Rearrangement with sodium iodide (ring opening) I'm not able to figure out the mechanism for this reaction. I don't see how sodium iodide in acetone could lead to rearrangement, and surprisingly ring opening!

What is this reaction called, if it has a special name? How does it occur?

P.S. In general, as far as I know, sodium iodide in acetone promotes excellent SN2 substitution, so all bromine atoms should be replaced by iodine. After this, because 1,2-diodo compounds (vicinal di-iodides) are unstable, elimination takes place (removal of iodine) and formation of alkene occurs (that's what I've seen happen, usually).

I think the final product will also contain some Bromine atoms attached to it. and the no. of double bonds will be one less than that of the compound given in the picture. I may be wrong but with all of my known possibility of Organic reactions, I can actually reach to the final product somewhat different from that given in the picture  and I don't think the final product given in the picture can ever be got from the given starting material. I will be happy if someone else can find flaws in the logic  given below.

$\ce{NaI}$ in acetone is the reagent used for Finkelstein reaction which is nothing but a standard substitution reaction with $\ce{I-}$ ions. But in presence of vicinal dibromides, or any special structure where there is a chance of elimination of hallide ions even further, the iodide ions can actually make the molecule undergo preferably an $E_2$ elimination (in most of the cases).

In this starting compound also, there are substitutable $\ce{Br}$ atoms present. So, the iodide ions first performs a substitution reaction, and then undergo an elimination via bond migration. This same procedure occurs for two times. and finally we are left with a product which has $\ce{Br}$ atoms attached to $sp^2$ carbon atoms, and therefore they are highly reluctant to undergo any substitution or elimination reaction. So, that's why, logically the end product should be, $\text{ (1E,3Z,5E,7E,9Z,11E)-2,5,8,11-tetrabromo-6a,12a-dihydrooctalene}$. Here is the structure of the possible end product.

$\textbf{Mechanism:-}$

As far as I see, there can not be any more further reaction to the final compound. According to me, the given compound in the picture should not be the final product if we take the starting material to be the correct one. Instead, the end product in the mechanism should be the one produced with higher yield at the end.


How surface passivation works? Surface passivation is done to deal with "dangling bonds" (for example, $\mathrm{SiO_2}$ layer on top of a $\mathrm{Si}$ layer). But isn't the new surface will have dangling bonds on its own at the new formed surface?

Surface passivation creates different chemistry on the surface than the chemistry of the bulk solid. Imagine bulk Si with a clean surface (presumably in a vacuum and freshly made). It can be thought of as having "dangling bonds", though this is not necessarily the best way to think about it. Think more generally that the structure on the surface is different to the bulk structure. There may be different elements present and the bonds may be "satisfied". Or they may be strained compared to the bulk material.

Expose the Si to, for example, O2 and the surface layer (your dangling bonds, if you like) will react to give some SiO2 just on the top layer. This new layer may be much less reactive than bulk silicon in normal air, hence can be said to passivate the surface.

If the chemical and physical properties of the new layer are right, the layer will protect the bulk from further reaction: this is the definition of passivation. A similar situation occurs naturally for aluminium, for example, where the Al2O3 layer that forms is both strong and relatively impermeable to air preventing further reaction (aluminium with the layer removed will react so rapidly with oxygen it gets warm and disintegrates). Iron, on the other hand, forms rust which doesn't protect the surface as it is flaky and weaker than the bulk metal. To passivate iron you need to add other elements which encourage the formation of stronger passivating layers (this is how many steels are made not to rust).

Chemical manipulation of the surface can create specific types of passivation that serve a particular purpose rather than relying on the natural reactions of air and the surface. This is common in the semiconductor industry and can also be used to generate the coloured layers on aluminium and titanium products where the goal is aesthetic as much as protective.


How to calculate molarity of a solution by solving the percentage? Molarity is number of moles per litre of solution and we calculate it by simply dividing moles by volume. But, if moles are not given and the w/w or w/v percentage is given then, how can we solve for the molarity?

For example, if we have 2 % w/v solution of $\ce{NaOH}$, how would we find its molarity? Do we solve by dividing 2 by 40 (molar mass of $\ce{NaOH}$)?

Both of them are pretty straightforward. Don't go by the formulae, but instead, internalize the steps required here.

$2\%$ w/v solution has $\pu{2g}$ of solute in $\pu{100 ml}$ solution. That implies $\pu{20g}$ of solute in $\pu{1L}$ solution. $\pu{20g}$ solute is 0.5 moles of $\ce{NaOH}$ (why?). So, molarity is simply moles divided by volume of solution.

The $2\%$ w/w solution case is trickier, and you'll also be given the density of solution in this case.  $2\%$ w/w solution has $\pu{2g}$ solute in $\pu{100g}$ solution, or $\pu{20g}$ solute in $\pu{1000g}$ solution. If the density is $\pu{2gm L^-1}$ (assumed), then the volume of solution would be $\ce{500 mL}$. Now, you can proceed by the same logic as above.

In fact, the conversion of $\pu{100ml}$ volume to $\pu{1L}$ in case 1 wasn't even necessary, but I only did that to make the step-by-step calculations more approachable for you.


Bipyridyl complex isomers What type of stereoisomers do these two complex compounds represent?

These are examples of enantiomerism or optical isomerism. If you imagine a mirror placed between the two images you will see that they are mirror images of each other. If you try to rotate them to match up (easiest to do with a model, either physical or virtual) you will find that they are non-superimposable and this is what gives them their isomeric properties. The two isomers rotate plane polarised light in opposite directions (this is where the name, optical isomerism, comes from).

The $\Delta$ and $\Lambda$ labels distinguish the two isomers by their absolute stereochemical configuration.

An decent explanation of this type of isomerism can be found here.


Partial charges in a covalent bond My professor stated the following statement:

  In a covalent bond, there is a presence of a partial charge in the atoms that combine to form a compound. For example in $\ce{H2O}$, since $\ce{O}$ is more electronegative, therefore there is a presence of a partial negative charge around $\ce{O}$. Similarly, as $\ce{H}$ is less electronegative, it has a presence of a partial positive charge around it. These partial charges are very feeble.

I found this statement to be little incomplete. So, I decided to find the reason of this partial charge. One of the video that talked about partial charges stated this:

  Since $\ce{O}$ is more electronegative than $\ce{H}$, therefore the shared electrons tend to spend more time around $\ce{O}$ than $\ce{H}$ and this is the reason of the partial charges.

Now, my questions are:

 If the shared electrons spend more time around $\ce{O}$ then it means that for some time $\ce{O}$ will be negatively charged and for some time $\ce{H}$. If this holds true, then why do we say that there is presence of  partial charges  around the atoms instead of saying that there is a presence of complete charge around the atoms but the charges keep on changing?

 What does my professor mean by 'feeble' charges? Is the charge reduced or something else?

If the question is unclear, ask me to edit it in comments.

We describe an object, which does not exist, but is very real, called and 'orbital'.  It is a region of space where an electron is 'likely' to be.  It is the result of charge field interactions between positively charged nuclei.  Normally we only discuss orbitals in the context of  1-2 Angstroms distance between atomic nuclei.  Some orbitals have a character which we call 'bonding'.  A bonding orbital, occupied by one or two electrons might be thought of as a 'bond'.  However, bonds also do not exist, even if they are very real.

It is somewhat nonsensical to describe where in an orbital the electron is.  If you must, the electron may be described as having a location of the entire orbital simultaneously, with vary weights to all of those locations, which are averaged.  This needs to be thought of in the concept of 'wavefunction', which is a mathematical function describing the distributions of these weights.  In the case of a polar covalent bond, the wavefunction more heavily favors locations near the more electron negative nucleus, and so the orbital 'concentrates' more of its charge near one end of the 'bond' and results in what we call a partial charge.

The reason we get squeamish about describing 'when' and 'where' is that the Heisenberg uncertainty principle renders the uncertainties energy and momentum of the system to be absurdly large when dealing with systems that are roughly the size of a chemical 'bond'.

At this point, nothing should be cleared up with regard to the original question.

The reality of the situation is that using when and where in this context is very likely to lead to predictions that will lead one astray when predicting physical phenomena.  As bizarre as quantum mechanics is, it has been shown to be highly predictive of the bizarre world of atoms and molecules, and so you have to try to figure out how to not ask questions from a classical  physics framework.


Reduction of epoxide Does $\ce{LiAlH4}$ and $\ce{NaBH4}$ reduce an isolated epoxide? I have two books where both are saying differently. One says it can and other says cannot. On searching the web there was no definite answer. Can they reduce epoxide to diol or something else ?
I am aware of the reduction of the other groups in the picture:

Sources:  NCERT: yes; JEE Advanced 2016 paper solution by Arihant: no; multiple coaching stuff: both yes/no

The correct question (taken from the 2016 JEE Paper is):

Lithium Aluminium Hydride reduces epoxides1, hence (A) would not give the required product. LAH also reduces ethers to primary alcohols, which is an easier way of eliminating the option (A). $\ce{BH3}$ in THF can reduce the acid, as explained in this question, hence the requisite product would not be obtained.
$\ce{NaBH4}$ cannot reduce esters, acids or epoxides in the absence of catalysts such as $\ce{PdCl2}$ or $\ce{AlCl3}$ and diglyme, hence (C) would be a correct option here.
Can you tell what Raney Ni would do here?
Raney Ni Reduces carboxylic acid and esters at elevated temperature (eg: $\pu{350 K}$); and no reaction with epoxide. So, it can be used for selective reduction of aldehyde at room temperature ($\pu{298 K}$).
Use this resource to help yourself out and as a reference for such reactions in the future.
References:

Murphy, David K., et al. “Reduction of Epoxides. III. Lithium Aluminum Hydride and Mixed Hydride Reduction of Some Secondary-Tertiary Epoxides.” Journal of the American Chemical Society, vol. 91, no. 10, May 1969, pp. 2649–53. doi:10.1021/ja01038a042.


What's the difference between demixing temperature and cloud point? I am learning something about polymer solution which has LCST phenomenon. But I met researchers used demixing temperature and cloud point to describe the transformation of this kind of solution from solutable into insolutable state. But for me, they seem the same. So I ask it here to learn what is the difference between these two concepts?

Cloud point typically refers to the LCST (or UCST) in solution. Demixing temperature usually refers to the LCST (or UCST) of a polymer mixture or copolymer in bulk.


Hydrogen cyanide - a Lewis acid? I am told that $\ce{HCN}$ is a Lewis acid.

From what I understand a Lewis acid possesses (an) empty orbital that is accessible to attacking nucleophiles.

Does $\ce{HCN}$ have any empty orbital(s)? So how can it behave as a Lewis acid? I know prototypical Lewis acids are for example boron trifluoride, and these have an empty p-orbital.

Low-lying, empty antibonding orbitals will suffice.  The reaction of carbon dioxide and water to form carbonic acid serves as a parallel example.  Water donates one of the lone pairs on oxygen towards bond formation, water is the Lewis base in this case.  Carbon dioxide has empty, low-lying $\ce{\pi}$* orbitals that serve to accept these electrons; carbon dioxide is a Lewis acid in this example.


Is it possible to measure pH without physically coming into contact with a solution? I've been researching ocean acidification, and the way that ocean pH is typically measured is by using sensors placed throughout all of the oceans. This is costly and time-consuming, and being able to measure ocean pH from space would be much more efficient as a global map could be created in a relatively short time and updated regularly.

However, when I researched pH measurement techniques, they all involve an instrument physically touching a solution, which would obviously not be possible from space. Ocean pH maps that have been created using satellite data measure pH indirectly by measuring ocean salinity, temperature, and other conditions, and they point out areas where it is "likely" that pH levels are low enough to harm ocean life. Is is possible to directly measure pH from a large distance? Would it be necessary to do so, that is to say, is measuring ocean pH levels indirectly accurate enough for environmental monitoring?

Interesting analytical chemistry problem. Currently, it is not possible to accurately determine pH in a remote fashion. Spectroscopy is the technique of choice for remote sensing, but the H$^+$ is not that spectroscopy friendly i.e., it is a quiet ion. It will respond to radiofrequency in a powerful magnetic field but the sensitivity is quite poor in an NMR experiment by current standards. You can Google "measurement of pH by NMR". In general what appears to be pure to an organic chemist by NMR experiment, analytical chemists call it a dirty sample because there a lot of minor impurities.

A quick literature search shows that that satellite data can be used to make an only an educated guess about the ocean pH. It not accurate but good enough for a ball-park estimate.

Remote sensing of ocean pH by R. Sabia, D. Fernández-Prieto, J. Shutler, C. Donlon, P. Land and N. Reul, "Remote sensing of surface ocean pH exploiting sea surface salinity satellite observations," 2015 IEEE International Geoscience and Remote Sensing Symposium (IGARSS), Milan, 2015, pp. 106-109.

  Within this context, the purpose of the ESA “Pathfinders-OA” project is to quantitatively and routinely estimate surface ocean pH by means of satellite observations in several ocean regions. Satellite Ocean Colour, Sea Surface Temperature and Sea Surface Salinity data (with an emphasis on the latter) will be exploited. A proper merging of these different datasets will allow to compute at least two independent proxies among the seawater carbonate system parameters and therefore obtain the best educated guess of the surface ocean pH.


What is the IUPAC nomenclature for the cyclic mono-alkene ions formed after proton or hydride abstraction? In this blog post, Steven Bachrach discusses the (anti-)aromaticity or lack thereof in the homologous series of anionic and cationic cyclic mono-alkenes formed respectively by deprotonation and hydride abstraction at a vicinal carbon:

$~~~\small\unicode{x201c}$$\small\textit{KassCPcmpds.png}$$\small\unicode{x201d}\textit{ by }$$\small\textit{Steven Bachrach}$$\small\textit{ is licensed under }$$\small\textit{CC BY ND 3.0}$$\small .$

Bachrach uses the construction -enyl anion/cation to name these species, which seems sensible to me.  However, I also vaguely recall seeing generalized use of the -ide and -ium suffixes for singly charged anions and cations. For example, I know that $\ce{NO+}$ is nitrosonium, but I seem to be recalling incorrectly that $\ce{NO-}$ is nitroxide. $($The NIST Webbook page only calls $\ce{NO-}$ the nitric oxide anion, and the Wikipedia article on nitroxide-mediated radical polymerization is discussing radical alkyl N-oxides $(\ce{R2NO^\bullet})$, not $\ce{NO-}$.$)$ On the other hand, $\ce{R3CO-}$ ions are unequivocally alkoxides.

So: What are the systematic IUPAC names for these compounds?

Anions

In general nomenclature, anions may be named by using functional class nomenclature; however, the resulting names are not the preferred IUPAC names. The corresponding subsection of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-72.2.1 Functional class nomenclature

  Functional class nomenclature can be used, in general nomenclature, to describe anionic compounds. An anion that can be considered as derived formally by adding an electron to a radical may also be named by adding the class name ‘anion’ as a separate word to the name of the substituent group. The names are formed by using the names of corresponding radicals (not necessarily the name of substituent groups) and the class name ‘anion’ as a separate word. The multiplying prefixes ‘di’, ‘tri’, etc., are added to the class name to denote multiple anions. This type of nomenclature is limited to anions having anionic centers in the same structure. Systematic names (see P-72.2.2) are preferred IUPAC names.

The functional class names for the anions that are given in the question are:

cycloprop-2-en-1-yl anion
cyclobut-2-en-1-yl anion
cyclopent-2-en-1-yl anion
cyclohex-2-en-1-yl anion

The preferred systematic names for anions that can be derived by deprotonation are formed according to P-72.2.2.1:

  P-72.2.2.1 Anions derived from parent hydrides and their derivatives

  An anion derived formally by the removal of one or more hydrons from any position of a neutral parent hydride is preferably named by using the suffix ‘-ide’, with elision of the final letter ‘e’ of the parent hydride, if any. Numerical prefixes ‘di’, ‘tri’, etc. are used to denote multiplicity; locants identify positions of the negative charges.

Therefore, the preferred IUPAC names for the anions that are given in the question are:

cycloprop-2-en-1-ide
cyclobut-2-en-1-ide
cyclopent-2-en-1-ide
cyclohex-2-en-1-ide

Cations

In general nomenclature, also cations may be named by using functional class nomenclature; but again, the resulting names are not the preferred IUPAC names.

  P-73.2.1 Functional class names

  Cationic compounds that can be considered as being derived formally by removal of electrons from the corresponding radical may be named by adding the class name ‘cation’ as a separate word after the name of the radical. Polycations are indicated by adding the multiplying prefixes ‘di’, ‘tri’, etc., as appropriate, to the class name. Systematic names formed by using the suffix ‘ylium’ are preferred IUPAC names (see P-73.2.2). When the charge is not localized, the structure is enclosed in square brackets.

The functional class names for the cations that are given in the question are:

cycloprop-2-en-1-yl cation
cyclobut-2-en-1-yl cation
cyclopent-2-en-1-yl cation
cyclohex-2-en-1-yl cation

The preferred systematic names for cations that can be derived by removal of a hydride ion are formed according to P-73.2.2.1.2:

  P-73.2.2.1.2 General method

  According to the general method, cations formally derived by the removal of one hydride ion, $\ce{H-}$, from any position of a parent hydride are named by adding the suffix ‘-ylium’ to the name of the parent hydride, with elision of the final 'e' in the name of the parent hydride, if present. Di- and polycations formally derived by the removal of two or more hydride ions from the parent hydride are named by using the suffix ‘ylium’ and the multiplying prefixes ‘bis’, ‘tris’, etc. (…)

Therefore, the preferred IUPAC names for the cations that are given in the question are:

cycloprop-2-en-1-ylium
cyclobut-2-en-1-ylium
cyclopent-2-en-1-ylium
cyclohex-2-en-1-ylium


confusion about pH of a substance Let's take an acid, $\ce{HCl}$;

Now what do you mean by the $p\ce{H}$ of $\ce{HCl}$? Does $\ce{HCl}$
 have free floating ions in it like water does?

Also, why can't the $p\ce{H}$ value of a substance be less than $0$? Like can't a substance have $10^1~\rm M$ of $\ce{H+}$ ions? In that case won't the $p\ce{H}$ be $-1$?

And when we talk about $p\ce{H}$ values, then are we talking about Arrhenius acids and bases or Brønsted–Lowry  acids and bases?

Dissociation of ions

Water, as you probably know, is highly polar and has a very high dielectric constant. When ionic compounds are added to the water, the electrostatic force of attraction between the ions get very weak and hence will result in dissociation of the compound.

The percentage of dissociation of a compound depends on various factors such as solvent, strength of ionic bond, etc.

$\ce{HCl}$ is a polar molecule and hence it dissociates into its constituent ions, namely, $\ce{H+}$ and $\ce{Cl-}$ ions.

Your initial question "what is the pH of HCl" is not quite right. More about it in the next section.

What is pH?$\space$ Can pH be negative?

pH by definition is a measure of hydrogen ion concentration.

$\mathrm{pH} = -\log[\ce{H+}]$ where $[\ce{H+}]$ denotes the active hydrogen ion concentration.

Hence, $\mathrm{pH}$ of $\ce{HCl}$ in an arbitary context won't make any sense since the $\mathrm{pH}$ of $\ce{HCl}$ would depend on its concentration.

The $\mathrm{pH}$ of a substance can be negative. $10\ \mathrm M$ solution of $\ce{HCl}$ would have a $\mathrm{pH}$ of $-1$.

The stronger the acid, the smaller is the value of $\mathrm{pH}$. $\ce{HCl}$ is a strong acid because it ungergoes almost 100 % dissociation.

Theory of acids and bases and their relation with pH

The theory of acids which you are refering to were developed to classify acids and bases rather not to measure the strength of the acids and bases.

However, one can arrive at the strength of the acid/base according to the theory by logical reasoning. For example, if someone were to ask you which one of the two compounds is an Lewis acid, you would investigate both the molecules and find out which one has a higher tendency to accept electrons to find the stronger acid.

Therefore, the theory of acids proposed by Arrhenius and Brønsted–Lowry are in no way related to $\mathrm{pH}$.

pH on the other hand, is a measure of strength of an acid and is indepdent of the above theories.

It is quite easy to compare $\mathrm{pH}$ with Arrhenius/Brønsted–Lowry theory of acids since according to the theory, the molecule which is willing to donate free $\ce{H+}$ ions will be considered an acid. A claim that higher the concentration of $\ce{H+}$ will be a stronger acid is reasonable, understandable and acceptable.

Bonus:
In case of bases, we have a strength measurement scale similar to the pH. We make use of hydroxide ion concentration in place of hydrogen ion concentration to find the strength of the base. We call it $\mathrm{pOH}$.

These two are related by the following formula when the acid/base are dissolved in water at room temperature.

$14 = \mathrm{pH} + \mathrm{pOH}$


What is the standard way of reporting/interpreting measurement uncertainty in scientific publications? As an undergraduate student I was often told in lab classes to report every single measured or calculated quantity with its associated uncertainty in lab classes, in the form $(10.1 \pm 0.1 )~\mathrm{g}$ for instance, where 0.1 is the standard deviation or estimated uncertainty. I'm starting to find, however, that this practice is seldom followed in scientific publications.

When reporting uncertainty in a scientific publication, is it good practice to report every measurement in the form $(300.1 \pm 0.1 )~\mathrm{K}$ ? I sometimes see this done in the body of scientific papers, but usually never in tables/lists of data. This worries me because, in my understanding, all measurements have an uncertainty associated with them, so I find these data hard to interpret unambiguously.

Is it usually assumed that numbers written without an explicit uncertainty are accurate to the last expressed digit? (i.e. that $10.5~\mathrm{g}$ implies $(10.5 \pm 0.1)~\mathrm{g}$ ?) sometimes this seems to be the case, but doesn't this negate the difference between $ \pm 0.1$ and $\pm 0.5$ for instance? Is this difference usually considered unimportant?

As a side question: Regardless of the usual practice in academic writing, do you consider that demanding every single number to be reported with an associated uncertainty is a good idea if teaching a lab class?

In my experience, this varies greatly with your field of study.

In organic chemistry, I have yet to experience proper uncertainty reporting in journal articles or lab reports with the notable exception of physical organic chemistry. Once in my undergraduate years, I noted while writing a report that $130~\mathrm{mmol}$ where actually $1.3 \times 10^2~\mathrm{mmol}$ and wrote it that way. Okay, I should have written $0.13~\mathrm{mol}$ in a more correct notation to signify the significant digits. But the way the TA marked the report clearly told me ‘I don’t give a damn about correct error margins or significant digits, just God damn write $130~\mathrm{mmol}$!’ Lesson learnt from that anecdote: synthetic organic chemistry does not care about minor details such as significant figures or uncertainty.

The closer you get to physical chemistry, the more your measurements actually represent data values to be cited in a table and thus the more exact the values determined become, the more people care about reporting error margins and uncertainty. And funnily enough, in biology this is also done much more regularly. I suspect that there are different underlying reasons: One side (physical chemistry/physics) does not want the negative outcome of ‘you pretended your results are more accurate than they are!’, while the other side, biology, knows that it is intrinsically dealing with very error-prone systems that require large repetition counts which can then be analysed statistically.

Although synthetic organic chemistry practically also deals with rather error prone systems, either the idea is that there are too many uncontrollable variables or the idea is $n=1$ (always!); in any case, error bars or uncertainty are deemed superfluous.


Correction formula for alcoholmeter I need to find the formula, if there is one, which would calculate the correction of alcoholmeter.

Basically convert observed %ABV (Alcohol by volume) to actual %ABV using the sample's temperature.

I know the correction tables, but I am looking for the formula or do I have to use tables?.

I have had the same question recently. And there is no simple answer. Tables (like ones from International critical tables of numerical data, physics, chemistry and technology or this) are still the 1st hand source. Formulas used are just approximating this data. Check formula here on page 8 (to get the temperature-corrected ABV you have to invert it by solving non-linear function). Alternatively check this recent study. To convert between ABV (alcohol by volume, %) and density (specific gravity SG, g/cm3) you may use this approximation:

$$\mathrm{SG} = -0.002\cdot\mathrm{ABV} + 1.0012$$

PS. There is an unconfirmed formula at vinolab:

  The hydrometer temperature correction for SG is performed with this expression:

  $$\text{corrected-reading} = r \times \frac{(1.00130346 - (0.000134722124 \cdot t) + (0.00000204052596 \cdot t^2)\ - (0.00000000232820948 \cdot t^3))} {(1.00130346 - (0.000134722124 \cdot c) + (0.00000204052596 \cdot c^2)\ - (0.00000000232820948 \cdot c^3))}$$

  where: $r$ = reading and $c$ = calibration temperature. This expression is based on °F, so the temperatures are first converted.

With some sample values it gave me results close to the table' ones (I have converted ABV to SG, amended as per temperature and formula above and finally converted back to ABV).


Why lithium gives flame coloration? Theoretically speaking,  beryllium and magnesium does not give flame test because their atoms are comparatively smaller and the valance electrons are strongly attached to the nucleus. Therefore their ionisation energies are high, so they need large amount of energy for the excitation of their valence electrons to higher energy level which is usually not available in the Bunsen flame. So they do not impart any colour to the flame.

Now, comparing the atomic radius of lithium and magnesium, both have nearly same atomic radius due to diagonal relationship.($\ce{152}$ pm and $\ce{160}$ pm respectively). Since both of them nearly have same atomic radius, why is lithium imparting flame coloration(crimson red) while magnesium is not? Both of the metals has to be supplied equal amount of energy to excite the electron which the burner is not able to supply and thus flame test should fail in this case but this is not happening.

I don't understand despite the reasoning for beryllium not giving flame coloration is quite correct, wikipedia claims beryllium  to give flame coloration(white).

I would love to find numbers, but alas, I have failed. If anyone can point me towards energy differences between the orbitals I am lacking, please feel free!

There is no a priori physical reason why lithium, sodium, potassium and calcium give flame colours but beryllium and magnesium would not. For all these elements (and hydrogen), the principal mechanism works in the same way: An electron is thermally excited into a higher orbital, and when it relaxes back to its ground state it releases a photon of exactly that wavelength. The flame colour is always due to gaseous atoms (not ions), so the electrons are always excited from an s-orbital.

The strongest lowest-energy excitation and relaxation should always be $n\mathrm{s} \rightarrow n\mathrm{p}$, i.e. from one shell’s s-subshell to that shell’s p-subshell. For sodium, this energy difference corresponds to $589~\mathrm{nm}$ or $2.10~\mathrm{eV}$, for lithium it is $671~\mathrm{nm}$ or $1.85~\mathrm{eV}$, for potassium $767~\mathrm{nm}$ and $1.61~\mathrm{eV}$ and for calcium $657~\mathrm{nm}$ and $1.89~\mathrm{eV}$.[1]

We can see that the energy difference rises from potassium to calcium, even though they are neighbours and the excitation is from the same orbital to the same different one. This is because calcium’s higher nuclear charge contracts the orbitals, stabilises them and thus lowers their energy — compare with the ionisation energy which also typically rises across a period. This effect is, of course, more pronounced when going across the second or third period when compared to the fourth; thus, I would expect an even greater energy difference when going from sodium to magnesium and even greater than that from lithium to beryllium.

The best explanation I have for the colourlessness of magnesium and beryllium is that their excitations are too close to ultraviolet to be properly observed due to strong contractions of the orbitals, but without numbers that is a weak explanation at best.

Notes:

[1]: I would be grateful for confirmation that these values I found are correct.


Volatile and non-volatile solutes in solution What is the difference between having a volatile versus a non-volatile solute at the boiling point of the solution?
I have been getting mixed answers online concerning the differences outlined above. This is what I have gathered so far:

A non-volatile solute does not produce vapour at the boiling point of the solution.
A volatile solute does produce vapour at the boiling point of the solution.

Can someone confirm these definitions?

A substance with higher vapour pressure vaporizes more readily than a substance with a lower vapour pressure. In case of a volatile solute this has a high vapour pressure and hence produces vapour. While in the case of a non-volatile solute due to its lower vapour pressure it does not produce vapour. The vapor pressure of a substance is the pressure at which its gas phase is in equilibrium with its condensed phases (liquid or solid).


Stereochemistry of Gabriel Synthesis I read that the primary amine product of Gabriel Synthesis is a racemic mixture. However, when analyzing the reaction the stereochemical step(s) are all $S_N2$. Thus, I figured if you are careful you can track the stereochemistry of the (or both) $S_N2$ reaction(s). Can anyone clarify this for me?

The lower part of the Dutch wikipedia about the Gabrielsynthese indeed displays an example where this reaction is carried out to yield a secondary amine:

The original publication (DOI: 10.1021/jm9804376, J. Med. Chem., 1999, 42, 593-600) equally includes similar conversions with racemic secondary substrates like (R/S)-($\pm$)-1-methylpentyl bromide to yield (R/S)-($\pm$)-N-(1-methylpentyl)phthalimide which in turn was then converted into (R/S)-($\pm$)-(1-methylpentyl)amine hydrochloride.


Sign of work done by gas during isothermal expansion $\pu{10 mol}$ of an ideal gas confined to a volume of $\pu{10 L}$ is released into atmosphere at $\pu{300 K}$ where the pressure is $\pu{1 bar}.$ The work done by the gas is:

Work done can be found as
$$W = -p_\mathrm{ext}(V_\mathrm{f} - V_\mathrm{i}).\tag{1}$$
$$V_\mathrm{f} = \frac{nRT}{p} = \frac{\pu{10 mol}\times\pu{0.083 L bar K^-1 mol^-1}\times\pu{300 K}}{\pu{1 bar}} = \pu{249 L}\tag{2},$$
$$W = -\pu{1 bar}\cdot(\pu{249 L} - \pu{10 L}) = \pu{-239 bar L}\tag{3}$$
As the question is asking us to find the work done by the gas, so shouldn't it be $\pu{+239 bar L},$ as work done by the gas is negative of the word done by external pressure?
But someone told me that the work done calculated is the work done by the gas and $\pu{-239 bar L}$ should be the answer. So, what mistake am I making and how do I determine if work done by the gas (and external pressure) is positive or negative under different conditions (expansion and compression)?
P.S.$\pu{-239 bar L}$ is the answer according to the answer key.

If we imagine our proverbial, frictionless piston containing some ideal gas, when $P_{surr} > P_{syst}$, the system is compressed by the piston and work is done on the system by the surroundings (i.e. work is positive). If the reverse is true ($P_{syst} > P_{surr}$), the gas in the piston will expand and work is done on the surroundings by the system (work is negative)
If you think of it in terms of internal energy, it might help. When work is done on the system, the internal energy increases for the ideal gas in the system (work is positive, internal energy increases). When the system does work on the surroundings (expansion), its ability to do work decreases (internal energy decreases, work is negative).
Hope this helps.


Finding change in activation energy from change in rate constant in catalysis I need to find by how much the activation energy decreases when the rate constant increases by a '$x$' factor when a catalyst is used in the reaction.
I'm just super confused about how to solve this. Using Arrhenius relation I have
$$\frac{k_2}{k_1} = \frac{A\mathrm e^{-E_\mathrm{a2}/(RT)}}{A\mathrm e^{-E_\mathrm{a1}/(RT)}}$$
$$x = \mathrm e^{(E_\mathrm{a1} - E_\mathrm{a2})/(RT)}$$
I took natural log on both sides to get rid of the exponent and then ended up finally with an equation I can't seem to get any relation from
$$RT\ln x = E_\mathrm{a1} - E_\mathrm{a2}$$
Now I need a ratio of the two activation energies, but I don't know how I get that from here as I have just a addition/subtraction relation between the two. Would be grateful if someone has a clue about how I could proceed.
The problem is I have come across a question in Atkins' Physical Chemistry [1, p. 841] and the answer is provided in percent, but I don't see how you could arrive at a solution like that with where I am:

Self-test 20D.4 Consider the decomposition of hydrogen peroxide, which can be catalysed in solution by iodide ion. By how much is the activation energy of the reaction reduced if the rate constant of reaction increases by a factor of 2000 at 298 K upon addition of the catalyst?
Answer: 25 per cent

References

Atkins, P. W.; De Paula, J. Physical Chemistry: Thermodynamics, Structure, and Change, 10th ed.; W.H. Freeman: New York, 2014. ISBN 978-1-4641-2452-5.

It appears that you are correct in your thought process.
Determining by how much the activation energy of the reaction is reduced requires you to basically find

$$Δ(\%) = \frac{E_\mathrm{a1} - E_\mathrm{a2}}{E_\mathrm{a1}} × 100\% = \left(1 - \frac{E_\mathrm{a2}}{E_\mathrm{a1}}\right) × 100\%$$

There is a system of three independent equations

$$
\begin{cases}
k_1 = A\mathrm e^{E_\mathrm{a1}/(RT)}\\
k_2 = A\mathrm e^{E_\mathrm{a2}/(RT)}\\
\frac{k_2}{k_1} = x
\end{cases}
$$

with five unknowns $k_1$, $k_2$, $E_\mathrm{a1}$, $E_\mathrm{a2}$, $A$, which requires either one more equation, or to know one more parameter of the system in order to find $E_\mathrm{a2}/E_\mathrm{a1}.$
The textbook provides you with activation energy in the previous illustration right next to your problem [1, p. 841], so I assume you are supposed to use $E_\mathrm{a1} = \pu{76 kJ mol−1}:$

At this point the problem can indeed be solved to a certain numerical value:

$$E_\mathrm{a1} - E_\mathrm{a2} = RT\ln x = \pu{8.3145 J mol-1 K-1} × \pu{298 K} × \ln{2000} \approx \pu{18.8 kJ mol-1}$$

$$Δ(\%) = \frac{\pu{76 kJ mol-1} - \pu{18.8 kJ mol-1}}{\pu{76 kJ mol-1}} × 100\% \approx 25\%$$

The question whether it's a textbook flaw, or authors assumed the reader would read the material more carefully still remains open.

References

Atkins, P. W.; De Paula, J. Physical Chemistry: Thermodynamics, Structure, and Change, 10th ed.; W.H. Freeman: New York, 2014. ISBN 978-1-4641-2452-5.


How can I balance this reaction using the algebraic method? How can I balance this reaction using the algebraic method?

$$\ce{K4Fe(CN)6 + KMnO4 + H2SO4 + H2SO4 -> KHSO4 + Fe2(SO4)3 + MnSO4 + HNO3 + CO2 + H2O}$$

My solution:

$$\ce{a K4Fe(CN)6 + b KMnO4 + c H2SO4 -> d KHSO4 + e Fe2(SO4)3 + f MnSO4 + g HNO3 + h CO2 + i H2O}$$

$\ce{K}{:~} 4a + b = d$
$\ce{Fe}{:~} a = 2e$
$\ce{C}{:~} 6a = h$
$\ce{N}{:~} 6a = g$
$\ce{Mn}{:~} b = f$
$\ce{O}{:~} 4b + 4c = 4d + 12e + 4f + 3g + 2h + i$
$\ce{H}{:~} 2c = d + g + 2i$
$\ce{S}{:~} c = d + 3e + f$

I couldn't solve further than that.

Now all you have to do is variable substitution, which is kinda tedious since it's such a large equation. Remember that solutions are defined up to multiplying everything by a constant factor, so we might as well pick

$a = 2$ (this guess is inspired by the equation for $\ce{Fe}$)
$\ce{Fe}{:~} e = 1$
$\ce{C}{:~} h = 12$
$\ce{N}{:~} g = 12$
Because $\ce{Mn}{:~} b = f$, we can rewrite $\ce{O}$ as $4c = 4d + 12e + 3g + 2h + i$
Subtract 4 times $\ce{S}$, and we get $0 = -4f + 3g + 2h + i$, so $4f = 60 + i$ (call this $\ce{O}'$)
Now rewrite everything in terms of $b$:

$\ce{K}{:~} 8 + b = d$, so $d = b + 8$
$\ce{Mn}{:~} f = b$
$\ce{O}'{:~} 4b = 60 + i$, so $i = 4b - 60$
$\ce{S}{:~} c = (b + 8) + 3 + b = 2b + 11$
$\ce{H}{:~} 2(2b + 11) = b + 8 + 12 + 2(4b - 60)$, so $4b + 22 = 9b - 100$, so $5b = 122$ and $b = 122/5$. This is a bit unfortunate since later on we have to multiply all variables by 5.

We now have $a=2$, $b=122/5$, $c=299/5$, $d=162/5$, $e=1$, $f=122/5$, $g=12$, $h=12$ and $i=188/5$.
By multiplying everything by 5 we get the final solution:

$a = 10$
$b = 122$
$c = 299$
$d = 162$
$e = 5$
$f = 122$
$g = 60$
$h = 60$
$i = 188$

You can check that these figures will make the equation balanced.


What does lowercase r-s notation mean? I came across a naming convention which I haven't seen before. I let ChemDraw name the following compound for me and got a name containing lowercase "r" and "s" configurations.

Can someone tell me what this means and something about the actual convention behind it? I think it must have something to do with rings and maybe with some kind of "pseudo-chirality" due to a certain conformation of the ethyl group which is attached to the ring, but I'm only guessing.

The name (1s,4s)-1-ethyl-4-methylcyclohexane is correct; it is even the preferred IUPAC name (PIN). It refers to the cis isomer of 1-ethyl-4-methylcyclohexane.

The preferred IUPAC name for the corresponding trans isomer is (1r,4r)-1-ethyl-4-methylcyclohexane.

In general nomenclature, according to Subsection P-93.5.1.2 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the stereodescriptors 'cis' and 'trans' may be used to denote the relative configuration of substituted monocyclic compounds.

  P-93.5.1.2 Relative configuration: the stereodescriptors 'cis' and 'trans'

  The stereodescriptors 'cis' and 'trans' are used to show the relationship between two ligands (atoms or groups) attached to separate atoms that are contained in a ring or a ring system. The two ligands are said to be located 'cis' to each other if they lie on the same side of a plane. If they are on opposite sides, their relative position is described as 'trans'. The appropriate reference plane of a ring or ring system (the ring being in a conformation, real or assumed, without reentrant angles at the two substituted atoms) is the mean plane of the ring(s). The stereodescriptors denote relative configuration; the absolute configuration must be described by CIP stereodescriptors such as 'R' and 'S'.

  (…)

In preferred IUPAC names, all stereogenic units shall be specified (unless an omission is explicitly allowed). The presence of stereogenic units does not require the molecule to be chiral.

According to Subsection P-93.5.1.1.1, the absolute configuration in substituted monocyclic compounds shall be described by Cahn-Ingold-Prelog (CIP) stereodescriptors.

  P-93.5.1.1.1 Absolute configuration

  In substituted monocyclic compounds the absolute configuration is specified in preferred IUPAC names by the CIP stereodescriptors such as 'R', 'S', 'r', and 's'.

This rule explicitly applies to achiral cyclic compounds, too.

  P-93.5.1.1.2 Achiral cyclic compounds

  The configuration of achiral cyclic molecules is also specified by CIP stereodescriptors in preferred IUPAC names.

  (…)

Therefore, this rule also applies to (1s,4s)-1-ethyl-4-methylcyclohexane, which is given in the question, and its isomer (1r,4r)-1-ethyl-4-methylcyclohexane. (The examples mentioned in the Blue Book include the similar compounds (1s,4s)-cyclohexane-1,4-diol, (1r,4r)-cyclohexane-1,4-diol, and (1r,4r)-4-chloro-4-methylcyclohexan-1-ol.)

The lower-case stereodescriptors 'r' and 's' are CIP stereodescriptors like 'R' and 'S'. The stereodescriptor 'r' or 's' is chosen by applying the CIP sequence rules as for the stereodescriptors 'R' and 'S'. However, the stereodescriptors 'r' and 's' are used to indicate the absolute configuration of pseudoasymmetric centers, whereas 'R' and 'S' are used to indicate the absolute configuration of real chirality centers.

  P-91.2.1.1 Cahn-Ingold-Prelog (CIP) stereodescriptors

  Some stereodescriptors described in the Cahn-Ingold-Prelog (CIP) priority system, called 'CIP stereodescriptors', are recommended to specify the configuration of organic compounds, (…), and in the nomenclature of natural products (…). The following stereodescriptors are used as preferred stereodescriptors (…):

  (a) 'R' and 'S', to designate the absolute configuration of tetracoordinate (quadriligant) chirality centers;

  (b) 'r' and 's', to designate the absolute configuration of pseudoasymmetric centers;

  (…)

The 'pseudoasymmetric' is defined in Subsection P-92.1.1 (d) as follows.

  Stereogenic units are called pseudoasymmetric (center, axis or plane) when they have distinguishable ligands 'a', 'b', 'c', 'd', two and only two of which are nonsuperposable mirror images of each other (enantiomorphic). (…). The 'r/s' (…) stereodescriptors describing a pseudoasymmetric stereogenic unit are invariant on reflection in a mirror (for example 'r' remains 'r', and 's' remains 's'), but are reversed by the exchange of any two ligands ('r' becomes 's', and 's' becomes 'r'). Lower case stereodescriptors are used to describe pseudoasymmetric stereogenic units.

For example, (2R,3r,4S)-2,3,4-trichloropentane contains two chirality centers (labeled 'R' and 'S', respectively) and one pseudoasymmetric center (labeled 'r').

The two 1-chloroethyl ligands of the pseudoasymmetric center are nonsuperposable mirror images of each other. On reflection of the entire molecule in a mirror, the configurations of the chirality centers are reversed ('R' becomes 'S', and 'S' becomes 'R'), but the configuration of the pseudoasymmetric center is invariant ('r' remains 'r').

The same principle applies to (1s,4s)-1-ethyl-4-methylcyclohexane, which is given in the question.

The two halves of the 1,4-disubstituted cyclohexane are nonsuperposable mirror images of each other. Therefore, the compound contains two pseudoasymmetric centers, which are invariant on reflection in a mirror ('1s,4s' remains '1s,4s').

Although this case is explicitly mentioned in the Blue Book, it may not be obvious how to choose the stereodescriptors 'r' or 's' for 1,4-disubstituted cyclohexanes. The ranking is a complex process and requires the use of auxiliary descriptors.

Generally, the stereodescriptors 'r' and 's' are assigned by applying the Cahn-Ingold-Prelog (CIP) Priority System as for the stereodescriptors 'R' and 'S'. The order of precedence is reached by the construction of a hierarchical digraph and the application of the Sequence Rules.

The constitutional formula of a cyclic molecule is transformed into an acyclic digraph according to the methodology that is described in Subsection P-92.1.4.3 and which has been recommended by Prelog and Helmchen.

In order to obtain the acyclic digraph of a stereogenic unit, the didentate ligand (i.e. the ring) is transformed into two monodentate ligands by leaving intact in each case one bond with the core of the stereogenic unit and cleaving the other bond. At the end of each of the two branches thus obtained, a duplicate atom of the core is attached. (Note that this methodology is similar to the case of multiple bonds.) For example, for the pseudoasymmetric stereogenic unit at the atom numbered '1' of (1s,4s)-1-ethyl-4-methylcyclohexane:

where $\ce{(C)}$ is a duplicate representation of the core of the pseudoasymmetric stereogenic unit.

As the digraph is acyclic, it is now possible to specify the stereodescriptor 'R' or 'S' for the stereogenic unit at the atom numbered '4', which is carrying the methyl group, in each branch. These are auxiliary descriptors that are different in each branch.

Since the auxiliary descriptors are different, they allow the branches to be ranked using Sequence Rule 5, which states that 'R' has priority over 'S'.

  P-92.1.3.5 Sequence Rule 5

  An atom or group with descriptor 'R', 'M', and 'seqCis' has priority over its enantiomorph 'S', 'P' or 'seq Trans'.

Therefore, the configuration of the pseudoasymmetric stereogenic unit at the atom numbered '1' of (1s,4s)-1-ethyl-4-methylcyclohexane can be determined to be 's'.

In the same way, the configuration of the pseudoasymmetric stereogenic unit at the atom numbered '4' of (1s,4s)-1-ethyl-4-methylcyclohexane can be determined as 's'.

Therefore, the complete configuration of (1s,4s)-1-ethyl-4-methylcyclohexane, which is given in the question, is indeed (1s,4s).


What are these "certain chemical compounds" denser than osmium? The Wikipedia entry for density states

  Osmium and iridium are the densest known elements at standard
  conditions for temperature and pressure but certain chemical compounds
  may be denser.

I have not been able to turn up what these chemical compounds are.  I can imagine a salt or some molecule in which dense atoms (e.g. of mercury) are packed in a conformation such that these atoms are held closer together than in the pure metal, so giving the molecule a higher density than the pure metal or of higher molecular weight metals like osmium.

I know there are elements with higher molecular weight than osmium and I have read about superdense exotic states of matter.  I am curious about this chemical (presumably comprised of atoms with lower MW that osmium) which is more dense than osmium - presumably at earthly temperatures and pressures.  Or an osmium-containing molecule more dense than metallic osmium would qualify.

If someone can answer this question he or she might also consider editing the Wikipedia article to clarify.

This text was added as part of a refactoring/cleanup of the page in this edit in March 2013, where it replaced the text "Osmium and iridium are the densest known elements at standard conditions for temperature and pressure but not the densest materials." This older text was tagged with the note "Which are denser, at standard conditions?" It in turn was added in this edit in May 2011 by an anonymous editor (the IP address used is apparently registered to the Wyoming Community College Commission). The original wording, "Osmium is [...] not the densest material", is ambiguous, but could be read as referring to exotic forms of matter such as neutronium, in spite of the earlier "standard temperature and pressure" caveat.

In short, this likely derives from a well-meaning but misguided editor attempting to note the existence of much higher-density, non-chemical substances in an inappropriate place.


Calculating the value of equilibrium constant in terms of pressure A mixture of $\pu{1 g}$ of $\ce{H2(g)}$ and $\pu{1.06 g}$ $\ce{H2S(g)}$ in a $\pu{0.5 L}$ flask come to equilibrium at $\pu{1670 K}.$ At equilibrium, there are $\pu{8E-6 mol}$ of $\ce{S2(g)}$ present. Determine $K_p.$

I set up an ICE table:
$$
\begin{array}{ccc}
\ce{& &2 H2 &+ &S2 &<=> &2 H2S} \\
\text{Initial} & &\pu{1 M} && 0 && \pu{0.062 M} \\
\text{Change} & &2x && x && 2x \\
\text{Equilibrium / Final} & &?  && \pu{1.6E-5 M} && ?
\end{array}
$$
From the $\ce{S2}$ column we see that $x = \pu{1.6E-5 M}$, so I calculated the equilibrium constant as follows:
$$K_p = \frac{(0.062 + 2\times\pu{1.6E-5})^2}{(\pu{1.6E-5})(1-2\times\pu{1.6E-5})^2} = 240$$
What is wrong with my solution?

The problem lies in that our information of the system is in terms of concentration, so you actually calculated $K_c$.
$$\ce{2H2 +S2<=>2H2S}$$
$$\begin{array}{|l|l|l|}
\ce{H2 & S2 & H2S} \\ \hline
0.992\ \mathrm{M}& 0\ \mathrm{M}& 0.062\ \mathrm{M} \\
2(1.6\times 10^{-5}) & (1.6\times 10^{-5}) & -2(1.6\times 10^{-5}) \\
0.992+2(1.6\times 10^{-5})  &  1.6\times10^{-5}\ \mathrm{M}   &  0.062-2(1.6\times 10^{-5})\end{array}$$
$$K_c=\frac{(0.062\ \mathrm{M}-3.2\times 10^{-5}\ \mathrm{M})^2}{(1.6\times10^{-5}\ \mathrm{M})\cdot(0.992\ \mathrm{M}+3.2\times 10^{-5}\ \mathrm{M})^2}=245\ \mathrm{M^{-1}}$$
which is numerically the same as the value you obtained, except that I used a more precise molecular mass for $\ce{H2}$.
Using the relation $$K_p=K_c(RT)^{\Delta n},$$
where $$\Delta n=(\sum \text{coefficient of gaseous products})-(\sum\text{coefficient of gaseous reactants}),$$ we can calculate $K_p$:
$$K_p=(245\ \mathrm{M^{-1}})\cdot(8.3144598\ \mathrm{J\ K^{-1}\ mol^{-1}}\cdot 1670\ \mathrm{K})^{-1}=0.0176\ \mathrm{kPa^{-1}}$$


Decarboxylation of THCA during CO2 Extraction of Cannabis We have a strange problem. We are performing CO2 extraction on cannabis and we are seeing considerable decarboxylation of THCA to THC.

We have high quality temperature sensors at the heat exchangers, the extractor vessels, and the separator. We have not observed any temperature spikes that would explain it.

Various parameters have been tried, i.e.

40 - 50C (104-122F) / 115 - 280 bar (1770 - 4050 psi)

We can't find any correlation between the temperatures / pressures and the level of decarboxylation.

The system has been run for 24 hours at 120bar/50C and we are getting approximately equal amounts of THC and THCA out.

The starting materials has been tested to being only slightly decarboxylated. Disturbingly, the spent material is also being decarboxylated.

The same equipment is used to test both the starting and spent material. The results on the starting material have been verified by a 3rd party lab. It seems unlikely to be a testing problem.

We have multiple extractors and we are seeing the same behaviour so it is unlikely to be a sensor issue. The temperature controllers also have their own thermostat that agrees with the extractor sensor. The extractors are all temperature controlled and all temperatures are logged every 30 seconds.

There is a research paper by Rovetto et al. that also observes significant decarboxylation but not to the extent that we are observing it:

https://www.researchgate.net/publication/315417305_Supercritical_carbon_dioxide_extraction_of_cannabinoids_from_Cannabis_sativa_L_plant_material

Is it possible that the normal decarboxylation threshold (e.g. 115C) is lowered when at high pressure?

I understand that THC extracts far more readily than THCA, however looking at the mass balance it seems that selective extraction can not explain it.

If anyone has any insights to share, I would very much appreciate it.

Thanks

If this problem has not already been solved by the author, then here: Supercritical extraction, by nature, breaks the carbon bond in the dissolution phase, which is why you are even seeing your biomass decarbed. It's not a heat thing, it's a function of SCF CO2.


Why does chloromethane have a larger dipole moment than chloroform? Why does $\ce{CH3Cl}$, methyl chloride, have a larger dipole moment than $\ce{CHCl3}$, chloroform?

Let us consider $\ce{C-Cl}$ bond moment to be $x$ and that of $\ce{C-H}$ bond to be $y$.

In $\ce{CCl4}$, the dipole moment of any three $\ce{C-Cl}$ atoms is balanced by the dipole moment of the fourth $\ce{C-Cl}$ bond dipole moment. This means the resultant of dipole moment of three $\ce{C-Cl}$ bonds in tetrahedral structure is equal to $x$, which is equal and opposite to the dipole moment $x$ of the fourth $\ce{C-Cl}$ bond.

Now, when we replace one of the chlorines with a hydrogen, the net dipole moment equals $x+y$, (as they are in the same directions). Similarly if we carry out the same procedure for $\ce{CH3Cl}$ it can be proved that its dipole moment is also $x+y$.

Where have I gone wrong?

Your conclusion is built on the assumption that the bond lengths will be constant. A bond’s dipole moment varies with bond length and thus for the vector addition to give the identical product, all $\ce{C-Cl}$ and all $\ce{C-H}$ bonds would have to be identical. They’re probably not.

Sutton and Brockway studied the bond lengths in chloromethane, dichloromethane and chloroform back in 1935 by electron diffraction.[1] Their results indicate that the $\ce{C-Cl}$ bond lengths don’t vary between the compounds in question as indicated in the table below. (Note that all chlorine atoms in each molecule are homotopic and thus equal bond lengths are theoretically expected.)

$$\begin{array}{lccc}\hline
\text{Compound} & \ce{CH3Cl} & \ce{CH2Cl2} & \ce{CHCl3}\\ \hline
d(\ce{C-Cl})/\mathrm{pm} & 177 & 177\phantom{^\circ} & 178\phantom{^\circ} \\
\angle(\ce{Cl-C-Cl}) & \text{n. a.} & 111^\circ & 111^\circ \\ \hline\end{array}$$

These results point towards the chlorine side of things not having an impact. But there is still the hydrogen side of things. Hydrogen atoms are generally very hard to locate in diffraction type experiments. X ray diffraction is observed if photons interact with electrons in a certain manner; hydrogen atoms do not have many electrons around them and are thus hardly seen. One needs to observe diffraction at very large angles to be able to locate hydrogens. I assume, although I don’t know, that it’s similar in electron diffraction. Indeed, the paper notes:

  […] and $l_{\ce{C-H}}$ was taken as $1.06~\mathrm{\overset{\circ}{A}}$.

The two carbons in chloromethane and chloroform are, however, very much not identical. One is surrounded by three electronegative chlorine atoms which will strongly remove electron density from it, polarising the $\ce{C-Cl}$ bonds towards each of the three chlorines. The other will have only one such withdrawing partner. The different electron densities around chlorine are well represented by the corresponding $\ce{^13C}$ NMR shifts: $\delta(\ce{CH3Cl}) \approx 22~\mathrm{ppm}$; $\delta(\ce{CHCl3}) \approx 77~\mathrm{ppm}$.[2] The hydrogen shifts are also different in similar magnitudes, showing that hydrogen in chloroform is much more polarised than in chloromethane.

These different electronics must have an influence on the $\ce{C-H}$ bond length of the two molecules. I can’t imagine two electronically such different systems to have equal bond lengths. My assumption is that $\ce{CH3Cl}$ should have the shorter $\ce{C-H}$ bond lengths, since carbon is more negatively charged while hydrogen remains positively charged; this should give a slightly greater attractive force and hydrogen, being such a small element, immediately feels these. Therefore, instead of the final dipole vector $\vec d$ being equal in both cases ($\vec d = \vec x + \vec y$), we have one lesser and one stronger vector.

$$\begin{gather}\vec d_{\ce{CH3Cl}} = \vec x + \vec y + \delta\vec y\tag{1}\\
\vec d_{\ce{CHCl3}} = \vec x + \vec y - \delta\vec y\tag{2}\end{gather}$$

References:

[1]: L. E. Sutton, L. O. Brockway, J. Am. Chem. Soc. 1935, 57, 473. DOI: 10.1021/ja01306a026.

[2]: Experimental spectra found for chloromethane/chloroform on SciFinder.


Schweizer's reagent from/with CuO? While attempting to make Schweizer's reagent in household conditions, using $\ce{CuSO4 + NaOH + 25\%  NH4OH}$ route (all at least 95% pure, either technical or lab grade purity), I've stumbled upon a problem. While trying to precipitate $\ce{Cu(OH)2}$, the temperature of the solutions rose to about 50 Celsius, and instead of expected blue precipitate, I got an almost black (with a hint of violet) one instead, which I assumed is $\ce{CuO}$. However, I was able to dissolve some of the precipitate in $\ce{NH4OH}$ slightly (a suspension was made from about 10 g of precipitate in 200 ml ammonia), resulting in dark blue solution with only traces of suspended black particles ($\ce{CuO}$?), appearing very like the expected Schweizer's reagent itself.

The resulting substance was able to dissolve regular cotton & cellulose samples overnight, however the dissolution was quite slow and unimpressive, with some of the sample material remaining almost intact; a chalk-coated paper was quite resistant to it (although it also dissolved somewhat eventually).

I don't have prior experience with Schweizer's reagent, so I don't know if the rate of dissolution I got here was the proper, expected behaviour here or a sign of dud.

Since, from what I know, $\ce{CuO}$ doesn't react nor dissolve in aq ammonia, does this experiment suggest a trace amounts of $\ce{Cu(OH)2}$ present in the (assumed) $\ce{CuO}$ precipitate, later forming a low-quality Schweizer's reagent, or am I missing something here?

As side questions:

was the precipitate really $\ce{CuO}$ (if so, how to avoid creating it, i.e. what are the necessary conditions for $\ce{Cu(OH)2}$ creation here), and what is the easiest way to check for $\ce{CuO}$ presence in general?
how efficient should be a properly done homemade Schweizer's reagent (using the aforementioned $\ce{CuSO4 + NaOH + 25\%  NH4OH}$ route) in dissolving various celluloses in practice?

Since, from what I know, $\ce{CuO}$ doesn't react nor dissolve in aq ammonia, does this experiment suggest a trace amounts of $\ce{Cu(OH)2}$ present in the (assumed) $\ce{CuO}$ precipitate, later forming a low-quality Schweizer's reagent, or am I missing something here?

Well in general this is true, but because the fresh precipitate can be partially hydrated and amorphous, which does allow the ammonium hydroxide to dissolve it.

  was the precipitate really $\ce{CuO}$ (if so, how to avoid creating it, i.e. what are the necessary conditions for $\ce{Cu(OH)2}$ creation here), and what is the easiest way to check for $\ce{CuO}$ presence in general?

Your precipitate was likely a hydrated copper oxide. Heat causes the copper hydroxide to lose its water.  To prevent $\ce{CuO}$ from forming, you want to keep the solution cool, perhaps with an ice bath. The easiest way to verify this is by visual observation. Copper oxide is black and copper hydroxide is blue, there is no false positive/negative to this for your preparation.

  how efficient should be a properly done homemade Schweizer's reagent (using the aforementioned $\ce{CuSO4 + NaOH + 25\%  NH4OH}$ route) in dissolving various celluloses in practice?

This is not easy to answer and depends greatly on the concentration of your reagents, temperature, pH and so on. I've seen videos on YouTube however showing about 2g of cotton balls dissolving in 500mL solution as a reference


Does isotretinoin affects contraception hormonal pills - Ethinylestradiolum, Drospirenonum Does isotretinoin affects contraception hormonal pills - Ethinylestradiolum, Drospirenonum? Asking as a pharmacologist not a patient.

My professional knowledge of vitamin A derivatives is limited to preparative aspects and some photochemical properties.

But it seems that the German Bundesinstitut für Arzneimittel und Medizinprodukte (BfArM) provides some information in german and english.

I didn't know whether a link to the search would survive the session, so the link above is just to their website and you would need to search yourself in case you're interested.

Note: I had a look at the Fachinformationen (RTF in German) among the documents "linked".

In this publication, it is very explicitly pointed out that Isotretinoin is a TERATOGEN and efficient contraception therefore is absolutely vital (in the truest sense of the word).

It is suggested that patients use two independent methods of contraception, of which one should be a "barrier method", e.g. the use of condoms.


Why is Density Functional Theory notoriously bad at describing oxygen molecules? I read in several papers that DFT is notoriously bad at describing free oxygen molecules but in none of them an explanation or reference concerning the problem has been provided. I guess this has something to do with its triplet ground state but I haven't come across a good explanation yet. Could you help me out?

Update: If someone wants to know the papers I'm refering to, here are the links and the passages that are most relevant (both groups use gradient-corrected (GGA) density functionals, so maybe this isn't a general problem of DFT but only of GGA- or LDA-functionals):

Rossmeisl et al. have devised a method for calculating the Gibbs Free Energy of the intermediate in the oxygen evolution reaction. The intermediates are treated with DFT but in order to calculate the Free Energy of the final products, including desorbed oxygen molecules, they use tabulated values.

  The free energy change of total
  reaction: $\ce{H2O -> 1/2 O2 + H2}$ is fixed at the experimentally found value of 2.46 eV per water molecule. This is done in order to avoid calculations of $\ce{O2}$, since this molecule has a complicated electronic structure, which is not described accurately with DFT.

There are other groups studying water splitting that use the experimental value too, instead of calculating it with DFT.
The paper by Bloechl includes calculations of the binding energies of a hydrogen molecule, which are in rather good agreement with experiment (4.338 eV (DFT) vs. 4.448 eV (exp.)), and of an oxygen molecule, where the DFT value is off by 0.795 eV.

  The binding energy of an oxygen molecule is calculated to be 5.912 eV, including the zero-point vibartion energy of 0.110 eV, which is comparable to the atomization energy of 5.906 eV, obtained with the Becke-Perdew-Wang gradient-corrected density functional. At 5.116 eV, the experimental binding energy is substantially smaller (by 0.795 eV) than the theoretical prediction.

The answer is most certainly the Self interaction error of an electron with itself in the exchange term.

This would probably be more helpful than what I can write here, as I have little time to make this answer more complete.

A. Droghetti PRB 2008

Also have a look at
Klupfel2012


Charged nucleophile and SN1/E1/SN2/E2 reactions How can we say that if a charged nucleophile/base is present we can safely rule out the possibility that the reaction can take place via SN1/E1 mechanism? I have been trying to figure this out since hours. Help would be appreciated. Thank you. Source

Strong nucleophiles tend to undergo SN1/E1 reactions whereas weak nucleophiles tend to undergo SN1/E1 reactions.

An increase in electron density will make a nucleophile stronger. This effect can be clearly seen when conjugate bases of common nucleophiles are stronger than the nuclephiles themselves. For example: $\ce{NH_2^- > NH_3} $.

In SN1/E1 reactions, the weak nucleophile can only attack the carbocation once the leaving group has left. This is precisely because it is a weak nucleophile.

IN SN2/E2 reactions, the strong nucleophile attacks the carbon before it can even form a carbocation. This forces the leaving group to leave. This is precisely because it is a strong nucleophile.


What is Inversion Temperature in terms of Joule Thomson Effect? Does Critical Temperature depend upon Inversion Temperature? I couldn't find a clear answer in my book and on the internet.
Please help! Your quick response will be appreciated.

This is half-part answer to your question:

Joule Thompson Inversion Temperature is the temperature when Joule Thomson coefficient $\mu_\mathrm{JT}$ is null.

$$\mu_\mathrm{JT} =  \left(\frac{\partial T}{\partial p}\right)_h$$

It means that the gas does not chill ($\mu_\mathrm{JT}>0$) or warm ($\mu_\mathrm{JT}<0$) when it experiences expansion. Pressurized Hydrogen has a negative Joule-Thomson coefficient, which must be taken in account if you plan to stock it and expand it, because it will significantly heat when expended.

This factor is influenced by gas molecular interactions (repulsive and attractive). Therefore, for an Ideal Gas implies that $\mu_\mathrm{JT} = 0$.

You will find detailed explanation in Atkins, Physical Chemistry, Chapter 2: First Law - The Joule-Thomson Effect.


Why are noble metals more electronegative then most metals? I was researching about electronegativity when I looked up what a graph of electronegativity within the periodic table is. And, this appeared. I scanned it, matching up everything I knew about the elements up to that table. Yep, Group 1 is all light colored; yea, the nonmetals are all really dark blue culminating in fluorine; and yes, the noble gases are mostly zero (with the exception of the traitors xenon and krypton, which coincidentally are also the only electronegative noble gases!)

But while I was scanning the page from Wikipedia on noble gas compounds, I suddenly realized that there was a compound called xenon hexafluoroplatinate, meaning that xenon could bond with...platinum? Checking the table, I was astonished to find that the entire noble metal group was actually more electronegative then the metals surrounding them! Why?! Aren't they extremely non-reactive? How then can they crave electrons even more then regular metals?

The noble metals are defined by a resistance to oxidation and corrosion, and this should not be interpreted as a lack of reactivity, but instead an aspect of their high EN.  So, there is no contradiction as you seem to be thinking.  Basically, they hold onto their electrons better than other metals, so it is harder for acids and oxygen to steal electrons from these metals.

The noble metals are generally considered to be ruthenium, rhodium, palladium, silver, osmium, iridium, platinum, and gold, so I'll mainly be addressing these particular elements. There are a few reasons why they should have a higher electronegativity:

The lanthanide contraction causes these atoms to have a higher than expected $Z_{\text{eff}}$. This means they hold on to their electrons tightly, so have higher electron affinities and ionisation energies which corresponds to a high EN. This doesn't carry into group 12 because accepting electrons for these elements would result in adding another energy level, so it's not as favorable.
The filling order: By period 6, the 6s, 4f, and 5d orbitals are all so close in energy that the filling order changes, which affects the properties and chemistry of the elements.  Notice that almost all of these metals, apart from Os and Ir, break the typical filling order. Adding electrons to these atoms therefore does not have the same effects that are dictated by normal periodic trends.
Relativistic effects - Somewhat related to filling order. By period 6, nuclei are so heavy that the core electrons are moving close to the speed of light.  This causes a contraction of the s orbitals, the inert s pair effect, and other things that have a big effect on the electronic structure and chemistry.

Also, the noble gases start becoming reactive around Kr, and more so for Xe, because they are so big with such high shielding that electronegative atoms are able to take electrons to form bonds.  Some people get confused about this because they think it's just Kr, and Xe being weird, but it's actually a trend and continues with Rn, but there isn't much data on Rn compounds, or much use for them, since Rn-222 is the longest lived isotope with a half-life of ~3 days.

Periodic Trends of Transition Metals could help explain some of this.


TEMPO Radical Analogue After investigating what makes the persistent radical TEMPO stable, I was wondering if another compound which is more feasible for my needs would be comparable to it.
Would oxidizing 2,2’-dimethyl-diphenylamine result in a compound similar to TEMPO? It will have an Aminoxyl group and have the steric hindrance of the methyl group next to it. By similar I mean if it will have a stable radical in the aminoxyl group.

Aminoxyl (trivial name: a nitroxyl radical or a nitroxide) denotes a radical functional group with general structure $\ce{R2N–O^•}$. It is well known that sterically unhindered aminoxyls baring $\alpha$‐hydrogens are unstable and undergo rapid disproportionation to nitrones and hydroxylamines (Ref.1: A new mechanism for the decomposition of $\alpha$‐hydrogen nitroxides is proposed). For example, decomposition of one of such compound is displayed below (The scheme is from Ref.1):

Sterically hindered aminoxyls without $\alpha$‐hydrogens such as TEMPO are stable radicals. However, it is not clear that steric hindrance at carbon $\alpha$ to nitrogen atom is important in the mechanism. Yet, the lone pair of electrons on nitrogen has played a significant role in the mechanism (Ref.1). Therefore, I speculate that 2,2'-dimethyldiphenylamine, when oxidized to relevant aminoxyl radical, may not be as effective as TEMPO, mainly due to the delocalization of the lone pair of electrons on nitrogen between two 2-methylphenyl groups.
Nonetheless, there are few available aminoxyl radicals, which are more efficient than TEMPO (Both following images are from Ref.2):

The research in Ref.2 has performed catalytic oxidation using few N-oxyl radicals, namely 2,2,6,6-tetramethylpiperidine-1-oxyl (TEMPO), 2-azaadamantane N-oxyl (AZADO), 1-methyl-AZADO (1-Me-AZADO), 9-azanoradamantane N-oxyl (Nor-AZADO), and 1,5-dimethyl-9-azanoradamantane N-oxyl (DMN-AZADO), in water at $\mathrm{pH} \  10$ on wood cellulose to determine the influence of different N-oxyl radical chemical structures on the reaction kinetics and structures of oxidized celluloses. The reaction time required for complete oxidation of wood cellulose using AZADO, 1-Me-AZADO, or Nor-AZADO was less than $\pu{20 min}$, while DMN-AZADO and TEMPO required $\pu{70 min}$ and $\pu{120 min}$, respectively.
There is also some comparative work has been done in Ref.3, which also showed the superiority of other N-oxyl radicals over TEMPO (Synthetic procedure to prepare Nor-AZADO has been given in the reference; see attached PDF).
References:

Aaron Nilsen,  Rebecca Braslau, "Nitroxide decomposition: Implications toward nitroxide design for applications in living free‐radical polymerization," J. Polymer Sci., Part A: Polymer Chem. 2006, 44(2), 697-717 (DOI:https://doi.org/10.1002/pola.21207).
Hiromasa Hondo, Tsuguyuki Saito, Akira Isogai, "Preparation of oxidized celluloses in a $\ce{NaBr/NaClO}$ system using 2-azaadamantane N-oxyl (AZADO) derivatives in water at $\mathrm{pH} \  10$," Cellulose 2019, 26, 1479–1487 (DOI:10.1007/s10570-018-2177-5).
Masaki Hayashi, Yusuke Sasano, Shota Nagasawa, Masatoshi Shibuya, Yoshiharu Iwabuchi, "9-Azanoradamantane N-oxyl (Nor-AZADO): a highly active organocatalyst for alcohol oxidation," Chem. Pharm. Bull. 2011, 59(12), 1570–1573 (DOI: https://doi.org/10.1248/cpb.59.1570)(PDF).


How durable is a silanized surface? I have two questions regarding silanization:

1) How durable is a silanized glass surface? (i.e. How long a silanized surface can be kept without any change in its wettability or contact angle?)

2) What are the optimum conditions for keeping a silanized surface while it's idle and dry? (e.g. maintenance temperature, humidity, etc. Should it be kept in a desiccator or not necessarily?)

If done properly, reactions with silanes or siloxanes are effectively permanent, unless removed by serious mechanical abrasion (e.g., scraping the material below the coating) or some sort of etching underneath.

The silanization process involves creating new covalent $\ce{Si-O}$ bonds, which are quite strong.

The silane surface itself should be durable. However, the contact angle may change as material adsorbs on top of it. For example, a colleague has shown that hydrocarbon contaminants change the wettability of many surfaces
I don't think any particular storage is necessary. We store in a vacuum desiccator to minimize the effect of contamination, but as long as you clean the surface, the properties should be consistent.


How to calculate the yield for an incomplete reaction? I had a doubt in part b) of the following question.

a) When 0.527 g of platinum is heated in fluorine, 0.732 g of a dark red volatile solid forms. What is the empirical formula of this product?

b) Suppose fluorine was a limiting reagent, such that 0.155 g of platinum remains after all the fluorine has been consumed. Calculate the percent yield of this reaction. How many grams of product would be produced?

I've given part a) also because it was told that we should use data from a).

This is my attempt:

The answer for b) is given as 70.6% and 0.517 g, but I am unable to find where I am wrong. Is my logic correct? (I've double checked that my calculations at least are correct.)

Well the problem doesn't say anything about how to get the % yield. If you start with 0.527 grams of platinum, but have 0.155 grams left, then 0.372 grams reacted.

$\frac{0.372}{0.527} * 100$% = $70.6$%

As for the yield

$\frac{0.372}{0.527} * 0.732 = 0.517$ grams


Local resolution of a crystal structure in PDB File (protein data bank) I am wondering if it is possible to find the local resolution of a crystal structure. Usually, the resolution of the PDB is assigned the resolution of the highest resolved part of the protein/nucleic acid. However, it is actually a range of resolution (ie. 2-8 A). I am wondering if it possible to extract the resolution of each atom from a PDB file?

The question has false statements:

  Usually, the resolution of the PDB is assigned the resolution of the highest resolved part of the protein/nucleic acid.

Usually (in ~90% of the PDB entries) the experimental method used is X-ray crystallography. In such case the resolution refers to a distance between crystal lattice planes: it is assigned the value of the smallest distance between lattice planes that was resolved in the diffraction pattern.

Crystallographic experiment measures reflection. Each reflection corresponds to a set of lattice planes in the crystal. The spacing between these planes, d, is the measure of resolution. "Resolution range 3 - 20 Angstroms" means that the reflections used in the refinement had d between 3 and 20 Angstroms.

The resolution is simply dmin - the smallest distance between crystal lattice planes that resulted in a reflection usable in the refinement. Each reflection contributes to each point in the electron density map, so while we can talk about local quality of the structure, the resolution cannot be local.

Unless the data comes from a Cryo-EM experiment, where the resolution is defined differently and people can talk about local resolution.


Difference in calculated pH and the real pH of a phosphate buffer? I'm from Colombia and I joined this site to ask a question that I could not solve yet. I am calculating the amounts to prepare a phosphate buffer with two salts. I want to prepare a buffer system with pH = 7.0 and a total concentration of Ct = 0.1 M. All my calculations are in the image attached. When I mix the calculated amounts, the obtained pH = 6.5 in the lab. I thought it was my mistake in calculations or an error when I weighed the amounts, but I repeated everything several times and still not get the pH = 7, and the value always was 6.4 or 6.5.

I don't know the reason but I would like to know this. Please help me. I think that If the concentration of the monobasic salt is greather than the dibasic, the pH tends to decrease, but in the calculations the relation [dibasic]/[monobasic] is always less than 1 and this again causes that [monobasic] > [dibasic]. Thanks, I really need a help.

The flaw in your calculations is the assumption that there exists a pK that is a constant independent of concentration.

In reality, pK is a function of ionic strength.

At infinite dilution (zero ionic strength) the pK is 7.199 as you assumed.

At 0.1M phosphate concentration, the pK is 6.81.

See Phosphate Buffers which cites to A. A. Green, "The Preparation of Acetate and Phosphate Buffer Solutions of Known PH and Ionic Strength", J. Am. Chem. Soc. 55, 2331, (1933).


Sign of the work done during a reversible compression The work done during a reversible compression - if $V_\mathrm{i} = V_1$ (initial) and $V_\mathrm{f} = \frac{1}{6}V_1$ (final) - of an ideal gas is?

I chose $1.8 p_1 V_1$. Is it right or should it be $-1.8 p_1 V_1$?

PV work is defined as
$$w = -p\Delta V,$$
where $p$ is pressure and $\Delta V$ is $V_\mathrm{final} - V_\mathrm{initial}$.

It has this form because in chemistry, work is defined from the perspective of the system - if it is compressed, we say that we did work on the system and so the sign of work is positive.

In this example, $V_\mathrm{f}$ is less than $V_\mathrm{i}$, so $\Delta V < 0$. If $\Delta V$ is negative, then $w$ must be positive.

This means your answer should have a positive sign. However, I get a different number from $1.8$ - you should double-check that.


Determining Half-lifes from a graph So I was studying for an exam I have thursday when I came across a question regarding half-life. I had previously thought that the definition of a half-life is the time it takes for the amount of material to half in its decay process. However, this question had a y-axis of Disintegration Rate, and used the same process.

They gave an answer of 8 days for the half life of this radioactive element(so they treated it as normal)

So my question is, what exactly does the half in half life represent? A halving of rate or an actual halving of decay of the nuclei?

At any time, the number of decays per reasonable$^1$ time unit is proportional to the amount of the isotope present. (This is expressed by the first-order rate equation for a one-decay process.) Therefore, it does not matter whether one looks at the amount or the number of decays. It may be easier to measure the number of decays by e.g. a Geiger counter and prefer it for that reason.

It now matters what you understand to be the rate: the number of decays per second or the percentage of nuclei decaying at any given time. The first becomes less over time, the second one is a constant for each isotope.

$^1$ You must have a reasonable number of decays within your time unit. If the half-life is long, it makes little sense to look at millisecond slices.


Lewis Structure For H2SO4 When I was drawing the lewis structure for $\ce{H2SO4}$ I got this configuration, now I know this isn't the best configuration but I don't see whats wrong with it. The number of valence electrons is right, the free electrons are on the most electronegative element and the formal charges are all 0. I'm not sure if there's a rule I'm missing.

As Jan mentions in the other answer, in general, a Lewis structure is preferred if it retains the octet rule. There can be more complicated bonding, and sulfur and phosphorous are two common examples (e.g., $\ce{SF6}$).

I find that many students can come up with "non-traditional" Lewis structures that, like yours, satisfy the number of valence electrons and minimize the formal charge. I usually consider these correct responses on an exam.

The "missing piece" is ring strain, which is not typically discussed until organic chemistry courses.

Note that your diagram as two O-S-O three-membered rings. These are extremely high in energy, because the O-O-S angle won't be anywhere near the expected 109.5°.

I tried a few quick calculations using the PM7 method and MOPAC. The best ring structure I could find looks like this (a distorted octahedral shape) and is estimated to have a $\Delta H_f^0$ = +33.05 kcal/mol. The O-O-S ring angles are ~65°.

The lowest energy geometry is the traditional $\ce{H2SO4}$ Lewis structure, estimated at $\Delta H_f^0$ = -177.88 kcal/mol. While PM7 isn't a highly accurate method, the difference in stability isn't close.


Are Organic Solvents used in Addition Reactions catalysts? Say for example you are performing a halogenation reaction with dichloromethane (DCM) as the solvent:

ethylene + chlorine $\stackrel{DCM}\longrightarrow$ 1,2-dichloroethane

I have 2 questions:

Is the DCM acting as a catalyst to the reaction, or just a solvent? And if just a solvent, why do we have to use an organic solvent, why not water?
Are solvents like DCM necessary if the reactants are liquids already, or only for gaseous or solid reactions?

Thank you

In the example you give, the dichloromethane is acting as a solvent. Solvents are usually chosen to be inert under the reaction conditions so that they do not react and produce unwanted products.

There are examples of reactions in which no solvent is used, or in which the solvent is a reagent. An example could be forming an ester from a carboxylic acid, using an excess of the alcohol as both solvent and reagent, in the presence of an acid catalyst.  That said, solvents are usually used as they severely effect the outcome of a reaction. Polar solvents (THF, diethyl ether) promote very different reactions to non polar ones (hexane) due to the stabilisation they provide for intermediates during the reaction, dilution can also play a role, deciding whether inter or intra-molecular reaction occurs.


What is the difference between the Joule and Joule–Thomson coefficients? Is there a difference between the Joule coefficient and the Joule–Thomson coefficient? Also, I am having a hard time understanding what they mean.

The Joule coefficient is $$\left(\frac{\partial T}{\partial V}\right)_U$$

Experimentally, Joule attempted to measure this value by expanding gas into an evacuated, insulated container, thus ensuring $U$ is constant.

The Joule–Thomson coefficient is $$\left(\frac{\partial T}{\partial P}\right)_H$$

Experimentally, this is realized by expanding a flow of gas in an insulted pipe from a high pressure upstream region to a lower pressure downstream region, the two regions being seperated by a porous frit.

For more information see THE JOULE AND JOULE-THOMSON EXPERIMENTS by Dr. J. B. Tatum, University of Victoria.


Finding vendor information programmatically I have a large collection of structures (>10,000) that I've gathered from literature searches and I would like to find out which of those can be obtained from vendors.  I know that eMolecules and ChemSpider provide such information, but getting that information for 10,000 structures in daunting.

On eMolecules, I can load a list of SMILES, search the databases, and download the results.  The download process is not very robust, though, and I have to reconstitute a lot of pieces of information.

ChemSpider provides additional information regarding vendors, and I've tried using the ChemSpiPy API, but I consistently get an error from the ChemSpider server.  Obviously something has changed with respect to the server that the API doesn't account for.

Are there other suggestions for where/how I might gather such vendor details for a very large collection of structures?

You can try molport, they are an intermediate. Just ask them directly, it will be done in a jiffy.
Or if you like you can do it in knime and use any API (like molport's API), which is much more fun, but for just a one-time thing a bit complicating.


Is cyclobutadiene antiaromatic? Cyclobutadiene is very unstable. But, some sources claim that this instability can be attributed to other factors such as ring and angle strain rather than antiaromaticity.

According to some, cyclobutadiene is simply non-aromatic (as opposed to antiaromatic) because it doesn't even have a fully conjugated pi system.

What is/are the real reason(s) for its instability?

If we go back to your earlier question on Frost diagrams (I've reproduced the key figure below), we see why simple molecular orbital theory or the "$4n+2$" rule suggests that benzene is aromatic while cyclobutadiene is antiaromatic.

Forming a planar, conjugated, 6-membered ring and placing 6 π-electrons in it creates a π-system that is energetically more stable than 3 conjugated C=C bonds (e.g. in hexa-1,3,5-triene). On the other hand, forming a planar, conjugated, 4-membered ring and placing 4 π-electrons in it creates a π-system that is energetically less stable than 2 conjugated C=C bonds (e.g. in butadiene).  For these reasons we say that the first system, benzene, is "aromatic", while the second system, cyclobutadiene, is "antiaromatic". Other measurements and physical phenomenon such as reactivity, bond lengths, ring currents, etc. support these conclusions.

Free cyclobutadiene has been observed as a transient intermediate.  Further theoretical analysis suggests that due to its lack of aromaticity it will distort from a square structure to a rectangular one with alternating single and double bonds (Jahn–Teller effect).

  sources claim that this instability can be attributed to other factors such as ring and angle strain rather than antiaromaticity

It turns out that other systems involving the cyclobutadiene skeleton have been prepared and studied.  Derivatives of both the cyclobutadiene dication[1] ($4n+2$, $n=0$) and the cyclobutadiene dianion[2] ($4n+2$, $n=1$) have been prepared and studied.  Their stability is much less than that of benzene, but much more than that of cyclobutadiene. Placing two charges in these small rings results in extremely high coulombic repulsions in both the dianion and dication, and this may be a significant part of the explanation as to  why they are less stable than say, benzene.  The fact that they exist as planar structures, with appropriate ring currents and are more stable than cyclobutadiene suggests that the instability observed with cyclobutadiene is not due to ring strain and angle strain alone.  Rather the aromatic stabilization - antiaromatic destabilization suggested by simple MO theory seems to be involved.

References

Olah, G. A.; Staral, J. S. Novel aromatic systems. 4. cyclobutadiene dications. J. Am. Chem. Soc. 1976, 98 (20), 6290–6304. DOI: 10.1021/ja00436a037.
Takanashi, K.; Inatomi, A.; Lee, V. Y.; Nakamoto, M.; Ichinohe, M.; Sekiguchi, A. Tetrakis(trimethylsilyl)cyclobutadiene dianion alkaline earth metal salts: new members of the 6π-electron aromatics family. Eur. J. Inorg. Chem. 2008, 2008 (11), 1752–1755. DOI: 10.1002/ejic.200800066.


Order of reactivity of nucleophilic addition at carbonyl group with Grignard reagent I thought the order would be $1 > 3 > 2$ as 1 is aldehyde and 2, 3 are ketones. As generally aldehydes are more reactive, I placed it first. I chose 3 over 2 as two alkyl groups would make cleaving of $\ce{C=O}$ difficult in 2.

But the answer is given $1 > 2 > 3$, even though phenyl is a bulky group and alkyl groups are electron releasing.

How to approach these type of questions?

All the compounds are planar, steric hinderance might have played a more significant role if the attack of nucleophile was in the plane. Nucleophilic addition at the carbonyl group takes place at an angle of 107° with the plane.!

Addition of grignard reagents at the carbonyl group are primarily governed by electronic effects. More the positive charge at the carbonyl carbon more will be the rate of reaction.

The two phenyl groups reduce the $\delta +$ most significantly by resonance, next the dimethyl ketone by two methyl groups which reduce the positive charge by hyper conjugation and inductive effects.

The order can be finalized as:

$$1>2>3$$

References

Clayden, J.; Greeves, N.; Warren, S. Organic Chemistry, 2nd ed.; Oxford UP: Oxford, U.K., 2012; pp. 130


Why is the ideal gas law so ubiquitous? Why is the ideal gas law usable in most gas problems?

For example, in a problem like this: "Suppose you have $1.00\: \mathrm{mol}$ of a gas at $0\:^\circ \mathrm{C}$, occupying a container which is $500\: \mathrm{mL}$ in size. What is the pressure of this gas in $\mathrm{atm}$?"

The ideal gas law is was applied to solve that problem and the answer was $44.8\: \mathrm{atm}$. But isn't the ideal gas law only applicable to gases which are in high temperatures and low pressure? Because the kinetic energy overcomes the intermolecular bonds and the molar volume is negligible, etc etc.

But the conditions for this "gas" are low temperature and consecutively high pressure which is basically a stamp for real gases so why isn't the real gas law applied to this problem instead?

You are right in saying that in your question, the gas is an non-ideal (real) gas. Basically, the reason why we use the ideal gas equation despite knowing that it is a real gas is simply because that the equation for real gases is too complicated and contain more variables which are hard to find given limited data.

The reason for this complicated formula is that several assumptions that we make for ideals gases don't hold for real gases. Recall that an ideal gas is considered to be a point mass - a particle so small that the volume of that particle is negligible. A real gas particle does have real volume. For an ideal gas, the collisions between gas particles was said to be "elastic" - no attractive or repulsive forces exist, and thus, no energy is exchanged during collisions. For a real gas, collisions are non-elastic. So, the ideal gas law must be corrected for this extra forces. There are several real gas laws, one is the van der Waals equation:

$$\left[P + a\left(\frac{n}{V}\right)^2\right](V-nb) = nRT$$

Notice how "corrections" are being made to the pressure term and the volume term. Since collisions of real gases are non-elastic, the term $a(n/V)^2$ is correcting for the interactions of these particles. Since real gas particles have real volume, the $nb$ term is correcting for the excluded volume. The values of $a$ and $b$ are constants, and must be determined experimentally for each gas.

As you can see from this equation that first the values of $a$ and $b$ must be known then you must solve this equation which can be complicated. Even then, when you do finally solve it, you will probably get a result which is similar to the result obtained if you used the ideal gas equation. This is why the ideal gas equation is continued to be used as in most cases it provides quite accurate results while bypassing complicated formulas.


What makes a compound basic? Aluminum chlorohydrate is often referred to as basic aluminum chloride.
The pH value of ACH is about 4. How is basicity in this context defined?

Aluminum chlorohydrate (ACH) is a group of specific aluminum salts having the general formula $\ce{Al_nCl_{(3n-m)}(OH)_m}$ (Wikipedia). It is used in cosmetics as an antiperspirant and as a coagulant in water purification. Specifically in wastewater treatment processes, ACH is used as a coagulant to remove dissolved organic matter and colloidal particles present in suspension. In his paper on Water Treatment Coagulants, Peter Gebbie describes what coagulants do (Ref.1):

  With aluminium-based coagulants, the metal ion is hydrolyzed to form aluminium hydroxide floc as well as hydrogen ions. The hydrogen ions will react with the alkalinity of the water and in the process, decrease the pH of the water as can be seen from Equation [...].  Similarly for ACH, which is described as being a pre-hydrolyzed coagulant, the following reaction takes place:
  $$\ce{Al2(OH)5Cl -> Al2(OH)5+ + Cl- + H2O -> 2Al(OH)3 + H+ + Cl- }$$
  Less hydrogen ions are produced with ACH, reflecting the hydroxylated nature of this compound.

There it says it all: ACH is more basic than other aluminum salts because it is pre-hydroxylated.

Aluminum chlorohydrate is one of the most common active ingredients in commercial antiperspirants. The variation most commonly used in deodorants and antiperspirants is $\ce{Al2Cl(OH)5}$, similar to the one used in water purification.

Note: Your statement of "The pH value of ACH is about 4" is evidently not correct. However, W. R. Grace & Company (producer of solution form of ACH) listed solubility of ACH as $\pu{1000 g/L}$ at $\pu{20 ^\circ C}$ and $\mathrm{pH} = 3.3$, the optimum condition.

Reference:

Peter Gebbie, “An Operator’s Guide to Water Treatment Coagulants,” 31st Annual Water Industry Workshop – Operations Skills; University Central Queensland Campus: Rockhampton, Australia, 4 to 6 July, 2006, pp. 14–20 (http://wioa.org.au/conference_papers/06_qld/documents/PeterGebbie.pdf).


Reaction of soap with water I've seen it many times... Why does soap become foam-like when it reacts with water (moisture) or when we touch it with wet hands.. What compound actually causes this physical change? And why does this physical change occur?

The simplest kinds of soap are just a fatty acid chain with something more ionic on the end - something like the ubiquitous sodium laureth sulphate:

When it's in bar form, it behaves a bit like a solid block of plastic, because chemically speaking, those long chains of carbon are quite a lot like a common plastic such as polythene. These chains attract each other and repel and exclude water, keeping the block together. The sulphate group on the end is hydrophilic, though - it's chemically stable surrounded by water. This means that when you add water, it infiltrates and teases apart the structure of strands, getting to the ionic ends on each chain and forming a "solvation shell" of waters around each one. There's no real chemical reaction here - as you say, it's a physical process, which is why rubbing the soap helps - you're applying shear to the structure and helping to separate the strands.

Once they're separated from the bulk, the soap forms tiny bubbles called "micelles", which have the chains pointed inwards towards each other and the ionic sulphate groups meeting the water at the outside.


How does iodide replace bromide when the latter is more basic? The following reaction follows the $S_N2$ pathway: $$\ce{CH3CH2CH2-Br + I- (presence of DMSO) -> CH3CH2CH2I + Br-}$$ I found this reaction in my textbook and according to me this reaction should not even take place. $\ce{Br-}$ is a stronger base than $\ce{I-}$ and keeping in mind, stronger the base, lesser is the leaving ability, why will $\ce{I-}$ replace $\ce{Br-}$ in the first place?

As you suggest "$\ce{Br−}$  is a stronger base than $\ce{I−}$ and keeping in mind, stronger the base, lesser is the leavability, $\ce{I−}$ should not replace $\ce{Br−}$ "

I actually agree with the logic and according to me, infact, the reverse reaction, i.e.

$$\ce{CH3CH2CH2I + Br- (presence of DMSO) -> CH3CH2CH2Br + I-}$$

should be much more favourable. Because we already know, that in the aprotic solvents, the nucleophilicity of halides is  $\ce{F− > Cl− > Br− > I−}$, so $\ce{I−}$ should be a better leaving group and $\ce{Br−}$ should be a better nucleophile .


How dangerous is a megacurie (MCi)? I was researching the Kyshtym (nuclear) Disaster, and it said that it released 20 MCi (20 megacuries) of radiation into the air in a dust cloud.
I read from this Yahoo Answers answer that:

The curie is the old unit of activity and a very big unit it was.
1 curie = 3.70 x 10^10 disintegrations per sec.
In today's units 1 curie is about 30 Giga Becquerel.
So a curie of any radioactivity (alpha, beta or gamma) would fry you.
A sealed microcurie source could be safely handled.

If a single curie is this powerful, then what would twenty million curies (a 20 MCi) be like? Could you give any examples of events that exemplify this scale of radiation and damage (if you can't think of any real ones, then maybe a theoretical example)?

There are a lot of factors at play here:

The curie and becquerel only tell you the number of disintegrations per unit time—they convey no information of the type or energy of radiation produced.
How dangerous a radioactive source is depends on the type of radiation produced and how one is exposed to it. External exposure to alpha particles is not very dangerous because alpha particles are easily stopped by skin, but inhaling alpha emitters can be very dangerous because there is direct access to sensitive alveolar cells.
How the radioisotopes are dispersed will greatly affect individual exposure to the radiation. A large amount of radioisotopes spread over a large area may increase the incidence of cancer slightly for a large number of people, but a smaller amount concentrated in one place may be very rapidly fatal to a single person.

20 MCi (740 PBq) is indeed a very large amount of radiation (~1/7 that released by the Chernobyl accident), but in a big cloud spread over thousands of square kilometres may not be acutely dangerous to people. That many radioisotopes is equivalent to 231,000 kg of cesium-137 and all that (magically) concentrated into a point source would produce a dose of .62 Gy/s for a person 1 m away—fatal in seconds.