Posted by: Mark Foreman | March 13, 2018

The Novichok agents

Dear Reader,

Things have become more complex, it has been claimed in the UK that the nerve agent attack involved a so-called novichok agent. Now I can not claim to know where these novichok agents came from, who did it or even if a novichok agent was used. I would rather not get sucked into a debate about who did it, I am mainly going to consider the chemistry.

However I can consider some of the chemistry, according to the internet the novichok agents were invented in the Soviet Union in an attempt to make chemical warfare agents even worse than sarin and VX. On Wikipedia it is claimed that methyl-(1-(diethylamino)ethylidene)phosphoramidofluoridate and 1-chloropropan-2-yl (E)-(((chlorofluoromethylene)amino)oxy)phosphonofluoridate are examples of novichok agents.

1-chloropropan-2-yl (E)-(((chlorofluoromethylene)amino)oxy)phosphonofluoridate

Part of the problem is that very little if anything has been written in the academic literature about novichok agents, well at least under that name. I did a search of the web of science and I did not find any mention of any paper with that word in the topic or title on the subject of chemical warfare. On the other hand I found almost 800 papers which have the word “sarin” in the title.

I also looked for methyl (E)-(1-(diethylamino)ethylidene)phosphoramidofluoridate in the organic chemistry literature using one common database, and I could not find any mention of this substance. Looking at it as a organophosphorus chemist I can see that it has the correct groups attached to a phosphorus atom to act as an acetylcholinesterase inhibitor. I also see an electron releasing group which would reduce the partial positive charge on the phosphorus atom when it is compared with sarin. This reduction in partial positive charge would reduce the rate at which the substance will react with water. This change would be likely to make the substance more able to persist in the environment.

However when you look for 1-chloropropan-2-yl (E)-(((chlorofluoromethylene)amino)oxy)phosphonofluoridate you can find plenty written on the subject in the Soviet chemical literature back in the 1960s and 1970s. By reacting dichlorofluoro(nitroso)methane with a phosphorus(III) compound a reaction similar to the Arbuzov reaction occurs which forms the product. Here is my best effort for the mechanism by which the compound is formed.


I hold the view that while the Soviets are thought to have developed the novichok compounds it would not be very hard for a competent (and well protected) phosphorus chemist to make a moderate amount of one of these rather exotic substances. So unlike polonium-210 (which is only produced) in very few nuclear reactors it is impossible to argue based on the identity of the nerve agent where it came from.

One of the problems in life is that sometimes bad people will choose tools or methods which are not normally used by people from their country in an attempt to confuse any investigator. For example the NKVD used German pistols to murder Polish officers at Katyn. This offers some degree of plausible deniability. Thus I think some other evidence other than the identity of the nerve agent is required before anyone can blame anyone for the horrible event which has just occured.

Posted by: Mark Foreman | November 25, 2016

How to get the energy of activation.

In the past, yesteryear we used to use graph paper to estimate the activation energy. We would plot graphs of ln(k) against 1/T (Kelvin -1).

Now a modern student wanted to do it without drawing a graph. Given data obtained at two different temperatures lets see how we can do it.

K1 = A exp (-Ea/RT1)


K2 = A exp (-Ea/RT2)


K1 / K2 = exp (-Ea/RT1) / exp (-Ea/RT2)


ln (K1 / K2) = ln (exp (-Ea/RT1)) – ln ( exp (-Ea/RT2))

is it a false start ?

No we can now write

ln (K1 / K2) =  (-Ea/RT1) –  (-Ea/RT2)

Which can be rearranged to

ln (K1 / K2) =  -Ea {(1/RT1) –  (1/RT2)}

Then to

ln (K1 / K2) =  -Ea/R {(1/T1) –  (1/T2)}

This gives us an equation which is very similar to the Clausius–Clapeyron equation.

Posted by: Mark Foreman | May 3, 2016

Cyclopropane chemistry

Dear Reader,

There is a deep problem with some chemistry text books, this is the writers of some books create questions (problems) for teaching using their own opinion about what the answer should be. The problem is often they seek to teach a given point in chemistry.

But they unwittingly ignore some other aspect of chemistry which causes the answer to be something other than the one which they expected. What I would suggest is that anyone who is writing a question for a chemistry book should make a quick literature search of the topic to make sure that they have got the right answer.

For example I have encountered one text book which claimed that the hydrides of the oxygen group increase in Bronsted acid strength as the central atom (O, S, Se, Te and maybe Po) get more and more heavy. Their reasoning was that the bond strength between hydrogen and the central atom would become weaker, thus the acid would get stronger.

In case you do not know HF and water are both much weaker acids than you might expect based on the electronegativity of the central atoms. This is because the bond strength is so high.

However in the case of the chalcogen hydrides while the bond strength decreases the electronegativity of the central atom also decreases. As a result hydrogen selenide and hydrogen telluride have similar pKa values when dissolved in water. Now that was quite painless, lets move onto a more complex case.

Now if we ignore that fact that interconversion of hydrogen atoms between the axial and equitorial sites in cyclohexane is quite slow we can reason that only one hydrogen environment exists in this molecule. If you cool it down you can get the proton NMR spectrum of this molecule to change from a sharp line to a broad line which then changes further to two sharp lines.

If we replace one of the hydrogen atoms with a chlorine atom then only one product (chlorocyclohexane) is possible. I am quite happy with this idea.

We could extend this idea for a moment to cyclopropane, this is a molecule where we have an even better case for arguing that all the hydrogens are the same. In this case the molecule can only adopt one conformation. It is stuck in single conformation.

We could reason that if we were to replace one of the hydrogens with a chlorine atom then we would have chlorocyclopropane.

The only problem was that the question suggested using chlorine gas under free radical chain reaction conditions to form this alkyl chloride. When we look in the literature we find that cyclopropane and chlorine gas forms a large amount of allyl chloride under these conditions.

What will happen is that a chlorine atom will abstract a hydrogen from a cyclopropane to form a free radical which will then rearrange in a ring opening reaction to form an allyl radical. The allyl radical will then react with chlorine to form allyl chloride which will then be likely to react with the chlroine to form 1,2,3-trichloropropane and a host of other products.

Posted by: Mark Foreman | February 13, 2016

Tin hydride reductions

Dear Reader,

Recently one of my readers asked about triphenyl tin hydride as a reducing agent, this is a reagent which will operate in a very different way to the anionic hydride reagents such as sodium borohydride which I have discussed in another post.

I would not expect triphenyl tin hydride or the more common tributyl tin hydride to be a nucelophile which is able to deliver a hydrogen atom to a carbonyl carbon, instead I would expect these reagents to act as reducing agents in free radical reactions. If we consider an alkyl halide then the following reaction mechanism can occur.

First a tin centred radical such as a tributyl tin radical will react with alkyl halide to form an alkyl radical and a tributyl tin halide. The alkyl radical will then react with the tributyl tin hydride to form an alkyl hydride (alkane) and a tin centred radical. Thus we can have a chain reaction. Now you have the words here is a diagram.

Reduction of a alkyl halide by a tin hydride reagent

Reduction of a alkyl halide by a tin hydride reagent

Posted by: Mark Foreman | February 11, 2016

Elimination from an alkyl halide

Dear Reader,

Something unexpected happened today when I was teaching, a student instead of drawing out the E2 mechanism did something else. Now normally the base attacks the hydrogen on the carbon next to the one bearing the leaving group. Here is the correct mechanism for you.

E2 reaction of a base with 1,1,1-trichloroethane ("methyl chloroform")

E2 reaction of a base with 1,1,1-trichloroethane (“methyl chloroform”)

Now instead of doing the easy E2 this student had drawn the formation of a carbene according to the following mechanism. Now the carbene synthesis is not a common reaction, I suspect that this is because the carbene has a much higher energy than that of the alkene. So I think that if a choice exists between a molecule doing an alkene synthesis by a E2 and a carbene synthesis then it is very likely that the alkene synthesis will occur instead of the carbene synthesis. Here for your delight is a carbene synthesis from chloroform.

Carbene synthesis from chloroform

Carbene synthesis from chloroform

It might look good but keep in mind that carbene formation is quite a rare event.

Posted by: Mark Foreman | February 4, 2016

Attack on carbonyl groups

Dear Reader,

The attack of a nueclophile on a carbonyl group is an important reaction, organic chemistry is littered with examples of carbonyl groups reacting with nucelophiles. It even pops up in interesting places such as the chemistry of the pyridines. The reaction of a nucleophile such as an amide ion with pyridine can be viewed as being like the reaction of a carbonyl. Here is the first stage of the Chichibabin reaction of pyridine and sodium amide.

The first stage of a Chichibabin reaction

The first stage of a Chichibabin reaction

Now to get us going we need to look at the list of groups, I have put the normal carboxylic acid derivatives in order of reactivity with a nucleophile. Here is the list.

List of carbonyl compounds in desending order of reactivity

List of carbonyl compounds in descending order of reactivity

Now the important thing to understand is that the pKa of the “leaving group” is increasing as we go down the list. At the top we have chloride which relates to the very strong acid hydrochloric acid while at the bottom we have an oxide group which corresponds to a very weak acid (hydroxide anion). In general if the nucelophile is the conjugate base of an acid with a higher pKa than the acid corresponding to the leaving group then the reaction is easy. If the reverse of this is true then the reaction is very hard or impossible. If the two pKas are the same then sometimes it is possible to do the reaction.

If by some miracle a chloride was to attack an ester on the carbonyl carbon then we would have the following tetrahedral intermediate.

Tetrahedral intermediate

Tetrahedral intermediate

When the negative charge on the oxygen is used to recreate the pi bond between the oxygen and the carbon (remaking the double bond) it will be much more likely that the chloride will be lost rather than the ethoxide as the chloride is far better leaving group. In general good leaving groups are the conjugate bases of strong acids, which makes them weak bases.

For example an acid chloride can react with an amine to form an amide with ease, but an amide can not react with hydrochloric acid to form an acid chloride. A second thing which contributes to the reduced reactivity of the compounds which are towards the bottom of the list is that the partial positive charge on the carbonyl carbon is very small. This small degree of positive charge on the carbonyl reduces the ability of the nucelophile to attack.

This change in the partial positive charge on the carbon explains the following reactivity series with nucleophiles such as borohydride and Grignard reagents.

Chart of carbonyl groups listed in order of reactivity

Chart of carbonyl groups listed in order of reactivity

In the near future I hope to post some more on this topic

Posted by: Mark Foreman | January 26, 2016

Spot the functional group

Dear Reader,

Some of you might be aware of a game played in UK newspapers called spot the ball, it is a photo of a sporting event where the ball has been painted out. The reader is invited to use their “skill and judgement” to work out where the ball is and then mark an X where it is. We will be doing something else which is “spot the functional group”. I am going to publish here a series of pictures of molecules. You need to spot the functional groups, name them and then consider what the answers are.

Before we get going it is important to understand that much of organic chemistry is about reactions of functional groups, but the functional groups can interact with each other. For example a hydroxyl group attached to a benzene ring is modified by the benzene ring. The pKa of the hydroxyl group goes down a lot, while the electron density in the pi cloud of the benzene ring increases a lot.

Now lets get started. Our first three molecules.


Lets look at them and search for the functional groups, before we move onto the next ones.


Now we will go through the different functional groups which we have spotted.

The alcohol is a sp3 oxygen atom which bears an aliphatic carbon and a hydrogen atom. The chemistry of the alcohol is similar to that of the phenols but the acidity of the alcohols is much lower. Normally alcohols groups are attached to saturated (sp3) carbons, but if an alcohol group was to be attached to a sp2 carbon in an alkene (such as vinyl alcohol) then it is normal for it to be converted by keto-enol tautomerism to an aldehyde or in some cases a ketone.

The phenol is like the alcohol but the carbon is part of a benzene ring, the pi system of the benzene ring withdraws electron density from the oxygen thus increasing the acidity of the OH group. In some ways the phenols can be regarded as a special case where the keto-enol tautomerism has been reversed. If we were to consider the keto form of phenol (cyclohexa-2,4-dien-1-one) then we can see how by converting the ketone into an “alkene” and a hydroxyl group the molecule can have an aromatic ring. We will get onto aromatic rings later. When I write “benzene ring”, it is important to understand that other aromatic systems such as naphthalene exist. A hydroxy naphthalene such as naphthol should be regarded as a phenol. Even while it has a different aromatic system.

The ether is a sp3 oxygen which is attached to two carbon groups, in common with the hydroxyl group it can donate electron density to a benzene ring. It is possible for both carbon groups to be aromatic, both aliphatic or even one of each. If the ether oxygen is attached to an aliphatic unsaturated group such as a vinyl then it is stable. The “keto-enol tautomerism” reaction will not convert it to something else.

The alkene is reactive towards electrophiles as it has a pi cloud in the form of the pi orbitals of the double bond. It is important to understand that a double bond is a sigma bond and a pi bond. We will attack this group later, at least in an intellectual manner rather as a hoard of electrophiles. I wounder what my students would be like if some miricle was to convert them into atoms / molecules. I think that they are better off as humans.

The benzene ring is not a simple collection of three alkenes, it has a special stability. Through resonance the bonds between the carbons in the ring have a bond order of 1.5 rather than 1 or 2. We need to pay special attention to the benzene ring as it is so common but it also can do many interesting things.

The last group we have is the amino group, this can be regarded as an ammonia which has had one or more hydrogen replaced with a carbon based group such as a aryl (aromatic) or alkyl (saturated aliphatic) group.

Posted by: Mark Foreman | August 12, 2015

Nulcear restart in Japan

Dear Reader,

It has come to my attention that a utility company in Japan have restarted a power reactor, now I know that a lot of people are very displeased about this saying that it is unsafe. But I have a question, after the Fukushima event the power reactors in Japan were shut down, now I imagine that some other fuel was used to provide the electric power while the reactors were shut down.

When people say that “nuclear is unsafe” do they consider the degree of harm caused by the alternative fuel or not ? A quick look at the EIA in the US suggests that after Fukushima that Japan was using coal, oil and gas to provide electrical power. As coal ash is often radioactive and as the oil / gas industry also are producers of radioactive waste. I have to ask the question of did the closure of the power reactors in Japan result in higher worker and public doses.

While the shutdown of a reactor might lower the doses of some people, the increased need for coal / oil might increase the doses of another group of people. What would be interesting is to find out what happened to the collective dose for the whole of the world’s population as a result of the closure of the Japanese power reactors.

I hold the view that the majority of the radioactivity which nuclear power plants generate is short-lived beta/gamma, while more activity (in Bq) might be produced per joule of electrical power delivered to the public than is handled during the use of a fossil fuel. If we take the view of how much activity will be released by the plant then the nuclear plant has a bit of an advantage that the radioactive waste is far better contained.

We also in our exposure assessment need to consider the fact that the longer half-lives of things like radium-226 in fly ash makes a release of it more serious than a release of the same activity of shorter lived fission products such as krypton-85. We also need to consider how the longer lived radionuclides from fossil fuel tend to be alpha emitters while the nuclear power fission products tend to be beta emitters, the alpha emitters might be short ranged but the internal threat they pose tends to be far higher than that posed by beta emitters.

Also the geochemistry and biochemistry of radium and its daughters increases the threat posed, much of the alpha activity in the back-end of the nuclear fuel cycle is plutonium. Plutonium is not very mobile in soil water and the human digestive system is not very able to absorb plutonium. On the other hand radium is quite mobile, radon is very mobile and the human digestive system is rather good at absorbing radium.

We also need to bear in mind that the use of fossil fuels can result in air pollution. Has this been considered in the judgement of nuclear as being unsafe ?

Also I know that MOX use has been demonized by some persons, but let us consider again the overall effect of MOX use. It is important to consider the whole rather than concentrating on a single part. For example it has been shown by one group in the USA that in terms of chemical toxicity to soil bacteria that plutonium is close to non toxic and that the alpha activity of the plutonium has no effect on the bacteria. They used both short-lived and long-lived plutonium to test the hypothesis that the plutonium only has a chemical effect on the bacteria.

If we were to use this study as our only source of information about plutonium we would come to the unreasonable conclusion that plutonium is harmless to humans. But back to the MOX question.

If MOX is used then the amount of alpha activity in the back end of the fuel cycle will increase, this alpha activity will be mostly in the form of plutonium, americium, curium and other long lived transuranium elements. A small increase in uranium-232 will occur but the alpha activity associated with this will be small compared with the transuranium elements. As the plutonium and the other transuraniums tend to be in high oxidation states they will tend to be not very mobile in soil water. The plutonium is likely to be as Pu(IV) while the americium / curium etc will be as Am(III) and Cm(III). These very highly charged cations will undergo plenty of hydrolysis and also tend to stick to mineral surfaces when the water is neutral or alkaline. Also once plutonium(IV) forms a solid it is very hard to redissolve, one of the enduring problems in the nuclear reprocessing sector is the fact that it is very hard to dissolve high plutonium content MOX fuels in nitric acid.

On the other hand the use of MOX will result in less uranium mining and less alpha emitting waste in the front end of the fuel cycle. The front end alpha waste will include the more mobile and toxic radium. The front end of the fuel cycle also is associated with human exposure to radon which is a problem which is less likely to appear in the back-end of the fuel cycle. But I note that inside the Chernobyl object shelter that the radon-220 (thoron) levels are rather high, so I would never say never. But compared with the other radioactive issues in a reprocessing plant the radon-222 and radon-220 will normally be small matters.

So while MOX use might form more alpha waste from a drinking water / ground water point of view the alpha waste problem could be mitigated by moving to MOX use.

Posted by: Mark Foreman | July 31, 2015

Reactions of Grignard reagents

Dear Reader,

I was recently asked on Quora about a reaction of a Grignard reagent with a ketone which instead of forming the alkoxide of a tertiary alcohol it forms the alkoxide from a secondary alcohol. In short the Grignard acts as a reducing agent which is a hydride donor rather than as a carbon nucleophile, we need to consider why.

The example I was asked about was a very hindered ketone which was reacting with a Grignard which had beta hydrogens. Now before those of you who have taken an organometallics class such as the second / third year ones which Tony Hill and David Widdowson gave to me about 20 years ago start to mention the beta hydride elimination, I have to say that I do not think that a magnesium hydride is responsible for this chemistry.

What is much more reasonable is the formation of an alkene from the Grignard and the nucelophilic attack of a hydrogen atom on the ketone. Here is the reaction with the curved arrows shown.

Reduction of a ketone by a Grignard reagent

Reduction of a ketone by a Grignard reagent

This reaction can be explained with steric effects, consider the steric repulsion between the isopropyl groups in 3-isopropyl-2,4-dimethylpentan-3-ol. I calculated this molecule in MM to have an energy of 26.5930 kCal per mole. When the same calculation is done for a molecule of propylene and a molecule of 2,4-dimethylpentan-3-ol then we get only 4.5838 kCal per mole. If the transition states for the two reactions are similar in energy to the products then this could explain why the Grignard acted as a reducing agent instead of as a carbon nucleophile.

Another reaction to watch out for is the Grignard acting as a base on a ketone to form an enolate. This is much more simple reaction. Here is the reaction with the curved arrows.

Grignard forming an enolate

Grignard forming an enolate


Posted by: Mark Foreman | February 25, 2014

Perylene II

Dear All,

I have looked up a paper which gives the C-C bond lengths in perylene

It is M.Botoshansky, F.H.Herbstein and M.Kapon, Helv.Chim.Acta, 2003, 86, 1113.

Here is a diagram of the values

C-C bond lenghts

C-C bond lenghts

Note the mean C-C distance for a single bond is 1.518 Å, all distances in the above diagram are in Å.

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