Categories
Uncategorized

Reaction of bromine / toluene and methyl benzoate and the nitration mixture

When we react bromine and toluene using the iron(III) chloride we have to have a step to form the true electrophile. We need a catalyst to make the reaction go. Here is a short film in which I had a solution of bromine in toluene. I added some steel staples and then I lifted them out with a magnet.

What you should notice is that when the steel was in the mixture the mixture generated bubbles of hydrogen bromide gas, when the steel was taken out the gas production stops. The surface of the steel reacts with a little of the bromine to form some ferric bromide. This ferric bromide is the catalyst for the reaction.

The first step is normally considered to be the reaction of the bromine molecule with the Lewis acid. Commonly this is thought (and taught) to be the reaction of bromine and the Lewis acid to form a pair of separated ions.

This will then be followed by the reaction with the aromatic ring. Now when the reaction occurs we have a choice of reacting in three different locations. We can react at the ortho, meta and para sites.

The rate of reaction at these three sites will not be equal. If we consider for a moment what the mechanism for each of the three possible reactions will be. If we use what we have found out from the nitration of benzene and the reaction of the anisole with the P4S10 then we should be able to make a set of three mechanisms like these.

We have to decide which ones of these will be faster, now to get to the cationic intermediate we have to pass through a transition state. Here is a graph of energy against the progress of the reaction for one molecule which I drew today.

Now as the transition state is similar to the cationic intermediate in energy we can reason that the lower the energy of the intermediate the lower the energy of the transition state will be. We have to overcome the energy barrier of TS1. Here is the diagram with the activation energy barrier marked on it.

Now if we have a second path to a different product, shown in red. Then I hope that you can understand that if the reaction is operating under kinetic control then we will not form the red product. The reason is that as the activation barrier of TS1(red) is larger the rate at which the intermediate for the red reaction is entered is smaller than the rate at which the black one is entered. Thus more of the black will be formed than the red one.

Now the key thing for us to understand is that the energy of the cationic intermediate is not dictated by only one resonance form. All the resonance forms have an effect upon the energy of the cation. If we imagine for a moment that all the resonance forms were stable cations then we can imagine how those which are formally drawn as secondary carbocations have higher energies than those which are formally drawn as tertiary carbocations.

But it is the combination of the resonance forms which determines the energy of the cation. If we ignore crazy high energy resonance forms like this one

and limit ourselves to the three lower energy forms which are commonly drawn then we can make some easy progress. If we assume for arguments sake that the three lower energy forms all contribute equally to the overall energy. Then if form A has an energy of 100, form B 70 and form C 100 then the average of these will be 90. So when one of the forms is more stable then we can understand how we lower the energy of the cation.

If we make more of the resonance forms lower in energy then the overall energy of the cation will be even lower. If we consider for a moment the reaction of bromine and 1,3,5-trimethyl benzene (mesitylene). Then we can see how all three of the resonance forms are more stable than the resonance forms which would exist if benzene was reacted.

So as a result if I was to react mesitylene with bromine then it should react faster than toluene under the same conditions. What I am going to do when I get the chance is to react some bromine with a mixture of mesitylene, chlorobenzene, toluene, tert-butyl benzene and some other aromatic compounds. If I get a GCMS trace of the mixture before and after adding the bromine we should be able to see the relative reactivity of the different compounds.

Now there is something which is an important difference between the reaction of the bromine and the P4S10. This is the size of the electrophilic reagent. In the case of bromine it is small while for P4S10 it is large. For large electrophiles we tend to see less reaction at the ortho sites as the electrophile is less able to reach the site. I think that the steric effect on the ortho / para ratio is a rather advanced idea which at Chalmers is not normally something we would expect a first year to recall / understand. But it is still an interesting thing.

The final stage of the bromination reaction is for the cation to lose the proton, the proton then reacts with the anionic iron bromo complex to reform the Lewis acid and form a molecule of HBr.

In the same way as a group attached to the ring can increase the stability of the cation, there are groups which lower the stability of the cation. If we consider a group which is able to withdraw electron density by both the inductive pull effect and the resonance effect for a moment. One such group would be a carboxylic acid or one of its derivatives. If we consider methyl benzoate for a moment. I have drawn some resonance forms which make it clear how the carbonyl group can pull electron density towards itself by means of the resonance effect.

There are at least two ways of thinking about it, one way is that if you are trying to increase the stability of the cationic intermediate by moving a positive charge onto a carbon which already is being used by another group in the same way. Then we will have contest for which group gets to put its positive charge onto the carbon. Here is an organic chemistry joke.

Q: What did the carbonyl group say to the Wheland intermediate (Arenium ion) ?

A: Clear off ! You are not putting your positive charge there ! I was here first !

Now we can consider it in a more serious way, if we take a more simple molecule (formaldehyde). we should understand that the oxygen is the most electronegative of the elements in the molecule. According to my quantum mechanical calculations we have charges on the different atoms.

Now if we repeat the calculation for acrylaldehyde (Acrolein, also known as propenal) we get a more electron rich oxygen. This indicates that the quantum mechanical calculation is indicating that the carbonyl group is withdrawing electron density from the alkene. This time I have put the partial charges on the non hydrogen atoms only.

Now we do this for fumaraldehyde, this is a trans alkene where there are two aldehyde groups which are playing “tug of war” with the electron density in the pi system.

What we should look at is the fact that when we have two aldehydes playing “tug o’ war” with the electron density that they get less electron density from the alkene. We can apply the same ideas to the cationic intermediate.

As the carbonyl group is trying to withdraw electron density from both the benzene ring and the cationic intermediate it will lower the amount of electron density in the benzene ring and it will also lower the stability of the cationic intermediate.

Now in organic chemistry a lot of things are dictated by kinetics, while a thermodynamic driving force might favor the formation of all three isomers of methyl nitrobenzoate the rates of formation of these things will be different.

There are some cases where the electrophilic substitution of aromatic things are reversible. But generally we run most of these reactions under conditions where the reverse reaction is very slow. As a result they do not reach equilibrium.

In our case we can regard the nitration to be irreversible, so which ever isomer is formed quickly will be the isomer which will be the major product. The ester group has two effects there is the inductive pull on the electron density in the ring. This occurs through the sigma bonds. There is also the resonance effect through which the ester group withdraws electron density.

While the reaction at all three sites is slowed down by the inductive removal of electron density by the ester group. The resonance effect only has an effect on the reaction at the ortho and para positions. So as a result the main product of the reaction is the meta isomer. The reaction at the meta site will be slower than the reaction of benzene, so if we were to make a 1:1 mixture of methyl benzoate and benzene and combine it with one equivalent of an electrophile then the majority of the electrophile would be consumed by the reaction with the benzene. You should recall how the young Mark Foreman reacted a mixture of xylene and ferrocene with an electrophile, when he did this the ferrocene reacted rather than the xylene.

When you are ready to consider the next aromatic reaction please click here, this will be phenol with nitric acid.

By Mark Foreman

I am a Swedish / British chemist who works in Sweden at Chalmers University of Technology as an Associate Professor. I am based in the Chemistry and Chemical Engineering Department and I work in the Nuclear Chemistry / Industrial Materials Recycling units. My academic interests include things from both nuclear chemistry, recycling and some other unrelated things.

I originally come from the UK (England). I have degrees from Imperial College (BSc + ARCS) and the University of Loughborough (PhD).

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s