6.1.1 (d,e,f,g) Electrophilic substitution

(d) The electrophilic substitution of aromatic compounds with:

(i) concentrated Nitric Acid in the presence of concentrated Sulphuric Acid

(ii) a Halogen in the presence of a Halogen-carrier

(iii) a Haloalkane or Acyl Chloride in the presence of a Halogen-carrier (Friedel–Crafts reaction) and its importance to synthesis by formation of a C–C bond to an aromatic ring

(e) The mechanism of electrophilic substitution in Arenes for Nitration and Halogenation

{For nitration mechanism, learners should include equations for formation of NO₂⁺.}

{Halogen carriers include Iron, Iron Halides and Aluminium Halides. For the Halogenation mechanism, the electrophile can be assumed to be X⁺}

(f) The explanation of the relative resistance to Bromination of Benzene, compared with Alkenes, in terms of the delocalised electron density of the pi-system in Benzene compared with the localised electron density of the pi-bond in Alkenes

(g) The interpretation of unfamiliar electrophilic substitution reactions of aromatic compounds, including prediction of mechanisms

{Extra information may be provided on exam papers.}

Benzene rings are energetically stable due to the delocalisation energy, addition reactions would involve breaking the ring but not re-forming it.

This is energetically unfavourable and so it is difficult to make aromatic compounds undergo addition.

However, substitution reactions allow the ring to reform.

So, substitution can happen if the conditions for breaking the ring are met.

You would think that electrophiles would be attracted strongly to all the electrons in the ring.

But this is not the case

Benzene rings don't polarise as much as Alkenes - each C-C bond has only 3 electrons rather than 4 in an alkene.

This makes them less attractive to normal electrophiles than an Alkene would be.

So, to make aromatic compounds undergo electrophilic substitution requires an electrophile with a positive charge.

You really have to learn this mechanism and be able to draw it accurately - it is the same for all electrophilic substitutions.

1. The first curly arrow must start on the ring - but it can start anywhere on the ring

2. The broken ring must cover 4 of the C-C bonds (not significantly more, not significantly less)

3. The examiners want to see the + charge on the broken ring away from Carbon 1 (people lose marks for this all the time!)

4. The second curly arrow must go back into the broken ring.

5. Don't forget that the displaced H atom must have a + charge.

Once you've learned the mechanism above you then have to learn how to make the positive electrophile for each different substitution.

In the case of Nitration (adding an NO₂ group) we need NO₂⁺(the Nitronium ion)

The obvious source of NO₂ would be Nitric Acid (HNO₃) but acids dissociate to make H⁺ ions and an anion.

HNO₃ --> H⁺ + NO₃^-

And an NO₃^- ion would never be able to draw electrons out of the ring.

So we need a substance that can stop Nitric Acid behaving in a typically acidic way.

In other words, a stronger acid.

If we use concentrated acids then there are few water molecules for the Nitric acid or the Sulphuric Acid to donate H+ ions to.

So the Sulphuric Acid donates to the Nitric Acid.

H₂SO₄ + HNO₃ --> HSO₄^- + H₂NO₃⁺

The unsightly looking protonated Nitric Acid then breaks up.

H₂NO₃⁺--> H₂O + NO₂⁺

After this, the mechanism proceeds as in the general mechanism shown first.

For Halogenation we need a Br⁺ or a Cl⁺ ion - neither of which are seen naturally.

A Halogen-carrier is required - a substance that will remove Br^- (or Cl^-) from Bromine (or Chlorine) and leave a Br⁺ or a Cl⁺ ion.

Steps 2 and 3 are common to all electrophilic substitutions - just drawn as double bonds rather than with the broken ring model.

So, the only thing to learn is how the Br⁺ was made in Step 1

Iron Chloride (FeCl₃) would be the equivalent halogen carrier for Chlorination, or we could use Aluminium Chloride (or Aluminium Bromide)

Note that examiners don't care if you write the mechanism showing Cl⁺ being made as Step 1, or if you draw it as above with a d+/d- arrangement on the halogen.

Note also that Iron metal is an acceptable halogen carrier to name - though not one you'll be able to use when drawing out the mechanism.

OCR examiners tend to avoid the term "Friedel-Crafts reaction".

But they do use Acylation and Alkylation - which have the same meaning

Acylation is considered to be very important because Benzene is an easy substance to make/extract but finding the correct substituted aromatic compound to make a pharmaceutical from would be harder.

Acylation allows us to form a C-C bond to a benzene ring, after which chemists can change the substituted group to suit their needs.