess HCl will then react with NaOH to determine the concentration of HCl and eventually Molar Mass of the provided carbonate substance.
NOTE: phenolphthalein is the indicator being used

What is the ph indicator for this experiment

Copper is a transition metal. As such, its ion will be a Lewis acid. The effect will, as with iron, be stronger with increased charge--so Fe(III) salts of strong acids give acidic solutions, more so than Fe(II) salts of the same acids.
This is because lone pairs can be donated into the empty d or hybrid orbitals from water. The structure produced in this way tends to lose hydrogen ions. This means that the acidity in your experiment has two sources for hydrogen ions.


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Combustion that generates no gas?
I'm learning on my own with Chemistry The Central Science, 13th ed. book. So far it's going well.
I've come across something I don't think I understand. I've googled and "stackechanged" with no avail.
How come a combustion generates no gas products? And in what circumstances?
This is the bit of the book that left me wondering:

For a noob like me, it's counterintuitive that something "burns brilliantly in air" and only generate solid products.

Well, the answer is: just like that. You ignite magnesium, it burns with a flash that leaves you blind for a minute or so, and produces nothing but $\ce{MgO}$ which has pretty high melting point (and even higher boiling point, of course). Who said a combustion must always generate gas products?

If $\ce{CO2}$ and $\ce{CO}$ weren't gaseous, then hydrocarbons would burn to form essentially all solid products, too.

I guess the exact definition of combustion collides a little bit with my natural experience. I'll work on that.

It should be noted that to the eye, there will be a lot of off gassing if you actually try this, but that is largely due to the extremely high temperature of magnesium burning igniting impurities present in the magnesium ribbon.  This is one case where the chemical prediction will differ from the pragmatic observation.  If you burned pure magnesium there would be no off-gassing.

I think the issue is that a kind of burning like this one could generate a smoke of solid particles. I guess that would fit with the "natural experience"; would it fit with the chemical issue at hand?

There are many reactions that do this. The simplest way is to substitute any other reactive metal for magnesium: aluminium, lithium, sodium, etc. All will burn in oxygen to give solid products / oxides.
This can be extended by using any of these metals and other reactive non metals, such as fluorine, chlorine, etc. in Group VII or sulphur, etc. in Group VI. Do not try it with $\ce{Se}$ or $\ce{Te}$ in combination with aluminium, however, as it is near-explosive and forms toxic gas on hydrolysis.
A simple experiment that does this is the reaction of sulphur with iron filings.

Welcome to Chem.SE!  Thanks for posting your answer here.  I've reformatted your post to make it more readable, and made some edits (removed element name capitalization, etc.) to bring it in conformity to site standards.

In our rocket club we used 1:1 mol/mol mix of zinc dust and flowers of sulphur as our basic rocket fuel.  The product is ZnS - no gas.  It burned with a bright green flame and produced a beautiful dense cloud & trail of white ZnS smoke.  The expansion was due simply to the extreme heating of the air in between the fuel/oxidizer particles.


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First usage of PEG (Poly ethylene glycol) on surfaces against biofouling
Is there something like a "famous initial case" when people used Poly Ethylene Glycol (Poly Ethylene Oxide, PEG) as passivating layer or so on surfaces? Especially against nonspecific Protein or Peptide binding?
I found one publication from 1971: "Treatment of controlled pore glass with poly(ethylene oxide) to prevent adsorption of rabies virus" (Hiatt et al), however this has been cited only very rarely.
On the other hand there is the famous precipitation method described by Polson et al from 1964.


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Why is it important to minimize impurities when doping a semiconductor?
I was reading "Electronic Devices and Circuits" by Boylestad, and in there the author talks about how the semiconductor industry seeks to add as little impurities as possible when manufacturing semiconductors.
What I don't understand is why it is so important to keep the material's purity? What characteristics are improved by doing that? If the objective of doping is improve the conductivity of the material (in a controlled manner), wouldn't reducing the ammount of doping material decrease that efficiency?

Semiconductors are based on impurities.  They just call it doping.  The addition of group 3 (Boron group) or group 5 (Nitrogen group) impurities create a hole or an extra electrno, respectively.  These form charge carriers which are necessary for a semiconductor to be used in a circuit.  However, the dopants need to be added in highly controlled and precise amounts to get the necessary mobility.  The silicon also needs to be highly pure for regular electronic response, for regular electronic properties over arbitrary distances.  Another important consideration is that the dopant allows some degree of tuning of the electrical conductivity of the semiconductor.
Also, highly pure crystalline silicon  fractures nicely along the crystal planes, meaning it can be cut with high precision.  Since most mass-produced electronic processors are made on 12- or 16-inch silicon wafers with dozens of devices manufactured at once, and initially all physically connected, the ability to precisely excise each tiny circuit is critical for high through put and yield in manufacturing.

If you add too much dopant to silicon the silicon will become conductive. In order other words it becomes a resistor, not a semiconductor.

When something becomes conductive it absolutely does not become a resistor.

@Lighthart :-( It wouldn't become an insulator, nor superconductive, so there is resistance. V=iR as I remember.

Does that mean, according to your language, semiconductors are not conductive either?


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Comparison of rate of reaction in bromination and chlorination
I did research, about this one but the data and other things prove chlorination is faster than bromination, because chlorine is more reactive. But, some data says reaction of chlorine is more exothermic then bromine, that is the reason. But, if this is the case, then according to me bromination should be faster. Because, less energy is needed to start bromination and less energy will be given out in formation hydrogen bromide.
And how reaction being more exothermic or less exothermic determines the rate of reaction?

The energy to start the reaction shouldn't be a key factor here;  one photon is enough to initiate the reaction for thousands of molecules. Also, you don't create hydrogen bromide in bromination.

If you are asking about free radical halogenation, see this earlier answer.

I don't understand.

Halogenation of what? An alkane? Alkene? Another functional group?

Alkane.
Free radical substitution.

@ron  Your link does not address rate.

What do you mean by link?

According to the Hammond postulate, a highly exothermic reaction is more likely to have a low transition state barrier than a less exothermic reaction.
Lower barriers lead to faster reactions.
This is a multistep mechanism, so for this to be relevant the rate-determining step.  According to this lecture the rate limiting step is the hydrogen abstraction from the alkane, forming a hydrogen-halide bond.
The bond dissociation energy for HCl is 431.8 kJ/mol and 365.7 kJ/mol for H-Br.  The formation of the H-Cl is this much more exothermic, and the reaction will be faster.

For the question why chlorination should be faster than bromination?
Answer should be
Because chlorination is more exothermic.
Not chlorine is more reactive.
Right?

Hammond's postulate says it is very difficult to separate those two circumstances.


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Why are silver halides insoluble in nitric acid
The silver halides $\ce{AgCl},~\ce{AgBr},~\ce{AgI}$ are insoluble in $\ce{HNO3}$. This is given in J.D.Lee's inorganic chemistry text. (The strength of nitric acid is not specified though).
But won't this reaction take place?
$$\ce{X- + HNO3 -> X2}$$
This example is taken from JD Lee's book itself:
$$\ce{6Br- + 8HNO3 -> 3Br2 + 2NO + 6NO3- + 4H2O}$$
because  concentrated Nitric acid is a powerful oxidizing agent.

What kind of reaction is that ? It's not balanced either in elements or in charges.

@Hippalectryon its a basic oxidation reaction.

Not all silver halides are insoluble in water. AgF is highly soluble in water.

Nobody is interested in the behavior of $\ce{AgF}$. Only $\ce{AgCl, AgBr, AgI}$ are studied and discussed.

Nitric acid (even concentrated) is essentially an aqeuous solution.
Effectively, silver halides will not dissolve in water (they  have incredibly low $K_\text{sp}$), so the reaction indicated is moot.... there is no appreciable concentration of bromide in solution to be oxidized.
$K_\text{sp}(\ce{AgBr}) = 5.0\times10^{-13} \Rightarrow [\ce{Br-}] \approx 700~\text{nM}$

Low but not incredible, Bi2S3 is almost incredible but even that is attacked by concentrated nitric acid [I think I have never tried it] All this seems to be a simple experiment for a high school lab [with good attention to safety]. Concentrated nitric is far from an aqueous solution. I have a friend who is in pain her whole life from a lab accident involving concentrated nitric acid.

The diagnostic test for Cl-, Br- and I- is reaction with Ag+ in a dilute nitric acid solution. The reason for the nitric acid is not because the silver halides are not attacked by nitric but that the dilute acid solution prevents the formation of other insoluble silver salts such as silver oxide, carbonate, phosphate, cyanide [careful here] and who knows what else. You cannot use sulfuric [perchloric will work but why even have it around] and acetic might work. Nitric is perfect. I have little doubt that the silver halides would be oxidized to the halogens or even higher oxidation states by hot concentrated nitric acid. You are welcome to try the next time you are in the lab.


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Synthesis from cyclic ketone to alcohol
How can I make the product in the picture below from cyclohexanone? I did Stork enamine synthesis, but it results in terminal primary alcohol. I need secondary alcohol with one more carbon atom at end. How to do that?

How about a Michael addition of the cyclohexanone enolate to methyl vinyl ketone; follow this with a reduction of both carbonyls.

@ron How do you prevent Robinson annulation?

@Loong Good point, so you'd use an enamine (like the OP suggested) instead of the enolate and run the Michael addition under neutral conditions.


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Why only the unstable resonating structure proceeds in the reaction?

For example, in the given image, I have starred the step where the relatively unstable resonating structure proceeds forward in the reaction.
Why?

These aren't two states, only two depictions of one.

Oh yeah. Let me clarify the question.

I would avoid the term 'resonance', since we're not dealing with two states that 'resonate' one between another. In reality, the electron densities are delocalised, making it that you can draw depictions of how this delocalized state would look like.

Indeed, in each case there is only one real structure, and the resonance variants are but artifacts of our representation. That being said, though, your observation might actually make some sense. A stable structure has no incentive to react, so to say.

Imagine a case where the resonance form with the negative charge localized on the oxygen went forward and attacked the second carbonyl:  That charge needs to go somewhere so the double bond is broken and the electrons go up to the second oxygen (the one that started as the second carbonyl). Those electrons would look to come back and reform the double bond, and in doing so would force one of the groups to leave. And the bond that leaves is the one you just made. This reaction happens with both atoms acting as a nucleophile except only one (the carbanion) goes forward and doesn't immediately leave again.

So you mean that the negative charge should be on Carbon, so that the electrons can be 'easily passed on' to next step in the reaction, otherwise Oxygen is more electronegative and won't lose 'pass on' electrons that easily? This seems acceptable, just confirm if you meant this.


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Determine rate equations for entire kinetic scheme
I am working with the following kinetic scheme that represents pyrolysis of a biomass particle:

The paper that discusses this scheme provides the following rate constants:

K1 for the pathway of $\text{Biomass} \rightarrow (\text{Volatiles + Gases})_1$
K2 for the pathway of $\text{Biomass} \rightarrow (\text{Char})_1$
K3 for the pathway of $(\text{Volatiles + Gases})_1 \rightarrow (\text{Volatiles + Gases})_2 + (\text{Char})_2$

The rate equation provided in the paper for the conversion of biomass is
$$\frac{\mathrm d\,B}{\mathrm dt} = -(K_1 + K_2)\,B$$
The rate equation for the char 1 component is given as
$$\frac{\mathrm d\,C_1}{\mathrm dt} = K_2\,B - K_3\,C_1$$
And finally the rate equation provided for the char 2 component is
$$\frac{\mathrm d\,C_2}{\mathrm dt} = \delta\,K_3\,C_1$$
where $\delta$ is a deposition coefficient reported as 1.45 in the article. Note that all components of the system are on a mass basis such as $\mathrm{kg/m^3}$.
It is fairly straightforward to calculate the concentrations of $B$, $C_1$, and $C_2$ at each time step given an initial concentration of $B$. However, I would like to determine the concentrations of the volatiles and gases in the system too. Since it doesn't seem possible to calculate the individual volatile and gas components from the given scheme, I would like to determine the concentration of the (Volatile + Gases) group which could be labeled as G. So how would I develop the overall rate equations for B, C, and G?
$$
\frac{\mathrm d\,B}{\mathrm dt} = -(K_1 + K_2)\,B \\
\frac{\mathrm d\,C}{\mathrm dt} = K_2\,B + ? \\
\frac{\mathrm d\,G}{\mathrm dt} = K_1\,B + ?
$$

(V+G)2 is the same as C2 and (V+G)1 is "straightforward" as you said.

@pH13 So the concentration of group (V+G)2 is the same as the concentration of C2; basically (V+G)2 = C2?

I'd say so, yes. As far as those subscripts are only indices.

@pH13 That makes sense to me but what I'm actually trying to do is figure out the individual components.  From (V+G)1 or (V+G)2 what is the concentration of V and G in each group?

As you don't know the amount of "Gases" I don't think that you can split it up further.

@pH13 Ok, well maybe you can help with the second part of my question regarding overall rate equations (see edited question above).

The Biomass-equation is:
$$\frac{\mathrm d B}{\mathrm d t} = -(k_1 + k_2) B$$
The (volatile + gases)$_1$ equation is:
$$\frac{\mathrm d VG_1}{\mathrm d t} = k_1 B$$
The (volatile + gases)$_2$ equation is equal to equation for (Char)$_2$:
$$\frac{\mathrm d VG_2}{\mathrm d t} = \frac{\mathrm d C_2}{\mathrm d t} = k_3 C_1$$
(It might be that it needs to be multiplied by 1/2.)
And the equation for (char)$_1$ is this:
$$\frac{\mathrm d C_1}{\mathrm d t} = k_2 B-k_3 C_1$$
The equations for the complete set of (char)s and (volatile+gases)s are then the sums of the solutions for $C_1$ and $C_2$ resp. $VG_1$ and $VG_2$.

For initial concentration of $B(0)=1$ and everything else $f(0)=0$, as well as $k_1=2$, $k_2=10$ and $k_3=3$, you should get something similar to this:

Thank you, I'll try this in my model and let you know if I need more help.  Also, what did you use to create the plot?  And what books would you recommend for learning about systems of kinetic reactions and how to compute their yields, i.e. something that discusses modern techniques or numerical methods for chemistry?

The calculation and the plot is made with Mathematica 10. Unfortunately I don't have any books to recommend as I'm not doing this very often.


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Synthesis of 3-Methylenecyclopentene from 1-methylCyclobutene
How do I make 3-Methylenecyclopentene

from 1-methylcyclobutene?

I thought to do ring opening reaction to add 1 carbon atoms but this won't add double bond in product. How to start please?

What sort of ring opening?

Here's a thought: 1) Simmon-Smith reaction, 2) bromine addition, 3) double dehydrobromination

I don't know  Simon-smith reaction. Can ring opening epoxide reaction work?

Here is my try.

Ozonolysis
Aldol condensation using a hindered base (like LDA)
Wittig reagent to convert $\ce{C=O}$ to $\ce{C=C}$

Step 1:

Step 2:
I used $\ce{LDA}$ which abstracts a proton from the least hindered carbon near to a carbonyl group. (So that the resulting carbanion is stable).
I am not drawing the enolate ion formation etc. You can find the reaction mechanism for aldol condensation here and the reaction using $\ce{LDA}$ here.

After protonation, you can heat the compound to remove $\ce{OH-}$ which is a simple dehydation. There are two possible alkenes. One of them (the one given below) is more stable because of conjugation.

Step 3:
We are almost done. To replace $\ce{C=O}$ with $\ce{C=C}$, use wittig reagent.

You'll need low temp for your intramolecualr aldol but otherwise, this is sold.

What are the chances of conjugate addition with a Wittig anion? Would Tebbe or Petasis be more likely to succeed? Well, if those things would be an issue, there's a ton of methylenation reactions.


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Why does increasing surface area of the electrode decrease resistance?

Which increases immediately if the surface area of the silver electrode is increased?

overall cell voltage
rate of change of $\ce{[Ag+]}$
mass of $\ce{Cu}$ electrode
change in ratio of electrode masses; ∆$(\frac{\text{ mass of Cu}}{\text{ mass of Ag}})$

The correct answer was (2), but I'm not sure why the reaction rate would go up if the surface area of the silver electrode increased. The rate of reaction is proportional to the current flowing through, which by Ohm's Law is $\frac{V}{R}$. This means that the resistance must have decreased if current increases. But resistance of the electrodes is only related to the length and cross sectional area. The problem does not specifically state any change in length or cross sectional area. It only says surface area, so what am I missing here?

For a galvanic cell (electrolytic cell with significant current) $\mathrm{R} \propto \dfrac{1}{\text{Area(electrode)}}$ because of the overpotential http://toruhara.page.tl/11-.--Electrochemistry-d--The-Basic-of-the-basic.htm

There is a greater point of contact with the larger electrodes allowing the current to flow at a faster rate.

Well actually in the circuit that you show, a high impedance voltmeter essentially allows no current to flow and you are just measuring the "theoretical" EMF of the cell.

Fluid flow is always an excellent analogy for understanding current flow.
Fluid flow rate -> electrical current
Pipe friction, valves and other throttlers -> electrical resistance
Fluid pressure -> voltage.
In this case, the amount of contact between the electrode and the solution is most likely the bottleneck of the circuit so increasing the area of contact should increase the current.  Similar to how cranking the valve on a faucet or outdoor spigot more and more open increases the flow rate of the water coming out.


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Current in a galvanic cell
I was looking through chemistry textbooks to find out how to determine how much current a galvanic cell should generate and what affects that current value. However, I did not find anything: textbooks just talk about the EMF of a cell.
I wanted to know what determines how much current a cell generates. Certainly, the amount of electrons produced externally contributes to the current produced by the cell. However, what about internally? Batteries have internal resistance. Is that true of cells too (it should be because batteries are made up of many cells)? What determines amount of internal resistance: does the type of electrolyte(s) affect it. If a cell uses a weak electrolyte (like acetic or carbonic acid) will it cause more internal resistance then if something like sulfuric acid (strong electrolyte) is used?
What if a cell used weak electrolyte as the cathode electrolyte and the strong electrolyte was the anode electrolyte, will that cause more internal resistance due to there being less ions in the cathode compartment as not all of the electrolyte has been dissociated into ions (thus, the cathode cation cannot be reduced)?

Although voltage can be calculated from the electromotive series, maximum current is harder to predict due to a number of factors, some of which you've mentioned.

Electrolyte conductivity, which varies with components, concentration, temperature, etc. As you state, weak electrolytes are less conductive. Resistance of a composite electrolyte would be the sum of its components.
Cell polarization, which is an effect at the electrode/electrolyte interfaces. The causes are buildup of gas bubbles, concentration gradients that develop in the electrolyte, etc.
Deterioration of the electrodes and/or electrolyte, particularly under high current drain.
Of course, as @MaxW states, a major factor is cell geometry: area of electrodes and separation due to insulators and electrolyte.

For these reasons, though the maximum current can be calculated from the resistances and EMF, in actual use, cells often fall far short of their theoretical potential (no pun intended).
The resistance of a battery is primarily the sum of that of the cells. Since cells are joined with metallic connections, the cell-to-cell resistance is usually negligible (though I've used poorly designed battery holders with high-resistance spring contacts that severely limited the current, and that damaged the plastic holder due to ohmic heating).

I think you missed a large factor which is electrode's surface area. A wire or a plate doesn't matter for EMF, but it has a large effect on current.

so if i have a salt electrolyte which is very dilute (i.e. 0.001M) will that be a poor conductor since their is not only less ions available but also because there is a lot of DI water (and DI water is a poor conductor)?

@MaxW is the electrode surface area the are that is in contact of the electrolyte or does it count the whole electrode (also the part that is not in contact with the electrolyte). Also, why will surface area have a large effect? Aren't most metal conductors (like aluminum and platinum) good conductors?

@user510 - Think... If you take the electrodes out of the solution how much current do you think will flow?!?

yes, no doubt that surface area will contribute to conductance, but is it that significant? I am saying this because when I made a cell and dipped more of my electrode in the electrolyte solution, the current went up only a little. Also is the surface area of the electrode just the area that is in contact? Specifically, when I am calculating resistance of the electrode, is the surface area the same as cross section area?

The explanation of your finding might be that the metering device has a large internal resistance or that the limiting factor is the electrolyte resistance. The current should be proportional to the surface area, given constant separation of the electrodes (for close geometry, the back surface provides little current).

what do you mean by metering device?

The device you are using to measure current.


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Why are there no hydrogen atoms in the crystal structure of a biomolecule?
I am taking a physics and chemistry at the nanoscale class at my university and for a project I must learn how to use Visual Molecular Dynamics (VMD). One of the models involved in this project is apo-core-streptavidin in complex with biotin at pH 4.5 (1SWE.pdb).  The crystal structure can be found protein data bank. There are no hydrogen atoms in this file/model and I was wondering why that is.  I can't find the answer to this anywhere online.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. With X-ray crystallography you measure electron density, for light elements the uncertainty is probably too high to depict their position accurately. Have a look at this file, maybe it helps you a little until you receive an answer.

Related: How to add hydrogens to a crystal structure?

Typically crystal structures are determined by X-ray diffraction off of electrons within a bond and around the nuclei of each atom. Since H only has 1 electron, and its often involved in a polar type bond (hence doesn't spend much time by the H nucleus when bonded) it is notoriously hard to "see" with X-rays.
In order to have X-rays diffract accurately off H atoms the sample crystal needs to be large and very regular; this is not usually the case with proteins. As a workaround information about the H position can be calculated by the other atoms hybridization. This is often unreliable especially with all the hydrogen bonding in proteins which pulls them out of their idealized positions.

Hydrogens don't diffract X-rays to the same extent (basically not at all) as compared to other atoms, so they do not show up in normal X-ray crystallography because their signal is so small.
This is because the electrons are actually responsible for diffraction and most other bio-elements have many more electrons than hydrogen, so their scattering signals are much larger than those for hydrogen.

Ok, this makes sense.  I never would have guessed it had to do with the measurement/ imaging of the molecule. Thanks for the answer.


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Chlorination by SOCl2 vs SO2Cl2
My textbook mentions that chlorination of alkanes by $\ce{SO2Cl2}$ is preferred over $\ce{SOCl2}$, but I didn't understand why.
Could someone explain which of the two reagents I should use in what situations, and briefly tell me why?
There is a similar question here:
How does SO2Cl2 react with alcohols?
However the answers there did not completely clarify my doubts.

The reactivity patterns of $\ce{SO2Cl2}$ and $\ce{SOCl2}$ are quite different. $\ce{SOCl2}$ is a good electrophile, and can be thought of as a source of $\ce{Cl-}$ ions.  These ions can go on to react in their typical nucleophilic fashion. $\ce{SO2Cl2}$ however is often a $\ce{Cl2}$ source, as it readily decomposes giving off sulfur dioxide. Usually much easier/safer to use this than measuring out (and getting into solution) chlorine gas. The chlorination of simple alkanes by $\ce{Cl2}$ gas (or something that makes it in solution) happens by a radical mechanism i.e. $\ce{Cl.}$ not $\ce{Cl-}$.

A solid answer.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I have updated your post with chemistry markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details.

Thanks a lot for the precise answer. So the major difference is socl2 reacts by nucleophilic substitution, but so2cl2 reacts by free radical mechanism, right?

I would say that the major difference is the oxidation state of the Cl formed. As stated in the answer, SOCl2 is a source of chloride ions whereas SO2Cl2 is a source of chlorine. That is a consequence of the different reactivity of the two species (SOCl2 is an electrophile and SO2Cl2 decomposes into SO2 + Cl2) I'm just restating the answer... my goodness.

Chlorination of alkanes by $\ce{SO2Cl2}$ is preferred over $\ce{SOCl2}$. Just pay little more attention on the word alkanes. So, the statement is absolutely correct.
Just go through it (link)!
Hence, For chlorination in alkanes, if you use $\ce{SO2Cl2}$, It undergoes a free radical substitution reaction and will get the product.

If we use $\ce{SOCl2}$, it undergoes a nucleophilic substitution reaction. Here, this reaction mechanism is explained.
So, You can see here that $\ce{SOCl2}$ doesn't produce a chlorine free radical. Hence, It is not possible to chlorinate alkanes using $\ce{SOCl2}$, which $\ce{SO2Cl2}$ does.

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center. Please visit [this page](http://chemistry.stackexchange.com/help/notation " latex="" at="" chem.se""="), this page and this one on how to format your future posts better with MathJax and Markdown.

Oh! That makes a lot of sense. A nucleophile wouldn't be able to attack an alkane, right? Thanks a lot!!


text:
Comparison of standard reducing potential of Co3+ and Mn3+
$\ce{Co^3+(aq) + e- → Co^2+(aq)}$   +1.81 V
$\ce{Mn^3+(aq) + e- → Mn^2+(aq)}$ +1.51 V
This is the data of Standard reducing potentials I have found on internet. Why is reducing potential of $\ce{Mn^3+}$ is less than that of $\ce{Co^3+}$. $\ce{Mn^3+}$ converts to $\ce{Mn^2+}$ which is half filled configuration and hence more stable. Am I missing something?

The answer is simple, you are overlooking the most important trend in the 3d block. Everybody likes to think about the fancy stuff like exchange energy, Jahn-Teller effects, blah blah blah but they forget that effective nuclear charge increases going across the 3d block.


text:
The difference between hydroxide and oxyhydroxide
I cant discern these two terminologies. Would someone be very kind to help me on it?
I think hydroxide mean M(OH)x where M is a metal ion, while oxyhydroxide means something like MO(OH). Is that right? Oxyhydroxide equals to oxide-hydroxide.

You may see this:-http://chemistry.stackexchange.com/questions/44159/what-is-the-structural-difference-between-mnooh-and-mnooh2

Thanks! Then I have a further question, is the OH in oxyhydroxide anion or it is charge balance like the OH in organics? I ask this because in the Chinese nomenclature of NiO(OH), they tend to the second one which means that the OH and NiO(I think it is the Ni in NiO) are bonding together by covalent bond.

In both the cases, the -OH group is covalently bonded to  manganese. So, I think in case of nickel  also the -OH group is covalently bonded. I am not too sure about it but don't worry. I will search  about it and tell you the answer.

Actually NiOOH follow a solid state chemistry. In page 3 of this link, it says that NiOOH  has various type of forms like beta-, gamma- etc. Here, the nickel oxide layer is stacked one above the other to form a lattice. But conventionally, NiOOH should be said nickel (III )oxyhydroxide.

Yes, you have correct idea.
Oxyhydroxide is any mixed oxide and hydroxide. For example Iron(III) oxide-hydroxide
While in hydroxide hydrogen and oxygen atom are joined with covalent bond.

Thanks Freddy! Then I have a further question, is the OH in oxyhydroxide anion or it is charge balance like the OH in organics? I ask this because in the Chinese nomenclature of  NiO(OH), they tend to the second one which means that the OH and NiO(I think it is the Ni in NiO) are bonding together by covalent bond.

To go back to the question given in the title, the difference between oxides and hydroxides is simply water $\ce{H2O}$.
Let's take the example of iron $\ce{Fe(III)}$. Its hydroxide is $\ce{Fe(OH)3}$. Its main oxyhydroxide is $\ce{FeOOH}$. And this oxyhydroxide is obtained by removing one water molecule from $\ce{Fe(OH)3}$ according to : $$\ce{Fe(OH)3 -> FeOOH + H2O}$$

Mate oxyhydride is -OH whereas hydroxide is HO+. Only difference is charge. I have heard somewhere that both can be used in writing chemical equations. However, this is wrong. You can use both OH and HO to signify presence of hydroxide. But oxyhydride is -OH whereas hydroxide is HO+. Hydroxide is an organic covalent bond. Oxyhydride is a mixed bond.

" oxyhydride is -OH whereas hydroxide is HO+" Reference, please

@Ian Bush. HO+ is a theoretical concept that does not exist in practice in the lab. OH- does exist. It is the anion hydroxide. This anion is part of metallic hydroxides. For example, OH- is contained in the famous sodium hydroxide $\ce{NaOH}$, which is one of the most important hydroxides in the world.

@Maurice I understand that entirely. But this answer to the question I found so ... surprising ...  I wanted to find where this, errr, assertion was derived from

@Ian Bush. Which assertion do you find surprising ?

@Maurice "oxyhydride is -OH whereas hydroxide is HO+" Hydroxide has never been OH+ in any of the 35 years since I got my chemistry degree, and nor has -OH been oxyhydride ( or oxyhydroxide for that matter ). I want to find out where this is being stated to be the case.

@Ian Bush this is not my words. I re checked and after consulting my science textbook and a quick search on Google I found this in Quora as well. It could completely be an error in my old textbook and you cannot trust Quora. Anyways, thanks a lot for correcting me.

Really sorry if I made a mistake. I'm always learning. Thank you for this information

The thing is I too thought how this happened. I recently came here from Quora. This is carry wierd. I just posted the thing I read on Quora and reddit.

@Maurice Where have I said that? I am just quoting the original poster. I am making no such claims.

True. You have not claimed that. I sincerely apologise for any misunderstanding I caused. You merely pointed out my mistake.

@Ian Bush. The word "oxyhydride" does not describe the ion OH−
This ion is called "hydroxide". Hydroxyde is the typical ion present in basic solutions in water. Hydroxyde is the anion OH−. It is not a cation.
It is produced by autoprotolysis in water, and by dissolving metallic hydroxides in water. It is not HO+.
!

@Maurice Last thing I will say. Please read what I said. I never claimed hydroxide is OH+. I never claimed oxyhydride is OH-. Note the inverted commas around my original comment. This implies I am quoting the above answer - it is not my own text, it is copy and pasted from the answer above it. I then ask for a reference to back up the assertion that I have quoted. Via this mechanism I am calling into question that very assertion - because IMO it is mistaken, but I want to give the person answering an opportunity to either correct or clarify themselves.

@Ian Bush. Thank you for your message. And sorry ! I had not really understood the meaning of the inverted commas around your original comment. .

@Maurice No problem - the internet is not always the easiest way to convey the intention behind your text, and unfortunately misunderstanding is rife. Glad we have it sorted it out.


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Delocalised electron in graphite
I have a question regarding the delocalised electron in graphite. There has been conflicting information between books.
Some books said the delocalised electron in graphite exists $\bf{between}$ layers.
Some said the delocalised electron in graphite exists $\bf{within}$ the hexagonal layer.
I support the second statement, because carbon has electronic configuration of 1s2 2s2 2p2. Graphite has sp2, therefore leaves 1 p orbital. And the giant pi bond should be extended over the layer, instead of between.
I am not sure my evidence is sound.
Any advice would be greatly appreciated. Thank you in advance.

It's just a matter of definition. If a "layer" is understood as an infinitely thin plane passing through the centers of carbon atoms, then the electron density of the huge delocalized $\pi$ orbital is located above and below that plane, i.e., technically between layers. If a layer is defined so as to include whole atoms, then the said orbital is within layer.
In any case, the delocalized electrons are more or less "free" to move along the planes and not perpendicular to them, which explains the high anisotropy in electrical conductance and other properties of graphite.

Is there a special pi bond which exists without forming a sigma bond?


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Trifluoroacetic acid, Formic acid and dichloromethane
I am working in a small lab (in a small company) with very poor conditions, which I have to improve (for example: no sink , I have to wash the glassware in the bathroom). My organic chemistry is very poor, I am new to using solvents, and I have several basic doubts:

How to correctly wash the glassware after use?
How to correctly dispose of wasted solvents?
Should I leave the chemical fume hood on after using it and, if so, for how long?

I basically only do electrospinning here, using the respective solvents:
- Trifluoroacetic acid
- Dichloromethane
- Formic acid

How to correctly wash the glassware after use?

The glassware must be rinsed with a non-toxic (or less-toxic) solvent such as acetone or isopropanol before washing to carry over the bulk of hazardous materials.  You should the wash the glassware with soap and water, rinse with water, followed by a rinse with DI water, isopropanol, or ethanol to eliminate water spots.

How to correctly dispose of wasted solvents?

Your solvents should go into an appropriate waste container and be marked as hazardous waste with the contents listed.  You also need to list the companies EPA number to dispose of the waste.  Have a reputable waste disposal service receive your waste and take it to a disposal facility.

Should I leave the chemical fume hood on after using it and, if so, for how long?

If anything hazardous is still in the hood you must leave it on. If there is nothing in the hood there is no reason you must leave it running.

Thank you so much for your answer, it is everything I needed to know! As soon as I get "enough rep to vote" I will come back here to upvote your answer.


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Why is LiOH a weaker base than NaOH
Why do LiOH is weak base as compared to NaOH even though it's hydration enthalpy is greater than Na?

Well, besides hydration energy, there's other factors at play. How do their lattice energies compare? Look them up, maybe that'll help you answer your own question.

http://link.aip.org/link/?JCPSA6/122/194509/1

$\ce{NaOH}$ and $\ce{LiOH}$ are both classified as ionic compounds, but as with most things in chemistry, this is always on a continuum scale, in this case between ionic and covalent bonds.  $\ce{NaOH}$ is 'more' ionic in character which means its $\ce{OH^-}$ is more frequently unaccompanied by its counter cation, $\ce{Na+}$ as compared to $\ce{LiOH}$ and its corresponding $\ce{Li+}$.
Without the cation to balance the character, $\ce{OH^-}$ will balance its charge via deprotonation, and hence it is a stronger base.

I'm confused Wikipedia has pKb as -0.63 for LiOH https://en.wikipedia.org/wiki/Lithium_hydroxide and a has pKb of +0.2 for NaOH. https://en.wikipedia.org/wiki/Sodium_hydroxide So isn't LiOH a stronger base?!?

See link above.


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Does the Dieckmann Reaction proceed via the conjugated enolate of the product?
In the following mechanism of the Dieckmann Condensation, the product has been formed in the fourth step, but still the reaction proceeds and $ \ce{RO^{-}}$deprotonates the product. Then $\ce{H^{+}}$ rejoins to form the same product. Won't this result in extra expenditure of energy? Also, if the reaction proceeds through this mechanism, shouldn't there be an infinite loop?

Can anyone explain why is there an extra step?

In the depicted mechanism there is consistently a charge for the $\ce{RO-}$ omitted.

The first four steps (i.e. the main reaction) are conducted under basic conditions due to the presence of alkoxide. When that reaction is complete, the result is the deprotonated ketoester--the formation of that stable species is what drives the reaction to completion. Note that the first 3 steps are shown as equilibria, whereas the fourth step is essentially irreversible.
Later, during workup (shown above as the fifth step of the process), acid is added to neutralize that anion and form the final product.

"Stable" means the ketoester anion is not a strong enough nucleophile to attack a ester or ketone irreversibly.

@Ishu This is why a slight stoichiometric excess amount of base is required, where a typical NAS reaction requires only a catalytic amount of base. Here, yes, the base is catalytic for the equilibrium steps, but an equivalent of that base is neutralized by the product immediately upon formation. Experimental data supports this, which lends evidence to this mechanism.

I would assume, that the enol-form would be predominant in this reaction. Just like it is observed for other compounds like for example here. Could you comment why the beta-ketoester should be the main product here? I don't think that the restriction of the ring is any indicator for a completely different behaviour. I also very much doubt that the first and the second step are in equilibrium, the C-C bond formation should be thermodynamically favoured over any other reaction involved.

@Martin-マーチン Tautomer distribution depends on the solvent, probably not keto-enol in DCM, but probably in THF, right? And the C-C bond formation is reversible, moreso than a beta-keto ester deprotonation by an alkoxide. That's a pKa difference of 15 or so in DMSO, but think about the malonic ester synthesis and the C-C bond fission there.

@DGS Reversible does not mean equilibrium. Protonation and deprotonation are usually equilibrium processes. As such, the keto-enol tautomerie is an equilibrium, too. I find it hard to believe, that under these conditions, the ring closing step is not the thermodynamic well, from an entropic perspective, this is very likely. I am also not convinced, how much the keto-enol tautomerie is dependent on the solvent. But I also don't want to worry about that any more.

@Martin-マーチン Well, then don't. But maybe you should take an NMR of acetylacetone in a non-H-bonding solvent like toluene or chloroform and one in something like THF or methanol, like I did when I took organic chem my freshman year of college. Then you don't have to worry. You'll just know because you'll see it.