Posted by: Mark Foreman | February 13, 2014

Cyclohexane compounds which can and can not do a E2 reaction

Dear Reader,

Here is a compound which can not do a E2 reaction, I have created a XYZ file which can be opened in ORTEP with ease.

30
XYZ file :   trans isomer of 1-bromo-4-tertbutyl cyclohexane
C -4.123 -0.017 -0.473
C -4.253 -1.394 0.174
C -2.923 -2.143 0.174
C -1.813 -1.320 0.821
C -1.683 0.056 0.174
C -3.012 0.806 0.174
C -5.442 0.727 -0.319
C -6.438 -0.152 0.425
C -5.212 2.010 0.469
C -5.996 1.068 -1.696
Br -0.124 -2.273 0.625
H -3.869 -0.161 -1.547
H -4.595 -1.267 1.226
H -4.987 -1.987 -0.415
H -3.045 -3.093 0.741
H -2.637 -2.344 -0.882
H -2.067 -1.177 1.896
H -1.340 -0.071 -0.877
H -0.948 0.650 0.764
H -2.890 1.756 -0.393
H -3.298 1.006 1.231
H -7.403 0.392 0.538
H -6.607 -1.090 -0.150
H -6.033 -0.401 1.432
H -6.176 2.555 0.582
H -4.484 2.653 -0.076
H -4.807 1.760 1.475
H -5.302 0.690 -2.480
H -6.993 0.590 -1.823
H -6.098 2.172 -1.792

Here is a cyclohexane ring which is locked into one chair with a tert butyl group which has a bromine atom arranged with the other atoms such that it can do a E2, I have changed the atom type of the hydrogens which would be removed in the E2 to be deutrium atoms.

30
XYZ file :   cis 1-bromo-4-tertbutyl cyclohexane
C -0.047 -0.171 -0.310
C 0.037 -1.198 0.839
C 1.277 -2.100 0.724
C 2.578 -1.308 0.658
C 2.504 -0.279 -0.464
C 1.270 0.633 -0.355
C -1.336 0.708 -0.296
C -2.601 -0.183 -0.356
C -1.439 1.598 0.961
C -1.381 1.627 -1.539
Br 3.036 -0.501 2.367
H -0.086 -0.764 -1.260
H 0.053 -0.672 1.820
H -0.859 -1.861 0.844
H 1.304 -2.829 1.568
D 1.184 -2.709 -0.208
H 3.428 -2.005 0.467
H 3.436 0.335 -0.497
D 2.457 -0.819 -1.441
H 1.361 1.268 0.554
H 1.285 1.320 -1.232
H -3.525 0.425 -0.475
H -2.559 -0.892 -1.214
H -2.747 -0.774 0.574
H -2.368 2.211 0.942
H -0.589 2.310 1.040
H -1.471 0.999 1.897
H -0.597 2.416 -1.512
H -1.254 1.046 -2.480
H -2.351 2.169 -1.615
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Posted by: Mark Foreman | February 4, 2014

Stereocentres

Dear Readers,

While some of you love stereochemistry I know that some of you are troubled by it, the diagrams on sheets of paper can be hard to see in 3D. I want to try and do something today to help you.

I suggest that you use the following to make a xyz file for ORTEP

8
XYZ file :   Halothane
C -0.687 -0.394 0.070
C 0.574 0.459 0.070
F -1.807 0.432 0.070
F -0.703 -1.197 -1.068
F -0.703 -1.198 1.206
Br 2.143 -0.698 0.070
Cl 0.596 1.495 -1.396
H 0.588 1.101 0.979

ORTEP can be downloaded and used for free for non commerical use from a very kind gentleman in Scotland named Louis Farrugia. In the tradition of crystalography he writes software for other crystalographers (and other people) and then he makes it avaiable for free to many people. This is a real deep and good service to the rest of the chemical community.

To use it you need to set some system variables in windows and then load the coordinates as XYZ files.

Now if you get the halothane coordinates into ORTEP (or some other viewing software) you can look at the arrangement of the atoms.

The key thing to do first is to work out which is the most heavy atom attached to the chiral centre, this is given the label 1, then the next heaviest is 2 etc etc until you do all four for a tetrahedral atom such as carbon.

What you also have to do is to rotate the molecule until the lightest atom is pointing away from you. For halothane here are four pictures taken as I turned it around. The bromine is brown, the chlorine is green and the fluorine is yellow/green.

First view of the molecule

First view of the molecule

Second view

Second view

Third view

Third view

Final view

Final view

You should see that the bromine is number 1, the chlorine is number 2 and the CF3 group is number 3. If you trace the path made by these atoms then it goes clockwise. Thus it is the R isomer.

You might want to try out some other files which I am showing below.

26

XYZ file : halogenated adamatane

C -3.965 -0.359 1.350
C -2.976 -0.544 0.187
C -4.872 -1.128 -1.348
C -5.872 -0.946 -0.193
C -3.660 -0.201 -1.148
C -4.142 1.260 -1.108
C -5.176 -1.285 1.136
C -5.139 1.455 0.049
C -6.342 0.519 -0.158
C -4.446 1.102 1.377
Br -7.403 -2.124 -0.455
Cl -2.498 -0.419 -2.500
I -5.806 3.497 0.096
F -3.341 -0.671 2.554
H -2.102 0.129 0.336
H -2.638 -1.604 0.161
H -4.526 -2.185 -1.372
H -5.373 -0.871 -2.308
H -3.268 1.932 -0.957
H -4.647 1.504 -2.069
H -5.896 -1.149 1.974
H -4.826 -2.341 1.105
H -7.062 0.654 0.680
H -6.835 0.767 -1.124
H -5.167 1.237 2.214
H -3.569 1.772 1.520

Here is another one a smaller molecule

5
XYZ file :   halon 1111
C 0 0 0
H 0.6 0.6 0.6
F 0.8 -0.8 -0.8
Cl -0.8 0.8 -0.8
Br -0.8 -0.8 0.8

And another one which is the other isomer of halon-1111

5
XYZ file :   halon 1111
C 0 0 0
H 0.6 0.6 0.6
F 0.8 -0.8 -0.8
Cl -0.8 -0.8 0.8
Br -0.8 0.8 -0.8

Well I hope that you have had fun with these molecules, my advice is “enjoy organic chemistry”.

Posted by: Mark Foreman | December 15, 2013

Titration I

Dear Student,

There appears to be no escape yet from equilibrium in aqueous media (maybe it is thermodynamically impossible to escape it), this in some ways is no bad thing as you need to learn these things in great depth to be good at physical chemistry.

Now imagine that we have 1 litre of 0.5 M acetic acid, what will the pH of this solution be ?

If we ignore the fact that the final acid concentration will be lower than the initial concentration, then we can use a very simple method.

KA = [H+][AcO]/[AcOH]

Then as [H+] = [AcO] rearrange to give

[H+]2 = KA[AcOH]

Then

[H+] = (KA[AcOH])½

Now if we mix the acetic acid solution with 500 ml of 0.5 M potassium hydroxide what will the pH be now.

We can attack the problem in a quick way or a more complex and better way.

Now the easy way is to use the buffer equation to deal with the problem.

pH = pKa + log10([AcO]/[AcOH])

For acetic acid pKa = 4.5

Now what is the pH of the mixture after 1000 ml of 0.5 M potassium hydroxide has been added.

The mixture is now 0.25 M potassium acetate

We have the equilibrium now of

AcO + H+ à AcOH

Which can be rewritten as

AcO + H2O à HO + AcOH

As

KW = KAKB

Then

pKW = pKA + pKB

So

pKW – pKA = pKB

Thus

pKB of acetate is 9.5

So KB = 3.16 x10-10

So if we assume that the final acetate concentration is equal to the initial then we can use

KB = [OH][AcOH]/[AcO]

Rearrange

[OH]2 = KB[AcO]

Next

[OH] = (KB[AcO])½

So for [OH] we get a value of (7.9 x 10-11) moles per litre

Which is 8.89 x 10-6 moles per litre

As

KW = [H+][HO]

Then we have a proton concentration of 1.125 x 10-9 moles per litre

This will give us a pH of 8.95

Posted by: Mark Foreman | December 6, 2013

Dilute solutions of weak acids

Now sometimes we want to calculate the ”proton” concentration in an aqueous solution of an acid from the initial concentration of the acid [HA]initial and the dissociation constant (KA) of the acid.

Now when the initial concentration is high and the dissociation constant is small then we can take a shortcut, here we will also ignore the autoioniziation of the water.

As

KA = [H+][A]/[HA]

Then

KA = [H+]2/[HA]

And then

[H+] = (KA[HA])½

Here we assume that [HA]initial = [HA]final

While this is OK for things like a 1 M solution of acetic acid it will fall down for many other things. One problem is that it predicts more than 100 % ioniziation of the acid. Here is a graph showing the bad prediction of how much of the acid will be ionised.

Prediction of the ionization of a weak acid by the very simple method

Prediction of the ionization of a weak acid by the very simple method

Now a slightly more rigorous method is to use the following mass balance equation

[HA]initial = [HA]final + [A]

And the charge balance equation

[H+] = [OH] + [A]

Together with the buffer equation

-log10[H+] = pH = pKa + log10([A]/[HA])

We will still ignore the autoionization of the water, so we assume that

[H+] = [A]

Now we need to make what I call a brute force attack on the problem, this is one which is quite simple in terms of maths but is one in which close to every possible combination is tried. It is like going through all the combinations which are possible on a safe. Given time you will get lucky and open the safe.

Now

[HA]final = KA / [H+]2

We can get [A] from the buffer equation, and we can then calculate for a series of proton concentrations the values of [A] and [HA]final. As the sum of [A] and [HA]final is [HA]initial then we can work backwards and make a graph of pH vs [HA]initial. Here is a graph of pH against the original concentration of the acid.

Graph of pH against original acid conc for two different methods of doing the calculation

Graph of pH against original acid conc for two different methods of doing the calculation

The good thing about the new model is that it does not predict more than 100 % ionization of the acid, well that is a good step forward !

Graph of percentage of ionization of the acid against the original concentration of the acid

Graph of percentage of ionization of the acid against the original concentration of the acid

So now we have a better model for the pH range around 4, but as the pH rises further and further this new model will become worse and worse.

So we should go back to our model and improve it further.

If

[H+] = [OH] + [A] and Kw = [H+][OH]

Then

[A] = [H+] – (KW/[H+])

We can use

[H+]3 + KA[H+]2 = (KW + KA[HA]initial) [H+] + KAKW

As we know KA and KW we can now create a table of values for (KW + KA[HA]initial) for different proton concentrations.

From this we can get the [HA]initial concentrations for a range of different proton concentrations.

From this we can get [A] and [HA]final to finish off our problem. Here is the graph of pH vs original acid concentration.

pH as a function of the original acid concentration, this is the best method with the most complex maths

pH as a function of the original acid concentration, this is the best method with the most complex maths

Posted by: Mark Foreman | November 13, 2013

Lets get physical, physical equilibrium

Dear Reader,

This week I was teaching physical equilibrium, my first problem which I needed to teach was about variation of vapour pressure as a function of temperature.  As I did it I wondered how best to make it relevant to chemical engineering students. Years ago as part of my teacher training I was taught by a man who teaches medical student how I can increase student motivation by explaining why they need to know something.

I choose to consider BLEVE accidents, now for those of you who do not know about BLEVE it is a class of explosion in which a sealed container of liquid heated above its normal boiling point breaks open. Here we clearly have an application of the Clausius-Clapeyron equation. As part of my lesson I talked about what will happen to a drinks can placed on a camp fire and what will happen to a propane tanker in a fire. For the later I showed the students a 2 minute film which was originally made as a teaching aid for firefighters.

This made me think of a question which might appeal to chemical eng students.

How about

A propane tank is a cylinder 2 meters long with a diameter of 1 meter. This tank is filled to 90 % of its capacity with propane at 273 K. The tank will burst open at a pressure of 960 psi (662 kPa).

If the boiling point of propane is 231 K and the enthalpy of vaporisation is 18.8 kJ mol-1, then calculate the pressure at 20 oC.

Next calculate the temperature at which the tank will burst open if it is heated in a fire.

If the fire which heats the tank delivers 20 kW to the tank then how long will it be before the tank bursts open ? For this we will need the heat capacity of liquid propane.

A second tank is fitted with a safety valve which limits the tank pressure to 500 kPa this valve vents the gas upwards and away from the tank. For the same fire calculate how long it will be between the start of the fire and when the valve opens, also calculate how many moles of propane will exit the tank via the safety valve per minute.

If the tank bursts open (due to the overheating of tank metal) when it is 30% full then calculate how long after the start of the fire does this second tank fail. To answer this question you will have to look in the literature for some missing data and also consider some other issues (hint check the critical temperature of propane).

Then I had to teach about melting point and vapour pressure depression of solvents which is caused by a solute. Now this is an interesting matter, while some text books have written off cryoscopy as a thing of the past I would like to point out that it is still relevant to some problems in chemistry.

While some people like the idea of using mass spectroscopy to get the molecular weight of everything, I would like to point out that for some things it cannot be used. For example polymers can be close to impossible to measure using a mass spectrometer, as an alternative cryoscopy, osmotic pressure measurement or viscosity can be used to deal with these macromolecules.

I would also like to point out that some species can exist in more than one form with different molecular masses. A favourite of mine is Lawesson’s reagent.

When Lawesson’s reagent is heated and bombarded with 70 eV electrons in a EI (electron impact) ionization source it gives a peak at with a m/q of 202 amu. While it would be unreasonable to assume that all Swedes are tall and blonde based on the observation of a single person it is equally unreasonable to assume that an electron impact mass spectrum gives a representative answer.

The method which the first people to make Lawesson’s reagent (long before Lawesson had thought of using it) was to measure the melting point of naphthalene when it had some Lawesson’s reagent dissolved in it.

The freezing point depression (ΔTf) is given by

ΔTf = kf  b

Where kf and b are a constant and the number of moles of solute per kilo of solvent (molality). For naphthalene the value of kf is 6.94 K Kg mol-1. If we take a data set for Lawesson’s reagent, I made this up with some slight random errors added, then we can solve the problem.

Mass of LR in g per kilo   of napthalene

ΔTf   (K)

4.0

0.069

8.9

0.153

10.9

0.187

14.9

0.257

 

We can then use this data to estimate that the Lawesson’s reagent has formula mass of XXX (you did not think I would blurt out the answer ?!)  when it is in solution. As the empirical formula of Lawesson’s reagent is {C7H7OPS2}n then we can work out how large is n. By the way the mass of C7H7OPS2 is 202 g mol-1.

The next major bit was when I had to teach about osmosis, now in one common text book you have the equation

Π = iRTc

Where the weird looking pi (Π ) is the osmotic pressure in Pa, I was dealing with this and then I noticed that something was horrible about this equation.

I used dimensional analysis on it then we notice something odd if we use the normal concentration in moles per litre for c,

Pa = Nm-2 ≠ J mol-1 K-1 . K . mol dm-3

J mol-1 K-1 . K . mol dm-3 = J dm-3

Now I cannot relate Pa to J dm-3 without multiplying everything by 1000 so something is wrong.

On the other hand

Pa = Nm-2 = J mol-1 K-1 . K . mol m-3

Which gives us

Pa = J m-3.

Now as

Pa = N m-2

and

J = N m

Then

Pa = N m m-3 = Nm-2

Success at last !

Now call me what you like but I think it is the duty of a person writing a book to draw attention to odd things like moles per cubic meters rather than the duties of the reader including the mitigation of the worst effects of bad writing.

A book should be written in a clear manner which educates and/or entertains the reader rather than bombarding them with a barrage of text. I recently read “Fashionable Nonsense” which is about the odd ideas that some people in European intellectual circles have had. For me the claim that the topology of a neurotic is a torus really took the biscuit, but after reading this chemistry text book I felt that maybe we should look at the way we write text books as well.

Posted by: Mark Foreman | November 4, 2013

Brayton cycle for a nuclear plant

Dear Reader,

As a result of an unforeseen event I have had to teach thermodynamics this year to the first years, while I cannot say that I normally experience the same ecstatic feelings when thinking about thermodynamics as I get when considering a decent bit of either organic or inorganic chemistry I saw something which interested me when I was reading a nuclear journal.

The problem is that after a SCRAM at a power reactor site the turbine will normally slow down within seconds and stop being a source of energy. Because of the decay heat of the used fuel the plant will need some power to operate pumps and other vital equipment to keep the core of a light water reactor covered with water.

While if one unit at a multireactor park like Ringhås was to shut down it would be possible to import electric power from another unit, if all the units were scrammed at the reactor park then electric power would need to be imported from elsewhere. If the power lines to the plant remained good then one simple method would be to take in energy from the grid.

Another option would be to use a diesel set to provide the power for vital services at the plant, some plants also have gas turbine sets as an alternative to running on diesel sets. What happened at Fukushima was that all the units which were running at the time of the earthquake scrammed (underwent an emergency shutdown which was commanded by a seismic sensor), but the earthquake and nasty big wave damaged the power lines, washed away the diesel fuel tank and soaked important electrical systems with water.

Now back to thermodynamics, a key thing in thermodynamics is heat engines which operate using the difference between a hot and a cold thing. One idea is to put a heat engine into the reactor which will use the steam from the SCRAMed reactor to heat up carbon dioxide.

The idea is to run a compressor using a turbine, the compressed gas is then heated using the steam from the reactor. The heated gas then expands through a turbine which shares a shaft with the compressor. The hot gas from the exit of the turbine then is cooled outside the reactor building before it is returned to the compressor.

The great thing is that the turbine generates more mechanical work than the compressor consumes, the difference could be used to run a generator or some other useful device. While the heat to electricity efficiency is poor this device is designed more to cool the steam coming from the core. The cooled steam will condense and then help to keep the core covered with water.

Posted by: Mark Foreman | October 27, 2013

The mechanism of a lithium aluminium hydride reduction

Dear Reader,

It is good to be back, while sadly I can not enter my classroom like “Eyeball Paul” in Kev and Perry go large (I have to use the door like everyone else) I can say that I have a treat for you deserving people.

Now recently one of my readers did not believe me regarding the mechanism of the lithium aluminium hydride reduction of a carbonyl, but I have dug up strong evidence which I hope will be of interest to you as it explains how our fiery friend LAH works.

Now I would like to make an observation about organic chemistry, organic chemists tend to draw curved arrows to explain mechanisms for everything but sometimes they draw a plausible mechanism without doing an experiment to prove it. However in the case of lithium aluminium hydride experiments have been done which prove what the mechanism is.

Now the late and great Ronald Snaith did a lot of very good chemistry where he showed that the binding of lithium to organic molecules is very important. I had the joy of meeting him once when I was a PhD student at an inorganic chemistry conference.

Now some years ago Karl E. Wiegers and Stanley G. Smith published a paper in the Journal of Organic Chemistry (1978, volume 43, page 1126) back in the days when I was in the infants section of primary school. I recall not long after this paper was published I picked up my first science textbook. It was an encyclopedia of science which seemed much more interesting to read rather pay attention to the storys which the deputy headteacher (my form teacher) was reading to the class.

One story which I read / was read about that time was about Marie Curie who was killed by overexposure to radiation, after reading about her untimely death I decided (as a young prechemist) that when I grew up to be a chemist I would do my experiments behind a wall of lead bricks in case I managed to make a radioactive chemical. A while later I discovered that I could not make radioactivity by a mere chemical reaction so I dropped the lead wall idea from my mind. I think that doing chemistry from behind a wall of lead would be much harder, I have had a go at using master slave grabbers of the type used for working with used nuclear fuel and trust me it is not easy to pick up a spanner or a bit of metal pipe with them.

Then a while later as a postdoc when I learnt about the joys of radioactivity from Jiri Hala I learnt about a radioisotope of beryllium whose decay rate is a function of its chemical environment. I will blog about this beryllium radioisotope another day.

Back to the paper of Karl E. Wiegers and Stanley G. Smith, they measured the rate of reaction of camphor with sodium and lithium aluminium hydrides in THF. They found that the sodium salt reacted more slowly which proves that the nature of the cation is important. The sodium cations are less able to bind to the carbonyl oxygen and increase the partial positive charge on the carbonyl carbon.

This provides us with some evidence to explain the reactivity series of lithium > Sodium for the aluminium and boron anionic hydrides.

Further evidence has been provided by the work of two French chemists (Pierre and Handel) who showed that when [2.l.l]cryptand is added to lithium aluminium hydride that it can no longer reduce many functional groups. This is in contrast to the reactivity of enolates where a strong lithium binding agent such as HMPA will make the enolate more reactive. For the enolate the more coordinated the lithium is by the solvent or the additive the more free the anionic enolate is to react with an electrophile. The same is true of many organometallic reagents such as Grignards and organolithiums, for these organometallics the more able the solvent mixture is to coordinate to lithium the more reactive the carbon nucleophile is.

The fact that lithium aluminium hydride shows the opposite trend when the solvent system is made more coordinating to lithium suggests that coordination of lithium cations to the carbonyl is as important (electrophilic nature of the lithium ion) as important as the nucleophilic part (the AlH4 anion) of lithium aluminium hydride.

Posted by: Mark Foreman | September 26, 2013

S orbitals and a H-H bond

Dear Reader,

I was going to entertaintorment you with vogon poetry but why wife has told me to be good and spare you the poetry. But lets get on with writing about orbitials.

One of the joys of chemistry is the shapes and forms of the atomic and molecular orbitals. The thing is that an electron can not be located, the uncertainty principle means that you can not know both the location and the energy of an electron. The better defined one is the worse the uncertainty with which the other is known.

We know the energies of electrons in atoms and molecules well through photoelectron spectroscopy (PES) and other spectroscopic methods so the locations of the electrons are poorly known. But there is hope. We can draw a surface around the shape which has a given probability of containing the electron. In the same way as the improbability powered space ship in the Hitch Hikers Guide to the Galaxy did lots of wonderful things like turning missiles into vases of flowers (and whales) the shapes of the volumes of space which have a given probability of containing an electron can do wonderful things for our chemistry.

Now to get going we will consider the s orbitals of a hydrogen atom, OK a hydrogen atom might be small and weedy but we need to understand these little things before we move onto big things. We need to start with the 1s orbitial which has a simple radial wavefunction of

Ψ = e-kx

The electron density (electrons per cubic Å) is given by the square of the wavefunction (Ψ)

This gives us a sphere shaped orbital which is like a ball, it is very symmetric. Here are two views of the electron density in the 1s orbital. First here is the bird’s eye view of a slice through the centre of the orbitial.

Bird's eye view of the electron density in a 1s orbitial

Bird’s eye view of the electron density in a 1s orbitial

Then here is the 3D view of the electron density in the orbiital

3D view of the graph of electron density as a function of distance from the centre

3D view of the graph of electron density as a function of distance from the centre

We have another s orbitial with a higher energy (2s) which has a slightly more complex radial function which is

This means it has an inner part with the opposite sign wavefunction. We will get onto that soon, I also hope to write a spreadsheet which will give me a better ability to draw oribitials.

Posted by: Mark Foreman | August 30, 2013

Epichlorohydrin

Dear Reader,

I have just escaped from a Swedish cray fish party, and now it is time to think about organic chemistry again. Now I am sure that all of you will use some plastic during your daily lives, for example we wear plastic (nylon and polyester), we store data on plastic (polycarbonate CDs), we use plastic electrical leads (PVC), we use fluorinated polymers (PVDF) for fuel lines for diesel systems and we use plastics for paints and coatings for surfaces.

One important class of polymers for paints / coatings and glue are the epoxy resins. These are typically made using bisphenol A  and a bifunctional (two functional group) molecule known as epichlorohyrin.

Here is a molecule of epichlorohydrin in all its glory.

Epichlorohydrin, note one of the hydrogens in the molecule is hidden from sight.

Epichlorohydrin, note one of the hydrogens in the molecule is hidden from sight.

The bisphenol A is made normally by an electrophilic substitution reaction from phenol and acetone. The traditional route to phenol generates acetone as a side product, so the production of bisphenol A makes some sense. If phenol production occurs by the cumene route then where there is phenol there will never be a shortage of acetone. Here is bisphenol A.

Bisphenol A

Bisphenol A

We will explain this later but the epichlorohydrin is made from glycerol or from allyl chloride. But the more interesting part of the chemistry of epichlorohydrin is the reaction with the bisphenol A this reaction is normally done by heating the biphenol A with sodium hydroxide in water with epichlorohydrin present.

Here is the solvent accessable surface of epichlorohydrin, the more positive an atom is the more red it is and the more nagative an atom the more blue it is.

The solvent accessable surface of epichlorohydrin

The solvent accessable surface of epichlorohydrin

The epichlorohydrin could react with our phenol in three ways, the phenol as the phenolate anion could attack the alkyl chloride in an SN2 reaction but this is unlikely as the carbon bearing the chlorine only has a weak partial positive charge (calculated by Huckel theory to be 0.028). But the two epoxide carbons have much higher partial positive charges. The carbon bearing the CH2Cl group has a partial charge of 0.316 while the other one has a charge of 0.219. I think that the carbon with the lower partial positive charge will be the site of the reaction. Here is what I think the neutral form of the initial product will look like this.

Stage one product

Stage one product

While the alternative product would be

Alternative and wrong product

Alternative and wrong product

According to a molecular mechanics calculation the product of reacting a phenol molecule with the middle carbon of epichlorohydrin has a steric energy of 7.8423 kCal per mol while the reaction product in which the phenoxy group is attached at the end carbon is 3.9212 kCal per mol. I suspect that the transition states for the reactions forming these products would also be different.

The transition state for the reaction reacting at the end will be lower, so there is both a kinetic and thermodynamic reason why the less substituted end of the epoxide reacts. This sets up the molecule to then lose chloride to form a new epoxide group which can react with a second phenolate ion to form a link between two phenol groups. Here is the mechanism for the reaction.

The reaction of expichlorohydrin with one phenolate

The reaction of expichlorohydrin with one phenolate

I think that you should be able to work out how the second phenolate reacts with the epoxide which is generated by the loss of the chloride. If this is repeated again and again then it will generate polymers if we use a diphenol.

Posted by: Mark Foreman | May 21, 2013

The banana bond

Bananas are fun things, those of you who might have thought that I have changed into Sean Banan can relax and breathe easy again. I am here to talk about the banana bond. This is the informal name for a bent bond, the banana bond I want to discuss now is a very bent bond which is found in diborane.

Now after we have seen how to draw the Lewis structure of a borohydride anion we can try to make a similar diagram for diborane using conventional two atom and two electron bonds. This is the best I can do and it is clear that it is a dismal failure, the borons only have six electrons around them and this falls short of the eight electrons which the atoms need to attain a noble gas like full outer shell.

Bad version of diborane, it has only six valence electrons around each boron.

Bad version of diborane, it has only six valence electrons around each boron.

Now what we have in this molecule are two three centred two electron bonds, these are bonds which involve three atoms and only have two electrons in them. With two sp3 and one s orbital we can draw out the three ways the orbitals can be combined to make three molecular orbitals. Here are the three molecular orbitals associated with the three atoms.

The three orbitals in the banana bond

The three orbitals in the banana bond

So after we back the electrons into the two bonding orbitals for the banana bonds, we can now get eight electrons around each boron atom. Thus even for these electron deficient molecules we can still have eight electrons around each boron.

Electrons in a diborane molecule

Electrons in a diborane molecule

We can move onto bigger and better things, for example if we replace the main group boron with a transition metal (such as copper) then to have a noble gas configuration we need 18 electrons around the transition metal. For example we can have a copper complex. How about this complex.

[Cu(BH4)(PPh3)2]

[Cu(BH4)(PPh3)2]

Now I hold the view that in organometallic chemistry that oxidation state can be a bit of fluid, at times it is not clear. It is possible to assign two different oxidation states to a metal in the same complex. Now if we assume that the copper in our complex has an oxidation state of zero then the zero valent copper has 11 electrons, ten of these electrons will stay around the copper in its atomic orbitals while one will enter one of the 3 centre 2 electron bonds. Two electrons from the bridging hydrogens will be in these banana bonds while one of the boron electrons will also be in a banana bond. The bonds to the phosphines contain two electrons from the phosphorus atoms (dative bonds) while the terminal hydrogen to boron bonds are simple normal two centre two electron bonds.

The copper borohydride complex with a zero valent copper and a zero valent boron

The copper borohydride complex with a zero valent copper and a zero valent boron

If we make the complex a copper(I) complex which has a boron in the (-I) oxidation state then the boron will have four electrons to start with. The complex will still have four electrons in the banana bonds, four in the dative bonds from the phosphines and four in the “normal” bonds to the terminal hydrides. Again we have 18 valence electrons around the copper and eight around the boron.

 

The copper borohydride complex with a copper(I) centre in it.

The copper borohydride complex with a copper(I) centre in it.

We will get onto another fun type of bond soon.

 

 

 

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