student carrying out practical in the lab

Electrophilic substitution-halogenation of benzene rings

Bromination of aromatic rings

Addition of bromine water to a substance is often used to test for unsaturation in a molecule, for example the addition of bromine water to an unsaturated alkene in a boiling tube will instantly result in the decolourisation of the reddish-brown bromine water. In the image below the left-hand test tube contains cyclohexane (C6H12); a saturated hydrocarbon while the right-hand test tube contains cyclohexene (C6H10); an unsaturated hydrocarbon.

the image shows the products of the addition of bromine water to cyclohexane and cyclohexene

If a small amount of bromine water is added to each test tube and then both test tubes are given a quick shake to mix the contents together, then the bromine water is instantly decolourised in the test tube containing the unsaturated cyclohexene while the bromine water stays orange in the test tube containing the saturated cyclohexane.

Switching layers

It is worth mentioning that the orange colour of the bromine will switch layers in the test tube containing the saturated cyclohexane. Halogens such as bromine are more soluble in organic solvents than in water, so when the contents of the test tube are shaken up, although no chemical reaction will occur immediately the majority of the bromine will move into the top less dense organic layer of cyclohexane while the aqueous layer which initially contained the bromine dissolved in it will turn clearer as the bromine dissolves in the organic cyclohexane layer, this is shown in the left hand test tube in the image above.

Electrophilic addition versus electrophilic substitution

As you may recall alkenes under electrophilic addition reactions. The mechanism below shows how bromine adds to the carbon carbon double bond in the unsaturated alkene ethene. You may recall that:


Reaction of bromine and ethene

The main product of the reaction of bromine water with the alkene ethene is a colourless solution of 2-bromoethanol. You may have expected the product to be 1,2-dibromoethane as was seen above when ethene was bubbled through bromine or a solution of bromine in an organic solvent. However if ethene is bubbled through an aqueous solution then the hydroxide ions (OH-) and also water molecules, both of which are better nucleophiles than bromide ions take part in the reaction and will attack the intermediate carbocation to form the substituted alcohol 2-bromoethanol. However 1,2-dibromoethane is also present in the solution but as a minor product. Equations for the two reactions taking place are shown below:
Word and symbolic equations for the reaction of bromine water with ethene.

Electrophilic substitution

However if the ethene is swapped for benzene or any another aromatic compound one difference in the reaction with bromine or bromine water is immediately obvious; aromatic rings will not react with bromine or other halogens, they are much less reactive than alkenes towards electrophiles. I am sure you are aware that while unsaturated molecules such as alkenes readily undergo electrophilic addition reactions aromatic molecules such as benzene will not undergo electrophilic addition reactions but instead undergo electrophilic substitution reactions.

In the image below bromine has been dissolved in an organic solvent such as chloroform or carbon tetrachloride, it is then added to a test tube containing benzene and another test tube containing the unsaturated hydrocarbon cyclohexene (C6H10). Both test tubes are stoppered and shaken. The results are shown in the image. Cyclohexene rapidly decolourises the orange bromine solution but in the test tube containing benzene the orange colour due to presence of bromine persists and does not fade.

Image shows the differences on the addition of bromine to liquid benzene and cyclohexene is boiling tubes.  Benzene does not readily react with bromine unless a catalyst is present.

Addition of a halogen carrier- a Lewis acid

As mentioned above if a solution of bromine in an organic solvent is added to a test tube containing benzene then no reaction occurs, the orange colour due to the presence of bromine persists. However if some iron filings or iron (III) bromide are added to the test tube then the orange colour due to the presence of bromine decolourises and white misty acidic fumes of hydrogen bromide are observed above the test tube. If a piece of moist pH paper or blue litmus paper is held above the test tube it turns red due to the acidic hydrogen bromide gas which is released. Equations for these reactions are shown below:

equations for the reaction of benzene with bromine

From the equations above I hope you can see that the bromine swaps or substitutes for one of the hydrogen atoms on the benzene ring. The reaction as you might expect from an aromatic ring is electrophilic substitution.

Electrophilic substitution versus electrophilic addition

cartoon style image stating that the delocalised pi electrons in aromatic rings adds extra stability Why then do unsaturated alkenes undergo electrophilic addition reactions but aromatic rings such as benzene undergo electrophilic substitution reactions instead? Well the pi(π) electrons in an alkene molecule are localised between two carbon atoms, whereas in an aromatic molecule such as benzene the pi(π) electrons are delocalised over all of the carbon atoms, this delocalisation makes benzene and other aromatic molecules very stable and means basically that in order for an electrophile to react with benzene or another aromatic molecule it needs to be an extremely "good electrophile".

Non-polar molecules such as bromine will NOT react with aromatic rings, simply because they are "not good" enough electrophiles. For this reason a catalyst or a halogen carrier as they are often referred to are needed to increase the ability of a species to act as an electrophile. The delocalisation of the π-electrons as mentioned results in an increase in the stability of the benzene ring and this will raise the activation energy for any reaction which will destroy this delocalisation energy.

Lewis acids and the halogenation of benzene rings

We have seen from the equations above that benzene will react with bromine or indeed chlorine and iodine in the presence of catalysts such as iron or even aluminium. These metals react with the halogens to form metal (III) halides such as aluminium chloride (AlCl3) or iron (III) bromide. This is shown in the equations below:

Iron(s) + bromine(g) iron (III) bromide(s)
2Fe(s) + 3Br2(g) 2FeBr3(s)

or in the case of chlorine we have:
Aluminium(s) + chlorine(g) aluminium chloride(s)
2Al(s) + 3Cl2(g) 2AlCl3(s)

The catalysts or halogen carriers are they are often called work by producing much more powerful electrophiles.


Lewis acids

3d representations and models of Lewis acid molecules using aluminium chloride. Lewis acids are substances which are able to accept a pair of electrons from another species; that is a Lewis base. In order to accept a pair of electrons the Lewis acid must have a space to put them! That is it must have empty orbitals available. Iron (III) ions (Fe3+) for example have empty d-orbitals which can accept an electron pair while aluminium chloride (AlCl3) has only 6 electrons in its valency shell so like iron(II) ions it has empty orbitals which can accept an electron pair.


Halogenation of aromatic rings

Iron (III) bromide(s) (FeBr3) is the Lewis acid catalyst or halogen carrier usually used to brominate aromatic rings while aluminium chloride (AlCl3) or iron (III) chloride (FeCl3) is usually used to chlorinate aromatic rings. The Lewis acid catalysts works by basically polarising the bromine molecule (Br2) or the chlorine molecule (Cl2) to form a much better electrophile. The image below shows how the Lewis acid or halogen carrier is able to polarise the halogen molecule.

It is the complex formed when the halogen makes a dative covalent bond to the Lewis acid or halogen carrier that acts as the electrophile. The formation of a dative covalent bond between the Lewis acid and the halogen molecule will induce a permanent dipole in the halogen molecule making it a much better electrophile.

FeBr3 acts as a Lewis acid catalyst and enables bromine to add 
to an aromatic ring

Self-check

Quickly check your understanding of a few of the key terms mentioned above by answering these two short questions:

1. What is aromaticity and how does it affects the reactions of benzene?
+
Poster showing the properties of aromatic rings

Aromaticity is a special stability found in certain cyclic compounds where:

Aromaticity makes molecules such as benzene very stable compared to alkenes. Because of this stability:


2. What is a halogen carrier (or Lewis acid) and explain how they work?
+
Atom pulling electrons from another atom

Lewis acids or halogen carriers are substances which are able to accept a pair of electrons from another species; that is a Lewis base. In order to accept a pair of electrons the Lewis acid must have a space to put them, that is it must have empty orbitals available.

The halogen carrier will form a dative covalent bond with the halogen molecule which will induce a permanent dipole in the halogen molecule; that is the halogen molecule will have partially charges ends; that is δ+ and δi ends which will make it a much better electrophile.


The mechanism for the halogenation of aromatic rings

An aromatic molecule such as benzene is able to use its delocalised pi electrons to attack the complex containing the polarised halogen molecule. The mechanism for the bromination of benzene is shown below. It is really looks no different to the other electrophilic substitution reactions that we have seen for other aromatic compounds. The reaction can be broken down into a number of key steps:

Mechanism for the bromination of benzene using Kekulé structures for benzene

Or if you prefer to usethe circle notation to represent the delocalised electrons in benzene we have:

Mechanism for the bromination of benzene using circle notation structures for benzene

The final step involves the attack of the FeBr4- ion, which helps to remove the hydrogen ion (H+) from the intermediate carbocation. The FeBr4- will react with the hydrogen ion (H+) to regenerate the Lewis acid catalyst- FeBr3 and also form the acidic gas hydrogen bromide (HBr). An equation for this reaction is shown below:

FeBr4- + H+ FeBr3 + HBr


Self-check- summary

Test your understanding of the main ideas covered above by filling in the blanks to complete the paragraphs below. The words to complete the blanks are shown in the yellow box and the drop down menus. Simply select the correct word or words from the drop-down menus to correctly complete the two short pargraghs.



Key Points

Image showing students in the chemistry lab

Practice questions and quick quiz

Check your understanding - Quick quiz on halogenation.

Check your understanding - Questions on halogenation

Check your understanding - Additional questions on halogenation

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