Conversion of molecular structure into Fischer projection for compounds having more than two chiral carbons Question

What is the Fischer projection of the given compound's enantiomer?

Answer

My attempt
I tried to solve this by first converting it into the Fischer projection of the given compound and then taking its mirror image since that would provide the Fischer projection of the enantiomer
This is what I got as the Fischer projection of the given molecule.

Why is this wrong?
How do you find the Fischer projection for a compound having more than two chiral centers?

To do this exercise, you must need to know how to assign stereo configuration according to Cahn–Ingold–Prelog priority rules, and what Fischer projection represent. First, get correct assignment of three stereo centers in the given molecule. That should be $(1R,2R,3S)$ as depicted in the image:

Now, sketch a Fischer projection with three stereo centers without assigning any groups but number the stereo centers as $1,2,3$. Put the least priory group of stereo center 1 (that would be $\ce{CH3}$ group) on top of the vertical line. By definition, top and bottom vertical lines are wedges-down (dashes) while all horizontal lines are wedges-up (solid wedges). If you put highest priority group ($\ce{Cl}$ group) on left horizontal line and second priority group ($\ce{OH}$ group) on right horizontal line, you get the correct clock-wise rotation for $(1R)$ configuration (See the image). You can follow same procedure or the one you are familiar to assign other stereo centers correctly. For example, for stereo center 3, the least priority group is $\ce{H}$ while the highest one is $\ce{F}$ group. The second priority group is substituted carbon chain, which is already in the projection (the last priority to consider is $\ce{CH3}$ group). Thus, I put $\ce{H}$ group on bottom vertical line (should be a dash) as did for stereo center 1. Then, to get correct counter-clock-wise rotation to give $(3S)$ configuration, you must put $\ce{F}$ group on right horizontal line so last priory group $\ce{CH3}$ would occupy the remaining left horizontal point.
Note: You can exchange any group at a stereo center with another group at same center to reverse the configuration, e.g., $(R) \rightarrow (S)$. If you repeat same with two different group, you get back the original configuration. Once I was done with assigning the groups, I have performed these operation twice to change the positions of three groups at stereo center 3 to my likeness. Still, the configuration remains same.
Finally, you may able to assign remaining two groups ($\ce{Br}$ and $\ce{F}$ groups) for stereo center 2 to get $(2R)$ configuration.
To get the enantiomer, you can sketch the mirror image of the compound, which is $(1S,2S,3R)$ as depicted in the image and also tally with the given answer for the corresponding enanthiomer.


As we go down a fluoride-alkali metal series, why doesn't the boiling point decrease? As you can see from the graph below, as we go down the blue fluoride-alkali metal series (alkali metal ion is varied from Lithium to Rubidium, which is represented by an increase in ionic mass on the graph), the boiling point increases then decreases.  This is contrary to my expectation that the boiling point should continuously decrease due to an increase in ionic radius.  The charge is obviously not a factor because it is consistent throughout.  What could be the factor that causes this deviation?

Thanks!

Polarizability of the fluoride ion by the alkali metal ion decreases going down the group, this decreases the covalent character in each ionic bond in question.

See the related questions and answers here:

Does covalent character increase or decrease bond strength?

Melting points of alkali metal halides


Selectivity in acetonide formation from a polyol In my total synthesis lecture notes the above example is given in which a polyol is reacted under acetonide forming conditions.

The notes point out that this product is the only product, no other actinides are observed. That said there possibly are two other products, both containing two acetonides, so why should just this one di-acetonide form?

The conditions are not very clear, but I assume all steps are reversible so what is actually being observed is thermodynamic product.

Acetonide formation (isoproylidine ketal formation, to be more correct) is indeed a thermodynamic process, and under acidic conditions acetonides will tend to migrate to form the lowest energy product.

If you look at the penta-ol you've been given, there are three possible permutations containing two acetonides (theres also the possibility of just forming a single acetonide, but these kinds of reactions are usually ran with an excess of the acetone/dimethoxypropane, so its likely to go on as many times as possible):

The green one (which is the one labelled A in your image) is the thermodynamic product, which becomes obvious if you consider the three-dimensional conformation of 1,3-syn and 1,3-anti acetonides (note that this analysis only holds up then there is no substituent at the 2-position. If there is a methyl group there, as is common in polyketide type motifs, then the acetonides don't necessarily adopt chair conformations):

In the syn acetonides, there are three equatorial and one axial groups (minimal strain), whilst in the anti acetonides (which both of the un-observed products contain one of), there are two equatorial and two axial groups, giving a 1,3-diaxial clash across the ring (hence raising the energy of these products).


Why is mass proportional to the strength of a dipole-dipole attraction? meaning more mass more strength I don't understand why the mass of a molecule make it attract more to other polar molecules.

Increasing mass increases the availability for molecules to attract through London dispersion forces (induced dipole-dipole). A longer chain has more atoms and hence more electrons and therefore experiences stronger bonding.

When you say 'polar' I assume you mean permenant dipole-dipole and hydrogen bonding which is actually not affected by the 'mass' or length of a carbon chain in a molecule. However if you take ethanol and pentanol as an example they both have hydrogen intermolecular bonding on account of the -OH group but pentanol also experiences greater London dispersion forces than ethanol which results in its significantly higher melting and boiling points.

So to recap increase mass is not a direct causal link to charge. It is in fact more mass meaning more atoms meaning more electrons meaning induced dipoles leading to greater intermolecular forces. I hope that's clear (:


How can I separate the vinegar from a 5% white vinegar to make a more concentrated solution. I was under the impression that if I put a semi full bottle of vinegar in the freezer the water would freeze but the vinegar would not until a lower temp.  Therefore I could remove the bottle after the water freezes and pour off the vinegar. Apparently the water and vinegar doesn't separate nicely before freezing so this didn't work so well.

I've seen boiling suggested but I'm afraid I'll lose some of the vinegar in the process.

What is the simplest way to concentrate the vinegar at home?

If you have access to a still you can distil the vinegar solution, however this can only go so far. You may find this section of the wikipedia article on azeotropy edifying - to summarise, adding ethyl acetate to a vinegar solution generates a water:ethyl acetate azetrope with a depressed boiling point that makes distillation more effective, however this may be beyond the pale for a home experiment and would obviously render the vinegar non-food-safe.


What does teflon give on decomposition? I have read that teflon coatings undergo decomposition at temperatures above 300 degree Celsius. What do they decompose to? And what other conditions apply for the decomposition? Is this decomposition harmful for humans as cooking goes beyond that temperature?

Like any other polymer decomposition process, the products of PTFE decomposition depends on the chemical species present while PTFE is undergoing the process and temperatures.

General process is this : Decomposition  is  initiated  by  random-chain
scission, followed by depolymerization. Termination is by
dis-proportionation. And all of this happens rapidly above 600 K (~326 C). There are no other conditions applied for the decomposition to take place hence can happen in dry or aqueous environments, which would give different by products.  Cooking below 200 C, it would be completely safe since the mass loss below 300 C is undetectable. Only above the glass transition temperature (~326 C) is the mass loss significant.

Decomposition products will depend on the environment and usually
Oxygen usually does not enter the cycle directly but through water, to give species like,

Carbonyl fluoride
Carbonyl difluoride

As expected other species like fluorinated alkanes and alkenes are obtained like,

the monomer, tetrafluoroethylene
Hexafluoroethane
Octafluorocyclobutane
Octafluoroisobutylene
Perfluoroisobutylene
Tetrafluoroethylene
and more.

However, the chemicals (species) listed above are for controlled lab experiments. What happens in real life cooking scenario is not (or cannot) be anticipated. But, it can be readily said that the above species can react with other chemicals in food to give fluorinated compounds that will be harmful for humans. As a precaution, one should not cook above 200 C on PTFE coated utensils to be completely safe.

Check this extremely detailed page for abstracts and toxicity remarks from research papers on PTFE decomposition on Fluoridealert.org where the conditions and level of toxicity is reported in an organized manner.

Polymer decompositions- check PTFE section for general comments.


Entropy - Negligible? In electrochemistry, the Nernst equation tells us the cell potential for non-standard conditions. I read that the factor $\frac{-RT}{nF}\ln{Q}$ accounts for difference in cell potential due to differing Gibbs Free Energy due to changing temperature and concentration.

This difference in Gibbs Free Energy is only due to entropy right? Because enthalpy is a constant? So obviously, entropy significantly interferes in the process of electrolysis.
And yet, it is ignored in virtually all spheres of chemistry. I'm beginning to doubt my chemical intuition because of entropy effects, and faith in it can only be restored if entropy really was negligible. So it is negligible or not? If not, why is it always ignored?

This difference in Gibbs Free Energy is only due to entropy right?

This is mostly true.  (I think it is 100% true when concentrations can be used in the reaction quotient $Q$, but am not sure if it is strictly true when solutions are not ideal and activities have to be used instead.)

  Because enthalpy is a constant?

Enthalpy is a "constant" only because of the way electrochemical standard states are defined.

  So obviously, entropy significantly
  interferes in the process of electrolysis.

Have no idea what this means.  How does it "interfere"?

  And yet, it is ignored in
  virtually all spheres of chemistry.

False.

  I'm beginning to doubt my chemical
  intuition because of entropy effects, and faith in it can only be
  restored if entropy really was negligible. So it is negligible or not?
  If not, why is it always ignored?

It isn't negligible and is not ignored.


For IUPAC nomenclature, does the number of multiple bonds take precedence over the chain length? I've received a few mixed answers to this question. Some sources say when determining the parent chain for unsaturated hydrocarbons, the longest chain should be considered before the chain with the most multiple bonds, for example:
https://www.acdlabs.com/iupac/nomenclature/79/r79_933.htm
While others say what appears to be the opposite, such as this excerpt from Wikipedia:

The steps for naming an organic compound are:
Identification of the parent hydrocarbon chain. This chain must obey the following rules, in order of precedence:

It should have the maximum number of substituents or branches cited as prefixes.
It should have the maximum number of substituents of the suffix functional group. By suffix, it is meant that the parent functional group should have a suffix, unlike halogen substituents. If more than one functional group is present, the one with highest precedence should be used.
It should have the maximum length.
It should have the maximum number of multiple bonds.
It should have the maximum number of single bonds.

Would be great if I were to receive clarification regarding this.

According to the new recommendations of IUPAC nomenclature rules, when choosing the senior parent structure, you have to first consider the senior parent structure, which has the maximum number of substituents corresponding to the principal characteristic group or senior parent hydride in accord with the seniority of classes. Then, if there is a choice, choose the principal chain, which has the greater number of skeletal atoms. Next in order is the chain, which has the greater number of multiple bonds, and then has the greater number of double bonds (This is according to Blue Book in 2013, which was displayed in 5 steps you have mentioned from Wikipedia).
Note that the new IUPAC recommendations (Blue Book in 2013) on order of chain length and number of multiple bonds have been changed (You can find more details here and here):

In acyclic parent structures the order of seniority between unsaturation and length of chain given in earlier recommendations is reversed. Thus, the first criterion to be considered in choosing a preferred parent acyclic chain is the length of the chain; unsaturation is now the second.

For example consider following structures:

In all first three structures, the longest chain has chosen over the multiple bonds (the old names are in orange). In the typical case, the principle group is $\ce{-CHO}$ and hence hence to hexanal is the principal structure which has $\ce{-OH}$ substituent and longest chain. Thus, the name is 2-ethynyl-5-hydroxyhexanal.


Nanoparticles synthesis I am following an article for synthesis of the nanoparticles.

There is a gold nanoparticles synthesis rout which is called UPD. In this rout, the underpotential deposition of Ag onto the surface of the Au particles is used. There are two metals (Au and Ag) in the obtained nanoparticles, but they aren't called bimetallic nanoparticles. Why is it called "gold nanoparticles synthesis" and not "gold-silver bimetallic nanoparticles synthesis"?

Its because the gold nanoparticles are the star of the show. The researchers were just looking at how surface modification using the silver would alter the morphologies of the gold nanoparticles. Probably the researchers wondered to themselves one day, "How could we possibly control the shape of these gold nanoparticles?" and decided to try with silver.
The bimetallic part would suggest that the researchers were aiming for gold-silver nanoparticles (agglomerated together, not one on the surface of another like in the research)


Acidity order of nitrophenols As in other posts I have seen and learnt that, if one is given four compounds, namely para-nitrophenol, ortho-nitrophenol, meta-nitrophenol and phenol and is told to arrange them in order of acidity, then one has to take into account the $\text{-R}$ and $\text{-I}$ effect.

Now, for the inductive effect, the greatest intensity is observed at the ortho-position, followed by meta-, and then para-.

For the resonating effect, the intensity is greatest for para and lowest for ortho, while meta has no such effect. Now keeping in mind, that $\text{-R}$ and $\text{-I}$ both tend to withdraw electrons away from the phenol group (and thus increases the acidity), and the $\text{R}$ effect has more contribution than the $\text{I}$ effect, we can thus deduce that para-nitrophenol is most acidic, followed by ortho-nitrophenol, then meta-nitrophenol and since phenol has no ring activating group, it comes out to be least acidic in the given compounds.

The facts are the proof of my theory, where the $\mathrm{p}K_\mathrm{a}$ for:
$$
\begin{array}{c|c}
\text{Compound} & \mathrm{p}K_\mathrm{a} \\
\hline
para\text{-nitrophenol} & 7.16\\
ortho\text{-nitrophenol} &7.2\\
meta\text{-nitrophenol} & 9.3\\
 \text{phenol}   & 9.95
\end{array}
$$

So far so good.

But one of my colleagues, when I asked the problem, tried to give the theory in different way. He said, in para-nitrophenol, hydrogen bonding is strongest, so it can't be most acidic, which in his view port should be least. I knew this is not the case, but I just can't explain or point out where he's been wrong. Can anyone of you make it little easy for me to understand it?

You have correctly identified that m-nitrophenol should be the weakest acid among the isomers as the resulting anion cannot be stabilised by the $\text{-M}$ effect of nitro group owing to its meta location.
The acidity order of the remaining two nitrophenols can be explained in this way:

Hydrogen bonding is possible in both of them. But note that:

$$\text{H-bonding} \propto \delta^+ \text{charge on $\ce{H}$ atom}$$
Inductive effect (here, $\text{-I}$) is distance dependent. So $\delta^+$ charge on the $\ce{H}$ of ortho-nitrophenol's -$\ce{OH}$ is going to be more than the $\delta^+$ charge the $\ce{H}$ atom of para-nitrophenol's -$\ce{OH}$.

Also,  quoting from this comment:

Usually the $\mathrm{p}K_\mathrm{a}$ of a substance is determined in dilute solution. So just due to dilution, there won't be many/any inter-molecular H bonds, but there can still be plenty of intra-molecular ones.

The answer above the comment relates the strength of these two types of hydrogen bonds to kinetic motion of molecules:

Inter-molecular interaction, such as inter-molecular hydrogen bonding only take place for a short time when two molecules are in each other's vicinity. Shortly after, the molecules separate and the interaction does no longer exist.
Intra-molecular interactions on the other hand are much stronger since the two atoms forming the hydrogen bond remain in each others proximity due to structure of the molecule. In other words, they exhibit strain.

These three factors help us conclude that:

Strength of intra-molecular hydrogen bonding in o-nitrophenol $\gt$ Strength of intermolecular hydrogen bonding in p-nitrophenol

Thus, p-nitrobenzoic acid is more acidic than the o-isomer.

...and since phenol has no ring activating group, it comes out to be least acidic in the given compounds.

This reasoning is incorrect. Phenol has no group that stabilises the negative charge on the resulting phenoxide ion through $\text{-M}$ or $\text{-I}$ effect (a deactivating group). Therefore, it's the weakest acid of all under investigation.


Which orbitals of the hydrogen atom are degenerate for n=3? Which of the following orbitals are degenerate in the hydrogen atom with $n = 3$?

  A. II and III only;
  B. I and IV only;
  C. I, II, and IV only;
  D. II, III, and IV only;
  E. all.

The answer says its E. All of them.

First of all isn't there only 1 electron in hydrogen? And how could the $\mathrm{s}$ orbital be degenerate? Doesn't degenerate mean there are multiple places pairs of orbitals can be?

First of all isn't there only 1 electron in hydrogen?

yes

  And how could the s orbital be degenerate? Doesn't degenerate mean there are multiple places pairs of orbitals can be?

"degenerate" means having the same energy.  "Degenerate" refers to a set of orbitals.  It doesn't make sense to say one orbital is degenerate.

Solving the non-relativistic Schrodinger equation, all the orbitals for a given "n" are degenerate.  Energy only depends upon n.

More complete consideration including relativity, spin and quantum electrodynamics shows that they are not all degenerate however.


How to calculate pH of a solution when H3PO4, NaH2PO4, Na2HPO4 and Na3PO4 are mixed together in certain amounts to form a solution? How to calculate pH of a solution when $\ce{H3PO4}$, $\ce{NaH2PO4}$, $\ce{Na2HPO4}$ and $\ce{Na3PO4}$ are mixed together in certain amounts to form a solution?

I know how to calculate pH when they are present individually but not when all of them together. I searched the net also but couldn't find any such example.

Usually doing calculations of this kind is not hard.  Roughly, you start from some idea of where the pH is going to end up.  For example, if you are only adding these phosphates, count the number of phosphate ions and the number of sodium ions.  Divide the number of sodium ions by the number of phosphate ions.  You (hopefully) get a number $x$ between $0$ and $3$, say $x=2.5$.  Almost all phosphates will then be either $\ce{NaH2PO4}$ or $\ce{Na2HPO4}$.  Pretend that these are the only ions and treat the problem like an ordinary two component buffer.  Then you are done.  This is an approximation but it will work rather well, as long as the $\mathrm{p}K_\mathrm{a}$ of the various acids are different enough.  If this is not good enough, you just treat this as a bunch of acids, with a bunch of $\mathrm{p}K_\mathrm{a}$ and use the acid-base equilibrium equation over and over, that is
$$\frac{[\ce{H}][\ce{B}_i]}{[\ce{A}_i]} = \mathrm{p}K_\mathrm{a}$$
where $[\ce{A}_i]$ is the concentration of an acid "$i$" and $[\ce{B}_i]$ that of the conjugate base.  Given the pH, this gives you the ratio of the concentration of each base to each acid.

But, really, for phosphate buffers, you are very well off if you just use one acid base pair.  For acids like terephthalic acid or isophthalic acid, which have two $\mathrm{p}K_\mathrm{a}$ values that are rather close to each other, that is they differ by less than $1.5$ or so, finding and solving such polynomial equations makes possible sense.  Otherwise, other approximations we routinely make are more important.

For an explicit example, suppose that we consider a compound $\ce{BH2}$ that can give up two $\ce{H+}$ ions, making both $\ce{BH-}$ and $\ce{B^2-}$.  Suppose that we have a total molar concentration of $\ce{B}$,
$$B_t=[\ce{BH2}]+[\ce{BH-}]+[\ce{B^2-}],$$
and that we have also added strong bases or strong acids to change the $[\ce{H+}],$ so that the total possible molar concentration of $[\ce{H+}]$ is
$$H_t=[\ce{H^+}]+[\ce{BH^-}]+2[\ce{BH_2}].$$

We then also have the relationships
\begin{align}
[\ce{H+}][\ce{BH-}] &= K_{\mathrm{a},\ce{BH_2}}[\ce{BH_2}]\\
[\ce{H+}][\ce{B^2-}] &= K_{\mathrm{a},\ce{BH-}}[\ce{BH-}]\\
\end{align}

If we now call $H=[\ce{H+}]$ and $B=[\ce{B^2-}]$ we find the equations
\begin{align}
H_t &= H + B (K_{\mathrm{a},\ce{BH-}}(H + 2K_{\mathrm{a},\ce{BH2}} H))\tag{1},\\
B_t &= B (1 + K_{\mathrm{a},\ce{BH^-}} H (1 + K_{\mathrm{a},\ce{BH2}} H))\tag{2}.\\
\end{align}

We need to solve these equations for $B$ and $H$. Equation $(2)$ is easy to solve for $B$ and we can then substitute into equation $(1)$ to find
$$H_t = H + B_t
  \frac{K_{\mathrm{a},\ce{BH-}}(H + 2K_{\mathrm{a},\ce{BH_2}}H)}
  {(1 + K_{\mathrm{a},\ce{BH-}} H(1 + K_{\mathrm{a},\ce{BH_2}}H))}\tag{3}.$$

We need to solve this formula for $H$, and (of course) $\ce{pH}=-\log_{10}(H)$.  Some algebra from this point gets us to a cubic equation for $H$.  Of course this has three possible solutions.  But, only one is sensible, provided we insist on "physical sense", e.g. all the equilibrium constants and concentrations are positive real numbers.  Somehow, this must be easy to see from the rules about roots of polynomials.  I am not currently willing to do that proof.

Physically, however, the total amount of hydrogen bound to the base will increase with increasing (positive) $H$.  So the right hand side of $(3)$ increases with increasing $H$ - from $H$ for $H=0$ to $2B_t + H$ as $H\to\infty$.  As the right hand side of $(3)$ increases with increasing $H$ there is exactly one positive solution.  However, this right hand side is also not a very "nice" function - the rational fraction is approximately constant except when we are near a "buffer" condition, where it changes quickly.  So, probably an effective way to solve the equation would be to use a computer and "regula falsi" or something like that.  A little thought will allow sensible guesses for any set of $H_t$ and $B_t$.  Then regula falsi will converge quickly.
Or, from a good enough initial guess even Newton-Ralphson.

Or, possibly it would be better in many cases to figure out which acid/base pair is closest to being in the "buffer" region.  And then, suppose that you have $H$ in that buffer region.  Solve for $H$ with only that acid-base pair.  Then, figure out what you did wrong - there can be $\ce{H+}$'s that are not bound to the base in the buffer pair you have chosen or $\ce{H+}$'s that are bound to stronger acids.  Assume that $H$ calculated from a single buffer pair is correct.  Calculate how many $\ce{H+}$'s are unexpectedly bound or unbound from your assumed buffer pairs.  Use this to calculate a new $H_t$.  If $H_t$ has not changed much, which will be the case unless there is another acid-base pair with very nearly the same $\mathrm{p}K_\mathrm{a}$ - differing I suppose by less than an integer from that of the pair you are considering, then use this value of $H_t$ to solve again for $H$, and recurse.

For the stated problem, all the $\mathrm{p}K_\mathrm{a}$ are very different so the $K_\mathrm{a}$ values differ by orders of magnitude.  And, really, all you need do to is figure out which buffer region you are near.  In principle, you can then find tiny errors using the recursion.  But, this is silly in most circumstances.  This will get a clear, continuous curve which always has the correct slope.  But, it will vastly overstate what you know about the problem.  The changes will be smaller than (I expect) lots of stuff: deviations from ideal solutions, errors in $K_\mathrm{a}$ values, etc..
).


What would be the effect of adding liquid beyond the calibration mark on a volumetric flask? If this happens in the course of preparing a volumetric solution, what should be done?

If you have plenty of solution and reagents, the solution should be prepared again, because the volume is not the nominal volume written on the flask. However, the proper trick to fill the volumetric flask is to use a dropper when you are about to reach the calibration line. Experienced hands can use a squeeze bottle until the end but the chances of this mistake are high.
However, if the solution was was just a single sample solution (provided by somebody else, or the compound is too expensive or rare), slightly overfilling the volumetric flask is not the end of the world. You can mark the liquid level after proceeding normally (dissolve and invert several times). With a help of a high quality volumetric syringe or gas tight syringe (e.g., 2.5 mL Hamilton used in gas chromatography injections) draw the excess liquid until the liquid level is back to the calibration line. Measure that volume in the syringe and add it to the nominal volume of the volumetric flask.
If the volumetric flask had a nominal volume = 100.0 mL and you had to withdraw 1.80 mL then the total volume would be 101.8 mL for molarity or other concentration calculation.


Does the temperature of water affect its ability to extinguish fires? Since water can be either hot or cold, does its temperature affect its ability to extinguish fires?

A fire needs 3 things as shown in the image:

Water reduces the heat as it turns into steam; and hot water turns into steam faster. But the most important thing water does, is to prevent oxygen getting to the fuel. So as long as there is liquid water, it does a great job.


Deciding the major product in this Wurtz reaction Today I was told:

  When 2-chlorobutane is reacted with 2-chloropropane in the presence of sodium and dry ether the following products are formed:

2,3-dimethylpentane
3,4-dimethylhexane
2,3-dimethylbutane

  3,4-dimethylhexane, being most symmetric, is the major product.

I'm not satisfied with the reason provided. Can anyone please explain the fact in a better manner?

What I'm unable to understand is that both 3,4-dimethylhexane and 2,3-dimethylbutane and equally symmetric since both have a plane of symmetry, don't they? Then how are we using symmetry to compare the yield?

I think that there should be a 'statistical product distribution', if I'm correct the products should be in the ratios 2:1:1.

You get to this conclusion as follows, call 2-chlorobutane A and 2-chloropropane B.

A can react with A to give AA, and
A can react with B to give AB, while
B can react with A to give BA, while
B can react with B to give BB.

When you realise that AB and BA are the same molecule you get AA:2AB:BB; that is, 1:2:1.


How do we know the size of atoms in body centered cubic system? I'm told that in the normal cubic system there are $\frac18$ of an atom at each corner and these atoms' radius are such that $a = 2r$ ($a$ is the lattice constant, $r$ radius of an atom).

In bbc (body centered cubic system) they say the atoms in the diagonals are touching, and as a consequence the atoms at each corner are not big enough to satisfy the above relation $a = 2r$

Why is this? Couldn't it be like the normal cubic system but stick an atom in the center to fill up the remaining space? How do we know that they touch along the cube's diagonal and not along the edges?

The easiest answer in my opinion; try it! If each atom has the same radius, it is simply not possible to fit one of them into the center of the cubic system (known as a cubic hole). It is, however, possible if the atom is sufficiently small. CsCl is an example of such a compound.

One can show this geometrically (as is done here), but you can also convince yourself that it must be the case using equally-sized tennis balls or something similar to try it out in practise.


Simple every-day-item hydrometer for distinguishing regular soda and diet sodas When ordering a diet soda (no sugar soda) in a cafe or restaurant I would like to be sure I am  getting a diet soda and not a regular sugar-infested soda. Because a regular soda, is through all the sugar, far more dense than a diet soda, my idea is to use a simple hydrometer. Ideally this hydrometer would be a simple everyday item that sinks in diet soda but floats in regular soda. I already tested this this idea with some simple items such as coins or small toys. However all tested items either sink in both cases or float in both cases. Anybody has a good idea for a suitable item that sinks in diet soda but floats in regular soda ?

UPDATE although it didn't really answer my question, the experiment suggested by matt_black was too simple not to try out. However what happened when I performed the experiment was that both the can of regular coke and zero coke sank. I discovered there are a lot of video's on youtube showing a can of diet coke floating and only the can of regular coke sinking. So I am quite surprised by my own findings.

Better than a hydrometer, use an Abbe refractometer to measure the index of refraction, which varies with sugar concentration. It requires just a drop of liquid, and that small sample is unlikely to have bubbles, reducing density. You can make a refractometer to measure the sugar concentration, or you can buy one for ~$US20.

That said, diet sodas have been linked to a number of health issues, such as diabetes and diabetic retinopathy. You might be better off with unsweetened products such as seltzer, club soda, tonic water or two cents plain.

On the other hand, if you want to explore brewing or candy making, that refractometer could be useful...

À votre santé!


Phenolphthalein as a indicator for the titration of benzoic acid Question: Justify the use of phenolphthalein $(\mathrm{p}K_\ce{in}=9.4)$ as a indicator for the titration of benzoic
  acid $(K_\mathrm{a}=6.3\times 10^{-5})$ with $\ce{NaOH}$.

My attempt:

I can calculate the $K_\mathrm{b}$ of $\ce{NaOH}$ from the above data.

And I also know the formula

$$\mathrm{pH} = \mathrm{p}K_\mathrm{b} + \log \frac{\ce{In}}{\ce{HIn}}$$

I know that this is the titration of a strong base and a weak acid. So the equivalent point will be at a $\mathrm{pH}>7$ and hence phenolphthalein is a good choice.

But how can I conclude mathematically that this is a correct choice?

You have to take two reactions into account.

First the protolysis of benzoic acid

$$\ce{HA + H2O <=> H3O+ + A-}$$
$$k_A = \frac{[\mathrm{H_3O^+}] [\mathrm{A^-}]}{[\mathrm{HA}]} \tag{1}$$

and second the (simplified) protolysis of phenolphthalein

$$\ce{HInd + H2O <=> H3O+ + Ind-}$$
$$k_{Ind} = \frac{[\mathrm{H_3O^+}] [\mathrm{Ind^-}]}{[\mathrm{HInd}]} \tag{2}$$

The color change is around the point where the concentrations $[\mathrm{HInd}]$ (colorless) and $[\mathrm{Ind^-}]$ (colored) are equal, i.e. $$k_{Ind} = [\mathrm{H_3O^+}] \tag{3}$$

If you substitute (3) into (1) and take the negative logarithm to base 10 you get

$$\mathrm{pK_A} - \mathrm{pK_{Ind}} = -\log\frac{[\mathrm{A^-}]}{[\mathrm{HA}]} = 4.4 - 9.4 = -5.2$$

what shows that the benzoic acid has been completely titrated at this point.


Why is Finkelstein reaction not conducted using NaF? I was studying the Finkelstein reaction and all sources through which I have studied this reaction indicate that the reaction uses $\ce{NaI},$ where $\ce{I-}$ acts as the nucleophile which will then replace $\ce{X}$ from the alkyl halide $\ce{R-X}.$

This is an $\mathrm{S_N2}$ reaction, hence we use a polar aprotic solvent (PAS, usually acetone) as PAS stabilizes the $\ce{Na+}$ cation. We know that while using a PAS the order of nucleophilicity is

$$\ce{F-} > \ce{Cl-} > \ce{Br-} > \ce{I-}.$$

The part I am struggling with is that even though $\ce{F-}$ is a better nucleophile than $\ce{I-}$ in a PAS, why do we still use $\ce{NaF}$ rather than $\ce{NaI}?$

Further to my comment - the solubility of Sodium Fluoride in Acetone is 2.4 microgrammes per 100g of solvent i.e. essentially insoluble source. There will be no F- in solution so the reaction will not occur.


Flexible & thermally conductive materials? Quick question: Are there any common or easily obtainable materials that are thermal conductors, but also flexible? By flexible, I mean along the lines of a plastic bag, with similar or greater durability.

There is a reasonable answer that does not involve aluminium. Thin silicone sheets are extremely available, but an even better solution to your problem would be a polyurethane elastomer with a thermal conductive filler however these are less readily available. Obviously they are not going to be as thermally conductive as metals but that is the sacrifice you make (currently) for flexibility.

Hopefully in the future graphene will be cheap and readily available but with the green energy crisis, I doubt it!


Is there a way to calculate how fast water will cool? Is there a way to calculate how fast water will cool if you know the water's starting temperature and temperature of the environment that you put it in, such as a freezer?

Check out Fourier's law. The rate of heat flux (transfer of thermal energy per unit time per unit surface area)  is proportional to the difference in temperature:

Since a change in temperature is directly proportional to the change in energy, you can form this into a fairly simple first-order differential equation. If we take one side (air in a room), at a temperature we'll call $T_{0}$ as a cold reservoir, we can say:

$$
\frac{d(cT_{1})}{dt}=c\frac{dT_{1}}{dt}=-kA\frac{T_{1}-T_{0}}{w}
$$

Where $T_{1}$ is the temperature of the water, $c$ is the water's specific heat, $A$ is the surface area, $k$ is the thermal conductivity of the material between them (i.e. styrofoam cup, say it has a styrofoam lid too so it surrounds the water completely), and $w$ is the thickness of the container.

You will notice that this equation leads to an exponential decay in the temperature difference.

Simply plug in the appropriate constants and solve this equation for any given $\Delta T$ and you can find how long it takes to get that close (it will never quite reach zero).

I'm a bit rusty, so please forgive any notational errors


Dropping Borosilicate glass mug I recently got a glass tea mug. It's has double wall construction with a vacuum insulation (advertised as such). I noticed on the label, it specifically mentioned it's made of borosilicate glass, and to "discard if dropped".

I understand the purpose of making it out of borosilicate glass (thermal expansion when adding hot tea). But what would be the reason for the directions to link the material to needing to discard the mug if dropped (and not cracked or broken)?

Does the fact that it's made of borosilicate mean it's structural or thermal properties might be compromised if it's dropped?

Pyrex glass labware can be dropped, scratched, even used if cracked (I've started filling a beaker, then noticing that there was a crack - but it still seemed OK).
Borosilicate glass for commercial (non-lab) use is often tempered, so that a scratch is a potential hazard - the item might fail unexpectedly. The issue is not so much the borosilicate but the tempering.


How is "F2/sig(F2)" the data quality in SXRD? When I recieve a dataset from single crystal X-ray diffraction, I can determine the quality from a graph in which F2/sig(F2) is displayed in dependence of the frame.
But I forgot the basics... I have found that the $R_{int}$ value is calculated as such:
$$R_{int}=\frac{\Sigma|F^2-F_{mean}^2|}{\Sigma|F^2|}$$
Reference: http://www.canadiancrystallography.ca/cccw/files/CIF-file%20and%20validation.pdf
But what exactly is F2/sig(F2) now?

$F^2/\sigma(F^2)$ is basically the signal-to-noice ratio, a simple measure of how well your crystal diffracts. It is hardly related to $R_{int}$ at all. Indeed, to calculate $R_{int}$ you need to have the whole dataset, otherwise it just would not make sense. As for $F^2/\sigma(F^2)$, you have it as soon as you've collected a little bit of data.


Why do ion cores not move in a metal? The delocalized electrons in a metal move about because of the electrostatic attractions between them and many different positive ion cores. But why does the same not occur with the ion cores themselves? Why are electrons, which bear the same magnitude of charge as nuclei, mobile, while the nuclei themselves are not?

The crystal structure of the metal holds the atom in place. BTW, since electrons are shared in a "sea" of electrons or conductivity band, no one atom is necessarily ionized. The crystal structure is also important for nuclear physics, e.g. using Mössbauer spectroscopy to measure gravitational effects.
Individual electrons don't actually move around the whole circuit, even in DC applications. Instead, think of it as a large stack of electrons; as one is pushed on the stack at one end, electromagnetic forces pop one off the other end. Though electrons do move a bit, the drift velocity is much less than the speed of light (in fact, it is somewhat slower than the last-placed snail in a race).
Compare the mass of a single electron and the mass of an atom, or the mass of an ion in a plasma. Note that acceleration of a particle is inversely dependent on its mass. This is important in fusion research, where confining electrons and ions is a bit tricky.

See this PPT presentation on metallic conduction.


Why isn't the side chain of arginine totally protonated at low pH? I was wondering why arginine cannot be protonated at $\mathrm{pH\ 0}$ on all the $\ce{N}$ atoms in the side chain, as they all have a free electron pair. As this image shows, it can only be protonated once.

Why don't the other two nitrogen atoms in the side chain get protonated? Does it have something to do with resonance with the $\ce{C=N}$ double bond?

It may be hard to see from a 2D-drawing, but there are simply no more available lone pairs on arginine than the ones protonated in form $\ce{A}$.

The guanidine group ($\ce{H2N-C(=NH)-NHR}$) is isolobal to a urea group, and indeed guanidine behaves much like urea except for the common differences between $\ce{C=O}$ and $\ce{C=NH}$. The lone pairs of $\ce{-NH2}$ and of $\ce{-NHR}$ are actively taking part in resonance with the $\ce{=NH}$ group’s double bond much like an amide structure. One can equivalently say that those lone pairs are delocalised into the $\ce{C=N}$ $\pi^*$ MO. Hence, only the $\mathrm{sp^2}$ hybridised lone pair on the $\ce{=NH}$ nitrogen is basic in any way and can be protonated.

Of course, you can further protonate arginine theoretically. However, it would take superacidic media.


Mercury vapor release from amalgam fillings What would cause vapour release from elemental mercury? Would it be pressure, friction... could one stop the vapour release through cooling the mercury?

All solids have a certain vapor pressure, and if this vapor pressure exceeds the atmospheric pressure, then evaporation of the solid occurs. Some of the solid turns into gas. More specifically, it's the uppermost layer of the solid that evaporates.

  What would cause vapour release from elemental mercury? Would it be
  pressure, friction...

Friction could definitely increase the kinetic energy of the mercury particles and increase the solid's vapor pressure. An increase in pressure could also increase the temperature and therefore kinetic energy of the solid's constituent mercury atoms.

  could one stop the vapour release through cooling the mercury?

Yes, that would decrease the kinetic energy of the particles and lower the solid's vapor pressure. Here's a chart relating temperature (in Kelvins) to vapor pressure. At physiological temperatures, we're still talking about an extremely low vapor pressure.

http://www.physics.rutgers.edu/~eandrei/389/NISTIR.6643.pdf


Theoretical basis behind orbital correlation diagram for pericyclic reactions When discussing pericyclic reactions, the Woodward-Hoffman rules allow one to predict whether a reaction will be thermally/photochemically allowed. This has been discussed in many questions on chem.SE, but essentially boils down to the following statement:

  A ground-state pericyclic change is symmetry allowed when the total number of $(4q+2)_\mathrm{s}$ and $(4r)_\mathrm{a}$ components is odd.

What the Woodward-Hoffman rules fail to explain however is why the reaction is allowed. Several explanations exist, such as the Huckel aromaticity of the transition states, however by far the most 'powerful' explanation is supposedly given by a well drawn orbital correlation diagram. Fleming devotes an entire section in his book on the topic of how to draw these diagrams:

"Correlation diagrams provide a compelling explanation, at least for those reactions that have well-defined elements of symmetry preserved throughout the reaction. The idea is to identify the symmetry elements maintained throughout the reaction, classify the orbitals undergoing change with respect to those symmetry elements, and then see how the orbitals of the starting materials connect with those of the product. The assumption is that an orbital in the starting material must feed into an orbital of the same symmetry in the product, preserving the symmetry throughout the reaction. Substituents, whether they technically break the symmetry or not, are treated as insignificant perturbations on the orbitals actually undergoing change."

Fig 1: Fleming, Molecular Orbitals and Organic Chemical Reactions- Reference Edition, 2010, Wiley

As far as I can find, and from recollection, Longuet-Higgins at Cambridge was the first person to use correlation diagrams to explain pericyclic processes (as per figure 2 below), but again, there's no explanation as to why this works.

Fig 2: Early (beautiful) example of a correlation diagram as applied to an electrocyclic reaction. J. Am. Chem. Soc. 1965, 87, 2045. (DOI)

To make an analogy to molecular orbital theory, clearly, correlation diagrams are simply 'mixing' things of the same symmetry (in this case A or S relative to some axis rather than the more expanded set of symmetry labels from group theory), but there seems to be little / no care in correlation diagrams as for the relative energies of the things being mixed. Is anyone able to explain the why of a correlation diagram?

Introduction

Instead of the usual cycloadditions (which has some extra complications in terms of the reacting orbitals), let's consider the concerted reaction $\ce{H2 + D2 -> 2HD}$. We'll come back to the cycloadditions at the end.

occurring via a square planar transition state. This can be analysed with the W–H rules and you'll find that it's a thermally forbidden process $(_\sigma 2_\mathrm{s} + {}_\sigma 2_\mathrm{s})$. So, hopefully that's what we find at the end of this discussion.

Ignoring the isotope labels,1 there is $D_\mathrm{2h}$ symmetry throughout the reaction, so we will categorise the orbitals by their symmetry in this $D_\mathrm{2h}$ point group.2 I've defined the coordinate axes such that the reaction takes place in the xy-plane, with the z-axis perpendicular to the plane of the paper.

There is one simplification which can be made: the symmetry of an orbital under the three mirror planes uniquely specifies the irreducible representation under which an orbital transforms. Hence, in order to identify the point group, we don't need to bother with the symmetry under the five other operations (identity, three rotations, and inversion); we simply need to find the symmetry under the three reflections.3

Reactant and product orbitals

The relevant MOs will be constructed from linear combinations of the 1s orbitals on each atom. First, we look at the orbitals of the reactants, labelled $\psi_1$ through $\psi_4$. We will overlap the 1s orbitals on adjacent atoms to produce $\sigma$ and $\sigma^*$ orbitals for both $\ce{H2}$ and $\ce{D2}$. However, since these MOs do not have definite symmetry under all the mirror planes, we need to take linear combinations of these MOs.

Hopefully the energy ordering of these MOs is clear (I actually listed them in increasing order of energy). It can be deduced from the extent of antibonding/bonding character between each pair of 1s orbitals.

The product MOs, labelled with $\phi_1$ through $\phi_4$, have a very similar form. It should not really be surprising, as the system is exactly the same, just rotated 90 degrees.

Orbital correlation diagrams

We can define a reaction coordinate $x$, such that $x = 0$ before the reaction (i.e. reactants) and $x = 1$ after the reaction (i.e. products). Over the course of the reaction, there will always be four MOs; and these four MOs will always transform as $\mathrm{A_g + B_{1g} + B_{2u} + B_{3u}}$, regardless of what the value of $x$ is. This has to be the case because the basis set (four 1s orbitals) transforms as a representation which can be reduced into that sum.

As the reaction proceeds, the positions of the nuclei vary continuously. Therefore, the Hamiltonian, and also the energies of these MOs $\varepsilon_i$ $(i = 1,2,3,4)$, will vary as the reaction proceeds, so $\varepsilon_i$ is a continuous function of $x$. The orbital correlation diagram simply plots how $\varepsilon_i$ varies with $x$. There will be four curves on the diagram: one for the MO with symmetry $\mathrm{A_g}$, one for the MO with symmetry $\mathrm{B_{1g}}$, and so on.

In theory, one could find the MOs and their energies at every value of $x$, and then you would find four curves, one for each $i$. However, for a qualitative understanding, this isn't necessary. We already know the approximate values of $\varepsilon_i(0)$ and $\varepsilon_i(1)$, and we also know the MOs to which these values correspond. So all we need to do is to figure out which reactant MO $\psi_i$ correlates with which product MO $\phi_j$.

From the symmetry labels, this mapping is quite self-evident, as the $\mathrm{A_g}$ reactant MO must become the $\mathrm{A_g}$ product MO, and so on.

In this example, it also makes physical sense that as the reaction proceeds, the reactant MO $\psi_2$ will simply become the product MO $\phi_3$ (pictured above). For other correlation diagrams, there is a subtlety, but it is a point we will return to later on.

State correlation diagrams

Orbital correlation diagrams are all fine and dandy, but they can be put on a much more solid footing by constructing the relevant state correlation diagrams. I have deliberately avoided using the word "state" up until now. Here, "state" refers to an electronic state of the system as a whole.4 The simplest expression for an electronic state is given by a Slater determinant, and we will simply abbreviate it to the usual form that the organic chemists do. For example, in the ground state of the reactants, we can write

$$\Psi_1 = \psi_1^2\psi_2^2$$

where the antisymmetrisation is implicit. The use of capital $\Psi$ instead of small $\psi$ emphasises that this is a total electronic state of a system. The subscript 1 indicates that it is the ground state.

The electronic states of the reactants are well-described by single Slater determinants. For a system with 4 MOs and 4 electrons, there are a total of $8!/4!4! = 70$ states, and so theoretically we need to look at $\Psi_1$ through $\Psi_{70}$. However, it suffices to consider the first three singlet states (the triplet states can be ignored5). These are

$$\Psi_1 = \psi_1^2\psi_2^2 \qquad \Psi_2 = \psi_1^2\psi_2\psi_3 \qquad \Psi_3 = \psi_1^2\psi_3^2$$

which transform as $\mathrm{A_g}$, $\mathrm{B_{1g}}$, and $\mathrm{A_g}$ respectively.6 Likewise we will consider the three product states

$$\Phi_1 = \phi_1^2\phi_2^2 \qquad \Phi_2 = \phi_1^2\phi_2\phi_3 \qquad \Phi_3 = \phi_1^2\phi_3^2$$

which transform as $\mathrm{A_g}$, $\mathrm{B_{1g}}$, and $\mathrm{A_g}$ respectively.

The state correlation diagram plots the energies of these states $E_i$ as they vary with the reaction coordinate $x$, so the question is again: which reactant state $\Psi_i$ correlates with which product state $\Phi_j$.

Let's first make the assumption that a single Slater determinant is sufficient to describe an electronic state, throughout the entire reaction coordinate. In this case, the assignment is very straightforward. Since $\psi_1$ and $\psi_2$ in the reactants smoothly become $\phi_1$ and $\phi_3$, it means that the reactant state $\Psi_1 = \psi_1^2\psi_2^2$ must smoothly become the product state $\Phi_3 = \phi_1^2\phi_3^2$ as the reaction coordinate increases. A similar consideration for the other states leads to the diagram below.

Let's pause for a moment and figure out what this diagram means. It means that this molecular Hamiltonian has three approximate solutions under consideration; we label these solutions as $S_1$, $S_2$, and $S_3$. The first approximate solution, $S_1$, can be described as a Slater determinant:

$$S_1(x) = f(x)^2g(x)^2$$

where $f$ and $g$ are MOs which vary with $x$. At the point $x = 0$, we have $f = \psi_1$, $g = \psi_2$, and hence $S_1 = \psi_1^2\psi_2^2 = \Psi_1$. As $x$ increases, $\psi_1$ continuously distorts towards $\phi_1$, eventually becoming $\phi_1$ when $x = 1$, and likewise $\psi_2$ continuously distorts towards $\phi_3$. Therefore the state $S_1$ continuously distorts from $\Psi_1$ to $\Phi_3$. Similar statements can be made of $S_2$ and $S_3$.

Now, we introduce the key assumption that the reaction must proceed adiabatically. This can be formalised very thoroughly in quantum mechanics, but essentially, it means that the state of the system cannot jump from one curve $S_i$ to another curve $S_{j \neq i}$. I'm not entirely sure about the justification for this, but Patterson has written about it in an article:7

  A W–H allowed reaction pathway conserves R (reactant) and P (product) nodal geometries, which in turn guarantees continuous covalent bonding along an adiabatic reaction pathway that takes the reactants through a transition state and on into products. A W–H forbidden pathway can either maintain this continuous bonding (interaction energy between R and P) along an adiabatic pathway and generate products in excited electronic states, or disrupt reactant nodal geometry or bonding to reach diabatically (or nonadiabatically) a lower-energy pathway to products. [...] Continuous bonding is energetically more favorable than either nonbonding or antibonding interactions.

Therefore, if we start the system in the state $\Psi_1$ and allow the reaction to proceed, it has to either end up in the product excited state $\Phi_3$ (as far as this treatment tells us); if it ends up in the product ground state $\Phi_1$, then there must be some disruption of continuous bonding. Both of these possibilities are energetically costly, and therefore the reaction is forbidden.

Configuration interaction

That gave us the correct result, but not for the correct reason. The problem is that this assumes that there is no interaction between $S_1$ and other states $S_n$. Mathematically speaking, this means that in the basis $\{S_1, S_2, S_3\}$ the Hamiltonian is assumed to be diagonal,8 such that

$$\langle S_i | \hat{H} | S_j \rangle = \delta_{ij}\langle S_i | \hat{H} | S_i \rangle = \delta_{ij}E_i$$

The question is, is this true? It turns out that, because the Hamiltonian transforms as the totally symmetric irreducible representation (TSIR), the matrix elements $H_{ij} = \langle S_i | \hat{H} | S_j \rangle$ generally only vanish if $S_i$ and $S_j$ are of different symmetry. Therefore, if two states $S_i$ and $S_j$ have the same symmetry, there will often be an off-diagonal element $H_{ij}$. This is the case with $S_1$ and $S_3$ in our example above.

As $x$ increases, the energies of $S_1$ and $S_3$ approach each other, and these two states mix into each other. The new electronic states are therefore linear combinations of our original solutions, $S_1$ and $S_3$. Since $S_1$ and $S_3$ are Slater determinants, the new electronic states are linear combinations of Slater determinants. This process of mixing different electronic states into the wavefunction is commonly known as configuration interaction (CI).

So much for the states; we're not really all that interested in them. The energies are more valuable. It turns out that the presence of this off-diagonal element leads to the non-crossing rule:9 two curves corresponding to states of the same symmetry cannot cross each other. Instead, they are repelled, and there is said to be an avoided crossing. The repulsion is most pronounced when the energies are close to each other, i.e. near the middle of the diagram. The state correlation diagram must be modified to reflect this. The coefficients $a$ and $b$ in the linear combinations must fulfil the criterion $|a|^2 + |b|^2 = 1$.

Within this framework, we can refine our argument for why the reaction is forbidden. In fact, the transition from $\Psi_1$ (reactant ground state) to $\Phi_1$ (product ground state) is adiabatic; however, there is an extremely large electronic activation energy to it, on the order of an electronic excitation energy, which cannot be provided by thermal means.

Real pericyclic reactions

You might notice that the orbital correlation diagram that we drew above is actually similar to that for the disrotatory ring-opening of cyclobutene. The principles are exactly the same, and the state correlation diagram can be constructed exactly analogously (and in fact, the state correlation diagram for this process is found in Fig. 2 of the question), and of course we find that this reaction is thermally forbidden. But, as I alluded to earlier, there is a slight subtlety. Here is the orbital correlation diagram:

Unlike in our first example, there is no obvious one-to-one correspondence between the form of the reactant MOs and the form of the product MOs. For example, it's not obvious why $\pi$ has to transform into $\phi_3$ and not $\phi_1$. It turns out that both $\phi_1$ and $\phi_3$ are actually linear combinations of what $\pi$ and $\sigma$ become in the new geometry.

In the reactant, cyclobutene, the $\pi$ and $\sigma$ MOs cannot mix by symmetry: there is an additional mirror plane (the plane in which the molecule lies), and $\pi$ is antisymmetric and $\sigma$ symmetric with respect to that mirror plane. However, when the reaction proceeds, the symmetry of the system is decreased, and these orbitals are allowed to mix and form linear combinations.

In our earlier example of $\ce{H2 + D2 -> HD + HD}$, the reactant MO $\psi_2$ does physically become the product MO $\phi_3$, as was fairly obvious from the nodal pattern. However, in this situation where there is orbital mixing, it cannot be said that the $\pi$ orbital physically becomes $\phi_3$ as the reaction proceeds. This point was emphasised by Patterson7 and was also described by Woodward and Hoffmann10 in their review of their rules (pp 797 onwards).

So, the correspondence should be determined using symmetry and the non-crossing rule (which also applies to MOs), without recourse to the form of the MOs.

Another example is that of the Diels–Alder reaction. If you look at the $\pi$ MO of ethylene (the second-lowest reactant MO), you find that it becomes the $\pi$ MO of the product cyclohexene. It makes no physical sense to claim that somehow the MO migrated through several carbons over the course of the reaction. Instead, it is more accurate to say that the three symmetric product MOs are linear combinations of what the three symmetric reactant MOs become over the course of the reaction.

Notes and references

If you're really bothered by it, you can change the reaction to $\ce{\color{red}{H}-\color{red}{H} + \color{blue}{H}-\color{blue}{H} -> \color{red}{H}-\color{blue}{H} + \color{red}{H}-\color{blue}{H}}$. But that's just confusing.
The square planar transition state has a higher symmetry of $D_\mathrm{4h}$. However, since $D_\mathrm{2h}$ is a subgroup of $D_\mathrm{4h}$, all the symmetry elements in $D_\mathrm{2h}$ will be preserved in $D_\mathrm{4h}$ symmetry, but not vice versa. For example, the transition state possesses a $C_4$ rotation axis but neither the reactants nor products do. Therefore, we cannot categorise the reactant or product MOs by their symmetry under a $C_4$ rotation; it only makes sense to categorise the orbitals using the more general $D_\mathrm{2h}$ point group. In organic chemistry parlance, this is what is meant by "identify the symmetry elements that are preserved throughout the reaction".
In organic chemistry, it is "conventional to be less formal with the  notation" (Atkins's words; Molecular Quantum Mechanics 4ed, p 400), and so the irrep symbols are replaced with a series of S's (for a character of +1) and A's (for a character of -1). For example, the $\mathrm{B_{1u}}$ irrep would be labelled as ASS since it has characters of $-1, +1, +1$ under the three mirror planes.
The reason why I have never referred to an MO as a "state of an electron" is because of quantum indistinguishability of the electrons, which necessitates that $\Psi$ be properly antisymmetrised. If the Slater determinant (eq. $(1)$) is expanded, it will become clear that the state of electron 1 is entangled with the state of all the other electrons. The wavefunction $\Psi$ cannot be factorised into a product of composite wavefunctions, and so, it is physically incorrect to speak of an electron being in an orbital.
The triplet states can be ignored because spin states are orthogonal. Therefore, for a singlet reactant to end up as a triplet product, there must be a non-adiabatic transition somewhere, and a key assumption is that the process is adiabatic.
The symmetry of a state is obtained by taking direct products of the populated MOs.
Patterson, R. T. An Improved Interpretation of the Woodward–Hoffmann Rules. J. Chem. Educ. 1999, 76 (7), 1002–1007. DOI: 10.1021/ed076p1002.
The diagonal entries of $\hat{H}$ are dependent on $x$, and hence the (approximate) eigenstates $S_i$ and (approximate) eigenvalues $E_i$ also vary with $x$. The basis set $\{S_1, S_2, S_3\}$ is also dependent on $x$, but we don't really need to worry about how it depends on $x$ are we are only looking for a qualitative description of the process.
In the case of two states mixing, the proof is simple: if you use the basis states $\{|a\rangle,|b\rangle\}$ the Hamiltonian is
$$\mathbf{H} = \begin{pmatrix}H_{aa} & H_{ab} \\ H_{ba} & H_{bb}\end{pmatrix}$$
and the eigenvalues are
$$E_\pm = \frac{(H_{aa} + H_{bb}) \pm \sqrt{(H_{aa} - H_{bb} + 4|H_{ab}|^2)}}{2}$$ since $H_{ba} = H_{ab}^*$. Then $E_+$ can only be equal to $E_-$ if $H_{ab} = 0$.

In general I don't want to prove it, because I don't know how to, but a useful reference for this is: Mead, C. A. The "noncrossing" rule for electronic potential energy surfaces: The role of time‐reversal invariance. J. Chem. Phys. 1979, 70 (5), 2276–2283. DOI: 10.1063/1.437733.
Woodward, R. B.; Hoffmann, R. The Conservation of Orbital Symmetry. Angew. Chem. Int. Ed. 1969, 8 (11), 781–853. DOI: 10.1002/anie.196907811


Am I using the Nernst equation correctly? From my understanding the Nernst equation is

$$E = E^\circ - \frac{RT}{nF}\ln Q$$

Where $Q$ is the reaction quotient.

So for a Daniell cell with copper and zinc (valence of 2) shouldn't the coefficient in front of $\ln Q$ be $0.0128$ at $25\ ^\circ\mathrm{C}$? Everywhere I'm looking the coefficient seems to be $0.0295$ but I can't work out where this number comes from. Could somebody explain this please?

I will show you why $\ce{0.295}$ is correct. :-)

Nernst equation can be written in different ways, I let you choose which you prefer. So for any reaction such as $$\ce{\nu_{Ox}Ox + n e^- \rightleftharpoons  \nu_{Red}Red}$$

One can show that at $\ce{25°C}$,

$$E_{Ox/Red} = E_{Ox/Red}^{°} + \frac{RT}{nF}\ln\left(\frac{a_{Ox}^{\nu_{Ox}}}{a_{Red}^{\nu_{Red}}}\right) \tag 1$$

Which can also be written as, see after how,

$$E_{Ox/Red} = E_{Ox/Red}^{°} + \frac{0.06}{n}\log\left(\frac{a_{Ox}^{\nu_{Ox}}}{a_{Red}^{\nu_{Red}}}\right) \tag 2$$

Where $\log$ is the logarithm in base $\ce{10}$.

So we have $$\frac{RT}{F}\ce{=\frac{8.314 \times 298}{96500}}$$
And we know that $\log(x)=\ln(x)/\ln(10)$ then the coefficient next to $\log$ is $$\frac{RT}{F}\times \ln(10)=\frac{8.314 \times 298}{96500} \times \ln(10) = 0.0591 \approx 0.06$$

To be a little more perfect I sometimes use $\ce{0.059}$ instead of $\ce{0.06}$

So for two electrons exchange the factor coefficient is $\ce{0.059/2}=0.295$


What is the official file extension for SMILES and SMARTS files I want to know what is the officially used file extension for files containing SMILES and SMARTS strings. Somewhere I see .SMI/.SMA and somewhere its the full .SMILES/.SMARTS.

I don't think I've ever seen anything proposed as "official," though I admit I don't use Daylight's official toolkit.

Open Babel accepts:

.smi
.smiles
.can (ideally canonical SMILES, but that matters more for writing, not reading)

I haven't seen many SMARTS files, but I would suspect both .sma and .smarts are out there "in the wild."

Incidentally, this is probably something to bring up with the OpenSMILES effort. I know there's an OpenSMARTS specification started as well.


How do metal's density affects the corrosion rate? Based on different formula that can be used to compute the corrosion rate of metals, all of them involves weight loss, time, surface area, and density. I thought that the weight loss, surface are, and time should definitely be considered. But, I got confused about the relevance of the metal's density. How does that affects metal's corrosion?

The reason you need density is to determine the mass lost by the system, assuming that the equations you are talking about look something like:

$$
m_{loss} = k\rho S_{area} t
$$

Where $k$ is a constant defining how fast the corrosion is occurring in your system (in this case k would be in units of [length]/[time]), $\rho$ is the density of the material, $S_{area}$ is the surface area, and $t$ is time: the equation outputs the total mass of material lost.

In more comprehensive corrosion equations $k$ would be substituted with equations describing the system causing the corrosion and may be dependent on (not a complete list): pH, temperature, surface activity, liquid flow, concentration of certain chemicals, passive layer growth, ect.


Probability of finding a molecule in the ground vibrational level I wanted to estimate the probability of finding a molecule in the ground vibrational level using the Boltzmann distribution:

$$
p_i = \frac{e^{-\epsilon_i/kT}}{\sum_{i=0}^{N}e^{-\epsilon_i/kT}}
$$

Using the quantum harmonic oscillator as a model for the energy

$$
\epsilon_i = h\nu (i+1/2) =/i=0/=\frac{h\nu}{2}
$$

In the Boltzmann distribution, we have the state of interest divided by the sum of all possible states. But how should I treat the denominator?

Searching a bit I found that the analytical expression for this geometric series is ($i$ not imaginary number)

$$
\sum_{i=0}^{N}e^{-i h\nu/kT} = \frac{1}{1 - e^{h\nu/kT}}
$$

However, is this using a shifted energy scale for the harmonic potential? In that the vibrational energies are $0$, $h\nu$, $2h\nu$, ..., and not $\frac{1}{2}h\nu$, $\frac{3}{2}h\nu$, $\frac{5}{2}h\nu$, ...? Should I make sure I use the same energy scale for the nominator and denominator in the Boltzmann distribution?

Doing what porphyrin suggested, I get

$$
\sum_{i=0}^{\infty} e^{-h\nu(i+\frac{1}{2})/kT} = e^{-h\nu/kT} \sum_{i=0}^{\infty} e^{- ih\nu/kT}
$$

Expanding the four first terms

$$
e^{-h\nu/kT} \sum_{i=0}^{\infty} e^{- ih\nu/kT} = (e^{-h\nu/kT} \cdot 1) + (e^{-h\nu/kT} \cdot e^{-h\nu/kT}) + (e^{-h\nu/kT} \cdot e^{-2h\nu/kT}) + (e^{-h\nu/kT} \cdot e^{-3h\nu/kT}) \\
= e^{-h\nu/kT} + e^{-2h\nu/kT} + e^{-3h\nu/kT} + e^{-4h\nu/kT} = \sum_{1}^{\infty}e^{-n\cdot h\nu/kT}
$$

which has an analytical expression for the converged value, right?

You're on the right track.

Also, using $i$ as an index can be confusing some times because it can be confused with the imaginary number; however, here it should not present a problem. As a matter of habbit however, I like to use $j$ or $n$ or something else..there are only so many letters in the alphabet.

The sum in the denominator is called the partition function, and has the form

$$Z = \sum_{j}e^{-\frac{\epsilon_j}{kT}}$$

For the harmonic, oscillator $\epsilon_j = (\frac{1}{2}+j)\hbar \omega$ for $j \in \{ 0,1,2.. \}$

Note that $\epsilon_0 \neq 0$ there exists a zero point energy.

Let's write out a few terms $$Z = e^{-\frac{\hbar \omega/2}{kT}} + e^{-\frac{\hbar \omega3/2}{kT}} + e^{-\frac{\hbar \omega5/2}{kT}} +..... $$

factoring out $e^{-\frac{\hbar \omega/2}{kT}}$

$$Z = e^{-\frac{\hbar \omega/2}{kT}} \left( 1+ e^{-\frac{\hbar \omega}{kT}} + e^{-\frac{2\hbar \omega}{kT}} +.....\right) $$

The sum in the bracket takes the form of a geometric series whose sum converges as shown below $$1+x+x^2+... = \frac{1}{(1-x)} $$
 herein, $ x \equiv e^{-\frac{\hbar\omega}{kT}} $

Putting all of this together

$$Z = \frac{e^{-\frac{\hbar \omega/2}{kT}}}{(1-e^{-\frac{\hbar\omega}{kT}})}$$

Now, $$p_0 = \frac{e^{-\epsilon_0/kT}}{Z} = \frac{e^{-\frac{\hbar \omega/2}{kT}}}{Z} = \frac{e^{-\frac{\hbar \omega/2}{kT}}}{\frac{e^{-\frac{\hbar \omega/2}{kT}}}{(1-e^{-\frac{\hbar\omega}{kT}})}} = (1-e^{-\frac{\hbar\omega}{kT}}) $$


Are there any molecules with the same spectrum? Are there any two distinct types of atom or molecule that have identical emission or absorption spectra?

In addition to the R/S chiral molecules mentioned by @user26143. There are also molecules with a distinct atomic composition that have experimentally indistinguishable spectra, even though theoretically there will be minor differences. One example, according to Rabitz et al. 2009 PRL (freely available for viewing) are flavin mononucleotide and riboflavine and in fact many other flavins. Due to there very similar structure you need unconventional techniques to be able to distinguish them.


Can the van der Waals coefficients be negative in the van der Waals equation for real gases? A gas described by van der Waals equation has the pressure that is lower than the pressure exerted by the same gas behaving ideally. True or false?

My approach

$$P(\text{ideal}) = P(\text{real}) + P(\text{changes due to intermolecular forces})$$

So the pressure of real gas can be more as well as less than that if a gas behaving ideally depending on whether intermolecular forces are repulsive or attractive in nature.

Questions

The answer to the question is given as True. So I want to understand what is wrong about my approach.
Since the intermolecular forces depend on $a$, for attractive nature $a$ is positive. Is $a$ negative for repulsive nature of forces?

This question requires a simplistic notion of real gas behavior.

The van der Waals equation was based on the notion that "real" gas particles occupy some volume, and have an attraction to each other. Thus the volume correction $b$ is negative in the equation and the pressure correction, $a$ is positive. The formula is

$$(P + a/V_\mathrm{m}^2)(V_\mathrm{m} -b) = RT$$

If you look at a table of van der Waals constants all the a and b terms are positive. Thus the volume calculated using the van der Waals equation will always be less than the volume calculated using the ideal gas equation.

The rest of the story...

The van der Waals equation isn't the best equation for corrections, particularly near the critical point for the gas. There are numerous other "real gas equations" which predict gas behavior better. (I'm not sure of what gas and what conditions, but there has to be a gas which has greater volume than would be predicted by ideal gas behavior.)


What is the product of a copper and vinegar reaction? I put a copper wire through a hole in the cap of a water bottle, filled the water bottle ~ half way full with vinegar, and left it over night.  The part of the wire inside the bottle that wasn't in vinegar turn black.  Really black.  Why is this?!

UPDATE: I wiped the black stuff off on paper, and it now looks green on the white paper.  Seems like it oxidized over night.  Why so fast?

If it looks green/blue on paper, you likely have copper (II) acetate. You likely have the anhydrous version $\ce{Cu(OAc)2}$ which is darker than the hydrous $\ce{Cu(OAc)4(H2O)2}$

Wikipedia Images include:

The rate of formation is a property inherent to the reactants and the products, but it also depends on temperature.


How would you go about synthesising amino sugars from sugars? How would you go about synthesising amino sugars from sugars? e.g. How would you synthesise the compound below from D-glucose?

Would adding ammonia ($\ce{NH3}$) to it work?

First, the anwer is “Would glucose and ammonia work?” is: no, definitely not. The $\ce{OH}$ group is a poor leaving group, and alcohol + ammonia definitely won't yield an amine. Moreover, in your case, you have many hydroxyl groups and you only want one transformed into an amine, which is tricky.

I don't know how the synthesis of the above compound can be achieved, but I know one way to help you start a literature search: your expected compound is called an amino sugar. In this particular case, it's the 6-amino-D-glucose.


What are the specific dangers associated with Grignard reagents? I would like to know if anyone has had any specific safety-related problems with Grignard reagents.  I have taught the year-long introductory Organic Chemistry sequence for about 15 years and have never had a student get any Grignard Reagent on their skin, but I've always questioned how safe it really is to have students work with these compounds (specifically PhMgBr).  I have never allowed my students to work with n-butyllithium (we don't even have any in our stockroom), so it seems an odd contradiction that I have much less reluctance to have them work with Grignard reagents.

The dangers of organolithium reagents, particularly t-BuLi and -- to a lesser extent -- n-BuLi are well documented.  It seems to me that the danger of these reagents largely arises from the flammability of the solvent (hexane) in which the compounds are dissolved as opposed to their extreme reactivity as bases.  The organolithium reagent as it reacts with moisture provides the energy needed to ignite the solvent. Since Grignard reagents are themselves extremely basic and dissolved in highly-flammable ethers, it would seem that the dangers of physical contact with a Grignard reagent would be equal to if not greater than those of using an organolithium reagent.  Just out of curiosity, has anyone actually had experience with any accidents and/or injuries involving Grignard reagents?

I have to disagree your position entirely.

Every chemical has associated hazards, with even the most unassuming chemicals having the potential to cause severe harm if not fatality.

However, the actual risk to the chemist is often severely minimised by taking sensible precautions, understanding the hazards, and working under a controlled set of conditions.

The whole purpose of organic chemistry education, especially the purpose of organic chemistry labs is to teach students how they should safely handle chemicals, not to tell them that they're too risky.


How to separate copper(II) chloride from 4-iodophthalic anhydride? I am planning a reaction in which I iodate phthalic anhydride. My reaction is I2, phthalic anhydride, and copper(II) chloride as a catalyst. The solvent is 99.9% isopropanol. How do I separate the product (4-Iodophthalic anhydride) from the copper(II) chloride and remaining iodine?
Edit: I planned to boil away the solvent, and sublimate away the iodine from the remaining solid. How do I separate my product from the resulting mix? At the moment, I only have isopropanol.

I just realized I can add sodium hydroxide, and filter off the resulting insoluble copper(II) hydroxide and sodium chloride, then boil down the result. If you can think of a better answer, feel free to suggest it to me.


Oxidation of Carbons In which of the reactions below is the first compound in the equation NOT oxidized?

The correct choice is (D), which can be seen through process of elimination, but I was wondering how to do this through the carbon oxidation number?

I feel that to know the oxidation number of nitrogen, I need the oxidation number of carbon, but to know the oxidation number of carbon, I need to know the oxidation number of nitrogen.

Or is the correct way to think about this problem that: (D) didn't gain any oxygen atoms, therefore, wasn't oxidized and leave it at that?

There are several ways to determine oxidation number/state. The most useful method for organic (covalent) compounds uses the following three rules, and works for most elements engaged in covalent bonding (just replace carbon with some other element):

Every bond between a carbon atom and a more electronegative atom increases the oxidation number of that carbon atom by 1.
Every bond between a carbon atom and a less electronegative atom (hydrogen is most common) decreases the oxidation number of that carbon atom by 1.
Every bond between a carbon atom and another carbon atom has no effect on the oxidation number of the carbon atom (although we could get more complicated since carbon atoms with different "hybridizations" have different electronegativities).

This approach is equivalent to assigning all of the electrons in a covalent bond to the more electronegative atom for purposes of determining oxidation number. This page on Master Organic Chemistry has a number of examples.

Here's an example from the Wikipedia article on the matter, using acetic acid:

Atom            Bonds to   Contribution to Ox#
H (methyl)       C         +1 (2.5 > 2.1)
          TOTAL Ox#        +1
C (methyl)       H         -1 (2.1 < 2.5)
                 H         -1
                 H         -1
                 C         +0 (same atom)
          TOTAL Ox#        -3
C (carboxyl)     O         +1 (3.5 > 2.5)
                 O         +1 (double bond counts as 2 bonds)
                 O         +1
                 C         +0
          TOTAL Ox#        +3
O (carbonyl)     C         -1 (2.5 < 3.5)
                 C         -1
          TOTAL Ox#        -2
O (hydroxyl)     C         -1
                 H         -1 (2.1 < 3.5)
          TOTAL Ox#        -2
H (hydroxyl)     O         +1 (3.5 > 2.1)
          TOTAL Ox#        +1


Phase stability of alcohols Tert-butyl alcohol seems unusual among alcohols in that its melting point is high (25°C) while its boiling point is also still low (82°C).  I am looking for more materials with phase-unstable liquid regions like this so I'm curious what makes the liquid phase so unstable relative to the solid and gas phases?

Do the methyl groups align and act like alkanes to stabilize the solid but the molecule is still small enough to have a low boiling point?

To my larger point, any advice on characteristics of materials that exhibit low liquid phase stability?

Highly symmetric molecules tightly fit into crystal lattices, give high melting points for their composition.  Their small surface areas minimize London dispersion forces and van der Walls forces, giving high volatility.  They tend to sublime.  Compare mp and bp data for cyclohexane and benzene versus their perfluorinated derivatives. Less exotic: n-octane, iso-octane, 2,2,4-trimethylpentane, 2,2,3,3-tetramethylbutane, and cubane.

http://en.wikipedia.org/wiki/Neopentane
n-Pentane and isopentane versus neopentane.

Chromium, molybdenum, and tungsetn metal versus their respective hexacarbonyls.  $\ce{-SiMe3}$ and $\ce{-CF3}$ plus symmetry confer remarkable volatility.  $\ce{I(CF3)7}$ melts and boils around 0 C.  That is molecular weight 609.95 - and it has no static molecular structure (Bartell mechanism).

http://www.patentgenius.com/patent/3992424.html
US Patent 3992424


What will methyl (2-oxocyclohexyl)acetate produce when reacted with sodium borohydride? What will methyl (2-oxocyclohexyl)acetate produce when reacted with sodium borohydride?

I feel the answer should be 2-(2-hydroxypropyl)cyclohexan-1-one:

Because the carbonyl carbon on the aliphatic chain should be more electrophilic than the one on the ring due to the presence of two neighbouring electronegative oxygen atoms. Correct me if I am wrong but I think the reaction begins with the hydride ion which acts as a nucleophile?

On the other hand the answer given in my book is 1-(2-hydroxycyclohexyl)propan-2-one:

Can someone please help me out here?

Both carbonyl carbons (and everything else in the molecule) are in aliphatic environments.  The six-membered ring you drew is not aromatic, it is saturated like a cycloalkane.

The carbonyl group in this ring is a ketone function, bonded only to other carbon atoms.  The carbonyl group in the chain is bonded to oxygen (did you miss that?) in an ester function, which is a less active electrophile than the ketone function.  Yes the additional oxygen atom is electronegative, but with its octet satisfied it can (to a degree) share a nonbonding pair with the carbonyl carbon, and that effect dominates making the ester carbon less electron deficient. The sodium borohydride reacts accordingly.


Are there ways to predict which chemicals will kill certain bacteria? Are there ways to compute what chemicals kill a give bacteria but not another given bacteria ?
Does that help much to make medication ?

The basic idea is sound, and is used to some degree; however, there are significant difficulties inherent in "tailoring" a chemical to selectively target certain organisms and not others.

There are three things that a chemical can do that make it toxic to cellular life:

Rip some vital chemical to shreds. Chlorine and other light halogens (fluorine, iodine) are huge here, because they have such high electron affinity that they can tear apart the otherwise relatively stable bonds within most organics, destroying that molecule. The most effective ones don't form any stable compounds; fluorine, for instance, is very effective at shredding organic compounds because it isn't happy just barging in on any one organic molecule and making itself at home; it'll deprotonate this one, split that one in half, before finally binding tightly and insolubly to calcium (robbing your body of that essential mineral).
Bind to a vital chemical and not let go. Carbon monoxide is a deadly danger to most aerobic organisms because the hemoglobin (or related compound) that loosely binds oxygen for transport to cells will bind more stably with CO, so the chemical can't transport oxygen and can't get rid of the CO. A lot of more selective poisons attach themselves to enzymes, not necessarily destroying them, but preventing the enzyme doing the work it's supposed to. Cyanide is poisonous specifically because it works like (well, it is) a nitrile group, of which there are very many in a lot of inorganic biochemicals, and specifically it targets a key enzyme involved in cell respiration and programmed cell death.
Mimic a chemical's interaction with a key receptor, causing it to go haywire. Many other "selective" poisons and antibiotics work this way; they fit into the receptor sites that a neurotransmitter or enzyme would, triggering some behavior of the cell, and by flooding the cellular environment with this chemical you cause the underlying programmed reaction to go haywire (or stop completely). A lot of insecticides and pesticides do this; they increase or shut down cell metabolism, or in creatures with a nervous system they mimic or trigger the release of massive amounts of neurotransmitters, basically causing cardiac or respiratory arrest because any normal message to control these actions is completely overridden.

The problem with all this comes in identifying which bacteria have a particular enzyme, receptor or other targeted chemical process, and thus would be susceptible to poisoning by the chemical. In addition, the chemical must not be toxic to human biochemistry (otherwise, no matter how beneficial it might be as a disinfecting agent, it will never see approval as an antibiotic medicine).

Antibiotics must have selective toxicity; while they do not affect eukaryotic cells (such as the ones that make up most plants and animals, thus allowing the chemical to be taken internally or directly applied), they have a much more pronounced effect on bacterial and archaeal cells. Usually, the chemical targets something common to a large class of bacteria (such as the thick peptidoglycan mesh forming the cell wall of Gram-positive bacteria), or most/all bacterial cells (such as the S30 robosomal subunit, key to bacterial protein synthesis and thus found in nearly all bacteria, but completely absent from eukaryote cells which use a different mechanism). However, even within an antibiotic's "target class", there are bacteria that are naturally resistant (perhaps they don't accept the chemical through the cell membrane due to some variation unique to the genus) or develop genetic resistance through overuse or misuse.

Also, bacteria are fickle; most are at least harmless if not beneficial in certain areas (E. coli, for instance, produces the essential vitamin K2 in your intestine), but lethal in others (ingestion of e-coli or introduction through an open cut can cause a fatal infection). Staph is relatively harmless on the skin, but fatal in the spine. Antibiotics don't stop and question bacteria; they kill any bacteria that comes across them, and so overuse of antibiotics to kill one harmful bacteria will kill the bacteria in beneficial places as well.

Lastly, not every infection is bacterial. Fungi and protozoa produce a significant portion of food-borne illnesses and poisonings, as well as other types of infection. These life forms are eukaryotes and unaffected by most antibacterial agents (in fact, penicillin was discovered when a fungus contaminated a bacterial culture; the fungus produces the antibiotic chemical as a natural defense).


Soxhlet extraction condenser - addition of compound to refluxing setup I've never used Soxhlet apparatus before and I have troubles understanding how to properly perform the following step from my protocol:

  To a gently refluxing suspension of 1 g Lithium aluminium hydride in 100 mL of diethyl ether, there was added 1.2 g of compound as a saturated diethyl ether solution with use of a Soxhlet extraction condenser which is modified allowing the continuous return of condensed solvent (through the thimble). After the addition was complete, the refluxing conditions were maintained."

Soo... How do I do it? I run RBF with condenser on top until reflux stabilizes and then I open it, add Soxhlet with compound and continue refluxing or I set up everything cold (RBF with ether + Soxlet with compound in ether + condenser) and then start refluxing?

Also - I'm not quite sure if I understand the thimble modification - can anyone shed some light on it for me?

In my opinion the intention is to have a very slow addition of the reagent to the hydride. The compound is dissolved in ether, so a Soxhlet is of no use since it is for solids.

What they are possibly referring to is a liquid-liquid extraction apparatus, specifically the one for extracting with solvents lighter than water:

The saturated solution of the compound is then placed in the thimble and when reflux starts, the solvent is condensed, falls into the tube and returns to the flask carrying over the compound. Normally the tube goes to the very bottom of the thimble. In fact, usually it is a loose tube with a wide opening on the top. How to know that the addition is complete is another story. If the compound is not colored, it will not be easy without opening the system.  In any case, maintaining reflux for a long time will ensure that everything goes into the flask.

I suppose that for some reason it is difficult to perform a slow addition through a funnel in this case. It would be, however, better to use a syringe pump but not everybody has one of those.


Was Wöhler’s urea synthesis carried out with or without oxygen? I learnt and was very impressed by the urea synthesis by Wöhler in 1828. The reaction was the heating of ammonium cyanate. One of my textbooks in the library said that ammonium cyanate was heated ‘in the absence of oxygen’.

But when I tried reading other organic chemistry textbooks, they also did mention the synthesis but never talk about whether it was anaerobic. Could anyone suggest a reliable source which clearly state about this matter?

Wöhler’s synthesis of urea is not from 1828 but from 1825. We’ll get to that in a second.

In cases like these, one should always rely on the original publications most. Unfortunately, publications of that era did not follow the somewhat stricter standards of experimental sections that nowadays’ publications do. Still, one can make very educated guesses based on the presence (or absence) of statements. The relevant quotation from the 1828 publication, in which Wöhler notes that the substance he created is indeed urea reads:[1]

  Ich habe schon früher angegeben, daſs man die oben erwähnte krystallisirte weiſse Substanz am besten erhält, wenn man cyansaures Silberoxyd durch Salmiak-Auflösung, oder cyansaures Bleioxyd durch flüssiges Ammoniak zersetzt.

Roughly translated:

  I have previously disclosed that the best method of obtaining the above-mentioned white crystalline substance is by decomposing cyanacidic silver oxide[2] with salmiak or cyanacidic lead oxide with liquid ammonia.

It took a while to locate the previous reference which — in accordance with common custom at the time — was given as ‘In einer früheren kleinen Notiz, die in dem III. Bande dieser Annalen abgedruckt ist, …’ (‘In a previous little note that was printed in the IIIrd volume of these annals …’). In the end, I located the most probable candidate to be the artice Ueber Cyan-Verbindungen (On cyan compounds) also published in the Annalen der Physik.[3] Therein, Wöhler reports two syntheses of a novel crystalline compound; the second one being the interesting one for us:

  Rein erhält man [die eigenthümliche kryſtalliſierte Materie], wenn cyanſaures Blei durch kauſtiſches Ammoniak, oder cyanſaures Silber durch Salmiak zerſetzt wird. Sie kryſtalliſirt in weiſsen durchſichtigen, ſtrahligen Kryſtallen, iſt leicht in Waſſer und Alkohol auflöslich.

Translated roughly:

  [The weird krystalline material] can be acquired in pure form if cyanacid lead is decomposed with caustic ammonia or cyanacidic silver with salmiak. It crystallises in white transparent spear-like crystals and is readily dissolved by water and ammonia.

Wöhler then goes on to describe how different substances interact (or rather: don’t) with what he newly synthesised as was common in the days.

Here, we cannot really make any educated guesses on the reaction procedure by what is written because the fragments are extremely short. However, we can make educated guesses by what is not written. Wöhler made no reference to using sealed glassware or any specific atmosphere. He does not not any specific purification of the reactants. And he does not indicate any special care that may be required. Therefore, it is reasonable to assume that all reactions were performed at ambient pressure in the presence of oxygen and water (although water content will have been neglectable in the case of liquid ammonia).

References and Note:
[1]: F. Wöhler, Ann. Phys. 1828, 88, 253–256. DOI: 10.1002/andp.18280880206.

[2]: My note: Wöhler (incorrectly) assumed the acid anion to be $\ce{(CN)2O}$ which meant that the other constituent must have been silver(I) oxide giving $\ce{[Ag2O][(CN)2O]}$ — which gives the correct sum of $\ce{Ag(OCN)}$.

[3]: F. Wöhler, Ann. Phys. 1825, 79, 177–182. DOI: 10.1002/andp.18250790208.


Can a Boc protecting group be removed with trifluoroacetic acid? I am trying to cleave a tert-butyloxycarbonyl (Boc) protecting group from an aminooxy group ($\ce{-O-NH-Boc}$) with trifluoroacetic acid (TFA) in $\ce{CH2Cl2}$ but i get minimum product and too much starting material.

I need the product to be a TFA salt and not an HCl salt so I need to use TFA. Any solutions?

If you do a literature search (SciFinder/Reaxys), it seems that HCl in polar aprotic solvents (dioxane, THF, ether) are the 'standard' conditions for this cleavage.

You mention that you require the TFA salt, not the HCl salt, but even if you use HCl to carry out the reaction, its fairly trivial to get rid of the HCl and then convert the free amine into whatever salt you desire.

(Though as vapid said, your reaction conditions are probably quite important, and you might find if you just push the TFA reaction a little harder you can force the Boc off).


Finding the color of crystal violet theoretically I am trying to find the color of an organic molecule, crystal violet, theoretically.

I suspect that the color is due to the large conjugated system. My question is, how would I go about finding the energy levels of such a system? I know how to solve the Schrodinger equation for a particle in a ring or a box (i.e. benzene, or a poly-diene) to solve for energy levels and find the lowest-wavelength transition. But how would I do it for this sort of molecule?

For a reliable and experimentally comparable answer, you really need to use a numerical method rather than a pencil and paper estimation. For an organic system this size you'll probably be safe using time-dependent density functional theory (TDDFT). Most computational chemistry packages have this built in. Off the top of my head, I know that Gaussian (commercial–standard for organic molecules like this), and GAMESS (free) both have TDDFT built in.

If you're intent on using a toy method you could use Hückel theory as orthocresol suggested. You could in principle tune the parameters of such a model until the excitation energy matches the experimental result, but I don't see what additional physical insight this would give you.


Symmetry element in the coversion of cyclopent-2-enyl anion to pentadienyl anion? I'm trying to draw the state correlation diagram for the reaction of cyclopent-2-enyl anion to pentadienyl anion but I don't know which symmetry element is preserved in this reaction. What could it be?

Let's look at the back reaction, the closure of the pentadienyl anion to the cyclopentenyl anion.  Once you understand this you'll see that the exact same reasoning applies to the reverse reaction that you asked about.  So here's a picture of the reaction we're talking about.

And here is a picture of the 5 molecular orbitals for the pentadienyl system

Since the pentadienyl anion has 6 pi electrons, the first 3 molecular orbitals (from the bottom up) are filled, each with two electrons.  That makes the third molecular orbital (the one in the middle of the diagram), let me call it $\ce{\psi}_3$, the highest occupied molecular orbital (HOMO).  The HOMO is so important because it is the orbital that will control our process.  If we look at the symmetry of the $\ce{P}$ orbitals on $\ce{C_1}$ and $\ce{C_5}$ in $\ce{\psi}_3$, we see that in order to get constructive overlap (overlapping the parts of the $\ce{P}$ orbitals with the same sign) as we rotate the $\ce{P}$ orbitals on these two carbons to form the bond that forms the cyclopentenyl ring, we must rotate the tops (or bottoms) of these two $\ce{P}$ orbitals towards each other.  If we were to rotate the top of one and the bottom of the other towards each other we would overlap a plus side of one $\ce{P}$ orbital with the minus side of the other $\ce{P}$ orbital and this would not result in constructive or bonding overlap.  The type of rotation that leads to bonding overlap in this example is termed "disrotatory".  It preserves a plane of symmetry bisecting the pentadienyl system (e.g. it contains $\ce{C_3}$) and is perpendicular to the plane of the screen.  This is the symmetry element that is preserved during the ring closure.  The exact same arguments also explain the ring opening of the cyclopentenyl anion to the pentadienyl anion.


Chemical reaction that can produce lots of heat from 2-3 simple liquid ingredients? Have an experiment I want to try out utilizing thermoelectric power and I want to generate the heat via a simple chemical reaction. Ideally I'd like to mix 2 liquids together (one being water or alcohol would be great) that can get to around boiling water temperature for a few minutes or longer. Byproducts need to be only other liquids that can be drained easily or gasses that aren't dangerous to humans. No solids wastes and non corrosive / carcinogen / caustic.

So far I've found one that's close to my needs but still produces a solid waste product and isn't particularly safe, which is Calcium Oxide + Water -> Calcium Hydroxide + Heat. This might work if there's another liquid I could add after the main reaction to turn the Calcium Hydroxide into a liquid for disposal.

The first thing that comes to mind, if you have access to a stock-room, is mixing two solutions: (1) NaOH; (2) HCl. This can release a lot of heat if your solutions are concentrated enough, and it forms salt-water if your NaOH and HCl are of the same molarity.

Edit: I'm on my lunch break, so I did some of the math...

The chemical reaction of interest in this case is:

$$\ce{OH- + H+ -> H2O}$$

The standard enthalpies of formation ($\Delta H_\mathrm f^\circ$) of these species are:

$$\begin{array}{lr}
\hline
\text{Species} & ΔH_\mathrm f^\circ/\pu{kJ mol-1} \\
\hline
\ce{OH-} & -229.99 \\
\ce{H+}  & 0.00       \\
\ce{H2O} & -288.83 \\
\hline
\end{array}$$

Thus, the change in enthalpy for the reaction is:

$$\Delta H^\circ = -288.83\ \mathrm{kJ/mol} - (-229.99\ \mathrm{kJ/mol}+0\ \mathrm{kJ/mol}) = -58.84\ \mathrm{kJ/mol}$$

Therefore, for $1\ \mathrm{mol}$ of $\ce{NaOH}$ + $1\ \mathrm{mol}$ of $\ce{HCl}$, you get $58.84\ \mathrm{kJ}$ of heat. Say you want to release enough heat to get the net solution up to $100\ \mathrm{^\circ C}$.

Water has a heat capacity of $4.18\ \mathrm{J/(g\ ^\circ C)}$. Say you have $1\ \mathrm L$ of $\ce{NaOH}$ + $1\ \mathrm L$ of $\ce{HCl}$, you'll need enough heat to raise the temperature of $2\ \mathrm L$ of water to $100\ \mathrm{^\circ C}$. I'll assume the water starts off at $25\ \mathrm{^\circ C}$, so you have $2\,000\ \mathrm g$ and $75\ \mathrm{^\circ C}$ to go.

$$4.18\ \mathrm{J/(g\ ^\circ C)} \cdot 2\,000\ \mathrm g \cdot 75\ \mathrm{^\circ C} = 62\,700\ \mathrm J = 62.7\ \mathrm{kJ}$$

How many moles of $\ce{NaOH}$ + $\ce{HCl}$ do you need for that much heat?

$$\frac{q}{\Delta H^\circ} = \frac{62.7\ \mathrm{kJ}}{58.84\ \mathrm{kJ/mol}} = 1.065\ \mathrm{mol}$$

That would mean that you can mix $1\ \mathrm L$ of $1.065\ \mathrm M$ $\ce{NaOH}$ + $1\ \mathrm L$ of $1.065\ \mathrm M$ of $\ce{HCl}$, and would theoretically expect to get a temperature close to $100\ \mathrm{^\circ C}$.

This might be off a bit because I've made some assumptions:

$\Delta H^\circ$ is constant with respect to temperature from $25\ \mathrm{^\circ C}$ to $100\ \mathrm{^\circ C}$. This may not be true.
the dissolved salts in water dont significantly affect its heat capacity
You have $2\,000\ \mathrm g$ of water in $1.065\ \mathrm M$ $\ce{NaOH}$ + $1.065\ \mathrm M$ $\ce{HCl}$

I think this could get you close though? I'm a bit surprised that the molarities aren't higher... It's a starting point at least.

Disclaimer of course: be careful with the $\ce{NaOH}$ + $\ce{HCl}$ solutions, they can be dangerous. Use proper chemistry hygeine protocols. The mixture should be benign, but you should confirm this with pH paper.


Solvation of uranyl formate/acetate Uranyl formate has a structure something like:

$\ce{UO2-(OCHO)2}$

So the structural formulae I've seen have two double-bonded oxygens directly attached to the U and each formate group is attached by single bonds to the oxygens. Like this.
The empirical formula for the dry powder is written as a monohydrate.

Can someone explain how this thing gets solvated in water and what the solvated ions are, exactly? I just realized I was assuming it dissociates into a 'uranyl' ion (whatever that looks like) and two formate ions$\ldots$but then I'm confused about where the formate's oxygen goes.

This is a uranyl ion, one of the most common forms of dissolved uranium:

Because the two U=O bonds can form a single delocalised region, there's a strong tendency towards linearity (c.f. carbon dioxide). It tends to dissolve in water by gaining equatorial water ligands (4 or 5), some of which it may then hydrolyse, depending on the pH:

The axial oxygen atoms are also valid targets for hydrogen bonding, typically gaining one hydrogen bond each.

The difficulty in dissolving uranyl formate in particular suggests that the formate ion is quite well-bound, and it seems that in this case you get a combination of water and formate ligands in the complex around the uranyl. I was expecting the formate to be a bi-dentate ligand, but according to the results of a combined computational and experimental study 1, it's uni-dentate (which incidentally allows it to form polymeric chains in solution).

Here's a picture from that study showing a computationally obtained structure of a mixed water/formate-ligated uranyl structure.

2013, Dalton Trans., Christian Lucks, André Rossberg, Satoru Tsushima, Harald Foerstendorf, Karim Fahmya, Gert Bernharda, DOI:10.1039/C3DT51711J