Dissociation of diatomic molecules At high enough temperature, diatomic molecules in a dilute gas should dissociate. Where do their electrons go? Will they dissociate into neutral atoms or into ions? Does it depend on the substance? What for $\ce{H2}$, $\ce{O2}$, $\ce{HCl}$, $\ce{HF}$? Which quantum mechanical technique can decide the question - Hartree-Fock?

Where do their electrons go?

IIRC, in the gas phase, the energy required for heterolysis is usually much greater than that required for homolysis. Consequently, usually, each of the atoms of a diatomic molecule retains one of the originally bonded electrons on dissociation.

  Which quantum mechanical technique can decide the question -
  Hartree-Fock?

In principle, one could scan the potential energy surface1) by elongating the bond using some quantum chemical method to look on what actually  happens. Beaware though, that the most used variant of the Hartree-Fock (HF) method, the so-called restricted Hartree-Fock (RHF) method is know to be ill-suited for such calculations: it is known to fail completely when the homolytical dissociation of a closed-shell molecule into two open-shell fragments is considered.

Such behaviour is usually refered to as the Hartree-Fock "dissociation catastrophe", although, strictly speaking, only the restricted variant of the Hartree-Fock method is a subject to it. The unrestricted Hartree-Fock (UHF) method does not suffer from the "dissociation catastrophe" and tends to give at least qualitatively correct dissociation limit. So, use UHF at least. Even better, use some correlated method.

1) Simply a curve in this case, since there is only single degree of freedom (bond length).


How can I tell whether or not a molecule is planar? I am doing several questions involving judgment on the planarity of a compound.

  Which of the following is not a planar molecule?

  $\ce{H_2C=C=CH_2}$
  $\ce{H_2C=C=C=CH_2}$
  $\ce{H_2C=C=O}$
  $\ce{NC-HC=CH-CN}$

I had the idea that the compound in with central atom has $\ce{sp}$ hybridisation is planar or the compound in which all the atoms has the same hybridization. But it is not working in this case. I am aware of finding out the hybridization of a atom in a compound but I feel trapped to decide the planarity of certain compounds. The question does not address this purpose.  Anyone has any idea to solve this and many related questions?

I think these general rules work:

If there is an $\ce{sp^3}$ hybridized carbon (or nitrogen), the molecular is NOT planar.

2) If there are no $\ce{sp^3}$ hybridized carbons (or nitrogens), and there is only one $\ce{sp^2}$ hybridized atom (carbon or nitrogen), it will be planar.

3) If there are no $\ce{sp^3}$ hybridized atoms, and there are 2 $\ce{sp^2}$ hybridized atoms (carbon or nitrogen) that are separated by an even number of double bonds and no single bonds, then the molecule will not be planar.

So a general simple rule is that:

  the molecule will not be planar if there is an $\ce{sp^3}$ hybridized carbon (or nitrogen) atom or two $\ce{sp^2}$ hybridized atoms of carbon/nitrogen which are separated by an even number of double bonds and no single bonds. Otherwise, its structure allows it to be planar.

Even though the molecule will have a structure that allows for it to exist in a planar conformation, there may be some/many that do not persist in a planar conformation due to steric effects, or complex three dimensional geometries.

In the problems you listed above, using this rule:

Not planar because there are no $\ce{sp^3}$ and the two $\ce{sp^2}$s are separated by an even number of double bonds.
Planar because there are two $\ce{sp^2}$s but they are separated by an odd number of double bonds (3) (and no single bonds)
Planar because there are no $\ce{sp^3}$s and only 1 $\ce{sp^2}$s that make 3 or more bonds (C or N).  The orbital geometry is NOT planar because the $\ce{sp^2}$ oxygen is separated from the $\ce{sp^2}$ carbon by an even number of double bonds.
Planar because 2 $\ce{sp^2}$s are separated by an odd number (1) of double bonds (and no single bonds)


Why starch (amylose) and cotton (cellulose) are so different? Both amylose and cellulose have the same "monomer" structure (glucose), so what makes them look/form so differently?

Your structures don't clearly show the different configurations between α- and β- linked glycopyranosides. Cellulose is a non-branching (poly) β-glycopyranoside. Amylose (a component of starch) is a non-branching (poly) α-glycopyranoside. The enzymes in your mouth and digestive system are good at breaking apart the α-glycopyranoside linkage, but not the β-glycopyranoside (it doesn't fit into the enzyme).
Amylopectin (the other component of starch) is a branched (poly) α-glycopyranoside. It branches with a 1-6 linkage. Glycogen is another branched (poly) α-glycopyranoside.
Cellulose is not the only (poly) β-glycopyranoside; there is also β-glucans.


Is strength of an acid proportional to concentration or dilution? I have a confusion between two seemingly contradictory factors.

Since pH =-log[H+], it suggests that more the concentration of the acid, lesser will be its pH, and therefore the stronger it is.
But, according to Ostwald's dilution law, the degree of dissociation of an electrolyte is directly proportional to the square root of dilution. Though this exact relationship is true only for weak electrolytes, I believe that the degree of dissociation of any electrolyte would increase with dilution, keeping the temperature constant. And, more the degree of dissociation, more would be the H+ions released, and stronger the acid.

Where am I wrong in my reasoning?
Thanks in advance.(Please give me an answer and not just a comment)

Is strength of an acid proportional to concentration or dilution?

This is an interesting question. It is sort of like being asked "What is your size?". If you are buying shoes then size means one thing, yet it is something different if you are buying pants.

So there are two different "strengths."

Strength can mean how much acid is present. So per unit volume of solution, a solution with 50% concentration has half the strength of the 100% concentration.

Strength can also mean how acidic, in terms of pH, would a specific solution be if the acid concentration were kept constant. In this sense hydrochloric acid is a stronger acid that acetic acid.

So the gist of your problem is that you're trying to draw a comparison between two different uses of the word strength when such a comparison is pointless.


Decomposition of iron(III) acetate? I recently asked about the synthesis of iron (III) acetate. I want a rather concentrated solution to use as a calligraphy ink (which will be dissolved in alcohol/water). Will it be okay to heat the solution of iron(III) acetate, hydrogen peroxide, acetic acid, and water to evaporate all of the other reagents? That is, will heat cause the decomposition of the iron acetate?

Iron(III) acetate will be stable under these conditions. According to this paper, thermal decomposition to iron oxide only starts at 423 K (150 °C). Heating iron and acetic acid together with an oxidation agent (oxygen from air or, like in your case, $\ce{H2O2}$) is actually a suitable way to prepare iron acetate dyes (source).


Hydrolysis of cyanohydrin The question is(See image):

I know that initially a cyanohydrin will be formed. On reaction with $\ce{H2SO4}$, the cyanohydrin will give a carboxylic acid/an amide.

Here's my doubt:

Since the medium is highly concentrated acid medium(95% $\ce {H2SO4}$), the quantity of water in the medium is less, which means lesser of the nucleophile(water in this case). Hence the cyanohydrin should undergo partial hydrolysis to give an amide(because of lesser water!).  Therefore, according to me the answer should be (D).(Had it been dilute acid medium I would have gone with the carboxylic acid product.) However, in this case my answer is not correct...:(

What is incorrect about my thinking?

So G is the expected cyanohydrim, but the second step looks very vigorous (concentrated sulfuric acid and heat).  I would expect two things to happen under these conditions, 1) hydrolysis of the nitrile group to the corresponding carboxylic acid (the reaction conditions are too strong to stop at the intermediate amide, rather the amide is further converted to the corresponding carboxylic acid; keep in mind that an equilibrium will initially exist between the amide and the carboxylic acid, but everytime the acid forms from the amide, ammonia is released and escapes from the system making the reaction irreversible, pushing this reaction further and further to the carboxylic acid side, ultimately only the carboxylic acid will be present.), and 2) acid catalyzed elimination of water from the alcohol portion of the molecule.  Taken together I would expect the final product to be the various cis-trans isomers of "A".


What will be the coordination number in Fe(0)? Fe(0) has its electronic configuration $\ce{3d^6 4s^2}$ in the ground state. In this case multiplicity will be either 1 or 3 during excitation. If multiplicity is 1, the vacant excited orbitals are $\ce{dsp^3}$. So, in this case it should be trigonal bipyramidal structure with the attachment of 5 ligands.

But in the case of multiplicity 3, the vacant excited orbitals will be $\ce{sp^3}$, excluding 2 singly occupied $\ce{3d}$ orbitals. If 4 ligands are there in attachment, what will be the structure -- tetrahedral or square planner?

Another additional information is also needed, is there any possibility to increase the number of ligand in this structure where multiplicity is 3? If yes, can you explain please?

This is another data point for my constant mantra that using hybridisation to explain coordination chemistry is not helpful in the slightest. And even ‘worse’: you are trying to invoke some strong meaning of the 18-electron guideline even if not explicitly. The 18-electron guideline is a guideline not a rule. Even the octet guideline (that I personally like to refer to as a rule) is not as strict a rule as proper physical rules.

As Ben points out, in diiron nonacarbonyl iron(0) has a coordination number of six, which immediately voids your ‘excitation hybridisation’ argument. I am unaware of any fully characterised tetracoordinate iron(0) species. However, I guess they exist due to bulky ligand design and that they will be either square planar or tetrahedral in nature depending on the type of ligand in question. The type of ligand and its interaction is much more important than a hypothetical ‘excitation hybridisation’ scheme.

Please forget any and all of the wrong things you learnt about hybridisation with respect to transition metal complexes. Please refer to molecular orbital theory to explain these instead.


How can you achieve negative pressure? Pressure is the magnitude (absolute value) of the force over the area, so how can that result in a negative value for pressure as would happen in certain circumstances?

The term “negative pressure” is often used in engineering to refer to a situation in which an enclosed volume has lower pressure than its surroundings. If the a region is surrounded by more pressurized area around it would cause substances to flow inwards and thus the term negative pressure.


What exactly is the "oxygen reduction reaction"? Is it

$\ce{O2(g) + 4H^+(aq) + 4e^- \rightarrow 2H_2O(l)}$

and are there multiple types of ORR?

See here for context: http://www.nature.com/srep/2014/140122/srep03821/full/srep03821.html

According to your cited paper the reaction they care most about is $$\ce{O2 + 4H+ + 4e- -> 2H2O}$$ as you are suggesting. This should also be the total reaction and the most interesting one for applications, e.g. in fuel cells.

There are various suggestions for the mechanistics of the Oxygen Reduction Reaction. A quite recent review was published by Nagappan Ramaswamy and Sanjeev Mukerjee. A very early general scheme can be found at electronicstructure.wikidot.com, where they cited Halina S. Wroblowa et.al..


Distinguish between glucose and Fructose samples I am looking for a simple way to distinguish between glucose and fructose samples. Currently, I am doing that by melting temperature method. Can anybody suggest me some easy way to do that?

One of the simplest ways would be bromine-water test. Bromine water oxidizes glucose to gluconic acid, hence decolorize the solution. Being a mild oxidizing agent, Bromine water is not capable of oxidizing fructose (ketone).


Salts that are more hydrophilic than NaCl What other, rather easy to obtain, salts are more hydrophilic than $\ce{NaCl}$? Is there a hydrophilic scale for substances like the Mohs scale of mineral hardness?

As per DavePhD's answer, the contact angle of NaCl with water is zero, so we need another measure.

I think the word you are looking for is hygroscopic. http://en.wikipedia.org/wiki/Hygroscopy Substances which are so hygroscopic that they dissolve in the water they absorb from the atmosphere are said to be deliquescent, although the property of deliquesence in itself does not directly indicate the strength of a substance as a drying agent.

The strength would best be measured as the equilibrium vapour pressure at a certain temperature, or alternatively, the temperature required to dehydrate: i.e. the temperature at which the vapour pressure of water vapour above the hydrated form of the substance becomes equal to atmospheric.

Cheap substances which are hygroscopic find applications as desiccants

Here is a list of some common ones http://en.wikipedia.org/wiki/List_of_desiccants. It includes the workhorse laboratory desiccants: silica gel for use in desiccators where products are dried under vacuum; molecular sieves, for drying solvents; CaCl2 for drying gases; and MgSO4 and CaSO4 in their anhydrous form for a quick drying of organic solutions. I would assume (though I'm not sure) that all of these would be stronger than NaCl.

Unfortunately the really strong ones are missing, so here's a list of mine. The weaker ones above tend to act by physical absorption, or by changing from anhydrous crystals to hydrated crystals. Stronger ones tend to react chemically.

Concentrated sulfuric acid Strong enough to dehydrate sugar to "charcoal." One of the reasons for the affinity is the energy released when H2SO4 proponates H2O, forming HSO4- + H3O+. More of a curiousity than a desiccant. Not widely used as a desiccant because of its acidity.

Phosphorous pentoxide Reacts with water to form various different polymeric phosphoric acids, ultimately ending up with 2P2O5 + 6H2O= 4H3PO4 though strength is gradually reduced as water is absorbed. Commonly used in the lab when a strong desiccant is required.

Calcium Oxide The reaction CaO + H2O = Ca(OH)2 has been proposed for large scale energy storage (the storage occurs in the reverse direction), but so far has been used only for smaller applications. As the calcium species are basic, exposure to air causes absorption of CO2, which then irreversibly degrades the material to bicarbonate and carbonate (unless heated to a much higher temperature.)

Sodium Metallic sodium, extruded into wire using a press to increase its surface area, is the best way to ensure absolutely dry solvents.


What would be an appropriate way to clean up equipment contaminated with ethidium bromide? I want to clean up some old equipment that was used to visualize nucleic acids with ethidium bromide. It's a UV table, built from glass and metal (stainless steel, I assume).

Ethidium bromide is generally considered scary in a biological lab (probably because most stuff there is pretty harmless), but there seems to be a lot of controversy about the actual danger.

What would be an appropriate way to clean this? I'd simply wipe the whole thing with paper towels and some ethanol and dispose of them in the chemical waste. Would that be sufficient?

As you stated in the question, there is some controversy surrounding the mutagenicity of ethidium bromide. However, a recently published opinion piece titled "The Myth of Ethidium Bromide" quickly dispels most of the uncertainty:

  Ethidium bromide, as far as can be told from the data, is not a human mutagen. It’s not a mouse mutagen or rat mutagen either. Nor apparently a mutagen in cows and other farm animals, where it’s used in veterinary medicine at concentrations one thousand times higher than the red solutions that are so feared in biology labs, seemingly with no bad effects.

This knowledge does however not give us clear directions as to how the clean-up and decontamination should proceed. For this we can turn to the Material Safety Data Sheet (MSDS).
The relevant sections are Hazards Identification, Handling and Storage, and Exposure Controls / Personal Protection, numbered 2, 7, and 8, respectively.

Hazards Identification tells us that it's fatal when inhaled (in its solid form). This means that when it's not in solution, you should be wearing a respiratory mask or have a fume hood panel separate you from the dust. The fume hood should be operational, of course.

Handling and Storage states that you should:

Avoid contact with skin and eyes.
Avoid formation of dust and aerosols.
Provide appropriate exhaust ventilation at places where dust is formed.

Finally, Personal Protection informs us that you should be wearing a lab coat and glasses (duh), but also gives us some data on the breakthrough time of the substance through nitrile gloves, which is 480 minutes. There is also a section on respiratory protection, but it is prefaced with "Where risk assessment shows [respirators] are appropriate," which is something that you must decide for yourself.

Thus, the answer is, as almost always, it depends:
In any case you will want to avoid brushing and dusting off the contaminated surface to reduce the formation of ethidium bromide dust. Personally, I would recommend to wipe all surfaces first with water (according to Wikipedia the solubility is 40 g / L) until clean using paper towels (and gloves, of course). Following that, one final wipe-down with ethanol may be appropriate, but that is left for you to decide.


Minimum resolution to acquire exact mass on a spectra from m/z peak I'm having a little difficulty with finding what the minimum resolution required to obtain the exact mass of a 400 $m/z$ peak in a mass spectrum. I would appreciate any help.

In mass spectrometry, resolution and mass accuracy are distinct yet related concepts.  Eliding the difference between the two often causes confusion.

  I'm having a little difficulty with finding what the minimum resolution required to obtain the exact mass of a 400 m/z peak on a mass spectrometry spectra is.

Resolution is a concept that always involves at least two chemical species that differ in $m/z$.  It doesn't make sense to talk about resolution if you are talking about a single ion.

Mass accuracy is a concept that more or less assumes there is a single chemical species at a single $m/z$.  It's very easy to define: how different is a measured $m/z$ from the correct $m/z$?

Now, due to the presence of multiple stable isotopes for most elements commonly seen in organic mass spectrometry, almost no compound will give rise to a single chemical species with a single $m/z$ value.  If your compound ionizes to give singly charged ions ($z=1$), these isotopologues will be separated by approximately 1 $m/z$.

A nice coincidence among commonly occurring elements in organic chemistry ($\ce{CHNOPS}$) is that the lightest stable isotope is by far the most abundant.  (This isn't true of $\ce{Cl}$, $\ce{Br}$, $\ce{Fe}$ though.)  For molecules as small as 400 Daltons, this means that the most abundant isotopologue will be (a) the lightest of all possible isotopologues, and (b) monoisotopic.

Thus, to get an accurate mass (really accurate $m/z$) measurement of an compound you'll need a resolution that is better than unit mass resolution.

However, you won't necessarily need a much greater resolution than that.  A lot depends on your sample.  Can you assume that there is only one chemical compound present that gives an $m/z$ of 400 Da?  If so, then even "low resolution" mass spectrometers can give very accurate masses.  However, if your sample is complex or contains other compounds that give ions very close to and $m/z$ of 400, then you will need more resolving power to separate the contaminating ions from your peak of interest.  In that case, it's impossible to say how much resolving power you need without specifying what the contaminants are.

For example, say your compound is $\ce{C29H37N}$, for example, maybe it's N-(3,3,3-triphenylpropyl)octan-1-amine.  In electrospray mass spectrometry it ionizes as $\ce{(C29H37N + H)+}$ at an $m/z$ of 400.3010.  If that was the only compound in your sample, then even old TOF instruments with resolutions as low as 2000 would suffice to obtain an accurate mass.  However, if there was also some $\ce{C26H41NS}$ in your sample, say 1,6-dimethyl-4-pentadecylquinoline-2-thione, its $\ce{(C26H41NS + H)+}$ ion would have an $m/z$ of 400.3043.  That would mean you would need a resolution of at least 122,000 to tell those two peaks apart, generally available only on Orbitrap or FT-ICR instruments.  But if there was no such contaminant in your sample, you wouldn't need that degree of resolution.


Heat transfer through layered wall An electrical heater is placed between two layers of insulation. The three layers (the heater and the two layers of insulation) have the same thickness. The figure below shows different temperature profiles at steady-state. The air temperature at the front- and backside of the heater is $\ce{T_{front}}$ resp. $\ce{T_{rear}}$. Heat transfer in the air film adjacent to the heater is negligible.

  a) Which profile corresponds to that of a turned on heater?

  b) Which profile corresponds to that of a turned off heater?

  c) Which profile is impossible?

  d) Which of the three layers has the lowest thermal conductivity? Which layer has the highest?

I have no answer for a) or b).

c) My reasoning is that since a heater is a conductor, the k-value needs to be high, resulting in a fairly straight line in the temp. profile. Hence, (C) is impossible. Is this correct thinking?

However, I don't know what it means when a temperature profile has that kind of curvature for one of the layers, such as in (A) and (B), so if someone could please explain that, I would appreciate it.

d) Based on the fact that low k-values give a steeper line, I would say that layer 3 has the lowest thermal conductivity. And because the heater is a conductor it should have the highest thermal conductivity.

Also, how can you tell if the heater is turned on or off from looking at different temperature profiles? I am having a very difficult time understanding this and would appreciate if someone could make this a bit clearer.

I apologize if these types of questions are considered off-topic, but I really want to know how to determine these things and we have not been provided with any  answers nor explanations.

I would start with guessing which one represents the "off" state. I would say (D), since the temperature slopes evenly from one air interface to the other, and is fairly constant in the heater layer. Based on this diagram you can also conclude as you did that layer 3 has the lower conductivity. And the slope in layer 2 indicates that it has a finite (not infinite) conductivity higher than that of 1 or 3.

Based on observations made in (D), you can conclude also that (B) represents the "on" state. As you explain, (C) is not sensible unless heating originates at the interface between layers 1 and 2. (A) also does not make sense, because it suggests that there is a heat sink between 1 and 2 (where is that heat going?).


Why is it lead(II) chromate and not lead(II) chromate(VI) tetraoxide? I am just beginning in chemistry and learning how to name the different molecules. I have this molecule: $\ce{PbCrO4}$.

I would give it the following name: lead(II) chromate(VI) tetraoxide. Lead and chrome have both multiple oxidation states. And as you have to specify about which ion you are speaking, it seems to me this would be the most correct name.

But this is the correct name apparently: lead(II) chromate.

Could someone please explain:

why you just specify the oxidation state for the first element and not the other? As they are both metals and both have multiple oxidation states.
why you don't say tetraoxide, although there is clearly a number 4 written?

Chromate ($\ce{CrO4^2-}$)is a poly atomic ion which is partially defined by the oxidation state of chromium.  Its like the sulfate ($\ce{SO4^2-}$), phosphate ($\ce{PO4^3-}$), nitrate ($\ce{NO3-}$), permanganate($\ce{MnO4-}$), and carbonate ($\ce{CO3-}$) polyatomic ions. For each ion the oxidation state of the center ion is defined therefore there is no reason to state it.  You could in principle call lead (ii) chromate lead(ii) chromium(vi) oxide or tetroxide, but this is less informative of the structure of the dissolved species.


Claisen Condensation vs. Trans-Esterification In the Claisen condensation, an alkoxide $\ce{RO-}$ acts as a base. However, in trans-esterification, an alkoxide $\ce{R'O-}$ attacks the carbonyl carbon of an ester as a nucleophile. Is this because a different 'R' group is used in trans-esterification?

In the Claisen condensation, notice that the alkoxide base usually matches the ester given. For example, if an ethyl ester is used, the base will be ethoxide. The reason for this is that trans-esterification is occuring, except that the product is the same as the reactant.

In general, the Claisen condensation is slow compared to trans-esterification. The rate limiting step of the Claisen is the C-C bond formation. Keep in mind that there will be very little of the active ester enolate nucleophile present, because the equilibrium of alkoxide deprotonating an ester's alpha-position will lie dramatically to the ester (by ~ 10 billion fold).

If one tried to run a Claisen condensation where the base does not match the ester (e.g., an ethyl ester with methoxide as the base), I would expect a mixture of methyl and ethyl esters as the products.


Thermodynamic and kinetic products - carbocation stability The professor wants to know which is the thermodynamic product and which is the kinetic product.

He says there are only two products - the two on the right that result from the resonance-stabilized complex. I can see a third product - the product on the far left with the question mark over an arrow.

The argument for the far right product being the kinetic product is that it is a tertiary carbocation and therefore the energy barrier to formation is probably lower than the energy barrier to formation of the secondary carbocation.

Alright, sure. However, could one also argue that the far left product is the kinetic product? It is after all the result of a tertiary carbocation. And it is the result of a "pure" tertiary carbocation - not a tertiary carbocation "diluted" by resonance; the far right product results from a resonance-stabilized complex that has both secondary and tertiary carbocation character. This was something I thought of at first. But then I realized I was probably taking resonance contributors too literally and that the resonance-stabilized complex would probably be more favored intermediate because the positive charge is "delocalized".

So what's your take?

Background

First off, let's discuss what we mean by "kinetic" and "thermodynamic" products.  The kinetic product is the first formed product, the one that forms the fastest, the one with the largest rate constant. The rate constant, k, can be described using the Arrhenius equation
$$\mathrm{k=Ae^{\frac{-Ea}{RT}}}$$
where A is the pre-exponential factor, a measure of the probability of a successful collision. Ea is the activation energy, the energy required to surmount the barrier to reaction, R is the gas constant and T is the temperature in Kelvin.

Usually, in a series of related compounds like we have here, the A factor (which is a very large number to begin with, so would need to change by a large amount to alter the course of a reaction) is assumed to remain relatively unchanged.  Consequently the activation energy is viewed as the controlling factor in terms of which reaction pathway is favored.

The thermodynamic product is the product with the lowest energy of formation, the most stable product.

In a reaction where several products can be formed the thermodynamic product could be the same as the kinetic product, or it could be different.

Two examples to further clarify kinetic and thermodynamic products can be found in this earlier answer describing the Diels-Alder reaction between cyclopentadiene and maleic anhydride. The endo isomer is formed first - it is the kinetic product, but if the reaction is run under conditions permitting equilibration, then ultimately the more stable exo product will be predominate. Another example can be found in the sulfonation of naphthalene where naphthalene-1-sulfonic acid is formed initially (the kinetic product), but upon heating will equilibrate and rearrange to the more stable 2-isomer.

A key point here is that in order for the kinetic product to rearrange to the thermodynamic product, the reaction must be reversible under the reaction conditions.

Back to your question

  could one also argue that the far left product is the kinetic product?

Technically, according to your diagram, the answer is "no". There are no equilibrium arrows in your reaction diagram, therefore if the far left product were the kinetic product it would be observed.

But maybe you inadvertently left out the equilibrium arrows, what would the answer be if you included them? All 3 possible products are rather analogous, so for starters we would expect similar A factors in the 3 pathways and look to differences in the activation energies to decide which is the preferred pathway (kinetic product). Your product on the left requires a tertiary carbocation as its intermediate, the other two products require a resonance-stabilized allylic carbocation which is further stabilized by a methyl substituent.  The latter carbocation is lower in energy (resonance stabilization plus inductive and hyperconjugative stabilization by the methyl group) and so the kinetic product will derive from it.

Of the two allyl resonance structures you've drawn, the one on the right will carry a bit more importance in terms of describing the actual allyl carbocation due to the methyl stabilization described above. Hence I would expect the kinetic product to involve attack at the methylated allyl carbon leading to your far right product being the kinetic product.

Attack at the other allyl terminus produces a product with a methyl group on the double bond. We know from heats of formation that alkylation of a double bond stabilizes the system (because electrons are moving from higher [$\ce{sp^3}$] to lower [$\ce{sp^2}$] energy). Hence, your middle product is a possible  thermodynamic product.  Your molecule on the far-left also contains a tri-substituted double bond, it may also appear as a thermodynamic product.

Whether or not equilibration between the kinetic and thermodynamic products occurs depends on the reaction conditions.  Assuming equilibration is possible, then short reaction times at low temperatures will favor formation of the kinetic product, while higher temperatures and longer reaction times will facilitate transformation of the kinetic product into the thermodynamic product.


What does the crystal field splitting diagram for trigonal planar complexes look like? For a trigonal planar complex (e.g. $\ce{[AgCl_3]^2-}$):

According to this image (source):

The $x^2 - y^2$ and $xy$ orbitals are on the same energy level. However, assuming that one of the $\ce{Cl}$ ligands lies on the $y$ axis (or the $x$ axis), wouldn't the $x^2 - y^2$ orbital be higher in energy? Thanks in advance.

Say you put a ligand in the trigonal complex (or an equatorial ligand in a trigonal bipyramidal complex) on the $x$-axis, where $z$ is the "axial" axis.  Then that ligand is directly aligned with an $x^2-y^2$ lobe and falls on a "seam" of the $xy$ orbital.  But the other two ligands in the plane are closer to an $xy$ lobe ($15°$ off) than to an $x^2-y^2$ lobe ($30°$ off).  These effects balance so that both in-plane $d$ orbital see the same total metal-ligand intetaction.


Why does the boiling and freezing-point of non-polar molecules rise with increasing molecular mass? In the section about fuels, my textbook mentions that the boiling and freezing-points of alkanes (which are non-polar) increase with an increase in molecular mass due to London dispersion forces.

Now, I understand what these are and why they become relevant when talking about non-polar molecules, but I can't figure out how they are connected to the boiling and freezing point of the matter. How do these London dispersion forces influence the boiling and freezing point?

In general, the larger the molecule, the larger the accessible surface area of the molecule and the more surface polarisable electron density. There are more atoms that can participate in dispersion interactions and the effect stacks up. To melt or boil a substance you need to overcome the intermolecular forces holding the molecules together and dispersion forces -  whilst weak even with respect to hydrogen bonding - can add up to large interaction energies. One of the more interesting examples of strong dispersion interactions is the gecko's foot, which permits the reptile to easily stick to sheer surfaces due to the exceptional surface area of the footpad, which is covered in many many tiny hairlike protrusions.


Terminology for "proton stripping"? This is going to probably be an easy question for an organic chemist, but I am looking for the proper terminology to describe the following reaction mechanism.

In particular, what is a more precise way than "hydrogen stripping" to describe the process where a photogenerated hole on an oxygen can pick off the hydrogen and its associated electron to create a radical that reforms to make formaldehyde?

Further, would I say something like "an HCO radical attacks an adsorbed methoxy to create methyl formate" or is there a more precise way to talk about it? The proper reaction names would be ideal.  Many thanks from a physical chemist who took orgo way too long ago!

EDIT: Thanks Orthocresol for the hint about hydrogen abstraction. Upon looking into things more, hydrogen abstraction aka hydrogen atom transfer (HAT), can be viewed as a subset of proton coupled electron transfer reactions (PCET). The difference is that HAT occurs with minimal solvent reorganization and between similar donors whereas PCET is more general where there can be heat transfer (i.e. it is diabatic versus adiabatic.) So I believe that PCET may be the proper name compared to "hydrogen/proton stripping".

One last question: does anyone know what the reaction between the HCO radical and the methoxy would be called?  Thanks again!

Photocatalysis - reactions caused by light. The term seems to be predominately related to free radicals.

This is obviously a misnomer. The photon is consumed which isn't true for a "real catalyst."

I know that lasers have been used to create excited molecular states which then react. I investigated this for reactions in gas phase of different atomic isotopes of a chemical.

I think if you get any more specific than that then it degenerates into "commonly accepted names" for certain reactions. Sort of like you just have to know what acetic acid is. There is no way to "figure out" what the name means if you don't just know.


Pressure in Ideal Gas Law In Boyle's law ($p$ inversely proportional to $V$ for constant $n$ and $T$), $p$ refers to the applied (external) pressure on the gas.

However, in the ideal gas law, my textbook says that $p$ refers to the pressure of the gas, which I think refers to the pressure applied by the gas. How is this possible, since the ideal gas law is derived from Boyle's law (and Charles' and Avogadro's laws)?

Shouldn't the $p$ in the ideal gas law then refer to the applied (external) pressure like in Boyle's law, not the pressure of (i.e. applied by) the gas?

At equilibrium (that is, if volume/temperature/mass is stable and not changing over time), the external pressure applied to the gas will equal the internal pressure of the gas.

This is (more or less) a consequence of Newton's third law. The external environment pushes on the gas with a certain force, and the gas pushes back with an equal and opposite force. At equilibrium, no work is being done on the gas or on the environment, so the forces have to balance. (This is very crude and hand-wavy - a more thorough description would require much more math.)

So (at equilibrium), there really isn't any difference between the two. In the Boyle's law case, the value is described as the external pressure, because the typical experimental setup in that case involves using something like a piston to control and vary the external pressure applied to the system. In the ideal gas case, you may not explicitly be controlling the external pressure, instead you're controlling temperature, volume, or particle numbers. In those case, you're normally measuring the internal pressure of the system, rather than (directly) controlling it or the external pressure. But (at equilibrium) the pressure of the gas pressing out is the same as the pressure of the container pushing in.

It's all just a matter of perspective, and how the system is set up.


NaN3 quantity for the generation of N2 You wish to design an airbag for a car. The gas that is released must be at a pressure higher than that of air – $\pu{110 kPa}$ at a temperature of $\pu{22 ^\circ C}$. The bag that you develop can expand to a volume of $\pu{45.5 L}$. What quantity of sodium azide $\ce{NaN3}$ should be used to generate the required mas of $\ce{N2}$ gas?

I've been given this question on my revision sheet. So far I've figured out using Boyle's law that I need 2 mol of $\ce{N2}$ gas. However, I am unsure of how to transfer this over to the $\ce{NaN3}$ and use the ratio or what not to find the number of moles there.

$$\ce{2NaN3 -> 2Na + 3N2}$$

As you already found the amount of $\ce{N2}$ required, I'll just give you a hint on how to get the amount of sodium azide required to produce that much nitrogen.

We know that 2 moles of sodium azide gives us 3 moles of nitrogen. So $x$ moles of sodium azide gives us $2$ moles of nitrogen. Solve for $x$. That's all!


How can I mix methylene chloride, methanol and hydrofluric acid to make a paint stripper? HF paint stripper on data sheets lists
Methylene chloride 70>85%
Methanol           10>25%
Hydrofluric acid.   5%
Previously I have found the chemical splits. Is there a particular method to properly blend all 3 parts?

It is not clear how reliable the concentrations given in the SDS are. It is not uncommon to give a range, but actually manufacture product at one end of the range.
I would take 25 parts of methanol and 70 parts of dichloromethane and add hydrofluoric acid till it becomes cloudy - then back off (add some more CH$_3$OH + CH$_2$Cl$_2$). You may be able to add 4.6 parts of hydrofluoric acid (round up to 5 for the SDS).
The solubility of dichloromethane in water is 1.60% at 20$^o$C and the solubility of water in dichloromethane is 0.24% at 20$^o$C. Ref 1  The solubility of dichloromethane in water increases as the temperature decreases. Ref 2
One unknown is the concentration of hydrofluoric acid and whether higher concentrations of HF are more easily dissolved in this mixture than lower concentrations. The SDS may not give you useful information, but I suggest trying to use the most concentrated HF available - you can always dilute it, and H$_2$O is no friend of this combination.
Methanol and water (and thus, probably also hydrofluoric acid) are miscible and so are methanol and dichloromethane. A recent question on StackExchange (Ref 3) gave a ternary diagram of water-methanol-dichloromethane:

The line separating the one-phase (M) from the two-phase regions (I) does not go all the way to the top, but you can see that there are homogeneous mixtures with a lot of water, methanol and dichloromethane.
The Safety Data Sheet is required to list hazardous ingredients, but not non-hazardous materials. Some stabilizer might be added, hidden in the 15% range of dichloromethane and 10% range of methanol, that could keep the mixture homogeneous. One compound that comes to mind is propylene glycol. If the commercial product is cloudy, it might be stabilized with a thickener like guar or xanthan gum. Wax has been used as an evaporation inhibitor in paint removers containing dichloromethane; it might also act as a thickener and stabilizer.
Ref 1. https://macro.lsu.edu/HowTo/solvents/Dichloromethane.htm
Ref 2. https://en.wikipedia.org/wiki/Dichloromethane
Ref 3. Mixture of water, methanol and dichloromethane


How can 30 ml of water be heated in less than 10 seconds? How is it possible to heat a tiny amount (30 ml)[1] of water to a high enough temperature to make a coffee, in less than 10 seconds and possibly instantly?

Most heaters that I know of heat water in no less than 90 seconds (induction heater) or 3 minutes which is way too much for my purposes, and a microwave oven takes me one minute and a half to heat.

[1] A single coffee is between 20 - 30 ml and my coffee machine makes 30 ml for each cup.

Well, let's do some math:

Assuming 30 mL of water is 30 g, and we want to heat our water from 20 °C to 90 °C, the energy required is:
$$\begin{align}E&=C_Pm\Delta K \\
&=\left(4.18 \mathrm{\frac{ J}{gK}}\right)(30\mathrm{\ g})(70\mathrm{\ K})\\
&=8.778\mathrm{\ kJ}\end{align}$$

So how much power do we need to do this in a given time? "Instant" doesn't really mean anything, so let's go with 10 seconds:

$$\begin{align}P&=\frac{E}{t}\\
&=\frac{8778\mathrm{\ J}}{10\mathrm{\ s}}\\
&=877.8 \mathrm{\ W}\end{align}$$

This is not an enormous amount of power, but the trick is that it all has to go into heating the water. A good microwave outputs a fair bit more power than this, but it generally doesn't all get absorbed by such a such a small amount of water in only 10 seconds. Your best bet is probably an electric heating element directly inserted into the liquid, though I don't know if you can get a ~1000 W one small enough to sit in that much water.

As Jon Custer notes, it's not necessary to produce all the heat at once. If you heat some kind of thermal reservoir and flow the liquid past/through it, it reduces the demands on your heat source.

Edit: Also, I just tried this with a 1200 W microwave and it only took 15 seconds. How fast do you really need this coffee?


What happens to the argon in the Haber-Bosch process? The Haber–Bosch process is used to synthesis ammonia from atmospheric nitrogen and hydrogen derived from a hydrocarbon such as methane. The process requires high temperatures and pressures and also requires repeated cycling of the gasses.
I would have thought that atmospheric argon and other noble gases would have slowly built up in the reactor. How are these inert gases removed?

Inert gaseous components such as methane and argon are indeed should be eliminated from the system in order not to lower the partial pressure of the reactants too much. Technically there is usually a standalone gas separation plant where the extraction of argon from the recycle gas is performed basically using modified Linde process [1 , pp. 428–430].

Suitable cryoprocesses [...] are available for production of noble gases from synthesis purge gas. Initially, extensive separation of nitrogen, argon, and methane is carried out by partial condensation. Purge argon is then recovered from the condensate in a two-stage condensation process. If helium is to be recovered it can be concentrated by liquefaction of the hydrogen in the hydrogen-rich gas phase, followed by purification. The heavier noble gases, krypton and xenon, pass into the methane fraction.
The purge gas at pressures of up to $\pu{70 bar}$ is ﬁrst introduced into the adsorber (a) where traces of water and ammonia are removed. It is then transferred to the heat exchangers (b1) and (b2) for cooling. The gas is then fed into separators (c1) and (c2) where separation of liquefied nitrogen, argon, and methane from gaseous hydrogen takes place, and dissolved hydrogen ﬂashes off. The liquid bottom product is fed into the fractionating column (d1) where methane (bottom) is separated from the nitrogen fraction (top).
A methane-free nitrogen – argon mixture (liquid) is withdrawn from the middle of this column and fed as reflux into the $\ce{Ar}$ purification column (d2), where nitrogen is separated from argon. The bottom product is argon of product purity and is transferred to a vacuum-insulated storage tank.
Both columns (d1) and (d2) are operated within an $\ce{N2}$ cycle in which cold is produced by expanding high- or medium-pressure nitrogen.
The $\ce{CH4}$ bottom stream from (d1), compressed by liquid pump (g), is evaporated against feed gas and normally led to battery limits as fuel gas. The higher boiling noble gases krypton and xenon are contained in this stream and, in principle, can be isolated.

Reference

Häussinger, P.; Glatthaar, R.; Rhode, W.; Kick, H.; Benkmann, C.; Weber, J.; Wunschel, H.-J.; Stenke, V.; Leicht, E.; Stenger, H. In Ullmann’s Encyclopedia of Industrial Chemistry; Wiley-VCH Verlag GmbH & Co. KGaA, Ed.: Weinheim, Germany, 2001. DOI: 10.1002/14356007.a17_485.


Phase diagram for first 10 elements I've been looking on the net for a reference which shows a phase diagram for elements at least from $\ce{H}$ up to $\ce{Ne}$. Specifically, I'm looking which of these elements can be solid or liquid in a vacuum.

I haven't really found anything, except some anecdotical comments here and there that hydrogen and helium evaporate in a vacuum at any temperature.

Any suggestions as to where these data might be published?

This may be what suffices you are looking for :

Periodic Table of Elements w. Selected Phase Transition Temperatures

This may get you started:

MatDL - The Materials Digital Library Synopsis
NIST SRD ( I would write them)
Java Applet: Phase Transition Temperatures (untested)

It is a very good question, which I came across as well. You could just google-image search phase diagram + ELEMENT to copy and past the most appealing charts into a table in your editor of choice.
If the intend is personal use, this might be the fastest route.

wolfram alpha should allow you to look up the state of the element at a given pressure as follows:

sodium phase at 0 bar

Helium itself is a particularly interesting subject.

I hope this helps for a start...


How do I calculate how much sodium hydroxide I need to add to get a new pH? A 100 ml $\ce{HCl}$ solution has a pH of $3.7$. You want the solution to be of pH 4.5. You have a solution of $10\ \mathrm M$ $\ce{NaOH}$. How much $\ce{NaOH}$ do you need to add to to the $100\ \mathrm{ml}$ solution of $\ce{HCl}$ to get a pH of 4.5?

I firstly calculated the concentration $\ce{HCl}$ in $100\ \mathrm{ml}$ solution = $10^{-3.7}$ = $2.0\times10^{-5}\ \mathrm{mol}$ $\ce{HCl}$ per $100\ \mathrm{ml}$. After this I calculated the $\left[\ce{HCl}\right]$ of the new solution: $10^{-4.5}=3.16\times10^{-6}\ \mathrm{mol}$ $\ce{HCl}$ per $100\ \mathrm{ml}$ of new solution. Now I'm stuck.

At the beginning, the amount of substance of hydronium ions is:
$$n(\ce{H3O+})=10^{-3.7}  \times 100 \times10^{-3}=10^{-4.7} \mathrm {mol}$$
After adding sodium hydroxide, the amount of substance of hydronium ions is:
$$n(\ce{H3O+})=10^{-4.5}  \times 100 \times10^{-3}=10^{-5.5} \mathrm {mol}$$

The difference represents the amount of substance of hydronium ions that have reacted with hydroxide ions.
$$n(\ce{H3O+})=10^{-4.7}-10^{-5.5}= n(\ce{OH-})$$
The volume of sodium hydroxide to add is:
$$v(\ce{OH-})=\frac {10^{-4.7}-10^{-5.5}}{10}= 1.68\times10^{-6}\mathrm{L}=1.68\,\mu \mathrm{L} $$

Assuming the volume of a droplet to be about 60 micro liter, it is much less than a droplet!


Unusual oxidation states of alkali metals When going through list of oxidation states on Wikipedia I encounterd there that sodium, potassium, rubidium  and caesium exhibit oxidation states of -1, but not lithium, even though its electronegativity is more than that of the others. How's that possible? Can some one provide an example.

You are talking about alkalides, salts where the anion is an alkali metal. There is a very brief overview on Wikipedia which also provides a couple of examples, including $\ce{[Na(\text{cryptand[2.2.2]})]+Na-}$. A good university level inorganic text book such as Greenwood and Earnshaw or Housecroft and Sharpe will provide more detail.

If you are interested in unusual oxidation states of alkali metals you might also like to know that in electrochemical experiments there is some evidence for $\ce{Cs^3+}$, which is isoelectronic with $\ce{Xe^2+}$. Again I think Greenwood and Earnshaw discusses this, but I don't have it to hand at the moment to confirm.


Why is sodium acetate a good nucleophile? As $\ce{CH3COONa}$ is bulky, I thought that $\ce{CH3COONa}$ is a weak nucleophile.
But I found out that it's a nucleophile good enough to perform $\mathrm{S_N2}$ reaction on a secondary carbon.

Why is it so?

The acetate anion is actually not particularly bulky. The negative charge is located on the two oxygens which are approximately at a $120^\circ$ angle away from the only residue that has the potential to be bulky. Whatever that residues does will be ‘behind’ the carbon and hardly noticeable. However, they are still rather bad nucleophiles since the negative charge is delocalised over two oxygen atoms in the π system.

I’m actually not sure which orbital attacks nucleophilically but it could equally be an oxygen lone pair rather than the π system. If that is the case, there is really very little steric hindrance that could inhibit the reaction from proceeding. Thus, acetate is able to attack nucleophilically, as it does for example in a Mitsunobu reaction.


Does boiling milk really make the nutrients in it less useful to the body when ingesting it? Does “denaturing” nutrients in milk by boiling it make them less useful, or actually more useful, as they undergo “denaturation” anyway during digestion?

I've read online that boiling milk "denatures" nutrients such as protein, calcium, and others contained in milk, rendering the milk less nutritious.

I've also read that because the nutrients in milk (and other foods) undergo "denaturation" in our body during digestion anyway, boiling it will not make milk less nutritionally valuable, and will actually help the body to more readily absord the nutrients in it into the bloodstream right after ingestion.

What is the truth in this matter? Thanks for any knowledgeable answers!  I put the word "denature" in quotes because it's a technical term and I'm not really sure in which ways it can correctly be used.

The term "denature" is generally used for proteins, and it is true that boiling milk will denature the proteins that are present. Calcium and other such elements would not be "denatured" per se. In fact, I can think of no adverse effect of boiling on the chemical nature of calcium, magnesium, etc. I'm not entirely sure what other minerals and/or nutrients are present in homogenized milk.

As for material being "denatured" during digestion, this is probably an inappropriate term since we generally use the term "metabolized" instead of "denatured". When proteins and other "nutrients" are metabolized, they result in useful products, as opposed to when they are "denatured". So, in summary, I'd say that boiling milk is not the same a metabolizing milk, and I'd say that boiling would likely render milk less "nutritious".


What is an appropriate way to write a recursive SMARTS that excludes trifluoromethyl groups as a potential R group? C[CX4&!$(C(F)(F)F)] is the smarts I've tried. It does not exclude the trifluoride in CC(F)(F)F.

I'm not sure if I correctly understand your intent.  My assumptions are

you have a set of molecules, e.g., as a list of SMILES strings (say FC(c1ccccc1)(F)F, CCCC(F)(F)F, CC(Cl)(Cl)Cl, and CCCO) in a file like probe.smi
you want to identify the molecules without the $\ce{-CF3}$ group

If so, I would i) search for the molecules with said group in a first step and then ii) remove this sub-set of positives from the set to know about the molecules without the $\ce{-CF3}$ group.

OpenBabel's positives: obabel -ismi probe.smi -s'C(F)(F)F' -osmi -O positives.smi.  In the present example, this yields only FC(c1ccccc1)(F)F and CCCC(F)(F)F, regardless of the different notation of the trifluormethyl group in the entries concerned.

The molecules lacking the $\ce{-CF3}$ group are just the complementary sub-set to the sub-set of the positives.  To start, one could opt for a Python script like:

#!/usr/bin/python3
"""Report molecules lacking the -CF3 group."""

def reporter():
    """Report .smi of molecules without -CF3 group."""
    # prepare lists about the molecules:
    all_mols = []  # all molecules, no discern
    positives = []  # molecules with -CF3
    negatives = []  # molecules without -CF3

    with open('probe.smi', mode='r') as source:
        for line in source:
            all_mols.append(line.strip())

    with open('positives.smi', mode='r') as source:
        for line in source:
            positives.append(line.strip())

    # truncate the set to yield the sub-set about molecules without -CF3 group:
    set_all_mols = set(all_mols)
    set_positives = set(positives)
    set_negatives = set_all_mols.difference(set_positives)

    # convert the sub-set (back) into a list of entries:
    negatives = list(set_negatives)

    with open('negatives.smi', mode='w') as newfile:
        newfile.write('\n'.join(negatives))

if __name__ == "__main__":
    reporter()

which in file negatives.smi correctly reports CC(Cl)(Cl)Cl and CCCO as molecules without the $\ce{-CF3}$ group.

Of course it would be nice to identify how to negate OpenBabel's search filter (-s'C(F)(F)F') to shorten the route traveled here.


Creating ethane through electrolysis of vinegar Some
 sources that I studied say that it is possible to produce ethane
through the electrolysis of ethanoic acid. Would this work with vinegar
(5-10% acid)? Also, other sources say that ethanoate salts should be used.
 I can easily obtain such salts (sodium hydrogen carbonate (baking soda)
 + ethanoic acid (vinegar)), and vinegar as well, so which one is better
for creating ethane? I don't care about the amount of carbon dioxide,
oxygen, hydrogen or any other gas produced.

Indeed that is quite possible. The reaction was first described by Hermann Kolbe and is hence called "Kolbe's Electrolysis"

It is a decarboxylative dimerisation of two carboxylate ions (carboxylate salts--either sodium, or potassium are used instead of carboxylic acids for precisely this reason as they disassociate more readily to furnish carboxylate ions).

The general reaction scheme is is a two-stage radical process: first, electrochemical decarboxylation gives a radical intermediate; then two such intermediates combine to form a covalent bond.

For sodium acetate (again would furnish acetate ions more readily than acetic acid) one would see the following reaction:

$\ce{2CH_3COO^-}\ce{Na^+} \rightleftharpoons  \ce{2CH_3COO^-}+ \ce{Na^+}$

At the anode: a) $\ce{2CH_3COO^-} \stackrel{-2e-}\rightarrow \ce{2CH_3COO.}$

Electrons shuffle around a bit in the acetate radical and we get

b)$ \ce{2CH_3COO.} \rightarrow \ce{2CH_3.} + \ce{CO_2} $

c)$ \ce{2CH_3.} \rightarrow \ce{CH_3CH_3} $

At the cathode: a) $\ce{H_2O} + \ce{e-}\rightarrow \ce{OH^-} + \ce{H.}$

and, finally $ 2\ce{H.} \rightarrow \ce{H_2}$

Now, here we observe that the solution is becomes more basic as the reaction proceeds since hydroxide ions are being produced. If we make the conditions mildly acidic, and draw on Le Chatelier's principle, we can shift the reaction to the right.

So, it would make sense to use sodium acetate, in the presence of ethanoic acid to perform this reaction.


Is there an established procedure for titanium tetrachloride titration? I’m attempting to do an aldol reaction with $\ce{TiCl4}$ as the catalyst. It worked well in the large scale ($100~\mathrm{mg}$) test system but not, so far, in my $5~\mathrm{mg}$ scale actual system. Our bottle of $\ce{TiCl4}$ is not the youngest, the seal has $\ce{TiO2}$ precipitates on the inner side, and when you finally manage to stick a needle from the argon line in, you almost get a geyser of hydrochloric acid steam — need I mention that the liquid is cloudy from precipitated $\ce{TiO2}$?

My supervisor said distillation is not an option due to unremovable residues of $\ce{TiO2}$ getting stuck in the distillation apparatus. But I would love to know what purity the $\ce{TiCl4}$ still has and thus how much I should upscale it when adding it to the reaction.

Are there any rather simple titration methods for $\ce{TiCl4}$ known or established like they are for n-butyllithium?

$\ce{TiCl4}$ is amenable to distillation if done carefully so as to exclude moisture. That doesn't mean it isn't a pain, however.

We do a lot of chemistry using $\ce{TiCl4}$ and other Ti Lewis acids (including some Ti aldol reactions on v. small scale), so end up having to distill it quite often (roughly once a quarter). Once distilled, it lasts in the fridge until it runs out so long as the flask is well sealed and everyone remembers to flush with argon before they're finished using it.

I'll quote from Purification of Laboratory chemicals for general information, but also note that we don't follow this procedure. (Mercury makes me uncomfortable)

  Ref: 'Purification of Laboratory Chemicals', 6 ed.

In practice, we simply distill the $\ce{TiCl4}$ under reduced pressure (high vacuum pump), discarding the first few mL as would be common procedure. This usually yields $\ce{TiCl4}$ that is clear in colour and sufficiently pure for our chemistry. If some colour remains, a second distillation from copper turnings is occasionally done, but this is rarely needed.

The key bit of information is to make sure that you quench all of the glassware and tubing used with a weak base (sodium hydrogen carbonate is sufficient), and to ensure that the pump exhausts into a good fume cupboard incase any fumes do escape. All of the glassware used should also be heavily dried prior to use.


How to identify a pseudo-first-order reaction? We can easily identify a pseudo first-order reaction with the integrated rate equation, but what if only the reaction is given? For example, what if I were given a question like the following on the hydrolysis of acetic anhydride?

Identify the type of this reaction: $$\ce{(CH3CO)2O + H2O -> 2CH3COOH}$$

A. Pseudo-first-order reaction
B. Pseudo-second-order reaction
C. Zero-order reaction
D. Third-order reaction

I would like to know how I could identify the correct answer if such a question were asked.

There are four steps:

Identify the mechanism.
Find the rate-determining step.
Write the corresponding rate equation according to the stoichiometric coefficients of that step.
Remove abundant reactants.

1. Identify the mechanism

Assuming that there is no acid to catalyze the reaction, it proceeds via a nucleophilic addition/elimination reaction.

An acetic anhydride molecule $(1)$ is attacked by the lone pairs on the water to form the complex $(2)$. This is the addition step.

The complex tautomerizes to give $(3)$, which is converted to our products, namely 2 acetic acid molecules $(4)$. This is the elimination step.

2. Find the rate-determining step.

Presumably, it is the first step.

3. Write the rate equation.

In the first step, one molecule of acetic anhydride reacts with one molecule of water. Therefore, it is a first-order reaction with respect to both of them:

$\text{Rate} = k [\ce{Ac2O}] [\ce{H2O}]$

4. Remove abundant reactants.

Water is abundant, so it is removed from the equation:

$\text{Rate} = k [\ce{Ac2O}]$

Conclusion

It is a pseudo-first-order reaction [A].


Understanding smoke chemistry of tobacco and/or marijuana In an effort to understand the effects of tobacco vs marijuana smoke on health, the chemistry of smoking plants, and remembering some of the chemistry I took in college, I came across
a 2007 study comparing mainstream and sidestream marijuana and tobacco cigarette smoke,  according to which:

  a number of chemicals were
  present in marijuana smoke at levels that were substantially higher
  than in tobacco smoke. For example, NO, NO x , hydrogen cyanide, and
  aromatic amines were present in marijuana smoke at levels 3-5 times
  higher than in mainstream tobacco smoke, while ammonia was present at
  levels 20 times higher than tobacco. Conversely, some compounds such
  as PAHs, formaldehyde, and acetaldehyde were found at moderately
  higher levels in tobacco.

I have mostly understood the study but I do need some things clarified.

Question 1: Could the combustion of any plant material whether it is an edible herb or in my particular example, tobacco or marijuana, potentially produce similar products or by-products (e.g HCN, ammonia), because of the same functional groups present in the different compounds of the tobacco or marijuana plant?

Question 2: Would all of the molecules in the figure below which are found in the tobacco plant, be independently reacting with oxygen and what is the reaction of nicotine producing HCN and/or an aromatic amine which if I remember correctly is a cyclic molecule?

Picture source

Question 1 is easy to answer: Yes.

The ‘burning’ of whichever plant you’re (or somebody else is) smoking is an incomplete combustion on the ember end of the cigarette/joint/whatever. In any incomplete combustion of nitrogen-containing organic compounds, $\ce{HCN}$ and ammonia can (and likely will) be a side product. Aromatic amines are probably also always generateable as are $\ce{NO_x}$. And PAHs, aldehydes etc. can also result from the incomplete combustion of practically all plant or animal material.

Question 2 is slightly harder. To the best of my knowledge, however, the answer is: We do not know.

While we may be able to say that all the molecules in your question would be independently reacting, the mechanisms are so many and so complex that I very much doubt they have been the target of investigations. I think I recall somebody once studying the reaction mechanism of the complete combustion of methane and coming up with tens if not hundreds of partial reactions and steps. That was complete combustion (i.e. enough oxygen present), so it all boiled down to:

$$\ce{CH4 + 2O2 -> CO2 +  2H2O}$$

In incomplete combustion, we cannot even be fully sure of the reaction equation (and the stoichiometries involved therein), let alone a mechanism.

Side note: the complete combustion of a cigarette in pure oxygen takes about seven seconds ($\pu{7s}$) and oxidises it entirely to $\ce{CO2}$ and $\ce{H2O}$ (plus something from the nitrogens and likely $\ce{SO3}$ from sulphur/$\ce{P4O10}$ from phosphorus)


Why is fluorine one of the most electronegative atoms even though it doesn't have the largest effective nuclear charge? Why doesn't electronegativity increase as the effective nuclear charge increases? Most atoms with a higher atomic number than fluorine have a higher effective nuclear charge. However, flourine is a more electronegative atom.

Fluorine is the most electronegative element because the definition of electronegativity makes it so. The electronengativity scales are defined based on experimentally determined properties of the elements. Fluorine has appropriate values for all of the common scales to ensure it has the highest electronegativity.

The Pauling scale, which is the first electronegativity scale proposed, relates the difference in electronegativity $\Delta\chi_{A,B}$ between two atoms $(A\ \& \ B)$ is related to the bond dissociation energies $(E_d)$ of the diatomic species $AA,\ BB,\ \& \ AB$:

$$\Delta \chi_{A,B}=|\chi_A - \chi_B| = (\text{eV})^{-1/2}\sqrt{E_d(AB)-[E_d(AA)+E_d(BB)]/2}$$

Since only differences can be determined, one element (hydrogen) was chosen to have an arbitrarily set value. On the Pauling Scale Hydrogen has a value of $2.20$. Since the $\ce{H-F}$ bond is much stronger $(E_d=586\ \text{kJ/mol})$ than the average of the  $\ce{H-H}$ bond $(E_d=436\ \text{kJ/mol})$ and the $\ce{F-F}$ bond $(E_d=157\ \text{kJ/mol})$. Data from here. In general, since fluorine forms stronger bonds to all other elements than it does to itself, fluorine will have the highest electronegativity.

The Mulliken scale relates electronegativity to two other properties: electron affinity $E_{ea}$ and ionization energy $E_i$:

$$\chi_A = \frac{E_{ea}(A)+E_i(A)}{2}$$

Fluorine has a high electron affinity (easy to gain an electron) and a high ionization energy (hard to lose an electron).


Standard Heats and Free Energies of Formation and Absolute Entropies of 2-Chloro-2-methylpropane Does 2-Chloro-2-methylpropane have standard heats and free energies of formation and absolute entropies values in a liquid state?
http://www.wiredchemist.com/chemistry/data/thermodynamic-data
This website only provides the values in gaseous state and I can't seem to find the values for the liquid state anywhere.

$\Delta H_\mathrm{f}^\circ = -211.3~\mathrm{kJmol^{-1}}$

Neither source provided entropy or free energy data though but this is a start.

http://courses.chem.indiana.edu/c360/documents/thermodynamicdata.pdf
    http://webbook.nist.gov/cgi/cbook.cgi?ID=C507200&Units=SI&Mask=2#Thermo-Condensed


Irreversible chemical reaction in a battery Inside an ideal battery, the chemical reaction is reversible (at equilibrium), thus the oxidation-reduction reaction $\ce{Zn + Cu^{2+} -> Zn^{2+} + Cu}$ is happening in both ways at the same rate right?

Now, for a real battery, there's resistivity inside it due to many factors, thus the chemical reaction is irreversible. My question is, during the irreversible reaction, is the reaction happening in both ways but at different rate due to resistivity inside the battery or it is just like the combustion reaction happening in one way?

The original poster has a serious misconception by equating a reversible chemical reaction with a reaction that is in equilibrium. The electrochemical reaction that allows the battery to deliver energy to the load (such as the copper/zinc reaction mentioned) is always reversible in theory by reversing the current flow. However in addition to the electrochemical reaction, there are other chemical reactions going on inside the battery which are not reversible as well as geometrical changes that may not allow the reactants to be close enough for the electrochemical reaction to run in reverse. In batteries designed to be recharged, these geometrical and irreversible chemical changes are minimized to allow as many recharge cycles as possible. Batteries not designed to be recharged therefore usually can't be recharged fully and have very few recharge cycles not to mention a high probability of gaseous or corrosive liquid leakage.

At equilibrium, both forward and reverse reactions are going at the same rate, but this is only discernable microscopically. At a macroscopic level you see nothing happening. So a battery is never at equilibrium while you are using it since at that point the battery is completely dead. Then (or preferably sooner) it is time to recharge the battery or throw it away.


Hill's system alphabetizes elements symbol or name? Take monosodium glutamate.
Its chemical formula is $$\ce{C5H8NO4Na or   C5H8NNaO4}$$
First way nitrogen, oxygen, sodium, with is in order of element's name.
Second way, N, Na, O, in order of element's symbol.
Have seen it both ways, hoping someone could clarify. Thanks.

Go straight to the original source, ON A SYSTEM OF INDEXING CHEMICAL LITERATURE; ADOPTED BY THE CLASSIFICATION DIVISION OF THE U. S. PATENT OFFICE, and you can see that Hill uses the element symbol to determine alphabetical order.

In your example $\ce{C5H8NNaO4}$ is correct.


Best Fuel for Potato Gun I have been using deodorant to propel potatoes (potato gun :D), but I want to switch over to pure butane. However, in order to do this safely I have to know how much butane is typically in deodorant (percentage). If you do know, please also include a source.

Secondly, for a potato gun, is it wiser to use butane or propane?

And could a spark ignite butane? Or is a real flame needed? Tested this already, it does.

How much butane is typically in deodorant (percentage)

Here's a link to an MSDS for a deodorant; it claims the butane range is 30-60% (they don't state if this is by volume or weight). It might be cheaper to just go to a hardware store and buy a small butane cylinder, like the ones used for outdoor cooking, then you know exactly how much butane you are using.

  For a potato gun, is it wiser to use butane or propane?

My first though was butane because it gives off a bit more heat per mole (or per unit volume) when it burns, than does propane.

$$\ce{C3H8 + 5 O2 -> 3 CO2 + 4 H2O + \pu{2219 kJ/mol}}$$
$$\ce{C4H10 + $\mathrm{6.5}$~ O2 -> 4 CO2 + 5 H2O + \pu{2878 kJ/mol}}$$

However, according to "Common Fuels for Combustion Spud Guns",

  The key value for comparing two fuels based on their heats of combustion is not the actual heats of combustion. Instead, the "Heat per mole Oxygen" should be used since the amount of energy in the combustion chamber is limited by the amount of oxygen present in the chamber. Fuel is added to match that amount of oxygen. As you can see from the table, there is relatively little difference between the various fuels based on their "Heat per mole Oxygen" values.

Both propane and butane give off $\pu{105 kcal/mol}$ of oxygen, so they should perform equivalently in a spud gun. The article goes on to note that, of common fuels, only hydrogen ($\pu{119 kcal/mol}$ of oxygen) and acetylene ($\pu{120 kcal/mol}$ of oxygen) produce significantly more energy under these conditions than other fuels.

Now going back to your first question, if you know the volume of your chamber, you can back-calculate the correct amount of air (oxygen) and butane.


Why doesn't hydrazine contain a double bond between two nitrogen atoms? I know that the Lewis structure of hydrazine contains $\mathrm{sp^3}$ hybridized nitrogen atom but on analyzing it's structure I realized there are many bond pair and lone pair repulsions. Instead of a single bond between the N atoms they could also have a double bond and a dative bond with h atoms.

When I asked my professor about this he said that the formal charge imparts much more instability. I was not satisfied with the argument. So, can anyone please provide a rigorous explanation of why I am wrong?

Very interesting question! Let's start with the nitrogens in hydrazine being approximately sp2-p hybridized, flat. The p orbitals on each nitrogen would overlap in a double bond configuration - but the "double bond" has four electrons, two from each nitrogen. If you consider a molecular orbital, the bonding AND the antibonding orbitals are filled, so the net bonding is zero. Actually worse than that, because the filled p orbitals on each nitrogen are aligned in the worst way - maximum repulsion.

The clever little nitrogens rehybridize to become pyramidal - perhaps not completely sp3, but not at all planar. This rearrangement hardly affects the N-H bonds, but allows the p orbitals to accept some s-character, which increases one lobe at the expense of the other. And of course, the prominent lobes find a way to point away from each other and reduce electrostatic repulsion of the electrons. You could say that the double bond character is reduced/eliminated, but the bonding loss is more than balanced by reducing the antibonding.

A useful comparison can be made with ethylene: planar, sp2-p, because there are only 2 electrons in the double bond.


Dilution terminology For thing like 8× dilution, that would be like 1 ml of a solution and then 7 ml of something like water to dilute right? Would 50 ml solution and 50 ml water be a 2× dilution? I’m a bit unsure about the terminology. If I an OD of 0.123 from a sample diluted 5×, would multiply 0.123 by 5 to get the real OD?

You are right.  The × in 8× refers to the fold-change in concentration between the concentrated source and diluted solution.

$$\mathrm{DF = \frac{C_{stock}}{C_{working}}}$$ where $\mathrm{C_{stock}}$ and $\mathrm{C_{working}}$ are the concentrations of a stock solution and the desired final concentration, respectively.

The concentration of the working solution can be expanded:

$$\mathrm{C_{working} = \frac{C_{stock} V_{stock}}{V_{working}} = \frac{C_{stock} V_{stock}}{V_{stock}+V_{diluent}}}$$ where $\mathrm{V_{stock}}$ is the volume of the stock solution and $\mathrm{V_{diluent}}$ is the volume of diluent (e.g. pure water) added.

Plugging back into the DF equation:

$$\mathrm{DF = \frac{C_{stock}}{\frac{C_{stock} V_{stock}}{V_{stock}+V_{diluent}}} = \frac{V_{stock}+V_{diluent}}{V_{stock}}}$$


Amine Inversion and Paramagnetism In amine inversion, there is a planar $sp^2$ intermediate.

The electron pair transferred from the blue lobe to the red one during the inversion. It is natural to assume, for a layman like me, that in the planar intermediate the electron pair is present in both lobes- one electron in each. That would impart paramagnetic character to the molecule. Does this really happen? I couldn't find any such phenomenon on internet.

Picture: http://chemistryjee.blogspot.in/2015/02/amines.html

For somebody who is not well-versed in quantum mechanics, it might be natural to guess that this is the case. But if I were you, I wouldn't assume so.

Both electrons share the same p-orbital, and are therefore still paired. They both occupy the entire volume of the p-orbital. What they do not do, is to divide the p orbital up into two half-orbitals, with each electron taking one half.

There is a somewhat related question here, which might address some issues: How do Electrons Cross Nodes in Orbitals?


How do scientists count the exact number of electrons, protons, neutrons in the atom? An atom is very small so it is probably difficult to focus exactly on a single atom. I wonder how scientists count the exact number of electrons, protons, neutrons in an atom of a certain element.

Could you explain it?

In other words, how do we know that for example, $\ce{^{19}_9F}$ has nine protons and ten neutrons and nine electrons?

For charge neutrality, the number of electrons has to equal the number of protons.  So, the question really is how to know how many protons and how many electrons. This is actually a very interesting question dating back to the early days (and even before) of nuclear physics and chemistry.

Now, one can do it fairly easily with an ion accelerator and a magnet. With the right set up, one can get a beam of mono-energetic ions out of the accelerator, send them through a stripper to remove electrons, and then send them through a magnet to separate the components out by velocity (which depends on the ion mass through the energy). From the magnet deflections one can get a charge-to-mass ratios, and get back to number of protons (charge on a fully stripped ion) and number of neutrons (from number of protons and the mass). This is the heart of mass spectrometry (although one can use time-of-flight techniques rather than, or in addition to, magnetic separation).


Is the compressibility factor smaller or greater than 1 at low temperature and high pressure? The compressibility factor of a gas is defined as $Z = pV/(nRT)$. If attractive intermolecular forces dominate then $Z$ tends to be smaller than 1, and vice versa if repulsive forces dominate.

In the case where temperature is low, attractive forces increase due to less kinetic energy, but at the same time repulsive forces also increase due to high pressure. How can I determine which force will dominate?

If we use the Van-der-Waals equation then a plot of Z vs reduced pressure can be constructed. By using reduced values, i.e. pressure and temperature relative to the critical values then a general plot can be made on which data from many different molecules can be superimposed. Look under sections dealing with the Law of Corresponding states in your textbook. From such a plot then it is possible to determine whether the attractive or repulsive part of the potential is the dominant one.

The figure shows such a plot of Z vs. reduced pressure $P_R$ at different reduced temperatures. The reduced temperature is $T_R=T/T_c$ where T is the actual temperature and $T_c$ the temperature at the critical point.

From the figure when $T_R \lt 2$ and $P_R \lt 6.5$ then $Z \lt 1$ and this indicates that a real gas (in so far as such is described by the Van-der-Waals eqn) has a lower pressure than an ideal gas and so this indicates that the molecules are more influenced by the attractive part of the potential than its repulsive part, which is due to their finite molecular volume. If the temperature is higher, above $T_R \approx 2$ (not shown), then Z is positive for all pressures, and similarly above $Z \approx 7$ for all temperatures.  In these cases repulsion dominates over attraction.


What's the composition of gun smoke and is it dangerous? Background

I've been conducting an informal (i.e. mental) risk assessment of visiting gun ranges. There are a couple of obvious sources of hazard:

Injury by bullet
Hearing loss
Hot casings flying around
Gun malfunction

The first one is addressed by the fact that you should assume every gun to be loaded until personally convinced otherwise, and to always keep it pointed at the target area when handling the weapon. The second and third points are addressed by safety measures such as wearing ear and eye protective equipment.
The fourth one is kind of a wildcard, because you would assume that people who go shoot guns in a range are professionals with training. This may not always be the case, and statistics of that should be taken into account in the final verdict of the risk assessment, but I will ignore it for the time being.

The Question

Suddenly, I was struck by the idea that guns also may pose a respiratory hazard. The smoke that is formed has a distinct smell, possibly combined from the lubricant on the gun and the combusted propellant of the bullet. What exactly is this "gun smoke" made of, and should I be concerned for my health in this regard?

An image to show what this gun smoke can look like (Source: Arizona Sonora News):

Bonus Question

There is an additional level of concern when handling even larger payloads than a simple 9 mm bullet, but the documentation may be restricted and as such I do not require an answer to the following question: Are there any requirements for military personnel operating cannons and other large artillery to protect their airways from the fumes?

Toxic smoke from propellant combustion has been known about for over 500 years..
The propellants used today are Nitrogen based, thus nitro glycerin, nitrocellulose and nitroguanidine . These produce gases containing oxides of nitrogen, carbon oxides, carbon monoxide, hydrogen cyanide, ammonia and several other gases..
The oxides of nitrogen can be inhaled deeply into the lung thus causing damage as deep as the Alveoli, thus causing bronchiolitis obliterins resulting in a "Pulmonary Edema". This can happen 5 years after exposure..
If you show signs within 24 hours of exposure then oxygen and Arterial Blood gas tests are the solution to treating the condition and knowning the effect. The symptons show just like influenza but no sign of virus.
This is known as an Acute Toxic Smoke Inhalation Injury.
WW1 and WW2 this was studied to reduce the effect on gun crews on both land and sea. It is still a problem today with Armoured vehicles and Artillery and one of the reason the Navies use unmanned turrets.
Look up this file and you will find quite a bit of information about this condition.

Occupational Health - The Soldier and the Industrial base

I had first hand experience of this injury in 1988 and am still alive to tell my story, I had a "Pulmonary Edema" in 1991 and survived that, so I am in the 2% of survivors from that period. In the mid 1990's they discovered how to treat a pulmonary edema thus it has increased the survival rate.

There is thousands of pages of information on this subject, just look for "Propellant exhaust gas exposure" or "Toxic gas inhalation" that will bring up more information.

The military threshold limits are higher than civilian but a tank will expose the crew to 1,300 the TLV for less than 5 minutes if the fume extractor is working. In my case it was not!

http://www.public.navy.mil/surfor/documents/p_5041.pdf  Chapter 10

http://www.cs.amedd.army.mil/FileDownloadpublic.aspx?docid=a1d49fb8-3a16-4489-9408-1089f7044b4c

https://permanent.access.gpo.gov/lps101453/www.bordeninstitute.army.mil/published_volumes/occ_health/OHch10.pdf

There are 3 links for you to go to...
Enjoy reading it...
Oh just as an idea, if you had 1,100 .308(7.62mm) rifles fired in a 100 m3 room, its would be the equivalent of 5x L7 105mm tank rounds...
Enjoy..


Why do single elements in a molecule have an oxidation state of zero? For example, take $\ce{O2}$.

The oxidation state of oxygen is -2, yet once its in a molecule its oxidation state becomes zero?

How is this so?

Oxidation states are something of an "accounting fiction", used to describe where electrons are spending most of their time in a molecule.  This is turn is dictated by electronegativity: the more electronegative, the more the element attracts electrons.

So when you say oxygen's oxidation state is -2, this is only with respect to other elements of lower electronegativity. Now in the case of oxygen, the only other element with higher electonegativity is fluorine. For example, the molecule Oxygen Difluoride ($\ce{OF2}$) has oxygen in the +2 oxidation state because fluorine is more electronegative!  For any other element in combination with oxygen, however, we assume the electrons spend their time with the oxygen filling its outer valence shell, thus the -2 oxidation state.

But two oxygens in a molecule of gas have the same electronegativity, and there is no way to say that the electrons want to spend more time with one or the other oxygen.  Therefore, the "average" electron state for an oxygen molecule is as for the individual atoms, and so the oxidation state is 0.


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

The classification we use to distinguish bonds in other chemical compounds will obviously be valid for the classification of oxides.

They can be classified as covalent or ionic.


How to calculate the concentration of a K₂Cr₂O₇ solution? How many ml of acidified $\ce{K2Cr2O7}$  solution, which includes 21.6 g of dichromate ions per Litre, are required for the oxidation of 44.4 g of  $\ce{CH3CH2CH(OH)CH3}$?

Obviously 44.4 g of $\ce{CH3CH2CH(OH)CH3}$ means 0.6 moles. So from the stoichiometry of the the oxidation we find out that 0.2 moles of $\ce{K2Cr2O7}$ are needed. I have trouble finding the concentration of $\ce{K2Cr2O7}$ though and thats why I need help.

First, you need a balanced equation for the oxidation.  The gives you the mole ratio of dichromate to substrate.  You know how many moles of substrate you have, you know how many moles of dichromate you need.  If you have 21.6 g of dichromate ions (as such)/liter, then you know how many moles/liter you have.  You know the number of moles you need, you know the number of moles/liter you have, you know the number of liters you need.

You will be doing your Jones oxidation in acid, of course.  Collins reagent is a sweeter deal for workup, since MEK is 27.5 g/100 ml water-soluble.


SN1 Reaction Involving Carboxylic Acids Which oxygen atom is more nucleophilic in carboxylic acids?

If we look at this page, it says the carbonyl oxygen is more nucleophilic:

However, I also learned in class that $\mathrm{sp^3}$ atoms are more reactive than $\mathrm{sp^2}$ atoms.

So, in an $\mathrm{S_n1}$ reaction where the solvent is acetic acid, which oxygen attacks?

The carbonyl oxygen is more nucleophilic. You can draw a resonance structure of the carboxylic acid that places a formal negative charge on that oxygen which rationalizes its greater nucleophilicity. Also, after it adds to the carbocation, the resulting intermediate will be resonance stabilized. If the other oxygen attacked, no resonance structures would be possible.


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

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

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

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


How is the Rate law of a reaction measured? What is the Rate law of the reaction below?
$$\ce{4 HBr + O2 -> 2 H2O + 2 Br2}$$

How do we approach finding the rate law of a reaction? Can this be found out only through experiments?

Preamble
For a reaction with no intermediate steps, you could derive the rate law from the reaction equation. For a generic equation
$$\ce{$a$A + $b$B -> C}$$
the rate is given by
$$r = k\cdot c^{a}(\ce{A})\cdot c^{b}(\ce{B}).$$
Rigorously, this is only possible for elementary reactions. A prominent example for a strictly unimolecular reaction is radioactive decay. Also quite commonly known is the photolytic or thermal cleavage of hydrogen or bromine:
\begin{align}
\ce{H2 &->[\Delta T] 2H. }\\
\ce{Br2 &->[h\nu] 2Br. }
\end{align}
An example for a bimolecular reaction would be
$$\ce{2NO2 -> NO3 + NO}$$
Reaction mechanism of the combustion of hydrogen bromide
This seems to be a reaction with a fairly complicated mechanism, involving multiple species, most of them radical. I would assume, that it is somewhat similar to the combustion of hydrogen, that I have described here.
A quick search on some forums will turn out a reaction mechanism, which I think is wrong. But I will show it here anyways, since it gives you a first clue, about how complicated this reaction might be. (Note: These are supposed to be gas phase reactions.)
\begin{align}\ce{
HBr + O2 &-> HOOBr\\
HOOBr + HBr &-> 2 HOBr\\
HOBr + HBr &-> H2O + Br\\
}\end{align}
The following mechanism is taken from Lilian G. S. Shum and Sidney W. Benson Int. J. Chem. Kin., 2004, 15 (4), 341-380. I think this is a much more suitable and credible approach to the problem. (Note: Most of the reactions are in equilibrium.)
\begin{align}\ce{
HBr + O2 &~<=> HO2 + Br\\
HO2 + HBr &~<=> H2O2 + Br\\
H2O2 + 2HBr &~->T[wall] 2H2O + Br2\\
Br + Br + M &~<=> Br2 + M\\\hline
4HBr + O2 &~<=> 2H2O + 2Br2\\
}\end{align}
Unfortunately I was unable to determine what $\ce{M}$ refers to in this case. It might be any inert gas used as a solvent in the reaction or the wall.
Rate law of the combustion of hydrogen bromide
As a consequence of the mechanism you cannot find the rate law from theory, but only from experiments. It is important to realize, that most of the reactions are in equilibrium and the whole system is dependent on the concentrations of each of the involved species. Therefore the kinetics of this reaction may be expressed as a complicated system of coupled differential equations.
However, Shum and Benson conclude, that the first step will be the rate-determining and the rate expression is given through

$$\frac{\mathrm{d}[\ce{H2O}]}{\mathrm{d}t} = 2k_g[\ce{HBr}][\ce{O2}].$$

Therefore it can formally be regarded as a bimolecular reaction.


How to accurately abbreviate solvents? IUPAC Red Book in section IR-4.4.4 Ligand abbreviations [1, p. 63] states the following:

  N.B. Abbreviations for solvents that behave as ligands should also be in lower case letters [e.g. dmso for dimethyl sulfoxide {(methylsulﬁnyl)methane}, thf for tetrahydrofuran]; the practice of capitalizing the abbreviation of a solvent when it does not behave as a ligand is strongly discouraged as an unnecessary distinction.

To my understanding, applying this rule to, lets say, $\ce{[CrCl2(C2H6OS)4]Cl * C2H6OS}$ complex [2] as example, leads to the following (correct me if I'm wrong):

\begin{array}{ll}
\ce{[CrCl2(DMSO)4]Cl * DMSO} & \color{red}{\text{wrong}} \\
\ce{[CrCl2(DMSO)4]Cl * dmso} & \color{red}{\text{wrong}} \\
\ce{[CrCl2(dmso)4]Cl * DMSO} & \color{red}{\text{wrong}} \\
\ce{[CrCl2(dmso)4]Cl * dmso} & \color{green}{\text{right}} \\
\end{array}

In the other hand, we are used to solvent abbreviations being capitalized, e.g. ACS Style Guide suggests abbreviating dimethyl sulfoxide as DMSO and it seems like it doesn't make any distinction between the ligand form and the free form [3, p. 177]:

  DMSO $\qquad$ dimethyl sulfoxide (also $\ce{Me2SO}$)

Does the rule from IUPAC Red Book also mean that when using in text (e.g. referring to a solvent as to reaction media) "dmso" is preferred over "DMSO", or this rule is only applicable when solvent acts as a ligand and/or solvate?

Bonus question: If both ways of capitalization are possible, then when creating list of abbreviations (e.g. using glossaries package with $\LaTeX$), do I need to separately list DMSO and dmso? If not, which one should I pick?

References

IUPAC “Red Book” Nomenclature of Inorganic Chemistry; Connelly, N. G., Royal Society of Chemistry (Great Britain), International Union of Pure and Applied Chemistry, Eds.; IUPAC Recommendations; Royal Society of Chemistry Publishing/IUPAC: Cambridge, UK, 2005.
Al-Najjar, N. A.; Al-Sudani, A.-R. H.; Shanshal, M. A.; Mousa, F. H.; Kariuki, B. M. Acta Cryst E 2013, 69 (7), m395–m396 DOI: 10.1107/S160053681301622X.
The ACS Style Guide: Effective Communication of Scientific Information, 3rd ed.; Coghill, A. M., Garson, L. R., Eds.; American Chemical Society; Oxford University Press: Washington, DC; Oxford; New York, 2006. DOI: 10.1021/bk-2006-STYG.fw001.

Ancient IUPAC literature

The IUPAC Interdivisional Committee on Nomenclature and Symbols states in recommendations from $1979$, Use of Abbreviations in the Chemical Literature,

  Since the style in which abbreviations are given (e.g., capital letters or small letters,
  with or without full stops) often depends upon well-established practices that vary widely
  from one journal to another, IUPAC does not recommend a specific style for general use.

  DOI: 10.1351/pac198052092229, archived version: here.

This means there is no one general guideline. You may compare this to units which are to be written, without exception, in upright Roman. (ACS Style guide does recommend Italic for molality $m = \ce{mol / kg solvent}$ but IUPAC disagrees.)

The IUPAC–IUB Combined Commission on Biochemical Nomenclature (CBN), in $1967$, set forth a document of revised tentative, i.e., provisional, rules. Abbreviations and Symbols for Chemical Names of Special Interest in Biological Chemistry does not list $\ce{DMSO}$ specifically, however does mention the similar $\ce{EDTA}$.

  7.$\ $Miscellaneous Compounds

  7.1 The following abbreviations are permitted; although they are fairly common, they should be defined in any paper if it is thought that readers might be unfamiliar with them. Some abbreviations are taken from the list published by Annual Review of Biochemistry.

  [---]

  $\ce{EDTA}$ – ethylenediaminetetraacetic acid (or acetate)

  DOI: 10.1111/j.1432-1033.1967.tb00070.x, archived version: here.

The above article mentions $\ce{EDTA}$ only as a separate compound, and is written once, exclusively in upper-case upright Roman.

Summary

Disclaimer: What follows is mostly personal opinion, and should be treated accordingly. It is a summary of my present understanding.

There really is not a recent IUPAC document discussing this topic.
IUPAC does not enforce a ubiquitous principle upon which rules of abbreviations are formed. Abbreviations should be used as rarely as possible, defined where possible, and the specifics depend on the field, journal, and purpose at hand.
That said, some rules exist. For instance, most new organic compound abbreviations should derive directly from acceptable full names set forth in IUPAC's $2013$ Blue Book.
If an abbreviation of a compound, such as $\ce{DMSO}$ or $\ce{EDTA}$, is applied in chemical formulae, they should be written in lower-case upright Roman. A non-exhaustive list of exceptions is listed in IUPAC's $2005$ Red Book, section IR-4.4.4, paragraph $\mathrm{(vi)}$.
The following cited quotation from the same paragraph is very confusing.

  Abbreviations for solvents that behave as ligands should also be in lower case letters [e.g. $\ce{dmso}$ for dimethyl sulfoxide {(methylsulfinyl)methane}, $\ce{thf}$ for tetrahydrofuran]; the practice of capitalizing the abbreviation of a solvent when it does not behave as a ligand is strongly discouraged as an unnecessary distinction.

I am fairly certain it means the following: If a collection of atoms attached to a central atom does not behave as a ligand, you should still use lower-case abbreviations when composing a chemical formula. Unless exception. This does not mean nor imply you ought to use lowercase when discussing a solvent on its own.
The formula $\ce{[CrCl2(dmso)4]Cl * dmso}$ seems correct to me. Not entirely sure if you could also use $\ce{[Cl2Cr(dmso)4]Cl * dmso}$.
I would include both options in the glossary but with a footnote. Mention that uppercase for $\ce{DMSO}$ is generally used the compound itself, lowercase is reserved for compiling a chemical formula. This might be overkill; if you prefer one, I would go with including only the upright version. But again, if space is not an issue, add both for maximum clarity.