Posted by: Mark Foreman | November 4, 2019

1,4 dicarbonyls

Dear Reader,

When I was a PhD student, now some of the PhD students of nuclear chemistry and even some of the seniors have noticed that this is a stock phrase but unless we learn from the past we are doomed to repeat the mistakes of the past without benefiting from the better bits.

Now I have already given my thoughts on 1,2 and 1,3 difunctional (not dysfunctional) compounds it is now time for 1,4. If we want to disconnect a 1,4 compound then I think we have two main choices. The top one requires a positive charged version of the enolate while the bottom one requires unpolung. Now unpolung is a funny sounding word which relates to the reversal of what is normal. In some ways both disconnections require it.

1 4 disconnection to synthons

Now if I had an aldehyde such as benzaldehyde then it can be regarded as a synthetic equivlent to a benzyl alcohol with a positive charge instead of one of the hydrogens on the benzylic carbon.

benzaldehyde normal synthetic equiv

What we need is a method of changing the benzaldehyde to make it act in a way which goes against its normal character. Thus we are reversing its polarity. There is a great wealth of reagents which will do this for us. The most simple one is to combine benzaldehyde with a little bit of sodium cyanide. This does the job for us in the following way.

PhCHO plus NaCN

The resonance stabalised carboanion on the right can react with a range of electrophiles including benzaldehyde and α,β-unsaturated compounds to form addition products. I would suggest that to increase the yield of the 1,4 compound that we should keep the concentration of the α,β-unsaturated compound nice and high and slowly add the aldehyde to the hot mxiture of sodium cyanide, water and alcohol. If we were to just make a strong solution of the aldehyde and add some cyanide to this we would increase the yield of 1,2-diphenyl-2-hydroxyethan-1-one which is formed by the action of the anion on a molecule of benzaldehyde.

Imagine in our case we use ethyl vinyl ketone as our α,β-unsaturated compound.

At the end of the reaction we recycle the cyanide by returning the negative charge to the oxygen which is attached to the carbon bearing the cyanide group. The carbon oxygen double bond reforms and the cyanide is lost from the molecule. Ready to continue its good work elsewhere.

end of the cyanide cat reaction

Another method is to react cyanide with an α,β-unsaturated compound to form an addition product. When then can convert the cyanide groups into carboxylic acid groups by boiling the living daylights out of the compound in concentrated hydrochloric acid. If you are going to do this be careful not to have a lot of residual sodium cyanide in the mixture when you add the acid. For example you can react cyanohexanone with ethyl cyanoacetate to form an α,β-unsaturated cyano ester which then reacts with sodium cyanide. Finally boil without mercy in hydrochloric acid and evaporate down to a mxiture of ammonium chloride and the diacid.

synthesis of a diacid from cyclohexanone which was reacted with ethyl cyanoacetate

Another method which is more expensive, but who cares about the cost of reagents when we are just thinking about chemistry, is to react the aldehyde with propane-1,3-dithiol and make a ring (1,3-dithiane). We can then deprotonate this with a strong base and then react it with a range of electrophiles. I am sure that this could be made to react with an unsaturated carbonyl compound. It is known that in HMPA that the separated ion pair form of the reagent will react in a 1,4 manner with cyclohex-2-en-1-one. But I suspect that like myself most chemists these days would rather not use HMPA. But it does not hurt to think about chemistry in HMPA.

A word of advice do not use HMPA if you have any choice, it is a nasty carcinogen which absorbs with ease through skin. To use it legally in Sweden you need a special license the last time I checked.

Posted by: Mark Foreman | November 4, 2019

1,3 compounds

Now the synthesis of 1,3-difunctional compounds offers us a vast array of things, I will admit that I am a little biased as I have made good use of this chemistry in my professional life. A reaction worked and did what it said on the tin before making something which results in publications for me always tends to give me a warmer feeling inside than a reaction which either failed to work or produced something which was less useful. For a 1,3-diketone we can disconnect it in two different ways, we can break a bond and form two synthons.

disconnection of 13 diketone

For this example which I used in Aberdeen (Scotland) there are two different disconnections, while both are equally valid. We have to choose which one we use. We have to select synthetic equivalents of the synthons. We can use an ethyl ester for the positively charged one and an enolate anion for the negatively charged one.

synthetic equiv for 13 diketone

We have to choose which one, in my experience when I form an enolate by the deprotonation of 2-acetylpyridine I tend to form more brown muck (from aldol reactions) than I do when I treat a less electron poor ketone with a base under similar conditions. So I would react the poetically named pinacolone with a base such as excess sodium ethoxide to form the enolate before adding the ethyl picolinate to it.

As the initially formed diketone is a stronger Bronsted acid than the ketone, an excess of base is needed. I recall how this reaction went like a dream, it processed along gracefully like an iridescent bird emitting a song of intense joy. Well please forgive my flowery English.

Sometimes we only want to add a single carbon atom and form a synthetic equivalent of something like 3-oxo-3-(pyridin-2-yl)propanal. Rather than tormenting myself with the unwanted condensation products of ethanal (acetaldehyde) I choose an alternative. I used a reagent which operates under less basic conditions and forms a masked aldehyde. The reagent is N,N-dimethyl formamide dimethyl acetal. This is one of my favourite reagents.

By boiling a mixture of 2-acetyl pyridine and N,N-dimethyl formamide dimethyl acetal we form 3-(dimethylamino)-1-(pyridin-2-yl)prop-2-en-1-one. This can be reacted with hydrazine to form a pyridyl pyrazole which I regard as a good friend. More than once in my professional life I have used 2-(1H-pyrazol-3-yl)pyridine as an intermidate for forming something else. While I was in Aberdeen I used it for making a capping ligand to reduce the crosslinking of my coordination polymers while in Sweden I used it as the basis of a nickel extraction agent.

reaction of acetyl pyridine with DMF dimethyl acetal and then hydrazine

Another method which involves thioethers is to use a base, carbon disulfide and methyl iodide. This combination of reagents can form 3,3-bis(methylthio)-1-(pyridin-2-yl)prop-2-en-1-one from 2-acetyl pyridine. This reagent can be used to form a range of interesting things. But sometimes it is a pain to have to remove the sulfur group later in the synthesis.

product of acetylpyridne and CS2 and methyl iodide

Regarding sulfur chemistry, I have been there, done it and brought the t-shirt ! But rather than falling into the macho nonsense which some chemists fall prey to before I do sulfur chemistry I always ask myself if I truly need to do it. For those of you who have never done both organo sulfur chemistry and radioactive methyl iodide (131I) chemistry. I can tell you that the two are very similar, when I make radioactive methyl iodide I feel myself channeling my inner sulfur chemist. The little sulfur chemist goblin inside me gets to come out to play !

My advice to a radiochemist about to do sulfur chemistry for the first time is to pretend you are working with something like sulfur-35, nickel-63 or a short lived alpha emitter and that the whole of the fumehood is crawling with radioactivity. My advice would be “do not touch anything inside the hood without gloves on” and to be very careful bringing anything out of the hood. For the radioactive case I would want to bag everything coming out while for the organic sulfur chemistry I would soak everything in chlorine bleach overnight before pulling it out of the hood.

It is best to separate the lightly contaminated things like liebig condensors which were used to boil things under reflux from the bulk stocks of organosulfur waste. Also keep in mind that the reaction between some sulfur compounds and chlorine bleach can be violent and exothermic. But it is nothing compared with phosphorus (III) compounds and chlorine bleach, as a PhD student I once added a couple of cubic centimeters of P,P-dichloro-N,N-diethylamino phosphine to some chlorine bleach. The reaction was exceptionally violent, for a moment the mixture of the two generated a flame. So in general when disposing of or decontaminating phosphine residues I would suggest using a more gentle oxidant.

One of my coworkers once made the mistake of using iodine in acetone to treat his waste, this was a big mistake as the combination of acetone, iodine and acid started to form the incredibly offensive iodoacetone. My advice is to use a combination of iodine and ethanol to decontaminate glass which has been used with things like P,P-diphenyl P-chloro phosphine.

While iodine-131 can be made to disappear by rendering it non volatile and putting it in a plastic bag before leaving it behind lead sheidling for about 80 day, organo sulfur compounds are everlasting. But they do not emit harmful rays which can pass through glass bottles and it is possible to send them for incineration with greater ease.

Posted by: Mark Foreman | November 3, 2019

1,2 difunctional compounds

Now in organic chemistry a common problem is knowing how to design the synthesis route for a substance. It is nice to be able to buy chemicals ready made but sometimes we need to be able to make things for ourselves. Sometimes the price of the readymade substance is too high or it might be impossible to find someone who will sell it.

The best way to plan the synthesis is often to look at the carbon framework of the molecule and look for ways in which we can use chemistry to cut up the molecule into more simple units. I generally advise you to look for bond breaking close to the functional groups.

If you see two carbonyl groups, two alcohols or a ketone and an alcohol next to each other then we can consider breaking the bond between the two carbons bearing these groups. If we look at the synthesis of the CyMe4 diketone used in BTP and BTBP chemistry we disconnect the molecule at this point to form a synthon.

A synthon is a entity which can be an imaginary one which can not exist in real life. We can break the molecule up to make an imaginary diradical.

disconnection to diradical

If we can think of a combination of reagents which reacts like the way that our synthon would then we can make some progress. In this case at Reading we used the diester as a synthetic equivalent of the diradical.

disconnection to diester

What we used was the reaction of the diester with sodium metal in toluene with trimethyl silyl chloride present. What happens in this reaction is that the ester groups accept an electron from the sodium metal to form radical anions which then dimerise as they are formed on the surface of the sodium metal.

mechanism of ester with sodium metal

After the dianion forms it will react with the chlorosilane to form the following type of compound.

diTMS product

What we can now do is one of three things, we can treat the compound with aqueous acid and then convert it into the 1,2-ketoalcohol. We can use hydrogen gas and a catalyst to hydrogenate the carbon-carbon double bond. This will give us a protected dialcohol. Finally we can oxidize it with bromine to form the 1,2-diketone which we want.

There are other reactions which form 1,2 difunctional molecules. For example we can react aromatic aldehydes with a small amount of cyanide to form 2-hydroxyketones. Also we can react ketones with magnesium amalgam to form 1,2-diols in the pinacol reaction.

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.

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